Hello students! Welcome to the fascinating world of Alkynes: acidity and important reactions! Get ready to unlock the secrets of these versatile hydrocarbons and master their unique reactivity for your exams and beyond.
Imagine a carbon atom not just forming single or double bonds, but a powerful, electron-rich triple bond! This is precisely what defines an alkyne, setting them apart from alkanes and alkenes. Alkynes are a class of unsaturated hydrocarbons containing at least one carbon-carbon triple bond, which grants them a distinct linear geometry around that bond and some truly exceptional chemical properties.
What makes alkynes so special? Beyond their unique structure, we'll delve into two pivotal aspects: their surprising acidity and their incredibly diverse range of important reactions. You might be accustomed to thinking of hydrocarbons as largely unreactive or non-acidic, but the presence of the triple bond introduces a fascinating twist. Specifically, the sp hybridization of the carbon atoms involved in a terminal alkyne (where the triple bond is at the end of the chain) dramatically increases the electronegativity of these carbons. This effect pulls electron density away from the attached hydrogen atom, making it unusually acidic for a hydrocarbon. This acidity isn't just a theoretical concept; it's a powerful tool in organic synthesis, allowing terminal alkynes to react with strong bases to form highly useful acetylide anions.
But the story doesn't end there! The electron-rich nature of the triple bond, combined with the possibility of forming acetylide anions, makes alkynes incredibly reactive and versatile synthetic intermediates. Think of them as "chemical chameleons", capable of transforming into a vast array of other functional groups. You'll learn about their characteristic reactions, including:
* Addition reactions: Where molecules like hydrogen, halogens, or hydrogen halides add across the triple bond.
* Oxidation reactions: Leading to the cleavage of the triple bond or the formation of diketones.
* Reduction reactions: Allowing selective conversion to alkenes (cis or trans) or even alkanes.
* Reactions involving acetylide anions: Opening doors to carbon-carbon bond formation and the synthesis of complex molecules.
For your JEE Main and Board examinations, a solid understanding of alkyne chemistry is fundamental. These concepts frequently appear in questions involving multi-step synthesis, reaction mechanisms, and distinguishing between different classes of organic compounds. Mastering the acidity and reactivity patterns of alkynes will equip you with essential tools to tackle challenging problems and build a strong foundation in organic chemistry.
So, get ready to explore the nuances of their structure, understand why they exhibit acidity, and unravel the exciting pathways of their diverse reactions. Let's begin this journey to unlock the full potential of alkynes!
Topic Explanations
🌐
Overview
📚
Fundamentals
Hey there, future chemists! Welcome to the exciting world of alkynes. You've already met alkanes (single bonds) and alkenes (double bonds). Now, get ready to dive into the chemistry of alkynes, those fascinating hydrocarbons with a carbon-carbon triple bond!
Today, we're going to explore two super important aspects of alkynes: their unique acidity and some of their key reactions. Understanding these fundamentals is crucial, not just for your exams, but for truly grasping organic chemistry. So, let's get started from scratch!
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### 1. What Makes Alkynes Special? A Quick Recap!
Remember how we categorize hydrocarbons?
* Alkanes: All C-C single bonds, sp3 hybridized carbons.
* Alkenes: At least one C=C double bond, sp2 hybridized carbons.
* Alkynes: At least one C≡C triple bond. The coolest thing about the carbons involved in this triple bond is their hybridization – they are sp hybridized!
This sp hybridization is the secret sauce behind many of alkyne's unique properties, especially its acidity. Each carbon in the triple bond forms one sigma bond and two pi bonds. The remaining two sp hybrid orbitals are used to form sigma bonds with other atoms (like hydrogen or another carbon).
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### 2. The Surprising Acidity of Terminal Alkynes: Why and How!
When we talk about 'acidity' in chemistry, we're generally referring to a compound's ability to donate a proton (H+). Acids taste sour, right? Well, you definitely shouldn't taste alkynes! But they do have an 'acidic' hydrogen.
Important Note: This acidity applies *only* to terminal alkynes.
A terminal alkyne is one where the triple bond is at the end of the carbon chain, meaning one of the triply bonded carbons is attached to a hydrogen atom. For example, ethyne (HC≡CH) or propyne (CH3-C≡CH).
Internal alkynes (like but-2-yne, CH3-C≡C-CH3) do not have this acidic hydrogen, so they are not acidic in the same way.
#### 2.1 The "Why": The Magic of sp Hybridization
So, why are these C-H bonds in terminal alkynes acidic when C-H bonds in alkanes and alkenes are generally considered non-acidic? The answer lies in the hybridization of the carbon atom bonded to that hydrogen!
Let's compare the hybridizations:
* Alkanes: sp3 hybridized carbon (25% s-character, 75% p-character)
* Alkenes: sp2 hybridized carbon (33% s-character, 67% p-character)
* Alkynes: sp hybridized carbon (50% s-character, 50% p-character)
What is 's-character' and why does it matter?
The 's' orbital is spherical and closer to the nucleus than 'p' orbitals. When an orbital has more 's-character', it means the electrons in that orbital spend more time closer to the positively charged nucleus. This makes the orbital, and thus the atom, effectively more electronegative.
Think of it like this:
* Imagine the nucleus as the "boss" of the atom.
* 's' orbitals are like employees who always sit right next to the boss – very close, strong influence.
* 'p' orbitals are like employees who sit a bit further away – less influence.
So, a carbon with 50% s-character (sp) is much better at pulling electron density towards itself than a carbon with 33% s-character (sp2) or 25% s-character (sp3).
How does this relate to acidity?
When a terminal alkyne loses its proton (H+), it forms a conjugate base called an acetylide ion (R-C≡C-).
R-C≡C-H ⇌ R-C≡C- + H+
The more stable this conjugate base (R-C≡C-) is, the more willing the original compound (alkyne) is to donate its proton, and thus, the stronger its acidity.
Because the sp hybridized carbon is highly electronegative, it can effectively stabilize the negative charge on the acetylide ion. It pulls the electron density of the lone pair closer to the nucleus, making the negative charge less "exposed" and thus more stable.
#### 2.2 Quantifying Acidity: pKa Values
Acidity is quantified by pKa values. A lower pKa value means a stronger acid. Let's compare some approximate pKa values:
As you can see, a terminal alkyne (pKa ~ 25) is significantly more acidic than an alkane (pKa ~ 50) or an alkene (pKa ~ 44). However, it's still less acidic than water (pKa ~ 15.7) or typical alcohols. This means you need a pretty strong base to deprotonate a terminal alkyne.
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### 3. Important Reactions of Alkynes: Building Blocks of Synthesis!
Alkynes are incredibly versatile in organic synthesis. Let's look at some of their fundamental reactions.
#### 3.1 Reactions Involving Acidity: Forming and Using Acetylide Ions
The acidic nature of terminal alkynes is not just a theoretical concept; it's a powerful tool in synthesis!
3.1.1 Formation of Acetylide Ions (Deprotonation)
Terminal alkynes react with very strong bases to form acetylide anions. Common strong bases include:
* Sodium amide (NaNH2) in liquid ammonia (NH3) - This is the most common and effective base.
* Sodium metal (Na)
* Grignard reagents (RMgX) (e.g., CH3MgBr)
* Alkyl lithium reagents (RLi) (e.g., BuLi)
R-C≡C-H + Base:- → R-C≡C- + Base-H
Example:
CH3-C≡C-H + NaNH2 (in liq. NH3) → CH3-C≡C-Na+ + NH3
(Propyne) (Sodium propynide)
This reaction works because the base (like NH2-) is stronger than the acetylide ion (R-C≡C-), meaning it has a stronger affinity for a proton. Since NH3 (pKa ~ 38) is a weaker acid than the alkyne (pKa ~ 25), NH2- is a stronger base than R-C≡C-, making the equilibrium favor the products.
3.1.2 Alkylation of Acetylide Ions (Carbon-Carbon Bond Formation!)
Once you have an acetylide ion, it's a powerful nucleophile (an electron-rich species that 'loves' positive centers). It can attack electrophiles, especially primary alkyl halides, in an SN2 reaction to form new carbon-carbon bonds! This is super useful for making longer carbon chains.
R-C≡C-Na+ + R'-CH2-X → R-C≡C-CH2-R' + NaX
(Acetylide ion) (Primary alkyl halide) (New internal alkyne)
Example:
Let's say we want to make pent-2-yne (CH3-C≡C-CH2CH3) from propyne.
1. Deprotonate propyne:
CH3-C≡C-H + NaNH2 → CH3-C≡C-Na+ + NH3
2. React with ethyl bromide (a primary alkyl halide):
CH3-C≡C-Na+ + CH3CH2-Br → CH3-C≡C-CH2CH3 + NaBr
(Sodium propynide) (Ethyl bromide) (Pent-2-yne)
This is a fantastic way to extend a carbon chain by adding an alkyl group to the terminal carbon of an alkyne. For JEE, this specific reaction is extremely important for synthesis problems!
#### 3.2 Addition Reactions: From Triple to Double to Single Bonds!
Just like alkenes, alkynes undergo addition reactions because of their reactive pi bonds. But since they have *two* pi bonds, they can add two equivalents of a reagent across the triple bond, potentially converting it to an alkene, and then an alkane.
3.2.1 Hydrogenation (Addition of Hydrogen, H2)
Adding hydrogen to an alkyne can lead to an alkane or an alkene, depending on the catalyst.
* Complete Hydrogenation (to Alkane):
R-C≡C-R' + 2H2 $xrightarrow{ ext{Ni, Pd, or Pt}}$ R-CH2-CH2-R'
This converts the alkyne completely into an alkane. The reaction is fast and generally not very selective if you want to stop at the alkene stage.
* Partial Hydrogenation (to Alkene): This is where it gets interesting and synthetically useful! You can selectively convert an alkyne into an alkene. And even better, you can control the *stereochemistry* (cis or trans)!
* Formation of *cis*-Alkene:
Using a "poisoned" or deactivated catalyst like Lindlar's catalyst (Pd deposited on CaCO3, poisoned with lead salts like lead acetate or quinoline), you can stop the hydrogenation at the alkene stage, and specifically get the cis-alkene.
R-C≡C-R' + H2 $xrightarrow{ ext{Lindlar's catalyst}}$ R
R'
(The hydrogen adds from the same side of the triple bond, resulting in *cis* geometry).
* Formation of *trans*-Alkene:
To get the trans-alkene, a different set of reagents is used: sodium (Na) or lithium (Li) in liquid ammonia (liq. NH3). This is called dissolving metal reduction.
R-C≡C-R' + Na/Li $xrightarrow{ ext{liq. NH}_3}$ R
R'
(The hydrogen adds from opposite sides, resulting in *trans* geometry).
For JEE, remembering Lindlar's catalyst for *cis* and Na/Li in liq. NH3 for *trans* is critical for synthesis and distinguishing reactions!
3.2.2 Hydrohalogenation (Addition of HX)
Alkynes react with hydrogen halides (HCl, HBr, HI). This reaction follows Markovnikov's rule, just like with alkenes. The hydrogen adds to the carbon with more hydrogens (or less substituted carbon), and the halogen adds to the more substituted carbon. Since there are two pi bonds, two equivalents of HX can add.
* One equivalent of HX: Forms a vinyl halide.
R-C≡C-H + H-X → R-C(X)=CH2
* Two equivalents of HX: Forms a geminal dihalide (both halogens on the same carbon).
R-C≡C-H + 2H-X → R-C(X)2-CH3
Example:
CH3-C≡CH + 2HBr → CH3-C(Br)2-CH3
(Propyne) (2,2-Dibromopropane)
In this case, both bromines end up on the central carbon because that carbon was the more substituted carbon in the intermediate vinyl bromide (CH3-C(Br)=CH2).
3.2.3 Hydration (Addition of Water, H2O)
Alkynes react with water in the presence of mercuric sulfate (HgSO4) and sulfuric acid (H2SO4) as catalysts. This reaction initially forms an enol (a compound with a double bond and an -OH group directly attached to it). Enols are typically unstable and rapidly tautomerize (rearrange) into more stable ketones or, if it's ethyne, an aldehyde.
R-C≡C-R' + H2O $xrightarrow{ ext{HgSO}_4, ext{H}_2 ext{SO}_4}$ [Enol] ⇌ Ketone (or Aldehyde for ethyne)
Example 1: Hydration of Ethyne (to Aldehyde)
HC≡CH + H2O $xrightarrow{ ext{HgSO}_4, ext{H}_2 ext{SO}_4}$ [CH2=CH-OH] → CH3-CHO
(Ethyne) (Vinyl alcohol - enol) (Acetaldehyde)
Example 2: Hydration of Propyne (to Ketone)
CH3-C≡CH + H2O $xrightarrow{ ext{HgSO}_4, ext{H}_2 ext{SO}_4}$ [CH3-C(OH)=CH2] → CH3-CO-CH3
(Propyne) (Propen-2-ol - enol) (Acetone/Propanone)
For JEE, the hydration of ethyne leading to acetaldehyde, and other terminal alkynes leading to ketones (following Markovnikov's rule for the initial enol formation) is a frequently tested concept.
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### 4. CBSE vs. JEE Focus: What to Emphasize
* CBSE: Focus on understanding *why* terminal alkynes are acidic (s-character explanation). Be able to write reactions for deprotonation and simple alkylation. Understand hydrogenation (complete and partial with Lindlar's/Na-liq. NH3), hydrohalogenation (Markovnikov), and hydration (enol-keto tautomerism, especially for ethyne).
* JEE Mains & Advanced: All the above, but with a stronger emphasis on synthetic applications. Expect multi-step synthesis problems where you need to use alkyne reactions (especially alkylation of acetylides, and selective partial hydrogenation) to build complex molecules. Be ready to predict products and propose reagents for reactions. Understanding reaction mechanisms (which will be covered in the 'Deep Dive' section) is also more important for advanced JEE.
---
And there you have it! The fundamentals of alkyne acidity and their crucial reactions. Alkynes are not just hydrocarbons; they're powerful tools in a chemist's synthetic toolbox. Keep practicing these reactions, and you'll soon be a pro at navigating the world of organic synthesis!
Today, we're going to explore two super important aspects of alkynes: their unique acidity and some of their key reactions. Understanding these fundamentals is crucial, not just for your exams, but for truly grasping organic chemistry. So, let's get started from scratch!
---
### 1. What Makes Alkynes Special? A Quick Recap!
Remember how we categorize hydrocarbons?
* Alkanes: All C-C single bonds, sp3 hybridized carbons.
* Alkenes: At least one C=C double bond, sp2 hybridized carbons.
* Alkynes: At least one C≡C triple bond. The coolest thing about the carbons involved in this triple bond is their hybridization – they are sp hybridized!
This sp hybridization is the secret sauce behind many of alkyne's unique properties, especially its acidity. Each carbon in the triple bond forms one sigma bond and two pi bonds. The remaining two sp hybrid orbitals are used to form sigma bonds with other atoms (like hydrogen or another carbon).
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### 2. The Surprising Acidity of Terminal Alkynes: Why and How!
When we talk about 'acidity' in chemistry, we're generally referring to a compound's ability to donate a proton (H+). Acids taste sour, right? Well, you definitely shouldn't taste alkynes! But they do have an 'acidic' hydrogen.
Important Note: This acidity applies *only* to terminal alkynes.
A terminal alkyne is one where the triple bond is at the end of the carbon chain, meaning one of the triply bonded carbons is attached to a hydrogen atom. For example, ethyne (HC≡CH) or propyne (CH3-C≡CH).
Internal alkynes (like but-2-yne, CH3-C≡C-CH3) do not have this acidic hydrogen, so they are not acidic in the same way.
#### 2.1 The "Why": The Magic of sp Hybridization
So, why are these C-H bonds in terminal alkynes acidic when C-H bonds in alkanes and alkenes are generally considered non-acidic? The answer lies in the hybridization of the carbon atom bonded to that hydrogen!
Let's compare the hybridizations:
* Alkanes: sp3 hybridized carbon (25% s-character, 75% p-character)
* Alkenes: sp2 hybridized carbon (33% s-character, 67% p-character)
* Alkynes: sp hybridized carbon (50% s-character, 50% p-character)
What is 's-character' and why does it matter?
The 's' orbital is spherical and closer to the nucleus than 'p' orbitals. When an orbital has more 's-character', it means the electrons in that orbital spend more time closer to the positively charged nucleus. This makes the orbital, and thus the atom, effectively more electronegative.
Think of it like this:
* Imagine the nucleus as the "boss" of the atom.
* 's' orbitals are like employees who always sit right next to the boss – very close, strong influence.
* 'p' orbitals are like employees who sit a bit further away – less influence.
So, a carbon with 50% s-character (sp) is much better at pulling electron density towards itself than a carbon with 33% s-character (sp2) or 25% s-character (sp3).
How does this relate to acidity?
When a terminal alkyne loses its proton (H+), it forms a conjugate base called an acetylide ion (R-C≡C-).
R-C≡C-H ⇌ R-C≡C- + H+
The more stable this conjugate base (R-C≡C-) is, the more willing the original compound (alkyne) is to donate its proton, and thus, the stronger its acidity.
Because the sp hybridized carbon is highly electronegative, it can effectively stabilize the negative charge on the acetylide ion. It pulls the electron density of the lone pair closer to the nucleus, making the negative charge less "exposed" and thus more stable.
#### 2.2 Quantifying Acidity: pKa Values
Acidity is quantified by pKa values. A lower pKa value means a stronger acid. Let's compare some approximate pKa values:
| Compound Type | Example | Hybridization of C-H carbon | s-character | Approximate pKa | Relative Acidity |
|---|---|---|---|---|---|
| Alkane | CH3-CH3 | sp3 | 25% | ~50 | Very Weak |
| Alkene | CH2=CH2 | sp2 | 33% | ~44 | Weak |
| Ammonia | NH3 | sp3 (N) | - | ~38 | Moderate |
| Terminal Alkyne | HC≡CH | sp | 50% | ~25 | Significantly more acidic than other hydrocarbons |
| Water | H2O | sp3 (O) | - | ~15.7 | Stronger than alkynes |
| Alcohols | CH3OH | sp3 (O) | - | ~16-18 | Stronger than alkynes |
As you can see, a terminal alkyne (pKa ~ 25) is significantly more acidic than an alkane (pKa ~ 50) or an alkene (pKa ~ 44). However, it's still less acidic than water (pKa ~ 15.7) or typical alcohols. This means you need a pretty strong base to deprotonate a terminal alkyne.
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### 3. Important Reactions of Alkynes: Building Blocks of Synthesis!
Alkynes are incredibly versatile in organic synthesis. Let's look at some of their fundamental reactions.
#### 3.1 Reactions Involving Acidity: Forming and Using Acetylide Ions
The acidic nature of terminal alkynes is not just a theoretical concept; it's a powerful tool in synthesis!
3.1.1 Formation of Acetylide Ions (Deprotonation)
Terminal alkynes react with very strong bases to form acetylide anions. Common strong bases include:
* Sodium amide (NaNH2) in liquid ammonia (NH3) - This is the most common and effective base.
* Sodium metal (Na)
* Grignard reagents (RMgX) (e.g., CH3MgBr)
* Alkyl lithium reagents (RLi) (e.g., BuLi)
R-C≡C-H + Base:- → R-C≡C- + Base-H
Example:
CH3-C≡C-H + NaNH2 (in liq. NH3) → CH3-C≡C-Na+ + NH3
(Propyne) (Sodium propynide)
This reaction works because the base (like NH2-) is stronger than the acetylide ion (R-C≡C-), meaning it has a stronger affinity for a proton. Since NH3 (pKa ~ 38) is a weaker acid than the alkyne (pKa ~ 25), NH2- is a stronger base than R-C≡C-, making the equilibrium favor the products.
3.1.2 Alkylation of Acetylide Ions (Carbon-Carbon Bond Formation!)
Once you have an acetylide ion, it's a powerful nucleophile (an electron-rich species that 'loves' positive centers). It can attack electrophiles, especially primary alkyl halides, in an SN2 reaction to form new carbon-carbon bonds! This is super useful for making longer carbon chains.
R-C≡C-Na+ + R'-CH2-X → R-C≡C-CH2-R' + NaX
(Acetylide ion) (Primary alkyl halide) (New internal alkyne)
Example:
Let's say we want to make pent-2-yne (CH3-C≡C-CH2CH3) from propyne.
1. Deprotonate propyne:
CH3-C≡C-H + NaNH2 → CH3-C≡C-Na+ + NH3
2. React with ethyl bromide (a primary alkyl halide):
CH3-C≡C-Na+ + CH3CH2-Br → CH3-C≡C-CH2CH3 + NaBr
(Sodium propynide) (Ethyl bromide) (Pent-2-yne)
This is a fantastic way to extend a carbon chain by adding an alkyl group to the terminal carbon of an alkyne. For JEE, this specific reaction is extremely important for synthesis problems!
#### 3.2 Addition Reactions: From Triple to Double to Single Bonds!
Just like alkenes, alkynes undergo addition reactions because of their reactive pi bonds. But since they have *two* pi bonds, they can add two equivalents of a reagent across the triple bond, potentially converting it to an alkene, and then an alkane.
3.2.1 Hydrogenation (Addition of Hydrogen, H2)
Adding hydrogen to an alkyne can lead to an alkane or an alkene, depending on the catalyst.
* Complete Hydrogenation (to Alkane):
R-C≡C-R' + 2H2 $xrightarrow{ ext{Ni, Pd, or Pt}}$ R-CH2-CH2-R'
This converts the alkyne completely into an alkane. The reaction is fast and generally not very selective if you want to stop at the alkene stage.
* Partial Hydrogenation (to Alkene): This is where it gets interesting and synthetically useful! You can selectively convert an alkyne into an alkene. And even better, you can control the *stereochemistry* (cis or trans)!
* Formation of *cis*-Alkene:
Using a "poisoned" or deactivated catalyst like Lindlar's catalyst (Pd deposited on CaCO3, poisoned with lead salts like lead acetate or quinoline), you can stop the hydrogenation at the alkene stage, and specifically get the cis-alkene.
R-C≡C-R' + H2 $xrightarrow{ ext{Lindlar's catalyst}}$ R
R'(The hydrogen adds from the same side of the triple bond, resulting in *cis* geometry).
* Formation of *trans*-Alkene:
To get the trans-alkene, a different set of reagents is used: sodium (Na) or lithium (Li) in liquid ammonia (liq. NH3). This is called dissolving metal reduction.
R-C≡C-R' + Na/Li $xrightarrow{ ext{liq. NH}_3}$ R
R'(The hydrogen adds from opposite sides, resulting in *trans* geometry).
For JEE, remembering Lindlar's catalyst for *cis* and Na/Li in liq. NH3 for *trans* is critical for synthesis and distinguishing reactions!
3.2.2 Hydrohalogenation (Addition of HX)
Alkynes react with hydrogen halides (HCl, HBr, HI). This reaction follows Markovnikov's rule, just like with alkenes. The hydrogen adds to the carbon with more hydrogens (or less substituted carbon), and the halogen adds to the more substituted carbon. Since there are two pi bonds, two equivalents of HX can add.
* One equivalent of HX: Forms a vinyl halide.
R-C≡C-H + H-X → R-C(X)=CH2
* Two equivalents of HX: Forms a geminal dihalide (both halogens on the same carbon).
R-C≡C-H + 2H-X → R-C(X)2-CH3
Example:
CH3-C≡CH + 2HBr → CH3-C(Br)2-CH3
(Propyne) (2,2-Dibromopropane)
In this case, both bromines end up on the central carbon because that carbon was the more substituted carbon in the intermediate vinyl bromide (CH3-C(Br)=CH2).
3.2.3 Hydration (Addition of Water, H2O)
Alkynes react with water in the presence of mercuric sulfate (HgSO4) and sulfuric acid (H2SO4) as catalysts. This reaction initially forms an enol (a compound with a double bond and an -OH group directly attached to it). Enols are typically unstable and rapidly tautomerize (rearrange) into more stable ketones or, if it's ethyne, an aldehyde.
R-C≡C-R' + H2O $xrightarrow{ ext{HgSO}_4, ext{H}_2 ext{SO}_4}$ [Enol] ⇌ Ketone (or Aldehyde for ethyne)
Example 1: Hydration of Ethyne (to Aldehyde)
HC≡CH + H2O $xrightarrow{ ext{HgSO}_4, ext{H}_2 ext{SO}_4}$ [CH2=CH-OH] → CH3-CHO
(Ethyne) (Vinyl alcohol - enol) (Acetaldehyde)
Example 2: Hydration of Propyne (to Ketone)
CH3-C≡CH + H2O $xrightarrow{ ext{HgSO}_4, ext{H}_2 ext{SO}_4}$ [CH3-C(OH)=CH2] → CH3-CO-CH3
(Propyne) (Propen-2-ol - enol) (Acetone/Propanone)
For JEE, the hydration of ethyne leading to acetaldehyde, and other terminal alkynes leading to ketones (following Markovnikov's rule for the initial enol formation) is a frequently tested concept.
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### 4. CBSE vs. JEE Focus: What to Emphasize
* CBSE: Focus on understanding *why* terminal alkynes are acidic (s-character explanation). Be able to write reactions for deprotonation and simple alkylation. Understand hydrogenation (complete and partial with Lindlar's/Na-liq. NH3), hydrohalogenation (Markovnikov), and hydration (enol-keto tautomerism, especially for ethyne).
* JEE Mains & Advanced: All the above, but with a stronger emphasis on synthetic applications. Expect multi-step synthesis problems where you need to use alkyne reactions (especially alkylation of acetylides, and selective partial hydrogenation) to build complex molecules. Be ready to predict products and propose reagents for reactions. Understanding reaction mechanisms (which will be covered in the 'Deep Dive' section) is also more important for advanced JEE.
---
And there you have it! The fundamentals of alkyne acidity and their crucial reactions. Alkynes are not just hydrocarbons; they're powerful tools in a chemist's synthetic toolbox. Keep practicing these reactions, and you'll soon be a pro at navigating the world of organic synthesis!
🔬
Deep Dive
Welcome, future engineers, to an in-depth exploration of Alkynes! Today, we're diving deep into two fascinating aspects of these hydrocarbons: their unique acidity and the plethora of important reactions they undergo. As you know, alkynes are characterized by the presence of a carbon-carbon triple bond, which bestows upon them distinct chemical properties.
### 1. Acidity of Terminal Alkynes
Let's begin by understanding why certain alkynes behave as acids. When we talk about acidity, we're essentially discussing the ability of a molecule to donate a proton (H$^+$). While hydrocarbons are generally considered non-acidic, terminal alkynes are a notable exception.
#### 1.1. The Role of Hybridization: The Heart of Acidity
The key to the acidity of terminal alkynes lies in the hybridization of the carbon atom involved in the triple bond.
* In alkanes, carbon atoms are sp$^3$ hybridized.
* In alkenes, carbon atoms are sp$^2$ hybridized.
* In alkynes, specifically the carbons involved in the triple bond, they are sp hybridized.
Let's compare the s-character in these hybrid orbitals:
* sp$^3$ orbital: 25% s-character (one s, three p orbitals)
* sp$^2$ orbital: 33% s-character (one s, two p orbitals)
* sp orbital: 50% s-character (one s, one p orbital)
Why is s-character important? The 's' orbital is spherical and closer to the nucleus than 'p' orbitals. Therefore, electrons in an 's' orbital are held more tightly by the positively charged nucleus. A hybrid orbital with a higher s-character means that the electrons in that orbital are, on average, closer to the nucleus and thus more strongly attracted to it.
In a terminal alkyne, the hydrogen atom is directly attached to an sp-hybridized carbon atom (C $equiv$ C-H). Due to the 50% s-character of the sp orbital, the carbon atom in the C-H bond of a terminal alkyne is significantly more electronegative than sp$^2$ or sp$^3$ carbons. This increased electronegativity pulls the shared electrons in the C-H bond closer to the carbon, making the hydrogen more electropositive and thus easier to remove as a proton.
The resulting conjugate base, known as an acetylide anion (RC$equiv$C$^ominus$), is stabilized because the negative charge resides on the highly electronegative sp carbon atom.
#### 1.2. Relative Acidity Comparison
Let's put the acidity of terminal alkynes into perspective by comparing it with other common acidic species:
JEE Focus: Notice that a terminal alkyne (pKa ~25) is more acidic than ammonia (pKa ~38) but less acidic than water (pKa 15.7) or alcohols (pKa 16-18). However, it is significantly more acidic than alkenes or alkanes. This makes it acidic enough to react with strong bases like sodium amide (NaNH$_2$) or Grignard reagents, but not with weaker bases like hydroxide (OH$^-$).
#### 1.3. Reactions Showing Acidity: Formation of Acetylides
Due to their acidity, terminal alkynes can react with strong bases to form metal acetylides:
1. Reaction with Sodium Metal (Na) or Sodium Amide (NaNH$_2$):
These are very strong bases capable of deprotonating terminal alkynes.
Example 1:
CH$_3$C$equiv$CH + NaNH$_2$ $longrightarrow$ CH$_3$C$equiv$C$^ominus$Na$^+$ + NH$_3$
(Propyne) (Sodium Amide) (Sodium Propynide) (Ammonia)
Here, sodium amide, a very strong base, abstracts the acidic proton from propyne, forming sodium propynide and ammonia. The equilibrium lies far to the right because ammonia (pKa ~38) is a much weaker acid than propyne (pKa ~25), meaning the amide ion (NH$_2$$^-$) is a stronger base than the propynide ion.
2RC$equiv$CH + 2Na $longrightarrow$ 2RC$equiv$C$^ominus$Na$^+$ + H$_2$
(Alkyne) (Sodium metal) (Sodium Alkynide) (Hydrogen gas)
2. Reaction with Grignard Reagents (RMgX):
Grignard reagents are strong bases and react with acidic protons.
Example 2:
CH$_3$C$equiv$CH + CH$_3$MgBr $longrightarrow$ CH$_3$C$equiv$C$^ominus$Mg$^+$Br + CH$_4$
(Propyne) (Methyl Magnesium Bromide) (Magnesium Propynide) (Methane)
This reaction is useful for synthesizing longer alkynes or other functional groups by subsequent reaction of the acetylide with an electrophile.
3. Reaction with Heavy Metal Ions (Tests for Terminal Alkynes):
Terminal alkynes react with certain heavy metal ions like Ag$^+$ (from ammoniacal silver nitrate, Tollens' reagent) or Cu$^+$ (from ammoniacal cuprous chloride) to form insoluble, often colored, metal acetylides. These reactions are used as qualitative tests to distinguish terminal alkynes from internal alkynes and alkenes.
* Tollens' Reagent (Ammoniacal Silver Nitrate, [Ag(NH$_3$)$_2$]OH):
RC$equiv$CH + [Ag(NH$_3$)$_2$]OH $longrightarrow$ RC$equiv$CAg$downarrow$ (White precipitate) + H$_2$O + 2NH$_3$
Example 3:
HC$equiv$CH + 2[Ag(NH$_3$)$_2$]OH $longrightarrow$ AgC$equiv$CAg$downarrow$ (White precipitate) + 2H$_2$O + 4NH$_3$
(Acetylene) (Silver Acetylide)
* Ammoniacal Cuprous Chloride (CuCl/NH$_4$OH):
RC$equiv$CH + [Cu(NH$_3$)$_2$]Cl $longrightarrow$ RC$equiv$CCu$downarrow$ (Red precipitate) + NH$_4$Cl + NH$_3$
Example 4:
CH$_3$C$equiv$CH + [Cu(NH$_3$)$_2$]Cl $longrightarrow$ CH$_3$C$equiv$CCu$downarrow$ (Red precipitate) + NH$_4$Cl + NH$_3$
(Propyne) (Cuprous Propynide)
Important Note for JEE: These metal acetylides (especially silver and copper acetylides) are often explosive when dry. They should be handled with extreme care and decomposed immediately after formation. Internal alkynes (e.g., CH$_3$C$equiv$CCH$_3$) do not have an acidic hydrogen and thus do not react with these reagents.
### 2. Important Reactions of Alkynes
Alkynes, due to their electron-rich triple bond (one sigma and two pi bonds), primarily undergo addition reactions. The pi bonds are relatively weak and can be easily broken to form new sigma bonds.
#### 2.1. Addition of Hydrogen (Hydrogenation/Reduction)
Hydrogenation involves the addition of hydrogen across the triple bond.
1. Complete Hydrogenation (to Alkanes):
In the presence of a metal catalyst (like Ni, Pt, or Pd) and excess hydrogen, alkynes are completely reduced to alkanes.
RC$equiv$CR' + 2H$_2$ $xrightarrow{ ext{Ni, Pt, or Pd}}$ RCH$_2$CH$_2$R'
Example 5:
CH$_3$C$equiv$CCH$_3$ + 2H$_2$ $xrightarrow{ ext{Pd/C}}$ CH$_3$CH$_2$CH$_2$CH$_3$
(2-Butyne) (n-Butane)
2. Partial Hydrogenation (to Alkenes):
This is a crucial reaction for synthesizing specific alkene isomers.
* Syn Addition (to *cis*-Alkenes): Lindlar's Catalyst
Lindlar's catalyst is a poisoned palladium catalyst (Pd/CaCO$_3$ with quinoline or sulfur added). The poisons reduce the activity of palladium, preventing further reduction of the alkene to an alkane. The hydrogen atoms add to the same face of the triple bond (syn addition), resulting in the formation of a cis-alkene.
RC$equiv$CR' + H$_2$ $xrightarrow{ ext{Lindlar's Catalyst}}$
(Imagine R and R' on the same side, and H and H on the same side)
Example 6:
CH$_3$C$equiv$CCH$_3$ + H$_2$ $xrightarrow{ ext{Lindlar's Catalyst}}$ *cis*-CH$_3$CH=CHCH$_3$
(2-Butyne) (*cis*-2-Butene)
* Anti Addition (to *trans*-Alkenes): Birch Reduction
The Birch reduction uses sodium or lithium metal in liquid ammonia (Na/Li in liq. NH$_3$) to reduce alkynes to trans-alkenes. This reaction proceeds via a radical anion mechanism, leading to anti addition of hydrogen.
RC$equiv$CR' $xrightarrow{ ext{Na/Li, liq. NH}_3}$
(Imagine R and R' on opposite sides, and H and H on opposite sides)
Example 7:
CH$_3$C$equiv$CCH$_3$ $xrightarrow{ ext{Na, liq. NH}_3}$ *trans*-CH$_3$CH=CHCH$_3$
(2-Butyne) (*trans*-2-Butene)
JEE Focus: Remember the stereoselectivity! Lindlar for *cis*, Birch for *trans*.
#### 2.2. Addition of Halogens (Halogenation)
Alkynes react with halogens (Cl$_2$, Br$_2$) to form tetrahaloalkanes via dihaloalkene intermediates. The reaction typically proceeds via anti-addition.
RC$equiv$CR' + X$_2$ $xrightarrow{ ext{CCl}_4}$ RC(X)=C(X)R' (dihaloalkene, predominantly *trans*)
RC(X)=C(X)R' + X$_2$ $xrightarrow{ ext{CCl}_4}$ RC(X)$_2$C(X)$_2$R' (tetrahaloalkane)
Example 8:
CH$_3$C$equiv$CH + Br$_2$ $xrightarrow{ ext{CCl}_4}$ CH$_3$C(Br)=CHBr (1,2-Dibromopropene)
CH$_3$C(Br)=CHBr + Br$_2$ $xrightarrow{ ext{CCl}_4}$ CH$_3$CBr$_2$CHBr$_2$ (1,1,2,2-Tetrabromopropane)
JEE Focus: Bromine water test: Alkynes, like alkenes, decolorize bromine water due to addition across the triple bond. This is a common test for unsaturation.
#### 2.3. Addition of Hydrogen Halides (Hydrohalogenation)
Alkynes react with HX (HCl, HBr, HI). The addition follows Markovnikov's rule, where the hydrogen atom adds to the carbon of the triple bond that already has more hydrogen atoms (or fewer alkyl substituents), and the halogen adds to the carbon with fewer hydrogens (or more alkyl substituents).
1. Single Addition (to Vinyl Halides):
RC$equiv$CH + HX $longrightarrow$ RC(X)=CH$_2$ (Vinyl Halide)
If the alkyne is symmetric, Markovnikov's rule still applies effectively.
Example 9:
CH$_3$C$equiv$CH + HBr $longrightarrow$ CH$_3$C(Br)=CH$_2$ (2-Bromopropene)
2. Double Addition (to Geminal Dihalides):
With excess HX, a second addition occurs, also following Markovnikov's rule, leading to geminal dihalides (both halogens on the same carbon).
RC(X)=CH$_2$ + HX $longrightarrow$ RC(X)$_2$CH$_3$ (Geminal Dihalide)
Example 10:
CH$_3$C$equiv$CH + 2HBr $longrightarrow$ CH$_3$CBr$_2$CH$_3$ (2,2-Dibromopropane)
3. Anti-Markovnikov Addition (only for HBr with Peroxides):
Similar to alkenes, the addition of HBr to terminal alkynes in the presence of peroxides follows an anti-Markovnikov pathway for the first addition only.
RC$equiv$CH + HBr $xrightarrow{ ext{Peroxides}}$ RCH=CHBr (vinyl bromide, anti-Markovnikov)
Example 11:
CH$_3$C$equiv$CH + HBr $xrightarrow{ ext{Peroxides}}$ CH$_3$CH=CHBr (1-Bromopropene)
JEE Focus: Remember that the peroxide effect is only observed with HBr, not HCl or HI, due to the radical mechanism. Also, it's typically for the first addition. The second addition usually follows Markovnikov's rule regardless of peroxides.
#### 2.4. Addition of Water (Hydration)
The addition of water to alkynes, typically catalyzed by mercuric sulfate (HgSO$_4$) and sulfuric acid (H$_2$SO$_4$), is a very important reaction.
RC$equiv$CR' + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$ R-C(OH)=CR' (Enol) $
ightleftharpoons$ R-CO-CH$_2$R' (Ketone)
This reaction follows Markovnikov's rule, meaning the hydroxyl group (-OH) adds to the more substituted carbon of the triple bond. The initial product is an enol (a compound with a hydroxyl group directly attached to an alkene carbon). Enols are generally unstable and rapidly undergo tautomerization to form more stable keto forms.
* For acetylene (HC$equiv$CH), hydration yields acetaldehyde:
HC$equiv$CH + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$ [CH$_2$=CH-OH] (Vinyl Alcohol, Enol) $
ightleftharpoons$ CH$_3$-CHO (Acetaldehyde)
* For terminal alkynes (RC$equiv$CH), hydration typically yields methyl ketones:
Example 12:
CH$_3$C$equiv$CH + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$ [CH$_3$C(OH)=CH$_2$] (Propen-2-ol, Enol) $
ightleftharpoons$ CH$_3$-CO-CH$_3$ (Acetone)
Notice how the -OH adds to the more substituted carbon (C2), leading to acetone.
* For internal alkynes (RC$equiv$CR'), if symmetric, they give a single ketone. If asymmetric, they can give a mixture of ketones, though one usually predominates according to Markovnikov's rule.
Example 13:
CH$_3$C$equiv$CCH$_2$CH$_3$ (2-Pentyne) + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$
Mixture of: CH$_3$COCH$_2$CH$_2$CH$_3$ (2-Pentanone, major) and CH$_3$CH$_2$COCH$_2$CH$_3$ (3-Pentanone, minor)
Mechanism Insight: The reaction proceeds via electrophilic attack by Hg$^{2+}$ on the triple bond, forming a mercurinium ion, followed by water attack and protonation, eventually leading to the enol. Tautomerism then converts the enol to the more stable keto form.
#### 2.5. Oxidation Reactions
Alkynes are susceptible to oxidation, which can lead to cleavage of the triple bond.
1. Ozonolysis (O$_3$):
Ozonolysis of alkynes cleaves the triple bond. The products depend on the work-up.
* Oxidative Work-up (H$_2$O$_2$): Leads to carboxylic acids. If a terminal alkyne, the terminal carbon is oxidized to CO$_2$.
RC$equiv$CR' $xrightarrow{ ext{1. O}_3 ext{ 2. H}_2 ext{O}_2}$ RCOOH + R'COOH
Example 14:
CH$_3$C$equiv$CCH$_3$ $xrightarrow{ ext{1. O}_3 ext{ 2. H}_2 ext{O}_2}$ 2 CH$_3$COOH (Acetic Acid)
CH$_3$C$equiv$CH $xrightarrow{ ext{1. O}_3 ext{ 2. H}_2 ext{O}_2}$ CH$_3$COOH + CO$_2$
* Reductive Work-up (Zn/H$_2$O or DMS): Less commonly used for alkynes in simple form. Can lead to dicarbonyl compounds if controlled, but often harsh oxidation leads to acids.
2. KMnO$_4$ Oxidation:
* Cold, Dilute, Alkaline KMnO$_4$ (Baeyer's Reagent): Under mild conditions, alkynes are oxidized to 1,2-diones.
RC$equiv$CR' $xrightarrow{ ext{Cold, dil. KMnO}_4}$ R-CO-CO-R' (1,2-Dione)
Example 15:
CH$_3$C$equiv$CCH$_3$ $xrightarrow{ ext{Cold, dil. KMnO}_4}$ CH$_3$CO-COCH$_3$ (2,3-Butanedione)
* Hot, Acidic KMnO$_4$: Under vigorous conditions, the triple bond is completely cleaved, yielding carboxylic acids. For terminal alkynes, the terminal carbon again forms CO$_2$.
RC$equiv$CR' $xrightarrow{ ext{Hot, acidic KMnO}_4}$ RCOOH + R'COOH
RC$equiv$CH $xrightarrow{ ext{Hot, acidic KMnO}_4}$ RCOOH + CO$_2$
Example 16:
CH$_3$CH$_2$C$equiv$CH $xrightarrow{ ext{Hot, acidic KMnO}_4}$ CH$_3$CH$_2$COOH + CO$_2$
(1-Butyne) (Propanoic acid)
#### 2.6. Polymerization Reactions
Alkynes can undergo polymerization, forming larger molecules.
1. Linear Polymerization: Acetylene, under specific conditions, can polymerize linearly to form polyacetylene, a polymer with alternating single and double bonds, which is a conducting polymer.
n HC$equiv$CH $longrightarrow$ (-CH=CH-)$_n$ (Polyacetylene)
2. Cyclic Polymerization (Trimerization):
When passed through a red-hot iron tube (or a copper tube), acetylene undergoes trimerization to form benzene. Other alkynes can also form substituted aromatic compounds.
3 HC$equiv$CH $xrightarrow{ ext{Red hot iron tube, 873K}}$ C$_6$H$_6$ (Benzene)
Example 17:
3 CH$_3$C$equiv$CH (Propyne) $xrightarrow{ ext{Red hot iron tube}}$ 1,3,5-trimethylbenzene (Mesitylene)
### 3. CBSE vs. JEE Focus
* CBSE/Board Level: You should know the concept of acidity of terminal alkynes, be able to write the reactions with NaNH$_2$, Na, and heavy metal ions (AgNO$_3$, CuCl). For reactions, focus on complete hydrogenation, Lindlar's catalyst (cis-alkenes), hydrohalogenation (Markovnikov and anti-Markovnikov for HBr), hydration (forming ketones/acetaldehyde), and the basic outcomes of ozonolysis/KMnO$_4$ oxidation. Cyclic polymerization of acetylene to benzene is also important.
* JEE Mains & Advanced: All the above are fundamental. Additionally, you need to understand the stereochemistry of partial hydrogenation (syn vs. anti), the mechanistic aspects of hydration (enol-keto tautomerism) and hydrohalogenation (carbocation vs. radical intermediates), the relative acidity comparisons with other compounds using pKa values, and the synthetic utility of these reactions in multi-step syntheses. Pay close attention to distinguishing tests for terminal vs. internal alkynes and the conditions required for each reaction. Questions often involve identifying unknown alkynes or predicting products in complex reaction sequences.
This deep dive into alkynes should provide you with a robust foundation for tackling both theoretical questions and problem-solving scenarios. Keep practicing with examples and mechanisms, and you'll master these fascinating compounds!
### 1. Acidity of Terminal Alkynes
Let's begin by understanding why certain alkynes behave as acids. When we talk about acidity, we're essentially discussing the ability of a molecule to donate a proton (H$^+$). While hydrocarbons are generally considered non-acidic, terminal alkynes are a notable exception.
#### 1.1. The Role of Hybridization: The Heart of Acidity
The key to the acidity of terminal alkynes lies in the hybridization of the carbon atom involved in the triple bond.
* In alkanes, carbon atoms are sp$^3$ hybridized.
* In alkenes, carbon atoms are sp$^2$ hybridized.
* In alkynes, specifically the carbons involved in the triple bond, they are sp hybridized.
Let's compare the s-character in these hybrid orbitals:
* sp$^3$ orbital: 25% s-character (one s, three p orbitals)
* sp$^2$ orbital: 33% s-character (one s, two p orbitals)
* sp orbital: 50% s-character (one s, one p orbital)
Why is s-character important? The 's' orbital is spherical and closer to the nucleus than 'p' orbitals. Therefore, electrons in an 's' orbital are held more tightly by the positively charged nucleus. A hybrid orbital with a higher s-character means that the electrons in that orbital are, on average, closer to the nucleus and thus more strongly attracted to it.
In a terminal alkyne, the hydrogen atom is directly attached to an sp-hybridized carbon atom (C $equiv$ C-H). Due to the 50% s-character of the sp orbital, the carbon atom in the C-H bond of a terminal alkyne is significantly more electronegative than sp$^2$ or sp$^3$ carbons. This increased electronegativity pulls the shared electrons in the C-H bond closer to the carbon, making the hydrogen more electropositive and thus easier to remove as a proton.
The resulting conjugate base, known as an acetylide anion (RC$equiv$C$^ominus$), is stabilized because the negative charge resides on the highly electronegative sp carbon atom.
#### 1.2. Relative Acidity Comparison
Let's put the acidity of terminal alkynes into perspective by comparing it with other common acidic species:
| Compound Type | Example | Approximate pKa | Acidity Comment |
|---|---|---|---|
| Alkanes | CH3CH3 | ~50 | Extremely weak acid, practically non-acidic. |
| Alkenes | CH2=CH2 | ~44 | Very weak acid. |
| Ammonia | NH3 | ~38 | Weak acid, stronger than alkenes. |
| Terminal Alkynes | RC≡CH | ~25 | Moderately acidic, stronger than water and alcohols. |
| Alcohols | CH3CH2OH | ~16-18 | Weak acid. |
| Water | H2O | 15.7 | Standard reference for weak acidity. |
| Carboxylic Acids | RCOOH | ~3-5 | Stronger organic acids. |
JEE Focus: Notice that a terminal alkyne (pKa ~25) is more acidic than ammonia (pKa ~38) but less acidic than water (pKa 15.7) or alcohols (pKa 16-18). However, it is significantly more acidic than alkenes or alkanes. This makes it acidic enough to react with strong bases like sodium amide (NaNH$_2$) or Grignard reagents, but not with weaker bases like hydroxide (OH$^-$).
#### 1.3. Reactions Showing Acidity: Formation of Acetylides
Due to their acidity, terminal alkynes can react with strong bases to form metal acetylides:
1. Reaction with Sodium Metal (Na) or Sodium Amide (NaNH$_2$):
These are very strong bases capable of deprotonating terminal alkynes.
Example 1:
CH$_3$C$equiv$CH + NaNH$_2$ $longrightarrow$ CH$_3$C$equiv$C$^ominus$Na$^+$ + NH$_3$
(Propyne) (Sodium Amide) (Sodium Propynide) (Ammonia)
Here, sodium amide, a very strong base, abstracts the acidic proton from propyne, forming sodium propynide and ammonia. The equilibrium lies far to the right because ammonia (pKa ~38) is a much weaker acid than propyne (pKa ~25), meaning the amide ion (NH$_2$$^-$) is a stronger base than the propynide ion.
2RC$equiv$CH + 2Na $longrightarrow$ 2RC$equiv$C$^ominus$Na$^+$ + H$_2$
(Alkyne) (Sodium metal) (Sodium Alkynide) (Hydrogen gas)
2. Reaction with Grignard Reagents (RMgX):
Grignard reagents are strong bases and react with acidic protons.
Example 2:
CH$_3$C$equiv$CH + CH$_3$MgBr $longrightarrow$ CH$_3$C$equiv$C$^ominus$Mg$^+$Br + CH$_4$
(Propyne) (Methyl Magnesium Bromide) (Magnesium Propynide) (Methane)
This reaction is useful for synthesizing longer alkynes or other functional groups by subsequent reaction of the acetylide with an electrophile.
3. Reaction with Heavy Metal Ions (Tests for Terminal Alkynes):
Terminal alkynes react with certain heavy metal ions like Ag$^+$ (from ammoniacal silver nitrate, Tollens' reagent) or Cu$^+$ (from ammoniacal cuprous chloride) to form insoluble, often colored, metal acetylides. These reactions are used as qualitative tests to distinguish terminal alkynes from internal alkynes and alkenes.
* Tollens' Reagent (Ammoniacal Silver Nitrate, [Ag(NH$_3$)$_2$]OH):
RC$equiv$CH + [Ag(NH$_3$)$_2$]OH $longrightarrow$ RC$equiv$CAg$downarrow$ (White precipitate) + H$_2$O + 2NH$_3$
Example 3:
HC$equiv$CH + 2[Ag(NH$_3$)$_2$]OH $longrightarrow$ AgC$equiv$CAg$downarrow$ (White precipitate) + 2H$_2$O + 4NH$_3$
(Acetylene) (Silver Acetylide)
* Ammoniacal Cuprous Chloride (CuCl/NH$_4$OH):
RC$equiv$CH + [Cu(NH$_3$)$_2$]Cl $longrightarrow$ RC$equiv$CCu$downarrow$ (Red precipitate) + NH$_4$Cl + NH$_3$
Example 4:
CH$_3$C$equiv$CH + [Cu(NH$_3$)$_2$]Cl $longrightarrow$ CH$_3$C$equiv$CCu$downarrow$ (Red precipitate) + NH$_4$Cl + NH$_3$
(Propyne) (Cuprous Propynide)
Important Note for JEE: These metal acetylides (especially silver and copper acetylides) are often explosive when dry. They should be handled with extreme care and decomposed immediately after formation. Internal alkynes (e.g., CH$_3$C$equiv$CCH$_3$) do not have an acidic hydrogen and thus do not react with these reagents.
### 2. Important Reactions of Alkynes
Alkynes, due to their electron-rich triple bond (one sigma and two pi bonds), primarily undergo addition reactions. The pi bonds are relatively weak and can be easily broken to form new sigma bonds.
#### 2.1. Addition of Hydrogen (Hydrogenation/Reduction)
Hydrogenation involves the addition of hydrogen across the triple bond.
1. Complete Hydrogenation (to Alkanes):
In the presence of a metal catalyst (like Ni, Pt, or Pd) and excess hydrogen, alkynes are completely reduced to alkanes.
RC$equiv$CR' + 2H$_2$ $xrightarrow{ ext{Ni, Pt, or Pd}}$ RCH$_2$CH$_2$R'
Example 5:
CH$_3$C$equiv$CCH$_3$ + 2H$_2$ $xrightarrow{ ext{Pd/C}}$ CH$_3$CH$_2$CH$_2$CH$_3$
(2-Butyne) (n-Butane)
2. Partial Hydrogenation (to Alkenes):
This is a crucial reaction for synthesizing specific alkene isomers.
* Syn Addition (to *cis*-Alkenes): Lindlar's Catalyst
Lindlar's catalyst is a poisoned palladium catalyst (Pd/CaCO$_3$ with quinoline or sulfur added). The poisons reduce the activity of palladium, preventing further reduction of the alkene to an alkane. The hydrogen atoms add to the same face of the triple bond (syn addition), resulting in the formation of a cis-alkene.
RC$equiv$CR' + H$_2$ $xrightarrow{ ext{Lindlar's Catalyst}}$
(Imagine R and R' on the same side, and H and H on the same side)Example 6:
CH$_3$C$equiv$CCH$_3$ + H$_2$ $xrightarrow{ ext{Lindlar's Catalyst}}$ *cis*-CH$_3$CH=CHCH$_3$
(2-Butyne) (*cis*-2-Butene)
* Anti Addition (to *trans*-Alkenes): Birch Reduction
The Birch reduction uses sodium or lithium metal in liquid ammonia (Na/Li in liq. NH$_3$) to reduce alkynes to trans-alkenes. This reaction proceeds via a radical anion mechanism, leading to anti addition of hydrogen.
RC$equiv$CR' $xrightarrow{ ext{Na/Li, liq. NH}_3}$
(Imagine R and R' on opposite sides, and H and H on opposite sides)Example 7:
CH$_3$C$equiv$CCH$_3$ $xrightarrow{ ext{Na, liq. NH}_3}$ *trans*-CH$_3$CH=CHCH$_3$
(2-Butyne) (*trans*-2-Butene)
JEE Focus: Remember the stereoselectivity! Lindlar for *cis*, Birch for *trans*.
#### 2.2. Addition of Halogens (Halogenation)
Alkynes react with halogens (Cl$_2$, Br$_2$) to form tetrahaloalkanes via dihaloalkene intermediates. The reaction typically proceeds via anti-addition.
RC$equiv$CR' + X$_2$ $xrightarrow{ ext{CCl}_4}$ RC(X)=C(X)R' (dihaloalkene, predominantly *trans*)
RC(X)=C(X)R' + X$_2$ $xrightarrow{ ext{CCl}_4}$ RC(X)$_2$C(X)$_2$R' (tetrahaloalkane)
Example 8:
CH$_3$C$equiv$CH + Br$_2$ $xrightarrow{ ext{CCl}_4}$ CH$_3$C(Br)=CHBr (1,2-Dibromopropene)
CH$_3$C(Br)=CHBr + Br$_2$ $xrightarrow{ ext{CCl}_4}$ CH$_3$CBr$_2$CHBr$_2$ (1,1,2,2-Tetrabromopropane)
JEE Focus: Bromine water test: Alkynes, like alkenes, decolorize bromine water due to addition across the triple bond. This is a common test for unsaturation.
#### 2.3. Addition of Hydrogen Halides (Hydrohalogenation)
Alkynes react with HX (HCl, HBr, HI). The addition follows Markovnikov's rule, where the hydrogen atom adds to the carbon of the triple bond that already has more hydrogen atoms (or fewer alkyl substituents), and the halogen adds to the carbon with fewer hydrogens (or more alkyl substituents).
1. Single Addition (to Vinyl Halides):
RC$equiv$CH + HX $longrightarrow$ RC(X)=CH$_2$ (Vinyl Halide)
If the alkyne is symmetric, Markovnikov's rule still applies effectively.
Example 9:
CH$_3$C$equiv$CH + HBr $longrightarrow$ CH$_3$C(Br)=CH$_2$ (2-Bromopropene)
2. Double Addition (to Geminal Dihalides):
With excess HX, a second addition occurs, also following Markovnikov's rule, leading to geminal dihalides (both halogens on the same carbon).
RC(X)=CH$_2$ + HX $longrightarrow$ RC(X)$_2$CH$_3$ (Geminal Dihalide)
Example 10:
CH$_3$C$equiv$CH + 2HBr $longrightarrow$ CH$_3$CBr$_2$CH$_3$ (2,2-Dibromopropane)
3. Anti-Markovnikov Addition (only for HBr with Peroxides):
Similar to alkenes, the addition of HBr to terminal alkynes in the presence of peroxides follows an anti-Markovnikov pathway for the first addition only.
RC$equiv$CH + HBr $xrightarrow{ ext{Peroxides}}$ RCH=CHBr (vinyl bromide, anti-Markovnikov)
Example 11:
CH$_3$C$equiv$CH + HBr $xrightarrow{ ext{Peroxides}}$ CH$_3$CH=CHBr (1-Bromopropene)
JEE Focus: Remember that the peroxide effect is only observed with HBr, not HCl or HI, due to the radical mechanism. Also, it's typically for the first addition. The second addition usually follows Markovnikov's rule regardless of peroxides.
#### 2.4. Addition of Water (Hydration)
The addition of water to alkynes, typically catalyzed by mercuric sulfate (HgSO$_4$) and sulfuric acid (H$_2$SO$_4$), is a very important reaction.
RC$equiv$CR' + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$ R-C(OH)=CR' (Enol) $
ightleftharpoons$ R-CO-CH$_2$R' (Ketone)
This reaction follows Markovnikov's rule, meaning the hydroxyl group (-OH) adds to the more substituted carbon of the triple bond. The initial product is an enol (a compound with a hydroxyl group directly attached to an alkene carbon). Enols are generally unstable and rapidly undergo tautomerization to form more stable keto forms.
* For acetylene (HC$equiv$CH), hydration yields acetaldehyde:
HC$equiv$CH + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$ [CH$_2$=CH-OH] (Vinyl Alcohol, Enol) $
ightleftharpoons$ CH$_3$-CHO (Acetaldehyde)
* For terminal alkynes (RC$equiv$CH), hydration typically yields methyl ketones:
Example 12:
CH$_3$C$equiv$CH + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$ [CH$_3$C(OH)=CH$_2$] (Propen-2-ol, Enol) $
ightleftharpoons$ CH$_3$-CO-CH$_3$ (Acetone)
Notice how the -OH adds to the more substituted carbon (C2), leading to acetone.
* For internal alkynes (RC$equiv$CR'), if symmetric, they give a single ketone. If asymmetric, they can give a mixture of ketones, though one usually predominates according to Markovnikov's rule.
Example 13:
CH$_3$C$equiv$CCH$_2$CH$_3$ (2-Pentyne) + H$_2$O $xrightarrow{ ext{H}_2 ext{SO}_4, ext{HgSO}_4}$
Mixture of: CH$_3$COCH$_2$CH$_2$CH$_3$ (2-Pentanone, major) and CH$_3$CH$_2$COCH$_2$CH$_3$ (3-Pentanone, minor)
Mechanism Insight: The reaction proceeds via electrophilic attack by Hg$^{2+}$ on the triple bond, forming a mercurinium ion, followed by water attack and protonation, eventually leading to the enol. Tautomerism then converts the enol to the more stable keto form.
#### 2.5. Oxidation Reactions
Alkynes are susceptible to oxidation, which can lead to cleavage of the triple bond.
1. Ozonolysis (O$_3$):
Ozonolysis of alkynes cleaves the triple bond. The products depend on the work-up.
* Oxidative Work-up (H$_2$O$_2$): Leads to carboxylic acids. If a terminal alkyne, the terminal carbon is oxidized to CO$_2$.
RC$equiv$CR' $xrightarrow{ ext{1. O}_3 ext{ 2. H}_2 ext{O}_2}$ RCOOH + R'COOH
Example 14:
CH$_3$C$equiv$CCH$_3$ $xrightarrow{ ext{1. O}_3 ext{ 2. H}_2 ext{O}_2}$ 2 CH$_3$COOH (Acetic Acid)
CH$_3$C$equiv$CH $xrightarrow{ ext{1. O}_3 ext{ 2. H}_2 ext{O}_2}$ CH$_3$COOH + CO$_2$
* Reductive Work-up (Zn/H$_2$O or DMS): Less commonly used for alkynes in simple form. Can lead to dicarbonyl compounds if controlled, but often harsh oxidation leads to acids.
2. KMnO$_4$ Oxidation:
* Cold, Dilute, Alkaline KMnO$_4$ (Baeyer's Reagent): Under mild conditions, alkynes are oxidized to 1,2-diones.
RC$equiv$CR' $xrightarrow{ ext{Cold, dil. KMnO}_4}$ R-CO-CO-R' (1,2-Dione)
Example 15:
CH$_3$C$equiv$CCH$_3$ $xrightarrow{ ext{Cold, dil. KMnO}_4}$ CH$_3$CO-COCH$_3$ (2,3-Butanedione)
* Hot, Acidic KMnO$_4$: Under vigorous conditions, the triple bond is completely cleaved, yielding carboxylic acids. For terminal alkynes, the terminal carbon again forms CO$_2$.
RC$equiv$CR' $xrightarrow{ ext{Hot, acidic KMnO}_4}$ RCOOH + R'COOH
RC$equiv$CH $xrightarrow{ ext{Hot, acidic KMnO}_4}$ RCOOH + CO$_2$
Example 16:
CH$_3$CH$_2$C$equiv$CH $xrightarrow{ ext{Hot, acidic KMnO}_4}$ CH$_3$CH$_2$COOH + CO$_2$
(1-Butyne) (Propanoic acid)
#### 2.6. Polymerization Reactions
Alkynes can undergo polymerization, forming larger molecules.
1. Linear Polymerization: Acetylene, under specific conditions, can polymerize linearly to form polyacetylene, a polymer with alternating single and double bonds, which is a conducting polymer.
n HC$equiv$CH $longrightarrow$ (-CH=CH-)$_n$ (Polyacetylene)
2. Cyclic Polymerization (Trimerization):
When passed through a red-hot iron tube (or a copper tube), acetylene undergoes trimerization to form benzene. Other alkynes can also form substituted aromatic compounds.
3 HC$equiv$CH $xrightarrow{ ext{Red hot iron tube, 873K}}$ C$_6$H$_6$ (Benzene)
Example 17:
3 CH$_3$C$equiv$CH (Propyne) $xrightarrow{ ext{Red hot iron tube}}$ 1,3,5-trimethylbenzene (Mesitylene)
### 3. CBSE vs. JEE Focus
* CBSE/Board Level: You should know the concept of acidity of terminal alkynes, be able to write the reactions with NaNH$_2$, Na, and heavy metal ions (AgNO$_3$, CuCl). For reactions, focus on complete hydrogenation, Lindlar's catalyst (cis-alkenes), hydrohalogenation (Markovnikov and anti-Markovnikov for HBr), hydration (forming ketones/acetaldehyde), and the basic outcomes of ozonolysis/KMnO$_4$ oxidation. Cyclic polymerization of acetylene to benzene is also important.
* JEE Mains & Advanced: All the above are fundamental. Additionally, you need to understand the stereochemistry of partial hydrogenation (syn vs. anti), the mechanistic aspects of hydration (enol-keto tautomerism) and hydrohalogenation (carbocation vs. radical intermediates), the relative acidity comparisons with other compounds using pKa values, and the synthetic utility of these reactions in multi-step syntheses. Pay close attention to distinguishing tests for terminal vs. internal alkynes and the conditions required for each reaction. Questions often involve identifying unknown alkynes or predicting products in complex reaction sequences.
This deep dive into alkynes should provide you with a robust foundation for tackling both theoretical questions and problem-solving scenarios. Keep practicing with examples and mechanisms, and you'll master these fascinating compounds!
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Shortcuts
Here are some useful mnemonics and shortcuts to help you remember the key concepts and reactions related to the acidity and important reactions of alkynes, crucial for both JEE and CBSE exams.
Mnemonics & Shortcuts for Alkynes
Mastering alkynes involves understanding their unique acidity and diverse reaction pathways. These mnemonics will help you quickly recall critical information.
1. Acidity of Terminal Alkynes
- Concept: Terminal alkynes (R-C≡C-H) are acidic because the sp-hybridized carbon atom has high s-character (50%), which makes the C-H bond stronger and the hydrogen more easily removable as a proton.
- Mnemonic: "S-P for Strong Proton"
- S-P hybridisation (of terminal alkyne carbon) means Stronger pull on electrons, making the Proton (H) more acidic.
- Alternatively: "50% S means Strong Acidity" (referring to 50% s-character).
- Reaction Shortcut: Terminal alkynes react with strong bases like NaNH₂, Grignard reagents (RMgX), or even Na metal to form acetylide anions (R-C≡C⁻), liberating H₂ or alkanes. Internal alkynes do not react in this way.
2. Important Reactions of Alkynes
A. Reduction (Hydrogenation)
The type of alkene formed depends on the catalyst used.
- Complete Hydrogenation (Alkane formation):
- Catalyst: H₂/Pd, Pt, or Ni.
- Shortcut: "All Metals, All Gone!" (All pi bonds are gone, forming an alkane).
- Partial Hydrogenation to *cis*-Alkene:
- Catalyst: Lindlar's Catalyst (Pd/CaCO₃, quinoline/BaSO₄).
- Mnemonic: "Lindlar's Loves Limited, Linked (Cis) Addition."
- The 'L' sound helps remember Lindlar and 'Linked' (cis).
- Partial Hydrogenation to *trans*-Alkene:
- Reagent: Birch Reduction (Na or Li in liquid NH₃).
- Mnemonic: "Birch Brings Broad (Trans) Products."
- Think of 'Birch' as a tree spreading its branches broadly, like the hydrogens on opposite sides of the double bond (trans).
B. Addition of HX (Hydrohalogenation)
- Rule: Follows Markovnikov's Rule for unsymmetrical alkynes.
- Mnemonic: "Markovnikov's Means More Hydrogens on the Hydrogen-rich carbon."
- The hydrogen atom of HX adds to the carbon atom of the triple bond that already has more hydrogen atoms.
- For alkynes, two moles of HX can add, leading to geminal dihalides (two halogens on the same carbon).
C. Hydration (Addition of H₂O)
- Reagents: H₂O, H₂SO₄, HgSO₄.
- Process: Forms an enol (unstable), which rapidly tautomerizes to a more stable ketone (or aldehyde for acetylene).
- Mnemonic: "Alkyne + Water-Mercury-Acid = Enol-Keto Tautomerization."
- Think of it as a direct route to the carbonyl compound.
- For acetylene (ethyne), it gives acetaldehyde (an aldehyde). For all other terminal alkynes, it gives methyl ketones.
D. Oxidation with KMnO₄
- Mild Oxidation (Cold, Dilute, Alkaline KMnO₄ - Baeyer's Reagent):
- Products: Typically diketones or diones via enol intermediates.
- Shortcut: "Cold KMnO₄ adds two 'O's (to make C=O, C=O bonds after rearrangement)."
- Strong Oxidation (Hot, Concentrated KMnO₄):
- Products: Cleavage of the triple bond, forming carboxylic acids. If a terminal alkyne, the terminal carbon becomes CO₂.
- Mnemonic: "Hot Concentrated KMnO₄ Cuts and Carboxylates."
- Cuts the C≡C bond and oxidizes each fragment to a carboxylic acid (or CO₂ for terminal H).
E. Ozonolysis (O₃, followed by H₂O)
- Products: Cleavage of the triple bond, forming carboxylic acids. For terminal alkynes, the terminal carbon forms CO₂.
- Shortcut: "Ozonolysis Clearly Cuts the triple bond and forms Carboxylic acids (or CO₂)."
- It's similar to strong KMnO₄ oxidation in terms of cleavage products.
Keep practicing these reactions with these memory aids, and you'll find recalling them much easier in your exams!
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Quick Tips
Here are some quick, exam-focused tips on the acidity and important reactions of alkynes, particularly useful for JEE and board exams:
1. Acidity of Terminal Alkynes
- Reason for Acidity: Terminal alkynes (R-C≡C-H) are acidic due to the sp-hybridization of the carbon atom bonded to hydrogen. sp-hybrid orbitals have 50% s-character, making them more electronegative than sp2 (33% s-character) or sp3 (25% s-character) orbitals. This greater electronegativity stabilizes the negative charge on the carbon atom of the acetylide anion (R-C≡C-), making the proton easier to remove.
- Acidity Order: Terminal Alkynes > Water > Alcohols > Alkenes > Alkanes. (Note: Terminal alkynes are weaker acids than carboxylic acids or phenols).
- Reactions Demonstrating Acidity (JEE Focus):
- React with strong bases like NaNH2 (sodamide), Na metal, or Grignard reagents to form acetylide anions (R-C≡C-Na+). This is a crucial step for subsequent alkylation.
- Form insoluble heavy metal acetylides (precipitates) with reagents like Tollens' reagent [Ag(NH3)2]+OH- (forms silver acetylide, a white precipitate) or ammoniacal cuprous chloride (CuCl/NH4OH) (forms red precipitate of cuprous acetylide). This reaction is a characteristic test to distinguish terminal alkynes from internal alkynes and other hydrocarbons.
2. Important Addition Reactions
- Catalytic Hydrogenation (Reduction):
- Complete Reduction: Alkyne + 2H2 (Ni, Pt, or Pd) → Alkane.
- Partial Reduction to Alkenes:
- cis-Alkene: Alkyne + H2 (Lindlar's Catalyst: Pd/CaCO3 poisoned with quinoline or BaSO4) → cis-Alkene.
- trans-Alkene: Alkyne + Na or Li (liquid NH3 at low temp) → trans-Alkene (known as Birch reduction for alkynes).
- Halogenation (e.g., with Br2/CCl4):
- Addition of 1 mole Br2 → trans-dihaloalkene.
- Addition of 2 moles Br2 → tetrahaloalkane. (Decolorizes bromine water, a test for unsaturation).
- Hydrohalogenation (Addition of HX):
- Follows Markovnikov's Rule. The H adds to the carbon with more hydrogens, and X to the carbon with fewer hydrogens.
- Addition of 1 mole HX → Vinyl halide.
- Addition of 2 moles HX → Geminal dihalide (both halogens on the same carbon).
- Anti-Markovnikov Addition (JEE Specific): With HBr in the presence of peroxides (HBr/ROOR), a terminal alkyne can give anti-Markovnikov addition, leading to a primary vinyl bromide.
- Hydration (Addition of H2O):
- Reagents: HgSO4/dil. H2SO4.
- Follows Markovnikov's Rule, forming an enol intermediate.
- The enol quickly undergoes tautomerization to form a more stable keto form (ketone) or aldehyde.
- Ethyne (acetylene) is unique: H-C≡C-H → H2C=CH-OH (enol) → CH3CHO (acetaldehyde).
- Higher alkynes (e.g., propyne) yield ketones (e.g., propyne → acetone).
3. Oxidation Reactions
- KMnO4 Oxidation:
- Cold, dilute, neutral/alkaline KMnO4 (Baeyer's Reagent): Forms 1,2-diones (vicinal dicarbonyl compounds).
- Hot, acidic KMnO4: Cleaves the triple bond, forming carboxylic acids. If the alkyne is terminal, the terminal carbon oxidizes to CO2 and H2O.
- Ozonolysis (O3):
- Oxidative workup (O3/H2O): Cleaves the triple bond, forming carboxylic acids. Terminal alkynes give CO2 from the terminal carbon.
- Reductive workup (O3/Zn/H2O): Less common for alkynes compared to alkenes, but can yield dicarbonyl compounds or carboxylic acids depending on the alkyne structure. Oxidative workup is more standard for alkynes.
4. Alkylation of Acetylides
- Acetylide anions (R-C≡C-) are strong nucleophiles.
- They react with primary alkyl halides (R'-CH2-X) via an SN2 mechanism to form new C-C bonds, yielding higher alkynes.
- Caution: Avoid secondary or tertiary alkyl halides, as these will primarily undergo E2 elimination due to the strong basicity of the acetylide anion.
5. Polymerization Reactions
- Linear Polymerization: Ethyne can polymerize to form polyacetylene, an electrically conductive polymer.
- Cyclic Polymerization: Three molecules of ethyne pass through a red-hot iron tube to form benzene. This is an important synthesis method for aromatic compounds.
Master these reactions and their specific reagents to confidently tackle alkyne-related questions in your exams!
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Intuitive Understanding
Welcome to the Intuitive Understanding section on Alkynes' acidity and important reactions! Here, we'll grasp the 'why' behind these chemical properties, making them easier to remember and apply.
Intuitive Understanding: Acidity of Terminal Alkynes
Terminal alkynes (those with a triple bond at the end of a carbon chain, e.g., R-C≡C-H) exhibit mild acidity, which is a unique property among hydrocarbons. But why?
- The 'Why' – Electronegativity of sp Carbon:
- Carbon atoms in a triple bond are sp hybridized. An sp orbital has 50% s-character and 50% p-character.
- Compare this to sp² hybridized carbons (alkenes, 33% s-character) and sp³ hybridized carbons (alkanes, 25% s-character).
- Intuitive Concept: The more 's' character an orbital has, the closer its electrons are, on average, to the nucleus. This makes the sp hybridized carbon behave like a more electronegative atom.
- Because the sp carbon is more electronegative, it pulls the shared electrons in the C-H bond towards itself more strongly. This weakens the C-H bond and makes the hydrogen relatively easier to remove as a proton (H+).
- Conjugate Base Stability:
- When a terminal alkyne loses its proton, it forms an acetylide ion (R-C≡C-). The negative charge resides on the sp hybridized carbon.
- Intuitive Concept: Due to the higher electronegativity of the sp carbon, it is better able to accommodate and stabilize this negative charge compared to sp² or sp³ carbons. A more stable conjugate base means a stronger acid.
- Comparison (JEE & CBSE Relevance):
Hydrocarbon Type
Hybridization of C-H carbon
s-character
Acidity
Alkyne (terminal)
sp
50%
Most acidic
Alkene
sp²
33%
Less acidic
Alkane
sp³
25%
Least acidic
This difference in acidity allows terminal alkynes to react with strong bases (like NaNH₂ or Na metal) to form acetylide salts, which are important in synthetic chemistry.
Intuitive Understanding: Important Reactions of Alkynes
Alkynes, with their electron-rich triple bond, undergo various addition reactions, but also have specific characteristics.
- Reactions Involving Acidity: Formation of Acetylides
- The 'Why': As discussed, the terminal hydrogen is acidic. Strong bases 'pluck' this proton away, forming a stable acetylide anion. This anion is a good nucleophile and can be used to form new C-C bonds (e.g., with alkyl halides), making longer carbon chains.
- Example: R-C≡C-H + NaNH₂ → R-C≡C-Na+ + NH₃ (acid-base reaction).
- Electrophilic Addition Reactions
- The 'Why': The triple bond has a high electron density (two pi bonds). Electrophiles (electron-seeking species like H+, Br+) are attracted to this electron richness, breaking one of the pi bonds to form new sigma bonds.
- Key Feature: Addition can happen twice, converting a triple bond to a double bond, and then to a single bond.
- Markovnikov's Rule (JEE & CBSE): In additions of unsymmetrical reagents (e.g., H-X, H-OH) to unsymmetrical alkynes, the positive part (H) adds to the carbon of the triple bond that already has more hydrogens. This happens because it leads to a more stable vinylic carbocation intermediate.
- Example (HBr addition): Propyne (CH₃-C≡C-H) + HBr → CH₃-C(Br)=CH₂ (following Markovnikov).
- Hydrogenation (Reduction)
- The 'Why': Breaking pi bonds and forming stronger sigma bonds with hydrogen is energetically favorable. Catalysts (Pd, Pt, Ni) lower the activation energy by adsorbing hydrogen and the alkyne, facilitating the addition.
- Controlled Reduction (JEE & CBSE):
- Complete Reduction: H₂/Pd (or Pt, Ni) converts alkyne → alkene → alkane. The catalyst is very active.
- Partial Reduction (to Alkene):
- Lindlar's Catalyst (H₂/Pd-BaSO₄/quinoline): A 'poisoned' catalyst that is less active, stopping the reduction at the alkene stage. Critically, it gives cis-alkene because both hydrogens add from the same face of the triple bond, adsorbed on the catalyst surface (syn addition).
- Na/Liquid NH₃: Also stops at the alkene stage, but gives trans-alkene. This reaction proceeds via radical anions, where the hydrogens add from opposite faces (anti addition).
- Oxidation Reactions
- The 'Why': Strong oxidizing agents (like KMnO₄, O₃) break the pi bonds, and sometimes even the sigma bond, replacing them with bonds to oxygen atoms. This leads to the formation of carboxylic acids or ketones, depending on the alkyne and conditions.
Understanding these underlying principles will help you predict reactions and apply concepts more effectively in exams!
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Real World Applications
Alkynes, with their unique acidity (especially terminal alkynes) and diverse reactivity, play a crucial role in various industrial and synthetic applications. Understanding these real-world uses helps to appreciate the practical significance of alkyne chemistry.
Real-World Applications of Alkynes
Synthesis of Complex Molecules (Leveraging Acidity):
Terminal alkynes are significantly acidic due to the s-character of the sp-hybridized carbon atom, allowing them to form acetylide anions. These anions are powerful nucleophiles and are extensively used in carbon-carbon bond forming reactions. This is vital in the synthesis of pharmaceuticals, agrochemicals, and natural products. For example, Sonogashira coupling reactions, which involve cross-coupling of terminal alkynes with aryl or vinyl halides, are indispensable for creating highly functionalized molecules, including precursors for drugs like certain anti-cancer agents and HIV protease inhibitors.
Industrial Production of Acetaldehyde (Hydration Reaction):
The hydration of acetylene (ethyne) is a classical industrial method for producing acetaldehyde. This reaction typically occurs in the presence of mercuric sulfate and sulfuric acid as catalysts. Acetaldehyde is a critical intermediate in the chemical industry, used in the manufacture of acetic acid, acetic anhydride, ethanol, and various resins. This showcases the utility of Markovnikov addition of water to alkynes.
Production of Electrically Conductive Polymers (Polymerization):
Acetylene can undergo linear polymerization to form polyacetylene, an organic polymer that exhibits electrical conductivity when doped. This unique property makes polyacetylene a subject of interest in materials science for applications such as organic light-emitting diodes (OLEDs), flexible electronics, sensors, and lightweight batteries. The discovery of conductive polymers, including polyacetylene, led to the Nobel Prize in Chemistry in 2000.
Synthesis of Vitamin A and Other Fine Chemicals (Partial Hydrogenation):
Controlled or partial hydrogenation of alkynes using specific catalysts like Lindlar's catalyst (palladium on calcium carbonate poisoned with lead acetate and quinoline) allows for the selective formation of cis-alkenes. This stereoselective reaction is crucial in the synthesis of compounds where the specific geometry of the double bond is critical, such as Vitamin A, pheromones, and various flavors and fragrances. These are high-value products in the fine chemical industry.
Oxyacetylene Welding (Combustion of Acetylene):
While not an "alkyne reaction" in the synthetic sense, the high exothermicity of acetylene's combustion is fundamental to its most well-known industrial application: oxyacetylene welding and cutting. The flame produced by mixing acetylene with oxygen reaches temperatures exceeding 3300 °C, sufficient to melt and join metals, making it an essential tool in fabrication and repair industries.
These examples highlight how the distinct chemical properties of alkynes, particularly their acidity and reactivity towards addition and polymerization, are leveraged across various sectors, from advanced materials to everyday industrial processes. For JEE and CBSE, while the detailed mechanisms are important, knowing these applications provides a broader perspective and context to the chemical principles studied.
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Common Analogies
Common Analogies for Alkynes: Acidity and Important Reactions
Understanding the unique properties of alkynes, particularly their acidity, can be made easier through simple analogies. These help to conceptualize why terminal alkynes behave differently from other hydrocarbons.
1. Acidity of Terminal Alkynes: The "Magnetic Pull" Analogy
The acidity of a terminal alkyne (R-C≡C-H) is its most distinguishing feature, stemming from the sp-hybridization of the carbon atom attached to the acidic hydrogen. Let's understand this using the "Magnetic Pull" analogy:
- Imagine the carbon atom's nucleus as a magnet, and the s-orbital electrons as having a stronger magnetic attraction compared to p-orbital electrons.
- Alkanes (sp3): Have 25% s-character. The carbon's "magnetic pull" on the C-H bond electrons is relatively weak (like a small magnet).
- Alkenes (sp2): Have 33% s-character. The carbon's "magnetic pull" is slightly stronger (a medium magnet).
- Terminal Alkynes (sp): Have 50% s-character. This carbon acts like a significantly stronger magnet. It pulls the shared electrons in the C-H bond much closer to itself.
Result: Because the sp-hybridized carbon pulls the C-H bond electrons so strongly, the hydrogen atom becomes relatively positive (electrophilic) and the C-H bond is weakened. It's like the hydrogen's electron is already halfway attracted to the carbon's "strong magnet," making it much easier for a strong base to "snatch away" the hydrogen as a proton (H+), leaving behind a stable acetylide anion (R-C≡C-). This enhanced stability of the negative charge on the sp-hybridized carbon is due to the electron-withdrawing nature of the s-character.
JEE Tip: This analogy helps explain why pKa values for terminal alkynes (~25) are significantly lower than alkenes (~44) and alkanes (~50), making them much stronger acids relative to other hydrocarbons.
2. Reaction with Strong Bases & Heavy Metal Ions: The "VIP Pass" Analogy
The acidic proton of terminal alkynes grants them a "Very Important Property" (VIP) pass, allowing them to participate in reactions that alkanes and alkenes cannot:
- Strong Bases (e.g., NaNH2, NaH): These are like bouncers at an exclusive club. They recognize the "VIP pass" (acidic proton) of terminal alkynes and immediately allow them entry by deprotonating them to form acetylide anions. Alkanes and alkenes lack this "pass" and are denied entry (do not react).
- Heavy Metal Ions (e.g., Ag+ from AgNO3, Cu+ from CuCl): Imagine these metal ions as "collectors" of unique and valuable items. The acetylide anion (formed from the acidic proton) is such a valuable item to them. They readily form insoluble, often colored, metal acetylides (precipitates). This reaction is a key distinguishing test for terminal alkynes in both CBSE and JEE practical organic chemistry. Alkanes and alkenes do not have this "collectible" feature.
This "VIP Pass" analogy helps to remember that the acidity is not just a theoretical concept but directly dictates unique and important reactions that are often tested in exams.
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Prerequisites
To effectively grasp the acidity and important reactions of alkynes, a strong foundation in several core chemistry concepts is essential. These prerequisites ensure you can understand the underlying principles and mechanisms, rather than just memorizing facts.
Prerequisites for Alkynes: Acidity and Important Reactions
- Basic Organic Nomenclature:
- Ability to name and draw structures of alkanes, alkenes, and simple alkynes. This forms the basis for identifying the compounds involved in reactions.
- Hybridization and Molecular Geometry:
- Understanding of sp, sp2, and sp3 hybridization. Specifically, the concept of sp hybridization in alkynes, leading to linear geometry around the triple bond.
- Knowledge of how hybridization affects bond length, bond strength, and especially s-character.
JEE Focus: A deeper understanding of s-character's implications on electronegativity is crucial for alkyne acidity.
- Electronegativity:
- Definition of electronegativity and how it varies across the periodic table.
- Its role in determining bond polarity and the stability of charges.
- Concepts of Acidity and Basicity:
- Brønsted-Lowry definition: Acids as proton donors, bases as proton acceptors.
- Understanding of pKa values and their relation to acid strength (lower pKa = stronger acid).
- Factors affecting the stability of conjugate bases (charge localization, electronegativity of the atom bearing the charge, inductive effects, resonance). This is fundamental to understanding terminal alkyne acidity.
- Fundamental Reaction Mechanisms:
- Electrophilic Addition: Basic understanding of how alkenes react with electrophiles (e.g., HBr, H2O in presence of acid). Alkynes undergo similar but distinct reactions.
- Nucleophilic Substitution (SN1/SN2) & Elimination (E1/E2): While not directly applied to alkyne reactions as primary mechanisms, a general understanding of these provides context for other organic reactions and understanding how functional groups are interconverted.
- Understanding of carbocations and carbanions as intermediates, and their relative stabilities.
- Inductive Effect:
- Basic understanding of electron-donating and electron-withdrawing groups and how they influence electron density in a molecule. This plays a role in stabilizing or destabilizing charged intermediates.
- Hydroboration-Oxidation (for Alkenes):
- A prior understanding of this anti-Markovnikov hydration for alkenes will make understanding its application to alkynes (leading to aldehydes) much easier.
By reviewing these foundational topics, you will be well-equipped to tackle the specific nuances of alkyne acidity and their diverse range of reactions, including those involving Markovnikov and anti-Markovnikov additions, and synthetic applications.
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Common Exam Traps
⚠ Common Exam Traps in Alkynes: Acidity & Reactions ⚠
Students often lose marks in alkyne-related questions due to common misconceptions and overlooking subtle details. Be vigilant about the following traps:
Trap 1: Confusing Terminal vs. Internal Alkyne Acidity.
- Mistake: Assuming all alkynes possess acidic properties.
- Correction: Only terminal alkynes (R-C≡C-H) are acidic. This is due to the high s-character (50%) of the sp-hybridized carbon, making the C-H bond more polar and the hydrogen more easily removable. Internal alkynes (R-C≡C-R') lack this acidic hydrogen and are non-acidic.
Trap 2: Incorrect Base Selection for Deprotonation.
- Mistake: Attempting to deprotonate terminal alkynes with weak bases like NaOH or NaHCO3.
- Correction: Terminal alkynes are very weak acids (pKa ~25). Stronger bases are required to form acetylides (alkynides). Common strong bases include:
- Sodium amide (NaNH2)
- Alkyl lithium (RLi)
- Grignard reagents (RMgX)
- Sodium hydride (NaH)
Trap 3: Ignoring Enol-Keto Tautomerism in Hydration.
- Mistake: Providing the enol as the final product for alkyne hydration (e.g., using H2SO4/HgSO4).
- Correction: The initial enol formed (e.g., vinyl alcohol) is unstable and rapidly undergoes keto-enol tautomerism to form the more stable ketone or aldehyde. Always show the keto form as the final product.
Example: R-C≡CH + H2O --(H2SO4/HgSO4)--> [R-C(OH)=CH2] (enol) --> R-CO-CH3 (ketone).
For ethyne: CH≡CH + H2O --(H2SO4/HgSO4)--> [CH2=CHOH] (enol) --> CH3CHO (acetaldehyde).
Trap 4: Misapplication of Markovnikov's Rule & Stereochemistry in Addition.
- Mistake (Hydrohalogenation, HX): Forgetting the number of equivalents or the anti-Markovnikov exception.
- Correction:
- 1 equivalent HX: Forms a vinyl halide (Markovnikov addition).
- 2 equivalents HX: Forms a geminal dihalide (Markovnikov for both additions, typically).
- HBr in presence of peroxides: Leads to anti-Markovnikov addition via a radical mechanism.
- Mistake (Hydrogenation, H2): Not distinguishing between catalysts for *cis* and *trans* alkene formation, or complete reduction.
- Correction:
Reagent
Product
H2/Pd, Pt, Ni Alkynes → Alkanes (complete reduction)
Lindlar's catalyst (Pd/CaCO3/quinoline) or P-2 catalyst (Ni2B) Alkynes → cis-Alkenes (partial, stereoselective)
Na/liquid NH3 (Birch reduction) Alkynes → trans-Alkenes (partial, stereoselective)
Trap 5: Incorrect Products for Oxidation Reactions.
- Mistake: Not differentiating between mild and strong oxidation conditions, or confusing products of terminal vs. internal alkynes.
- Correction:
- Hot Acidified KMnO4 or Hot K2Cr2O7 (Strong Oxidation): Causes cleavage of the triple bond.
- Terminal Alkynes (R-C≡CH): Yields a carboxylic acid (R-COOH) and CO2 (from oxidation of formic acid).
- Internal Alkynes (R-C≡C-R'): Yields two carboxylic acids (R-COOH + R'-COOH).
- Ozonolysis (O3 then H2O): Similar to strong oxidation, producing carboxylic acids.
- Terminal Alkynes (R-C≡CH): R-COOH + HCOOH (formic acid, often further oxidizes to CO2).
- Internal Alkynes (R-C≡C-R'): R-COOH + R'-COOH.
- Hot Acidified KMnO4 or Hot K2Cr2O7 (Strong Oxidation): Causes cleavage of the triple bond.
Trap 6: Misidentifying Reagents for Distinguishing Alkynes.
- Mistake: Assuming Tollens' reagent or ammoniacal cuprous chloride can react with internal alkynes.
- Correction: These reagents are specific tests for terminal alkynes due to their acidic hydrogen. They form insoluble metal acetylides (e.g., white precipitate with Ag(NH3)2OH, red precipitate with ammoniacal CuCl). Internal alkynes do not react as they lack the acidic terminal hydrogen.
💪 Practice with varying reagents and conditions to master these distinctions and avoid common pitfalls!
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Key Takeaways
Key Takeaways: Alkynes - Acidity and Important Reactions
This section summarizes the critical concepts regarding alkyne acidity and their most significant reactions, crucial for both JEE and board exams.
1. Acidity of Terminal Alkynes
- Reason for Acidity: Terminal alkynes (R-C≡C-H) are significantly more acidic than alkanes and alkenes due to the sp-hybridization of the carbon atom bonded to hydrogen. The sp-hybrid orbital has 50% s-character, which is higher than sp2 (33%) and sp3 (25%). This higher s-character means the valence electrons are held closer to the nucleus, making the sp-hybridized carbon more electronegative. Consequently, the C-H bond in terminal alkynes is polarized, and the hydrogen can be removed as a proton more readily.
- pKa Value: The pKa of a terminal alkyne is approximately 25, making it a weak acid but strong enough to react with strong bases.
- Reagents for Deprotonation: Strong bases like sodium amide (NaNH2), sodium hydride (NaH), or alkyl lithiums (RLi) are required to deprotonate terminal alkynes, forming acetylide anions (R-C≡C-).
JEE Tip: This deprotonation is a key step for carbon-carbon bond formation via nucleophilic attack of the acetylide anion.
2. Important Reactions of Alkynes
Alkynes undergo a variety of reactions, primarily addition reactions across the triple bond and reactions involving their acidic hydrogen.
| Reaction Type | Reaction/Reagents | Product(s) | Key Features / JEE Specifics |
|---|---|---|---|
| Hydrogenation | 1. H2 / Ni, Pt, Pd (excess) 2. H2 / Lindlar's catalyst (Pd-CaCO3/Pb(OAc)2/quinoline) (1 eq.) 3. Na / liquid NH3 (Birch reduction) (1 eq.) | 1. Alkane 2. cis-Alkene 3. trans-Alkene |
|
| Electrophilic Addition | 1. HX (X = Cl, Br, I) (Markovnikov's) 2. X2 (X = Cl, Br) (anti-addition) 3. H2O / H2SO4, HgSO4 (Hydration) | 1. Vinyl halide (1 eq.), Geminal dihalide (2 eq.) 2. Dihaloalkene (1 eq.), Tetrahaloalkane (2 eq.) 3. Enol (intermediate) → Ketone (terminal alkyne gives methyl ketone, except acetylene gives acetaldehyde) |
|
| Oxidation | 1. Cold, dilute, alkaline KMnO4 2. Hot, concentrated KMnO4 or Ozonolysis (O3, then H2O) | 1. Vicinal diones 2. Carboxylic acids (internal), Carboxylic acid + CO2 (terminal) |
|
| C-C Bond Formation | 1. R-C≡C-Na+ + R'-X (SN2 reaction) 2. Oxidative Coupling (Glaser coupling): 2R-C≡C-H + O2 / CuCl, NH4Cl 3. Cyclic Polymerization: CH≡CH / Red hot iron tube | 1. Substituted alkyne (R-C≡C-R') 2. Diyne (R-C≡C-C≡C-R) 3. Benzene, substituted benzenes |
|
Mastering these reactions and the concept of alkyne acidity is fundamental for solving hydrocarbon-related problems in competitive exams. Keep practicing reaction mechanisms and product predictions!
🧩
Problem Solving Approach
Problem Solving Approach: Alkynes - Acidity & Important Reactions
A systematic approach is crucial for mastering alkyne reactions and acidity concepts, especially for competitive exams like JEE. This section outlines a step-by-step methodology to tackle common problems.
1. Approach to Alkyne Acidity Problems
Problems related to alkyne acidity often involve comparing acid strengths or utilizing the acidic proton for synthesis.
- Step 1: Identify the Type of Alkyne:
- Is it a terminal alkyne (RC≡CH)? Only terminal alkynes possess an acidic proton directly attached to the sp-hybridized carbon.
- Is it an internal alkyne (RC≡CR')? Internal alkynes do not have acidic protons and thus do not show acidic properties.
- Step 2: Recall the Reason for Acidity:
- The acidity of terminal alkynes is due to the high s-character (50%) of the sp-hybridized carbon atom. This makes the sp carbon more electronegative than sp² or sp³ carbons.
- This high electronegativity stabilizes the negative charge on the acetylide ion (RC≡C⁻) formed after deprotonation.
- JEE Focus: Be ready to compare acidity: R-C≡CH > R-OH > H₂O > R-NH₂ > R-CH=CH₂ > R-CH₂-CH₃.
- Step 3: Predict Reactions Based on Acidity:
- Terminal alkynes react with strong bases like NaNH₂, Na, or organometallic reagents (e.g., Grignard reagents) to form metal acetylides (RC≡C⁻Na⁺).
- These acetylide ions are strong nucleophiles and can be used in C-C bond forming reactions, particularly in alkylation (with primary alkyl halides) or addition to aldehydes/ketones.
2. Approach to Alkyne Reaction Problems
Solving alkyne reaction problems requires identifying the starting material, reagents, and predicting the product while considering regioselectivity and stereoselectivity.
- Step 1: Analyze the Alkyne Structure:
- Is it terminal or internal? This determines if the acidic proton is available and influences regioselectivity in addition reactions.
- Is it symmetrical or unsymmetrical? This affects the number of possible products, especially in addition reactions (e.g., hydration, hydrohalogenation).
- Step 2: Identify the Reagents and Reaction Type:
The reagent provides the primary clue about the reaction mechanism and product.
Reagent Category
Key Reagents
Resulting Transformation
Reduction
H₂/Lindlar's catalyst; Na/Li in liq. NH₃
Alkene (cis- or trans-); complete reduction to alkane (H₂/Pd, Pt, Ni)
Electrophilic Addition
HX (X=Cl, Br, I); H₂O/HgSO₄/H₂SO₄; X₂/CCl₄
Vinyl halide/dihaloalkane (Markovnikov); Ketone (via enol tautomerism); Vicinal dihalide/tetrahalide
Hydroboration-Oxidation
BH₃-THF/H₂O₂, OH⁻
Aldehyde (terminal alkyne); Ketone (internal alkyne), Anti-Markovnikov addition
Oxidation
KMnO₄ (cold, dilute); KMnO₄ (hot, conc.); Ozonolysis (O₃/H₂O)
Diketone (internal); Carboxylic acids (terminal), CO₂ (terminal C); Carboxylic acids/CO₂
Dimerization/Cyclization
CuCl/NH₄Cl; Red hot iron tube
Vinylacetylene; Benzene (trimerization of ethyne)
- Step 3: Predict Regioselectivity and Stereoselectivity:
- Regioselectivity (where the new atoms attach):
- Most electrophilic additions to unsymmetrical alkynes follow Markovnikov's rule (e.g., HX, H₂O/HgSO₄).
- Hydroboration-oxidation follows Anti-Markovnikov's rule.
- Stereoselectivity (orientation in space):
- Lindlar's catalyst gives cis-alkene (syn addition).
- Na or Li in liquid NH₃ gives trans-alkene (anti addition).
- Halogenation (X₂) usually involves anti addition.
- Regioselectivity (where the new atoms attach):
- Step 4: Draw the Final Product(s):
Always draw the full structural formula, paying attention to all atoms and bonds, especially in multi-step reactions. Consider tautomerism for hydration products.
Example: Propose a synthesis for butanal from 1-butyne.
Approach: Butanal is an aldehyde. Aldehydes can be formed from terminal alkynes via anti-Markovnikov hydration.
- 1-butyne (terminal alkyne) needs to be hydrated in an anti-Markovnikov fashion.
- The suitable reagent for anti-Markovnikov hydration of an alkyne is Hydroboration-Oxidation.
- Reaction: CH₃CH₂C≡CH (1-butyne) $xrightarrow{ ext{1. BH}_3 cdot ext{THF}}$ CH₃CH₂C(BH₂)═CH₂ $xrightarrow{ ext{2. H}_2 ext{O}_2, ext{OH}^-}$ CH₃CH₂CH═CHOH (enol) $xrightarrow{ ext{Tautomerism}}$ CH₃CH₂CH₂CHO (Butanal).
"Consistent practice and a clear understanding of reaction mechanisms will significantly improve your problem-solving speed and accuracy in Hydrocarbons."
📝
CBSE Focus Areas
CBSE Focus Areas: Alkynes - Acidity and Important Reactions
For CBSE Board examinations, the focus on alkynes typically revolves around understanding their acidic nature, especially for terminal alkynes, and a select set of characteristic reactions. Emphasis is placed on reaction conditions, products formed, and distinguishing features from other hydrocarbons.
1. Acidity of Terminal Alkynes
- Reason for Acidity: Terminal alkynes (R-C≡C-H) are acidic due to the sp-hybridized carbon atom directly bonded to hydrogen. The sp-hybrid orbital has 50% s-character, which makes it more electronegative than sp² (alkenes) or sp³ (alkanes) hybridized carbons. This increased electronegativity pulls the shared electron pair closer to carbon, making the C-H bond weaker and the hydrogen more easily removable as a proton.
- CBSE Relevance: This concept is crucial for explaining why terminal alkynes react with strong bases to form alkynides (acetylides) and for distinguishing them from non-terminal alkynes, alkenes, and alkanes.
- Key Reactions Demonstrating Acidity:
- Reaction with active metals (e.g., Sodium):
R-C≡C-H + Na → R-C≡C-Na+ + ½ H₂ - Reaction with strong bases (e.g., Sodamide, NaNH₂):
R-C≡C-H + NaNH₂ → R-C≡C-Na+ + NH₃
(Note: Sodamide is a powerful base often used to deprotonate terminal alkynes). - Reaction with Grignard reagents (R'MgX):
R-C≡C-H + R'MgX → R-C≡C-MgX + R'H
- Reaction with active metals (e.g., Sodium):
2. Important Reactions of Alkynes
CBSE students should know the conditions and products for the following reactions:
- Catalytic Hydrogenation:
- Complete Hydrogenation: Alkyne → Alkane (using Ni, Pt, or Pd catalyst).
R-C≡C-R' + 2H₂ (Ni/Pt/Pd) → R-CH₂-CH₂-R' - Partial Hydrogenation (Selective):
- cis-Alkene formation: Using Lindlar's catalyst (Pd/BaSO₄ poisoned with quinoline or sulfur).
R-C≡C-R' + H₂ (Lindlar's catalyst) → cis-R-CH=CH-R' - trans-Alkene formation: Using Na/liquid NH₃ (Birch Reduction).
R-C≡C-R' + 2Na (liq. NH₃) → trans-R-CH=CH-R'
- cis-Alkene formation: Using Lindlar's catalyst (Pd/BaSO₄ poisoned with quinoline or sulfur).
- Complete Hydrogenation: Alkyne → Alkane (using Ni, Pt, or Pd catalyst).
- Addition of Water (Hydration):
- Alkynes react with water in the presence of H₂SO₄ and HgSO₄ to form carbonyl compounds. This proceeds via an enol intermediate which rapidly tautomerizes to a more stable keto form.
- Example: Ethyne (Acetylene): Forms ethanal (acetaldehyde).
HC≡CH + H₂O (H₂SO₄/HgSO₄) → [CH₂=CH-OH (enol)] → CH₃-CHO (ethanal) - Example: Propyne: Forms propanone (acetone) following Markovnikov's rule.
CH₃-C≡CH + H₂O (H₂SO₄/HgSO₄) → [CH₃-C(OH)=CH₂ (enol)] → CH₃-CO-CH₃ (propanone)
- Addition of Halogen Acids (Hydrohalogenation):
- Follows Markovnikov's rule for unsymmetrical alkynes. Adds in two stages to form gem-dihaloalkanes.
R-C≡C-H + HX → R-C(X)=CH₂ (vinyl halide)
R-C(X)=CH₂ + HX → R-C(X)₂-CH₃ (gem-dihaloalkane)
- Follows Markovnikov's rule for unsymmetrical alkynes. Adds in two stages to form gem-dihaloalkanes.
- Cyclic Polymerization:
- When ethyne (acetylene) is passed through a red-hot iron tube, it undergoes cyclic trimerization to form benzene. This is a significant reaction for the synthesis of aromatic compounds.
3 HC≡CH (Red hot Fe tube) → C₆H₆ (Benzene)
- When ethyne (acetylene) is passed through a red-hot iron tube, it undergoes cyclic trimerization to form benzene. This is a significant reaction for the synthesis of aromatic compounds.
CBSE Tip: Pay close attention to the specific reagents and catalysts used, as they dictate the product formed (e.g., Lindlar's vs. Na/liq. NH₃ for stereochemistry, and HgSO₄/H₂SO₄ for hydration).
🎓
JEE Focus Areas
For JEE Main & Advanced, understanding alkynes goes beyond simply memorizing reactions. A deep grasp of their unique acidity and the various reaction mechanisms is crucial for problem-solving.
I. Acidity of Terminal Alkynes
- Reason for Acidity: Terminal alkynes (R-C≡C-H) possess an acidic hydrogen atom directly attached to a triply bonded carbon. This carbon is sp hybridized.
- s-Character: An sp hybridized carbon has 50% s-character. This high s-character means the electrons in the C-H bond are held more closely to the carbon nucleus, making the hydrogen relatively more positive and thus easier to abstract as a proton.
- Acidity Order (JEE Focus): This is a very common comparative question. The acidity order is crucial:
- HC≡CH > H2O > ROH > HC≡CR > NH3 > Alkene > Alkane
- pKa values (approximate): Acetylene (~25), Water (~15.7), Alcohols (~16-18), Terminal Alkynes (~25), Ammonia (~38), Alkenes (>40), Alkanes (>50).
- Reagents for Deprotonation: To remove the acidic proton, a strong base is required. Common reagents include:
- NaNH2 (Sodium Amide): Forms sodium acetylides (R-C≡C-Na+). This is a strong base commonly used.
- Na (Sodium Metal): Can also deprotonate, forming sodium acetylides and H2 gas.
- Grignard Reagents (RMgX): Act as bases, deprotonating terminal alkynes to form alkynyl magnesium halides.
- NaH (Sodium Hydride): Another strong base.
- Importance: The resulting acetylide ion (R-C≡C-) is a strong nucleophile and can be used to form new carbon-carbon bonds, e.g., in alkylation reactions with primary alkyl halides (SN2) or in reactions with carbonyl compounds.
II. Important Reactions of Alkynes
A. Addition Reactions: Alkynes undergo electrophilic and nucleophilic addition reactions across the triple bond.
- Hydrogenation:
- Complete Hydrogenation (JEE): Alkyne + H2 (excess), Pd/C (or Pt, Ni) → Alkane. All double bonds are reduced.
- Partial Hydrogenation (Stereoselective - JEE Advanced):
- cis-Alkene formation: Alkyne + H2, Lindlar's Catalyst (Pd/CaCO3 + quinoline/BaSO4) → cis-Alkene. (Catalyst poisons prevent further reduction).
- trans-Alkene formation: Alkyne + Na/Li in liquid NH3 (Birch reduction conditions) → trans-Alkene. (Involves a radical mechanism).
- Halogenation:
- Alkyne + X2 (1 eq), CCl4 → dihaloalkene. (Usually trans-addition).
- Alkyne + X2 (2 eq), CCl4 → tetrahaloalkane.
- Test for Unsaturation: Decolorizes Br2/CCl4 or Br2/water.
- Hydrohalogenation (Markovnikov Rule - JEE):
- Alkyne + HX (1 eq) → vinyl halide (adhering to Markovnikov's rule).
- Alkyne + HX (2 eq) → geminal dihalide (both halogens on the same carbon, Markovnikov addition).
- Anti-Markovnikov addition: Only with HBr in presence of peroxides → bromoalkenes with anti-Markovnikov regioselectivity.
- Hydration (Keto-Enol Tautomerism - JEE Advanced):
- Alkyne + H2O, dil. H2SO4, HgSO4 → Enol → Ketone (via tautomerism). This follows Markovnikov's rule, the -OH adds to the more substituted carbon.
- Specific Case: Acetylene (ethyne) upon hydration gives acetaldehyde (an aldehyde). All other terminal alkynes (e.g., propyne, but-1-yne) give ketones.
- Hydroboration-Oxidation: An alternative for hydration that yields an aldehyde from a terminal alkyne (anti-Markovnikov addition of water, followed by tautomerism from enol). Reagents: (i) BH3 or disiamylborane; (ii) H2O2, OH-.
B. Oxidation Reactions:
- Mild Oxidation (Baeyer's Reagent - JEE):
- Alkyne + Cold, dilute, alkaline KMnO4 → vicinal diketone (after tautomerism from diols). This decolorizes Baeyer's reagent.
- Strong Oxidation/Oxidative Cleavage (JEE Advanced):
- Alkyne + Hot, concentrated KMnO4 or Ozonolysis (O3, then H2O) → The triple bond breaks.
- Internal alkyne: Produces two carboxylic acids.
- Terminal alkyne: Produces one carboxylic acid and CO2 (from the terminal -C≡CH group).
- Alkyne + Hot, concentrated KMnO4 or Ozonolysis (O3, then H2O) → The triple bond breaks.
C. Polymerization/Cyclization:
- Cyclic Polymerization (JEE):
- Three molecules of acetylene + Red hot iron tube (or Cu tube) at 873K → Benzene. This is a crucial reaction for aromatic synthesis.
These focus areas cover the essential concepts and reactions of alkynes for JEE. Pay close attention to the reagents and their specific outcomes, especially regarding stereochemistry and regioselectivity.
🌐
Overview
Terminal alkynes are relatively acidic (pKa ≈ 25) due to sp-carbon high s-character. They form acetylide anions with strong bases, enabling C–C bond formation. Key reactions: hydrogenation (to alkenes/alkanes), halogenation, hydrohalogenation, hydration (ketones), oxidative cleavage, and Na/NH3 reduction (trans-alkenes).
📚
Fundamentals
• sp C has 50% s-character → higher acidity for terminal alkynes.
• Acetylide formation requires strong base; SN2 works best on 1° halides.
• Hydration of terminal alkyne → methyl ketone after tautomerism (Markovnikov).
• Acetylide formation requires strong base; SN2 works best on 1° halides.
• Hydration of terminal alkyne → methyl ketone after tautomerism (Markovnikov).
🔬
Deep Dive
pKa rationalization via hybridization; mechanism of Hg2+-catalyzed hydration; subtlety of anti-Markovnikov hydration via hydroboration-oxidation route.
🎯
Shortcuts
“sp is Specially acidic”; “Lindlar = Lovely cis; Na/NH3 = Notably trans.”
💡
Quick Tips
• Avoid SN1/E1 with acetylides; prefer primary halides.
• Watch for over-reduction with H2/Pd; use Lindlar for cis-alkene.
• Hydration often needs Hg2+ catalysis for good rates.
• Watch for over-reduction with H2/Pd; use Lindlar for cis-alkene.
• Hydration often needs Hg2+ catalysis for good rates.
🧠
Intuitive Understanding
Greater s-character holds the negative charge closer to the nucleus, stabilizing the conjugate base—hence terminal alkynes can be deprotonated by strong bases.
🌍
Real World Applications
Synthesis of larger molecules via acetylide alkylation; generation of internal alkynes; partial hydrogenation to trans-alkenes (Birch/Na/NH3).
🔄
Common Analogies
Think of sp as a tighter “grip” on electrons versus sp2/sp3; the tighter grip stabilizes charge, making the acidic H easier to remove.
📋
Prerequisites
pKa and acidity; hybridization and s-character; strong bases (NaNH2, organolithiums); Markovnikov vs anti-Markovnikov additions (context).
⚠️
Common Exam Traps
• Trying SN2 on tertiary halides (fails).
• Forgetting tautomerization to ketones after hydration.
• Confusing cis/trans outcomes of hydrogenation conditions.
• Forgetting tautomerization to ketones after hydration.
• Confusing cis/trans outcomes of hydrogenation conditions.
⭐
Key Takeaways
• Terminal ≠ internal: only terminal deprotonates readily.
• Choice of reagents controls cis/trans hydrogenation.
• Acetylide alkylation is a powerful C–C forming step.
• Choice of reagents controls cis/trans hydrogenation.
• Acetylide alkylation is a powerful C–C forming step.
🧩
Problem Solving Approach
Identify if terminal; choose base (NaNH2) for acetylide; consider SN2 scope; pick addition conditions for desired product (Markovnikov, anti-Markovnikov, cis/trans).
📝
CBSE Focus Areas
Relative acidity trend; key reaction list; simple mechanism sketches for acetylide formation and additions.
🎓
JEE Focus Areas
Synthesis design using acetylides; regio-/stereoselectivity in additions; oxidative cleavage outcomes.
CBSE 12th Board Problems (12)
Problem 255
Easy
1 Mark
Complete the following reaction: <br> CH≡CH + H₂O ( xrightarrow{ ext{HgSO₄/H₂SO₄}} ) ?
Show Solution
1. Ethyne undergoes hydration in the presence of HgSO₄/H₂SO₄. <br> 2. This is an addition reaction where water adds across the triple bond. <br> 3. Initially, an enol (vinyl alcohol) is formed. <br> 4. The enol tautomerizes to a more stable keto form (acetaldehyde).
Final Answer: CH₃CHO (Ethanal/Acetaldehyde)
Problem 255
Easy
2 Marks
How will you distinguish between Propyne and Propene using a chemical test?
Show Solution
1. Identify the functional groups: Propyne is a terminal alkyne, Propene is an alkene. <br> 2. Terminal alkynes have an acidic hydrogen atom. <br> 3. Use a reagent that reacts specifically with acidic hydrogen of terminal alkynes but not with alkenes. <br> 4. Ammoniacal silver nitrate (Tollens' reagent) or ammoniacal cuprous chloride can be used.
Final Answer: Ammoniacal Silver Nitrate (Tollens' Reagent) Test.
Problem 255
Easy
1 Mark
Convert Propyne to Propane.
Show Solution
1. Identify the change: A triple bond needs to be completely reduced to a single bond. <br> 2. This requires complete hydrogenation. <br> 3. Use hydrogen gas in the presence of a strong catalyst.
Final Answer: Hydrogenation with H₂/Ni or H₂/Pt or H₂/Pd.
Problem 255
Easy
2 Marks
Which of the following will react with ammoniacal cuprous chloride solution: But-1-yne or But-2-yne? Give reason.
Show Solution
1. Recall the conditions for reaction with ammoniacal cuprous chloride. <br> 2. It reacts with terminal alkynes due to their acidic hydrogen. <br> 3. Identify which of the given compounds is a terminal alkyne.
Final Answer: But-1-yne will react. Reason: It is a terminal alkyne with an acidic hydrogen.
Problem 255
Easy
1 Mark
Write the product formed when propyne reacts with HBr (1 equivalent).
Show Solution
1. This is an addition reaction of HBr across a triple bond. <br> 2. The addition follows Markovnikov's rule. <br> 3. The hydrogen atom adds to the carbon with more hydrogen atoms, and the bromine adds to the carbon with fewer hydrogen atoms (among the sp-hybridized carbons).
Final Answer: CH₃-C(Br)=CH₂ (2-Bromopropene)
Problem 255
Easy
1 Mark
Name the catalyst used for the partial hydrogenation of alkynes to cis-alkenes.
Show Solution
1. Recall types of hydrogenation: complete and partial. <br> 2. Recall methods for partial hydrogenation: one gives trans-alkene, the other gives cis-alkene. <br> 3. Identify the catalyst specific for cis-alkene formation.
Final Answer: Lindlar's catalyst (Palladium supported on BaSO₄ poisoned with quinoline or sulfur compounds).
Problem 255
Medium
2 Marks
How will you convert Propyne to Propanone?
Show Solution
1. Hydration of Propyne using HgSO4/H2SO4 results in the formation of an enol (prop-1-en-2-ol). 2. The enol undergoes tautomerism to yield Propanone, a more stable keto form.
Final Answer: CH3C≡CH + H2O <span style='color: blue;'>→</span>[HgSO4/H2SO4] CH3C(OH)=CH2 (enol) <span style='color: blue;'>→</span> CH3COCH3 (Propanone)
Problem 255
Medium
2 Marks
Give a chemical test to distinguish between 1-Butyne and 2-Butyne.
Show Solution
1. Use Tollen's reagent (ammoniacal silver nitrate solution). 2. 1-Butyne, being a terminal alkyne, will react with Tollen's reagent to form a white precipitate of silver butynide. 3. 2-Butyne, being an internal alkyne, will not react.
Final Answer: Tollen's reagent test: 1-Butyne forms a white precipitate, 2-Butyne does not react.
Problem 255
Medium
3 Marks
Identify A, B, and C in the following reaction sequence: CH3CH2C≡CH --(NaNH2)--> A --(CH3I)--> B --(H2/Lindlar's Cat.)--> C.
Show Solution
1. 1-Butyne reacts with NaNH2 (a strong base) to form sodium butynide (A), an alkynide. 2. Sodium butynide (A) reacts with methyl iodide (CH3I) via SN2 reaction to form 2-pentyne (B). 3. 2-Pentyne (B) undergoes catalytic hydrogenation with Lindlar's catalyst to give cis-2-pentene (C).
Final Answer: A = CH3CH2C≡CNa, B = CH3CH2C≡CCH3, C = CH3CH2CH=CHCH3 (cis isomer)
Problem 255
Medium
2 Marks
Predict the major product(s) when 2-Butyne reacts with HBr (excess).
Show Solution
1. The first molecule of HBr adds to the triple bond, following Markovnikov's rule, to form 2-bromo-2-butene. 2. The second molecule of HBr adds to the double bond, again following Markovnikov's rule, to form a gem-dihalide.
Final Answer: 2,2-Dibromobutane (CH3CBr2CH2CH3)
Problem 255
Medium
2 Marks
How can Ethyne be converted to Benzene?
Show Solution
Ethyne undergoes cyclic polymerization when passed through a red hot iron tube, resulting in the formation of Benzene.
Final Answer: 3 HC≡CH <span style='color: blue;'>→</span>[Red hot iron tube] C6H6 (Benzene)
Problem 255
Medium
2 Marks
Complete the following reaction: CH3C≡CCH3 + Na/Liq. NH3 → ?
Show Solution
2-Butyne undergoes Birch reduction (reduction by Na in liquid NH3) to form a trans-alkene.
Final Answer: CH3CH=CHCH3 (trans-2-Butene)
IIT-JEE Main Problems (12)
Problem 255
Easy
4 Marks
How many acidic hydrogen atoms are present in the molecule of propyne?
Show Solution
1. Identify the structure of propyne: CH₃-C≡CH.
2. Recall that only hydrogens attached to sp hybridized carbon atoms in terminal alkynes are acidic due to the high electronegativity of sp hybridized carbon, which stabilizes the resulting acetylide anion.
3. In propyne, only the hydrogen atom attached to the terminal sp hybridized carbon is acidic.
Final Answer: 1
Problem 255
Easy
4 Marks
One mole of propyne is treated with excess sodium amide (NaNH₂). How many moles of the resulting sodium acetylide derivative are formed?
Show Solution
1. Write the reaction of propyne with sodium amide.
CH₃-C≡CH + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃
2. The reaction shows a 1:1 molar ratio between propyne and the sodium acetylide derivative.
3. Since 1 mole of propyne is used, 1 mole of the derivative will be formed.
Final Answer: 1
Problem 255
Easy
4 Marks
How many different organic products (excluding stereoisomers) are formed when but-1-yne reacts with H₂SO₄/HgSO₄?
Show Solution
1. Identify the reaction type: HgSO₄/H₂SO₄ is for the Markovnikov hydration of alkynes.
2. Write the structure of but-1-yne: CH₃-CH₂-C≡CH.
3. Apply Markovnikov's rule: The -OH group adds to the more substituted carbon of the triple bond, and -H adds to the less substituted carbon.
4. This forms an enol: CH₃-CH₂-C(OH)=CH₂.
5. Enols are unstable and tautomerize to the more stable keto form.
6. The keto form will be CH₃-CH₂-CO-CH₃ (Butan-2-one). This is the only major organic product.
Final Answer: 1
Problem 255
Easy
4 Marks
How many different organic products (excluding stereoisomers) are formed when but-1-yne undergoes hydroboration-oxidation (i) BH₃.THF, (ii) H₂O₂/NaOH?
Show Solution
1. Identify the reaction type: Hydroboration-oxidation of alkynes is an anti-Markovnikov hydration.
2. Write the structure of but-1-yne: CH₃-CH₂-C≡CH.
3. Apply anti-Markovnikov's rule: The -OH group effectively adds to the less substituted carbon of the triple bond, and -H adds to the more substituted carbon.
4. This forms an enol: CH₃-CH₂-CH=CH-OH.
5. Enols are unstable and tautomerize to the more stable aldehyde form.
6. The aldehyde form will be CH₃-CH₂-CH₂-CHO (Butanal). This is the only major organic product.
Final Answer: 1
Problem 255
Easy
4 Marks
How many moles of hydrogen gas (H₂) are required for the complete hydrogenation of 1 mole of but-2-yne to form butane using a Ni catalyst?
Show Solution
1. Write the structure of but-2-yne: CH₃-C≡C-CH₃.
2. Complete hydrogenation means converting the triple bond to a single bond (an alkane).
3. Each π bond requires 1 mole of H₂ for reduction.
4. A triple bond consists of one σ bond and two π bonds.
5. Therefore, 2 moles of H₂ are required to reduce the two π bonds.
Final Answer: 2
Problem 255
Easy
4 Marks
When propyne undergoes ozonolysis followed by hydrolysis (O₃/H₂O), what is the total number of carbon atoms present in the organic products formed? (Consider only the main stable organic products)
Show Solution
1. Write the structure of propyne: CH₃-C≡CH.
2. Ozonolysis (O₃/H₂O) of an alkyne cleaves the triple bond oxidatively.
3. CH₃-C≡CH will break into CH₃-COOH (acetic acid) from the internal part and H-COOH (formic acid) from the terminal part.
4. Under oxidative conditions (O₃/H₂O), formic acid (H-COOH) is further oxidized to CO₂ and H₂O.
5. Therefore, the main stable organic product is acetic acid (CH₃-COOH).
6. Acetic acid has 2 carbon atoms.
Final Answer: 2
Problem 255
Hard
4 Marks
Identify the final major product 'C' in the following sequence of reactions:
Show Solution
<ul><li><strong>Step 1: Formation of A (Alkyne)</strong><br>1,2-Dibromobutane (CH3CH2CH(Br)CH2Br) undergoes double dehydrohalogenation with excess NaNH2 in liquid NH3 to form but-1-yne (CH3CH2C≡CH). This is product 'A'.</li><li><strong>Step 2: Alkylation to form B</strong><br>But-1-yne (A), being a terminal alkyne, is deprotonated by NaNH2 to form the but-1-ynide anion (CH3CH2C≡C⁻Na⁺). This anion then reacts with methyl iodide (CH3I) via an SN2 reaction to form pent-2-yne (CH3CH2C≡CCH3). This is product 'B'.</li><li><strong>Step 3: Stereospecific Reduction to form C</strong><br>Pent-2-yne (B) is an internal alkyne. Reduction with H2 in the presence of Lindlar's catalyst (Pd-BaSO4/quinoline) is a stereospecific cis-addition of hydrogen. This yields (Z)-pent-2-ene. This is product 'C'.</li></ul>
Final Answer: (Z)-Pent-2-ene
Problem 255
Hard
4 Marks
An organic compound 'A' (C6H10) on reaction with ammoniacal cuprous chloride gives a red precipitate. When 'A' is treated with excess HBr, it forms a geminal dibromide 'B' which upon dehydrohalogenation with strong base (NaNH2) regenerates 'A'. What is the structure of 'A'?
Show Solution
<ul><li><strong>Clue 1: Reaction with Ammoniacal Cuprous Chloride</strong><br>The reaction of compound 'A' with ammoniacal cuprous chloride to give a red precipitate indicates that 'A' is a terminal alkyne.</li><li><strong>Clue 2: Molecular Formula (C6H10) and Terminal Alkyne</strong><br>Based on C6H10 and being a terminal alkyne, possible structures are Hex-1-yne, 4-methylpent-1-yne, 3-methylpent-1-yne, or 3,3-dimethylbut-1-yne.</li><li><strong>Clue 3: Reaction with Excess HBr</strong><br>Terminal alkynes react with excess HBr via Markovnikov's rule to form geminal dibromides (two bromine atoms on the same carbon). For instance, Hex-1-yne (CH3CH2CH2CH2C≡CH) reacts with excess HBr to form 2,2-dibromohexane (CH3CH2CH2CH2CBr2CH3).</li><li><strong>Clue 4: Dehydrohalogenation of 'B' regenerates 'A'</strong><br>Geminal dibromides, upon treatment with a strong base like NaNH2 (excess), undergo double dehydrohalogenation to regenerate the alkyne. This confirms the sequence. All the possible terminal C6H10 alkynes would form corresponding geminal dibromides and then regenerate the alkyne. However, the simplest unbranched terminal alkyne of formula C6H10 is Hex-1-yne. In JEE problems, if not specified for isomers, the straight-chain unbranched structure is often implied.</li></ul>
Final Answer: Hex-1-yne
Problem 255
Hard
4 Marks
How many distinct organic products are formed when propyne is reacted sequentially with (i) sodium amide in liquid ammonia, and then (ii) 1,3-dibromopropane?
Show Solution
<ul><li><strong>Step 1: Formation of Alkynide Anion</strong><br>Propyne (CH3C≡CH) is a terminal alkyne and reacts with a strong base like sodium amide (NaNH2) in liquid ammonia to form the prop-1-ynide anion (CH3C≡C⁻Na⁺). This is an acid-base reaction.</li><li><strong>Step 2: Alkylation with 1,3-Dibromopropane</strong><br>1,3-Dibromopropane has two electrophilic sites (primary alkyl bromides). The prop-1-ynide anion, being a strong nucleophile, can attack these sites via SN2 reaction.<ul><li><strong>Monosubstitution:</strong> The prop-1-ynide anion can attack one of the primary bromine atoms in 1,3-dibromopropane, displacing Br⁻. This results in the formation of 1-bromohex-3-yne (CH3C≡C-CH2CH2CH2-Br). This is one distinct product.</li><li><strong>Disubstitution:</strong> If sufficient prop-1-ynide anion is available (or if the reaction is driven to completion with excess reagents, though 'sequentially' and '1,3-dibromopropane' usually implies possibility), a second prop-1-ynide anion can attack the remaining bromine atom. This leads to the formation of hepta-2,5-diyne (CH3C≡C-CH2CH2CH2-C≡CCH3). This is a second distinct product.</li></ul><li><strong>Stereoisomers:</strong> Neither 1-bromohex-3-yne nor hepta-2,5-diyne possess chiral centers or geometric isomerism around the triple bonds. Therefore, no stereoisomers are formed.</li></ul>Given 'sequentially' and '1,3-dibromopropane' (not specified as 'one equivalent of 1,3-dibromopropane'), both monosubstituted and disubstituted products are plausible. If the question implies addition of the second alkynide, it would be 'excess' alkynide, but in the context of 'distinct products', it implies all possible stable products from the given reagents.
Final Answer: 2
Problem 255
Hard
4 Marks
Identify the reagents P, Q, and R in the following synthetic sequence:
Show Solution
<ul><li><strong>Reaction 1: Propyne to 2,2-Dibromopropane</strong><br>Propyne (CH3C≡CH) reacts with an excess of HBr to undergo Markovnikov's addition twice. The two bromine atoms add to the more substituted carbon, resulting in 2,2-dibromopropane (CH3CBr2CH3). Therefore, <strong>P = HBr (excess)</strong>.</li><li><strong>Reaction 2: 2,2-Dibromopropane to Propyne</strong><br>2,2-Dibromopropane (a geminal dibromide) undergoes double dehydrohalogenation to form an alkyne (Propyne). This reaction requires a strong base, typically sodium amide (NaNH2) in excess. Therefore, <strong>Q = NaNH2 (excess) in liquid NH3</strong>.</li><li><strong>Reaction 3: Propyne to (E)-propene</strong><br>Propyne (a terminal alkyne) is reduced to (E)-propene (trans-propene). This is a stereospecific reduction using sodium metal in liquid ammonia (Birch reduction conditions). Therefore, <strong>R = Na/liquid NH3</strong>.</li></ul>
Final Answer: P = HBr (excess), Q = NaNH2 (excess)/liquid NH3, R = Na/liquid NH3
Problem 255
Hard
4 Marks
Identify the major product 'X' formed in the following reaction sequence:
Show Solution
<ul><li><strong>Step 1: Reaction with Zn dust/EtOH</strong><br>1,1,2,2-Tetrabromopropane (Br2CH-CBr2-CH3) undergoes reductive debromination with excess zinc dust in ethanol. This removes all four bromine atoms and forms a triple bond, yielding propyne (CH≡C-CH3). This is product 'A'.</li><li><strong>Step 2: Reaction with NaNH2 (excess)/liquid NH3</strong><br>Propyne (A) is a terminal alkyne. It reacts with excess sodium amide to form the prop-1-ynide anion (CH3C≡C⁻Na⁺). The excess NaNH2 ensures complete deprotonation.</li><li><strong>Step 3: Alkylation with CH3CH2Br</strong><br>The prop-1-ynide anion, being a strong nucleophile, reacts with ethyl bromide (CH3CH2Br) via an SN2 reaction to form pent-2-yne (CH3C≡CCH2CH3). This is product 'B'.</li><li><strong>Step 4: Reduction with Na/liquid NH3</strong><br>Pent-2-yne (B) is an internal alkyne. Reduction with sodium metal in liquid ammonia (Birch reduction conditions) is a stereospecific anti-addition of hydrogen. This yields (E)-pent-2-ene. This is product 'X'.</li></ul>
Final Answer: (E)-Pent-2-ene
Problem 255
Hard
4 Marks
How many distinct organic products (including stereoisomers) are formed when 3-methylbut-1-yne is treated sequentially with (i) one equivalent of sodium amide in liquid ammonia, (ii) 1-iodopropane, and (iii) H2/Pd-BaSO4/quinoline?
Show Solution
<ul><li><strong>Step 1: Deprotonation</strong><br>3-methylbut-1-yne (CH≡C-CH(CH3)2) is a terminal alkyne. It reacts with one equivalent of sodium amide (NaNH2) in liquid ammonia to form its corresponding alkynide anion: Na⁺⁻C≡C-CH(CH3)2.</li><li><strong>Step 2: Alkylation</strong><br>This alkynide anion then reacts with 1-iodopropane (CH3CH2CH2I) via an SN2 reaction. The prop-1-yl group is attached to the terminal carbon of the alkyne, yielding 6-methylhept-3-yne (CH3CH2CH2-C≡C-CH(CH3)2).</li><li><strong>Step 3: Stereospecific Reduction</strong><br>6-methylhept-3-yne is an internal alkyne. Reduction with H2 in the presence of Lindlar's catalyst (Pd-BaSO4/quinoline) results in a stereospecific cis-addition of hydrogen to the triple bond. This yields (Z)-6-methylhept-3-ene.<ul><li>Let's analyze the structure of (Z)-6-methylhept-3-ene: The double bond is between C3 and C4. C3 is bonded to H and CH2CH2CH3. C4 is bonded to H and CH(CH3)2. Since the groups on each carbon of the double bond are different (H vs alkyl), geometric isomerism (cis/trans) is possible. Lindlar's catalyst specifically yields the (Z) or cis-isomer.</li><li>Is the product chiral? The molecule is CH3CH2CH2-CH=CH-CH(CH3)2. There are no chiral carbons in this structure. The isopropyl group -CH(CH3)2 is achiral. The double bond does not create chirality here.</li></ul><li><strong>Distinct Products:</strong> Since only the (Z)-isomer is formed and there are no chiral centers, only one distinct organic product, (Z)-6-methylhept-3-ene, is formed.</li></ul>
Final Answer: 1
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Important Formulas (6)
Acidity of Terminal Alkynes (Formation of Acetylide Ion)
R-Cequiv C-H + B^- longrightarrow R-Cequiv C^- + B-H
Text: R-C≡C-H + Base (e.g., NaNH2) → R-C≡C⁻Na⁺ + NH₃
Terminal alkynes (those with a hydrogen directly attached to a triple bond carbon) are acidic due to the sp-hybridized carbon's high electronegativity stabilizing the resulting carbanion. This allows them to react with strong bases to form acetylide ions. Common bases include <strong>sodium amide (NaNH₂)</strong>, sodium metal (Na), or Grignard reagents.
Variables: To form carbon-carbon bonds via nucleophilic attack of the acetylide ion, or to differentiate terminal alkynes from internal alkynes or alkenes/alkanes.
Hydration of Alkynes (Markovnikov's Rule)
R-Cequiv C-R' xrightarrow{H_2SO_4, HgSO_4} [enol] longrightarrow keto
Text: R-C≡C-H + H₂O (H₂SO₄, HgSO₄) → [Enol] → Ketone (for terminal alkyne, propyne gives acetone)
Acid-catalyzed hydration of alkynes, usually with <span style='color: #FF0000;'>mercuric sulfate (HgSO₄) and sulfuric acid (H₂SO₄)</span>, follows Markovnikov's rule, adding H to the less substituted carbon and OH to the more substituted carbon. The initial product is an enol, which rapidly tautomerizes into a more stable ketone (or aldehyde for acetylene). For terminal alkynes, except acetylene, it yields ketones.
Variables: To synthesize ketones from alkynes. <span style='color: #0000FF;'>JEE Note: Acetylene gives acetaldehyde.</span>
Hydroboration-Oxidation of Alkynes (Anti-Markovnikov)
R-Cequiv C-H xrightarrow{1. BH_3 ext{ or } (Sia)_2BH, 2. H_2O_2, OH^-} [enol] longrightarrow aldehyde
Text: R-C≡C-H + BH₃ (or disiamylborane) then H₂O₂, OH⁻ → [Enol] → Aldehyde
Hydroboration-oxidation adds H and OH across the triple bond in an <span style='color: #FF0000;'>anti-Markovnikov fashion</span>. This reaction is regioselective and proceeds via syn-addition. The resulting enol tautomerizes to an aldehyde (for terminal alkynes) or ketone (for internal alkynes if asymmetric). Disiamylborane [(Sia)₂BH] is often preferred to prevent double addition.
Variables: To synthesize aldehydes from terminal alkynes, or ketones from internal alkynes, specifically achieving anti-Markovnikov hydration.
Partial Hydrogenation (Lindlar's Catalyst)
R-Cequiv C-R' xrightarrow{H_2, ext{ Lindlar's Catalyst}} R-CH=CH-R' ext{ (cis-alkene)}
Text: R-C≡C-R' + H₂ (Pd-CaCO₃/BaSO₄ + Quinoline/S) → R-CH=CH-R' (cis-alkene)
Hydrogenation of alkynes using <span style='color: #FF0000;'>Lindlar's catalyst</span> (palladium on calcium carbonate poisoned with lead acetate/quinoline or sulfur) selectively reduces the alkyne to a <strong>cis-alkene</strong>. The catalyst's activity is reduced to prevent over-reduction to an alkane.
Variables: To synthesize cis-alkenes from alkynes.
Partial Hydrogenation (Birch Reduction of Alkynes)
R-Cequiv C-R' xrightarrow{Na ext{ or } Li, ext{ liq. NH}_3} R-CH=CH-R' ext{ (trans-alkene)}
Text: R-C≡C-R' + Na (or Li) in liquid NH₃ → R-CH=CH-R' (trans-alkene)
Reduction of alkynes with <span style='color: #FF0000;'>sodium (or lithium) in liquid ammonia</span> converts alkynes into <strong>trans-alkenes</strong>. This reaction proceeds via a radical anion mechanism and is highly stereoselective for trans-products.
Variables: To synthesize trans-alkenes from alkynes.
Oxidative Cleavage of Alkynes
R-Cequiv C-R' xrightarrow{1. O_3, 2. H_2O} R-COOH + HOOC-R' ext{ (or } CO_2 ext{ for terminal H)}
Text: R-C≡C-R' + KMnO₄ (hot, conc.) or O₃ then H₂O → Carboxylic acids (or CO₂ for terminal H)
Strong oxidizing agents like <span style='color: #FF0000;'>hot, concentrated KMnO₄</span> or <span style='color: #FF0000;'>ozonolysis (O₃ followed by water workup)</span> cleave the triple bond, forming carboxylic acids. If the alkyne is terminal (R-C≡C-H), the terminal carbon is oxidized to CO₂.
Variables: To determine the position of the triple bond in an unknown alkyne or to synthesize carboxylic acids.
References & Further Reading (10)
Book
Organic Chemistry
A well-regarded textbook with a strong emphasis on reaction mechanisms and synthesis. Provides in-depth coverage of alkynes, their synthesis, reactivity, acidity, and key reactions like hydroboration, hydration, and various coupling reactions.
Note: Ideal for students seeking a deeper mechanistic understanding, crucial for JEE Advanced. Covers a broad spectrum of alkyne reactions and synthetic strategies.
Book
Website
Reactions of Alkynes
Part of a comprehensive online organic chemistry textbook, this section details various reactions of alkynes, including catalytic hydrogenation, hydrohalogenation, halogenation, hydration (both mercuric ion-catalyzed and hydroboration-oxidation), ozonolysis, and reactions of terminal alkynes with strong bases.
Note: Provides a well-organized summary of alkyne reactions with mechanisms. Good for reinforcing concepts and understanding the scope of reactions relevant for all exam levels.
Website
PDF
Alkynes - Structure, Acidity, and Reactions
Comprehensive lecture notes from IIT Bombay, specifically focusing on the structure, nomenclature, acidity, and important reactions of alkynes. Covers electrophilic additions, reduction, oxidation, and reactions of terminal alkynes, with a strong emphasis on mechanisms and applications.
Note: Directly aligned with the level and depth expected for JEE Main and Advanced, given its origin from an IIT. Highly recommended for thorough understanding and problem-solving. Provides concise and accurate information.
PDF
Article
A Review on the Application of Alkyne Reactions in Organic Synthesis
A review article summarizing the synthetic utility of various alkyne reactions, including additions, cyclizations, and transformations involving acetylide anions, highlighting their importance in constructing complex organic molecules.
Note: While a general review on synthetic applications might be slightly advanced, it helps in understanding the broader context and importance of alkyne reactions beyond basic mechanisms, useful for advanced JEE perspective.
Article
Research_Paper
Sonogashira Coupling: A Versatile Tool in Modern Organic Synthesis
This is a review article summarizing the Sonogashira coupling reaction, a crucial cross-coupling reaction involving terminal alkynes, showcasing its mechanism, scope, and synthetic applications in various fields of organic chemistry.
Note: While a research review, it covers the Sonogashira coupling, which is an important reaction involving terminal alkynes and their acidity-derived reactivity. Highly relevant for JEE Advanced level synthesis questions and understanding modern synthetic methods.
Research_Paper
Common Mistakes to Avoid (62)
Minor
Other
❌ Confusion in Stereochemistry of Alkyne Reduction
Students often correctly identify the reagents for partial reduction of alkynes to alkenes but frequently confuse the specific stereochemical outcomes (cis vs. trans) associated with Lindlar's catalyst and dissolving metal reduction (Na/NH3). This leads to incorrect product stereoisomers in final answers.
💭 Why This Happens:
This mistake primarily stems from the similar nature of these two partial reduction reactions, often taught consecutively. Under exam pressure, students might mix up which reagent yields the cis product and which yields the trans product, especially if their understanding is based on rote memorization rather than the underlying mechanism. The subtle mechanistic differences leading to distinct stereoisomers are sometimes overlooked.
✅ Correct Approach:
To avoid this, it's crucial to understand the mechanism behind each reduction. Lindlar's catalyst (Pd/CaCO3, quinoline, PbO) facilitates a syn-addition of hydrogen, meaning both hydrogen atoms add to the same face of the alkyne, consistently yielding the cis-alkene. In contrast, dissolving metal reduction (Na or Li in liquid NH3) proceeds via a radical anion mechanism involving anti-addition, which sterically favors the more stable trans-alkene.
📝 Examples:
❌ Wrong:
Consider the reaction:
CH3-C≡C-CH3 ---> Product?
[Lindlar's Catalyst]Wrong Answer: The student draws (E)-But-2-ene (trans-But-2-ene) as the product.
✅ Correct:
For the same reaction:
CH3-C≡C-CH3 ---> Product?
[Lindlar's Catalyst]Correct Answer: The product is (Z)-But-2-ene (cis-But-2-ene).
Similarly, for dissolving metal reduction:
CH3-C≡C-CH3 ---> Product?
[Na / liq. NH3]Correct Answer: The product is (E)-But-2-ene (trans-But-2-ene).
💡 Prevention Tips:
- Create a Comparison Table: Systematically list reagents, conditions, mechanism type (syn/anti), and stereochemical outcome for partial alkyne reductions.
- Visualize Mechanisms: Understand why Lindlar's gives cis (surface adsorption, syn-addition) and why Na/NH3 gives trans (radical anion intermediates, anti-addition).
- Practice with Examples: Solve various problems involving different alkynes and reagents, drawing out the correct stereoisomers. This is crucial for both CBSE and JEE Advanced, where stereochemistry is frequently tested.
JEE_Advanced
Minor
Conceptual
❌ <p>Confusing Acidity of Terminal vs. Internal Alkynes</p>
A common mistake is the incorrect generalization of acidity to all alkynes. Students often fail to recognize that only terminal alkynes (those with a hydrogen atom directly attached to the sp-hybridized carbon) exhibit significant acidic character, capable of reacting with strong bases.
💭 Why This Happens:
This misconception stems from an incomplete understanding of the structural requirements for alkyne acidity. The high s-character (50%) of the sp-hybridized carbon in terminal alkynes makes it more electronegative than sp2 or sp3 carbons, thus polarizing the C-H bond and allowing deprotonation. Internal alkynes lack such a C-H bond.
✅ Correct Approach:
Always identify the position of the triple bond. If the triple bond is at the end of the carbon chain (e.g., R-C≡CH), it's a terminal alkyne and acidic. If the triple bond is within the carbon chain (e.g., R-C≡C-R'), it's an internal alkyne and non-acidic.
- JEE Pointer: The acidity of terminal alkynes (pKa ≈ 25) is significant enough to react with strong bases like NaNH2, Grignard reagents, and even alkali metals.
📝 Examples:
❌ Wrong:
Assuming 2-butyne (CH3-C≡C-CH3) will react with sodium amide (NaNH2) to form a sodium acetylide salt.
✅ Correct:
Propyne (CH3-C≡CH) reacts with sodium amide (NaNH2) to yield sodium propynide (CH3-C≡C-Na+) and ammonia (NH3), as propyne is a terminal alkyne.
CH3-C≡CH + NaNH2 → CH3-C≡C-Na+ + NH3
💡 Prevention Tips:
- Visual Check: Always look for a hydrogen directly bonded to the sp-hybridized carbon of the triple bond.
- Acidity Trend: Remember the relative acidity: Alkyne (terminal) > Alkene > Alkane. This distinction is crucial for understanding reaction feasibility with bases.
- Avoid Generalization: Do not assume all compounds containing a triple bond are acidic.
JEE_Main
Minor
Calculation
❌ Stoichiometric Errors with Acidic Alkynes
Students frequently miscalculate the required moles of base or the expected number of products when terminal alkynes react with strong bases. This error typically arises when multiple acidic protons are present or the stoichiometry of the base is misunderstood.
💭 Why This Happens:
- Incomplete Proton Identification: Failure to correctly identify all acidic terminal alkyne protons in a molecule (e.g., overlooking a second terminal alkyne in a di-yne).
- Ignoring Stoichiometry: Not paying close attention to the number of equivalents of base provided in the problem statement.
✅ Correct Approach:
To avoid these errors, a systematic approach is essential:
- Identify Acidic Protons: Always draw the full structure of the alkyne and clearly mark all terminal alkyne protons, recognizing their pKa value (around 25) makes them acidic enough to react with strong bases like NaNH2.
- Count Sites & Match Base: Determine the total number of acidic terminal alkyne protons available. The moles of strong base (e.g., NaNH2) consumed will directly correspond to the moles of acidic protons abstracted. For instance, a di-terminal alkyne requires two equivalents of base for complete deprotonation.
📝 Examples:
❌ Wrong:
Wrong Assumption: Assuming that when but-1,3-diyne (H-C≡C-C≡C-H), a di-terminal alkyne, reacts with only 1 equivalent of NaNH2, both terminal alkynes will be deprotonated to form a di-anion (Na+-C≡C-C≡C-Na+). This outcome incorrectly implies that 1 equivalent of base can deprotonate two sites; it actually requires 2 equivalents of base.
✅ Correct:
Correct Scenario: Consider but-1,3-diyne (H-C≡C-C≡C-H), which possesses two acidic terminal alkyne protons.
- If reacted with 1 equivalent of NaNH2, only one proton will be abstracted, yielding H-C≡C-C≡C-Na+.
- If reacted with 2 equivalents of NaNH2, both protons will be abstracted, resulting in the formation of Na+-C≡C-C≡C-Na+.
The key for JEE Main is to correctly match the equivalents of base with the number of acidic protons available for reaction.
💡 Prevention Tips:
- Visual Inspection: Always draw the complete structural formula to unambiguously identify all terminal alkyne protons.
- Stoichiometric Check: Carefully read the number of equivalents of reagents provided in the problem statement, as this dictates the extent of the reaction.
- Practice Di-alkynes: Pay particular attention to reactions involving di-terminal alkynes, as they have two distinct acidic protons that can react.
- Conceptual Clarity: Solidify your understanding that a strong base like NaNH2 deprotonates terminal alkynes, and the reaction proceeds stoichiometrically based on the moles of base added.
JEE_Main
Minor
Formula
❌ Confusing Acidity-Related Reagents with Addition Reagents for Terminal Alkynes
Students frequently misidentify the specific role of reagents when dealing with terminal alkynes. They might mistakenly use reagents meant for addition reactions (e.g., HBr, H2O/H2SO4) to abstract the acidic proton, or conversely, use strong bases (e.g., NaNH2) hoping for an addition across the triple bond.
💭 Why This Happens:
This error stems from an incomplete understanding of the unique acidic nature of terminal alkynes (due to the sp-hybridized carbon) and a failure to differentiate between electrophilic attack on the triple bond and nucleophilic attack by a strong base on the acidic hydrogen. Rote memorization of reagents without understanding their specific mechanisms also contributes to this confusion.
✅ Correct Approach:
Always remember that terminal alkynes possess an acidic hydrogen atom (pKa ≈ 25). Reactions targeting this acidity require strong bases like NaNH2, Grignard reagents (RMgX), or alkali metals (Na). These form nucleophilic acetylide anions. Addition reactions, on the other hand, involve the direct electrophilic attack on the π-bonds of the alkyne and require different reagents such as HX, X2, H2O/H2SO4, etc.
📝 Examples:
❌ Wrong:
Attempting to convert propyne (CH3-C≡CH) to propargyl bromide (CH3-C≡C-Br) by reacting it directly with HBr. This will primarily lead to addition products across the triple bond, not substitution of the acidic hydrogen.
✅ Correct:
To react with the acidic hydrogen of propyne (CH3-C≡CH), you would use a strong base:
1. CH3-C≡CH + NaNH2 → CH3-C≡C-Na+ + NH3 (Acetylide anion formation)
This acetylide anion can then be used in subsequent reactions, for example, alkylation:
2. CH3-C≡C-Na+ + CH3I → CH3-C≡C-CH3 + NaI
💡 Prevention Tips:
- JEE Tip: Categorize alkyne reactions into 'Acidity-based reactions' and 'Addition reactions' in your notes.
- Focus on understanding the function of each reagent. Is it a strong base? An electrophile? A nucleophile?
- Practice identifying terminal vs. internal alkynes as only terminal alkynes exhibit significant acidity.
- Pay close attention to reaction conditions (e.g., solvent, temperature) as they often provide clues about the intended reaction pathway.
JEE_Main
Minor
Unit Conversion
❌ Inconsistent Energy Units (J vs. kJ) in Reaction Calculations
Students frequently overlook the conversion between Joules (J) and kilojoules (kJ) when dealing with energy values in problems related to alkyne reactions (e.g., hydrogenation, combustion, or energy changes associated with acidity). This often leads to numerically incorrect answers, even if the underlying chemical principles are understood correctly.
💭 Why This Happens:
- Lack of rigorous unit checking for all given numerical values.
- Assuming that all energy data provided in a problem statement are already in a consistent unit.
- Forgetting the crucial conversion factor: 1 kJ = 1000 J.
- Overemphasis on the chemical reaction mechanism or acidity concept, leading to reduced attention on numerical and unit details.
✅ Correct Approach:
Before commencing any calculation involving energy terms, make it a habit to explicitly write down the units for each value. Convert all energy values to a common, consistent unit (e.g., all to kJ/mol or all to J/mol) at the very beginning of the problem. This ensures that all subsequent additions, subtractions, or comparisons are performed with compatible units.
📝 Examples:
❌ Wrong:
Consider a problem asking for the net energy change of an alkyne reaction. If one energy input is given as 500 J/mol and an energy output is 200 kJ/mol, a common mistake is to simply calculate 200 - 500 = -300. This is incorrect because the units are different.
✅ Correct:
To correctly calculate the net energy change from the above example:
First, convert 500 J/mol to kJ/mol: 500 J/mol = 0.5 kJ/mol.
Then, perform the calculation with consistent units: 200 kJ/mol - 0.5 kJ/mol = 199.5 kJ/mol.
Similarly, if the final answer needs to be in Joules, convert 200 kJ/mol to 200,000 J/mol before calculation.
First, convert 500 J/mol to kJ/mol: 500 J/mol = 0.5 kJ/mol.
Then, perform the calculation with consistent units: 200 kJ/mol - 0.5 kJ/mol = 199.5 kJ/mol.
Similarly, if the final answer needs to be in Joules, convert 200 kJ/mol to 200,000 J/mol before calculation.
💡 Prevention Tips:
- Always include units with every numerical quantity throughout your problem-solving steps.
- Before starting calculations, identify the target unit for the final answer and ensure all given data are converted accordingly.
- For JEE Main, be aware that answer choices might include values resulting from common unit conversion errors.
- Practice problems that intentionally mix different energy units to sharpen your attention to detail.
JEE_Main
Minor
Sign Error
❌ Misinterpreting Polarity of Terminal Alkynes and Acetylide Anion Charge
Students frequently make 'sign errors' by incorrectly perceiving the terminal hydrogen in an alkyne as having a partial negative charge (δ⁻) or assuming the resulting acetylide anion (R-C≡C⁻) is positively charged or neutral. This fundamental error leads to incorrect predictions regarding alkyne acidity and their subsequent nucleophilic reactions.
💭 Why This Happens:
This misconception stems from an incomplete understanding of how hybridization affects electronegativity and the stability of carbanions. Students might generalize from other organic compounds without considering the unique properties of sp-hybridized carbon, which possesses significantly higher s-character (50%) compared to sp2 (33%) or sp3 (25%) carbons. This higher s-character makes the sp-hybridized carbon more electronegative, causing it to withdraw electron density from the hydrogen.
✅ Correct Approach:
The sp-hybridized carbon in a terminal alkyne is highly electronegative, making the C-H bond polar. The carbon carries a partial negative charge (δ⁻) and the hydrogen a partial positive charge (δ⁺), making the hydrogen acidic. Upon deprotonation, a stable acetylide anion (R-C≡C⁻) is formed, which carries a full negative charge on the carbon atom. This anion acts as a strong nucleophile or a strong base in subsequent reactions.
📝 Examples:
❌ Wrong:
Predicting that the terminal carbon in R-C≡CH is δ⁺ and the H is δ⁻, or attempting to react R-C≡C⁻ as an electrophile rather than a nucleophile.
✅ Correct:
Consider the deprotonation of propyne and its subsequent reaction:
Here, the acetylide anion (CH₃-C≡C⁻) has a negative charge on the terminal carbon. This anion then acts as a nucleophile, for instance, in an SN2 reaction:
The negatively charged carbon of the acetylide attacks the electrophilic carbon of methyl bromide.
CH₃-C≡CH + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃Here, the acetylide anion (CH₃-C≡C⁻) has a negative charge on the terminal carbon. This anion then acts as a nucleophile, for instance, in an SN2 reaction:
CH₃-C≡C⁻Na⁺ + CH₃-Br → CH₃-C≡C-CH₃ + NaBrThe negatively charged carbon of the acetylide attacks the electrophilic carbon of methyl bromide.
💡 Prevention Tips:
- Reinforce Hybridization Effects: Understand that sp-hybridized carbon is more electronegative, making the terminal C-H bond acidic.
- Always Assign Charges Correctly: Terminal hydrogens are δ⁺, and the resulting acetylide carbon is always negatively charged (C⁻).
- Identify Reactivity: A negatively charged species (anion) acts as a nucleophile or a base, not an electrophile.
- Practice Electron Flow: Trace the movement of electrons during deprotonation and nucleophilic attack to solidify charge understanding.
JEE_Main
Minor
Approximation
❌ <span style='color: #FF0000;'>Overgeneralizing "Strong Base" for Alkyne Deprotonation</span>
Students often incorrectly assume that common strong bases like NaOH (sodium hydroxide) or NaOR (sodium alkoxide) are sufficient to quantitatively deprotonate a terminal alkyne and form an acetylide anion.
💭 Why This Happens:
This mistake stems from a general understanding that terminal alkynes are acidic and that bases like NaOH are considered 'strong' in many contexts. However, students frequently overlook the crucial aspect of relative pKa values. Terminal alkynes (with a pKa value of approximately 25) are significantly weaker acids than water (pKa ~15.7) or alcohols (pKa ~16-18). Consequently, bases such as NaOH or NaOR are not strong enough to shift the deprotonation equilibrium fully towards the formation of the acetylide, resulting in very little product.
✅ Correct Approach:
To effectively deprotonate a terminal alkyne, a base that is significantly stronger than the conjugate base of the alkyne is required. The pKa of the conjugate acid of the chosen base must be substantially higher than the pKa of the alkyne (ideally by at least 10 units for a nearly complete reaction). Therefore, bases like sodium amide (NaNH₂), sodium hydride (NaH), or organolithium reagents (e.g., BuLi) are necessary. These bases have conjugate acids (NH₃ pKa ~38, H₂ pKa ~35, alkanes pKa ~50) with pKa values much higher than 25, ensuring the equilibrium lies far to the right, yielding quantitative acetylide formation.
📝 Examples:
❌ Wrong:
CH≡CH + NaOH → X (Acetylene is not significantly deprotonated by NaOH)
✅ Correct:
CH≡CH + NaNH₂ → CH≡C⁻Na⁺ + NH₃ (Quantitative formation of sodium acetylide)💡 Prevention Tips:
- Always compare the pKa of the terminal alkyne (approx. 25) with the pKa of the conjugate acid of the base being used.
- For effective and quantitative deprotonation, the pKa of the conjugate acid of the base must be significantly higher (by 10 or more units) than the pKa of the alkyne.
- Remember the hierarchy for alkyne deprotonation in JEE: NaNH₂, NaH, BuLi are strong enough; NaOH, NaOR are generally not.
- (JEE Main Tip) While the exact pKa values might not be asked, understanding their relative magnitudes is crucial for predicting reaction feasibility.
JEE_Main
Minor
Other
❌ <span style='color: #FF0000;'>Confusing Terminal Alkynes with Internal Alkynes</span>
Students often fail to distinguish between terminal alkynes (R-C≡C-H) and internal alkynes (R-C≡C-R'), treating them interchangeably. This is critical because the acidic C-H proton in terminal alkynes dictates a different reactivity, especially with strong bases. Ignoring this leads to incorrect predictions in synthesis.
💭 Why This Happens:
Over-generalization: Assuming all triple bonds react identically, regardless of their position.
Lack of structural analysis: Not closely examining the position of the triple bond and the presence of hydrogen atoms directly attached to it.
Underestimating acidity: Overlooking the significance of the relatively acidic nature of the terminal alkyne's proton.
✅ Correct Approach:
Always, as a first step, identify whether the given alkyne is terminal or internal. This crucial distinction dictates the reaction pathway:
This differentiation is crucial for predicting reaction pathways and products accurately.
Terminal Alkynes: Possess an acidic hydrogen (pKa ~25). They react with strong bases (e.g., NaNH₂, Grignard reagents) to form nucleophilic acetylide anions (R-C≡C-Na+), which are excellent nucleophiles for C-C bond formation (e.g., alkylation).
Internal Alkynes: Lack acidic hydrogens at the triple bond. They do not react with strong bases via deprotonation and cannot form acetylide anions for subsequent alkylation or similar reactions.
This differentiation is crucial for predicting reaction pathways and products accurately.
📝 Examples:
❌ Wrong:
Assuming 2-butyne (CH₃-C≡C-CH₃) will react with NaNH₂ followed by CH₃I to yield 2-pentyne:
CH₃-C≡C-CH₃ + NaNH₂ (no acidic proton) ⟶ No reactionReasoning: Internal alkynes do not have a proton attached to the sp-hybridized carbon, so they cannot be deprotonated by NaNH₂.
✅ Correct:
Considering the reaction of 1-butyne (CH₃-CH₂-C≡C-H) with NaNH₂ followed by CH₃I:
1. CH₃-CH₂-C≡C-H + NaNH₂ ⟶ CH₃-CH₂-C≡C-Na+ + NH₃
2. CH₃-CH₂-C≡C-Na+ + CH₃I ⟶ CH₃-CH₂-C≡C-CH₃ (2-pentyne) + NaIReasoning: The terminal proton is acidic, allowing deprotonation by NaNH₂ to form an acetylide anion, which then acts as a nucleophile in an SN2 reaction with methyl iodide.
💡 Prevention Tips:
Always draw the structure: Visually confirm the position of the triple bond and any attached hydrogens.
Check for acidic protons: Before assuming deprotonation, verify the presence of a C-H bond directly on an sp-hybridized carbon.
Practice specific reactions: Work through problems that explicitly require distinguishing between terminal and internal alkyne reactivity.
JEE Specific: This distinction is frequently tested in JEE Main, especially in multi-step synthesis or matching type questions.
JEE_Main
Minor
Other
❌ Ignoring Tautomerism and Specific Catalysts in Alkyne Hydration
Students frequently overlook the crucial role of specific catalysts (HgSO₄/H₂SO₄) for alkyne hydration and fail to correctly identify the final product due to neglecting the tautomerization of the unstable enol intermediate to a more stable carbonyl compound (aldehyde or ketone).
💭 Why This Happens:
This mistake often stems from a lack of attention to reaction mechanism details and specific reagent requirements. Students might incorrectly stop at the enol stage or simply add water across the triple bond without considering the subsequent rearrangement, especially for unsymmetrical alkynes.
✅ Correct Approach:
Always remember that alkyne hydration is catalyzed by mercuric sulfate in dilute sulfuric acid. The initial product is an enol, which is unstable and rapidly undergoes keto-enol tautomerization to form a more stable aldehyde (from ethyne) or ketone (from other alkynes, following Markovnikov's rule).
📝 Examples:
❌ Wrong:
Reaction: Propyne + H₂O
Incorrect Product: Propen-2-ol (the enol intermediate) or Propan-1-ol (incorrect addition).
✅ Correct:
Reaction: Propyne + H₂O (HgSO₄/H₂SO₄)
Correct Product: Propan-2-one (Acetone)
Explanation: Water adds to propyne according to Markovnikov's rule, forming an enol (Propen-2-ol). This enol then tautomerizes to the more stable ketone, Propan-2-one.
💡 Prevention Tips:
- Memorize Catalysts: Always associate alkyne hydration with the catalyst combination of HgSO₄/H₂SO₄.
- Understand Tautomerism: Recognize that the formation of an enol is always followed by its rapid conversion to a more stable keto or aldehyde form. This is a fundamental concept for these reactions.
- Apply Markovnikov's Rule: For unsymmetrical alkynes, the 'H' of H₂O goes to the carbon with more hydrogens, and the 'OH' goes to the carbon with fewer hydrogens, leading to the formation of a ketone, except for ethyne which forms ethanal (acetaldehyde).
CBSE_12th
Minor
Approximation
❌ Misjudging the Relative Acidity of Terminal Alkynes
Students often correctly identify terminal alkynes as acidic due to the high s-character of the sp-hybridized carbon. However, a common minor error is to over-approximate their acidity, treating them as strong acids or equating their strength with carboxylic acids. This leads to incorrect predictions about reactions with weaker bases like water or alcohols.
💭 Why This Happens:
- Insufficient emphasis on 'relative acidity' compared to common reagents. While CBSE emphasizes the concept of acidity, JEE often requires deeper understanding of relative strengths (pKa values).
- Lack of clear comparison with other acidic compounds (e.g., water, alcohols, carboxylic acids) in terms of required base strength for deprotonation.
- Generalization from 'more acidic than alkenes/alkanes' to simply 'acidic enough for any base.'
✅ Correct Approach:
Understand that while terminal alkynes are acidic (pKa around 25), they are weak acids. They are stronger than ammonia (pKa ~38) but significantly weaker than water (pKa ~15.7), alcohols (pKa ~16-18), or carboxylic acids (pKa ~4-5). They react only with strong bases whose conjugate acids have a pKa higher than 25 (e.g., NaNH₂, Grignard reagents, organolithium reagents).
📝 Examples:
❌ Wrong:
Predicting a reaction between propyne and NaOH(aq):
This reaction does not proceed significantly because NaOH is not a strong enough base to deprotonate propyne effectively (pKa of H₂O ~15.7, propyne ~25).
CH≡C-CH₃ + NaOH(aq) → Na⁺⁻C≡C-CH₃ + H₂O
This reaction does not proceed significantly because NaOH is not a strong enough base to deprotonate propyne effectively (pKa of H₂O ~15.7, propyne ~25).
✅ Correct:
Predicting a reaction between propyne and sodium amide (NaNH₂) in liquid ammonia:
This reaction proceeds effectively because NaNH₂ is a very strong base (pKa of NH₃ ~38, propyne ~25).
CH≡C-CH₃ + NaNH₂ (liq. NH₃) → Na⁺⁻C≡C-CH₃ + NH₃
This reaction proceeds effectively because NaNH₂ is a very strong base (pKa of NH₃ ~38, propyne ~25).
💡 Prevention Tips:
- Memorize key pKa benchmarks: Carboxylic acids (~4-5), Water (~15.7), Alcohols (~16-18), Terminal Alkynes (~25), Ammonia (~38).
- Apply Acid-Base equilibrium principle: A reaction favors the formation of the weaker acid and weaker base. For deprotonation to occur, the base used must be stronger than the conjugate base of the alkyne.
- Do not assume all 'acidic' compounds react with all 'basic' compounds. Always contextualize the relative strength of the acid and base involved.
CBSE_12th
Minor
Sign Error
❌ Incorrect Charge Assignment and Role of Acetylide Anions
Students frequently understand that terminal alkynes are acidic due to the s-character of the sp-hybridized carbon, which makes the C-H bond more polar. However, a common 'sign error' arises when representing the products of deprotonation or their subsequent reactions. They might correctly remove a proton (H⁺) but then fail to assign the negative charge correctly to the carbon atom in the resulting acetylide anion (R-C≡C:⁻), or they misinterpret its role as a nucleophile by not showing the electron pair attacking an electrophile.
💭 Why This Happens:
This mistake often occurs due to a lack of detailed mechanistic understanding of acid-base reactions and nucleophilic substitutions. Students focus on the net reaction but overlook the precise electron movement. They might remember 'H is removed' but forget that the bonding electron pair remains with the carbon, leading to a negative charge. Additionally, confusion with other functional groups or overlooking the carbanion's strong nucleophilic nature contributes to incorrect representations.
✅ Correct Approach:
Always remember that when a terminal alkyne loses its proton (H⁺), the electron pair from the C-H bond remains on the carbon atom, forming a carbanion called an acetylide anion (R-C≡C:⁻). This acetylide anion is a strong nucleophile and its negative charge allows it to readily attack electrophilic centers (like alkyl halides in SN2 reactions) to form new carbon-carbon bonds. Always explicitly draw the lone pair and the negative charge on the carbon.
📝 Examples:
❌ Wrong:
| Incorrect Representation | Reason for Error |
|---|---|
R-C≡C-H + NaNH₂ → R-C≡C + Na⁺ + NH₃ | The carbon atom in R-C≡C is shown without a negative charge, implying a radical or a cation, which is incorrect. |
R-C≡CH + R'-X → R-C≡CR' + HX (without showing intermediate steps and charges) | This omits the nucleophilic attack by the acetylide anion, potentially leading to a misunderstanding of the mechanism and charges involved. |
✅ Correct:
| Correct Representation | Explanation |
|---|---|
R-C≡C-H + NaNH₂ → R-C≡C:⁻ + Na⁺ + NH₃ | Shows the correct deprotonation, with the negative charge and lone pair on the sp-hybridized carbon. |
R-C≡C:⁻ + R'-X → R-C≡CR' + X⁻(Arrow from lone pair on C to R', arrow from R'-X bond to X) | Clearly depicts the acetylide anion as a nucleophile, attacking the electrophilic carbon of the alkyl halide, leading to a new C-C bond formation. |
💡 Prevention Tips:
- Draw Mechanisms: Practice drawing the full reaction mechanisms, including electron-pushing arrows, for deprotonation and subsequent nucleophilic attacks.
- Label Charges: Always explicitly write the negative charge on the carbon of the acetylide anion.
- Identify Nucleophiles/Electrophiles: Clearly identify the acetylide anion as a nucleophile and the species it attacks as an electrophile.
- JEE Specific: In multi-step synthesis problems, correctly identifying the charged intermediate is crucial for planning the next step.
- CBSE Specific: Even for direct reactions, showing the correct charges in intermediates can fetch valuable marks.
CBSE_12th
Minor
Unit Conversion
❌ Inconsistent Volume Units in Stoichiometric Calculations
Students often use volumes in milliliters (mL) directly in stoichiometric calculations involving molarity (M), which is defined as moles per liter (mol/L), without converting mL to liters (L). This oversight leads to incorrect calculations for moles of reactants or products, impacting the overall answer.
💭 Why This Happens:
This mistake commonly occurs due to a lack of meticulous attention to unit consistency. Under exam pressure, students might recall the formula (moles = Molarity × Volume) but neglect the crucial detail that the volume (V) must be expressed in liters for molarity to be in mol/L. It's a fundamental unit conversion often overlooked in haste.
✅ Correct Approach:
Always ensure all units are consistent before performing calculations. If molarity is given in mol/L, then any volume used in the calculation must also be in liters. The conversion from milliliters to liters is straightforward: divide the volume in mL by 1000 (since 1 L = 1000 mL).
📝 Examples:
❌ Wrong:
Consider a reaction where 50 mL of a 0.2 M 1-butyne solution is used. To find the moles of 1-butyne:
Incorrect calculation: Moles = 0.2 mol/L × 50 mL = 10 moles. This result is absurdly high and physically impossible for such a small volume and concentration.
Incorrect calculation: Moles = 0.2 mol/L × 50 mL = 10 moles. This result is absurdly high and physically impossible for such a small volume and concentration.
✅ Correct:
Using the same scenario: 50 mL of a 0.2 M 1-butyne solution.
Correct calculation:
1. Convert volume to liters: 50 mL = 50 / 1000 L = 0.050 L.
2. Calculate moles: Moles = Molarity × Volume (in L)
= 0.2 mol/L × 0.050 L
= 0.01 moles.
This is the correct amount of 1-butyne.
Correct calculation:
1. Convert volume to liters: 50 mL = 50 / 1000 L = 0.050 L.
2. Calculate moles: Moles = Molarity × Volume (in L)
= 0.2 mol/L × 0.050 L
= 0.01 moles.
This is the correct amount of 1-butyne.
💡 Prevention Tips:
- Explicitly write units: Always include units (e.g., mol, L, mL) with numerical values throughout your calculations. This helps in verifying unit cancellation and consistency.
- Pre-calculation check: Before starting complex numerical problems, list all given quantities and their units. Identify any necessary unit conversions and perform them first.
- Molarity rule: For any calculation involving molarity (M), remember that volume (V) must always be in liters (L) for the unit 'mol/L' to be correctly applied.
- CBSE vs. JEE: While this is a common error in both, JEE problems often involve more complex multi-step calculations where such a mistake can cascade and lead to completely wrong final answers, making unit vigilance even more critical.
CBSE_12th
Minor
Formula
❌ <b>Confusion in Reagents for Partial Hydrogenation of Alkynes</b>
Students often confuse the specific reagents used for partial hydrogenation of alkynes, leading to incorrect stereochemistry (cis vs. trans) of the alkene product. This directly impacts the reaction's outcome and product identification.
💭 Why This Happens:
The similarity in the goal (partial hydrogenation) but distinct reagents and stereochemical results (Lindlar's catalyst for cis; Na in liquid NH₃ for trans) often leads to mix-ups. Lack of explicit memorization of these reagent-product pairs is a key reason.
✅ Correct Approach:
- Lindlar's catalyst (Pd/CaCO₃, quinoline, or BaSO₄) promotes syn-addition of hydrogen, yielding a cis-alkene.
- Sodium in liquid ammonia (Birch reduction) promotes anti-addition of hydrogen, yielding a trans-alkene.
- Explicitly link each reagent system to its specific stereochemical outcome.
📝 Examples:
❌ Wrong:
(Incorrect usage of Na/Liq. NH₃ to obtain a cis product from an alkyne.)
CH₃-C≡C-CH₃ + H₂ Na/Liq. NH₃--> CH₃-CH=CH-CH₃ (cis-2-butene)
(Incorrect usage of Na/Liq. NH₃ to obtain a cis product from an alkyne.)
✅ Correct:
| Reagent for Partial Hydrogenation | Expected Product Stereochemistry (e.g., from 2-Butyne) |
|---|---|
| H₂/Lindlar's catalyst | CH₃-CH=CH-CH₃ (cis-2-butene) |
| Na in liq. NH₃ | CH₃-CH=CH-CH₃ (trans-2-butene) |
💡 Prevention Tips:
- Create flashcards or a comparison table for Lindlar's catalyst and Birch reduction, explicitly noting their reagents and stereochemical outcomes.
- Focus on understanding the specific role and mechanism of each reagent system.
- This distinction is vital for JEE for accurate product prediction; for CBSE, knowing the reagents and their general stereochemical impact is important.
CBSE_12th
Minor
Calculation
❌ Miscounting Acidic Hydrogens and Incorrect Reagent Stoichiometry
Students frequently make minor 'calculation' errors by incorrectly counting the number of acidic hydrogens in a given alkyne structure or by failing to use the correct stoichiometric amount (equivalents) of base required for complete deprotonation. This leads to predicting partially reacted products or an incorrect final product, especially when dealing with terminal alkynes and strong bases.
💭 Why This Happens:
- Lack of Attention to Detail: Students might quickly scan a structure and miss the presence of multiple terminal alkyne groups (less common in CBSE, but possible in diynes).
- Confusing Bonds with Hydrogens: Mistaking the number of C≡C bonds for the number of acidic hydrogens.
- Ignoring Stoichiometry: Overlooking the mole ratio requirement for complete deprotonation (e.g., assuming 1 equivalent of base is sufficient for all acidic hydrogens, regardless of how many there are).
✅ Correct Approach:
To avoid this error, follow these steps:
- Identify All Terminal Alkynes: Carefully examine the alkyne structure for all -C≡CH groups. Each such group contains one acidic hydrogen.
- Count Acidic Hydrogens: Determine the total number of hydrogens directly attached to sp-hybridized carbon atoms.
- Match Reagent Equivalents: For complete deprotonation (e.g., using strong bases like NaNH₂), the number of equivalents of base must match the number of acidic hydrogens identified. Each acidic hydrogen requires one equivalent of a strong base.
📝 Examples:
❌ Wrong:
Reactant: But-1,3-diyne (HC≡C-C≡CH)
Incorrect Reaction: HC≡C-C≡CH + 1 equivalent of NaNH₂ → Na⁺⁻C≡C-C≡CH + NH₃
Reason for Error: The student only reacted one of the two terminal alkyne groups, incorrectly assuming only one acidic hydrogen or insufficient base was provided for complete deprotonation.
✅ Correct:
Reactant: But-1,3-diyne (HC≡C-C≡CH)
Correct Reaction: HC≡C-C≡CH + 2 equivalents of NaNH₂ → Na⁺⁻C≡C-C≡C⁻Na⁺ + 2NH₃
Explanation: But-1,3-diyne has two terminal alkyne groups, each with an acidic hydrogen. Therefore, two equivalents of a strong base like NaNH₂ are required for complete deprotonation to form the disodium salt.
💡 Prevention Tips:
- Always draw the full structure: Don't just rely on the name. A drawn structure helps visualize all terminal hydrogens.
- Count carefully: Explicitly count the 'H' atoms attached to 'sp' hybridized carbons.
- Verify stoichiometry: For reactions involving strong bases with terminal alkynes, ensure the equivalents of base match the number of acidic hydrogens to achieve the desired product (monosubstituted vs. disubstituted).
- CBSE vs. JEE: In CBSE, simple terminal alkynes (like propyne) are more common, having only one acidic hydrogen. However, be vigilant for diynes or other poly-functional molecules in advanced problems (JEE scope).
CBSE_12th
Minor
Conceptual
❌ Confusing Acidity of Terminal vs. Non-Terminal Alkynes
Students frequently assume all alkynes are acidic or incorrectly attribute acidic properties to internal (non-terminal) alkynes. This leads to errors in predicting reactions with strong bases or active metals.
💭 Why This Happens:
This misconception arises from a lack of understanding of the structural requirement for alkyne acidity. Only terminal alkynes (RC≡CH) possess an acidic hydrogen atom directly attached to an sp-hybridized carbon. The high electronegativity of the sp-hybridized carbon stabilizes the resulting carbanion (acetylide ion), making the C-H bond weak enough to be deprotonated by strong bases. Internal alkynes (RC≡CR') lack such a hydrogen atom.
✅ Correct Approach:
Always identify if the alkyne is terminal (has a hydrogen directly on the triple bond) or internal (no hydrogen on the triple bond carbons). Remember that only terminal alkynes show acidic character. They react with very strong bases like sodamide (NaNH₂) or active metals like sodium to form acetylides. This concept is crucial for both CBSE and JEE.
📝 Examples:
❌ Wrong:
A student might write:
CH₃−C≡C−CH₃ + Na → No reaction, but sometimes students incorrectly predict reaction.
✅ Correct:
The correct understanding is:
CH₃−C≡C−CH + Na → CH₃−C≡C⁻Na⁺ + ½ H₂
Or with a stronger base:
CH₃−C≡C−H + NaNH₂ → CH₃−C≡C⁻Na⁺ + NH₃
💡 Prevention Tips:
- Visualize the structure: Before predicting reactivity, always draw the structure and check if a hydrogen atom is directly bonded to an sp-hybridized carbon.
- Key definition: A terminal alkyne has at least one hydrogen attached to an sp-hybridized carbon. Internal alkynes do not.
- Reagent association: Reactions with NaNH₂ (sodamide) or Na metal are characteristic tests/reactions for terminal alkynes only, due to their acidic hydrogen.
CBSE_12th
Minor
Approximation
❌ <span style='color: #FF0000;'>Approximating Terminal Alkyne Acidity</span>
Students often correctly identify terminal alkynes as acidic due to the sp-hybridized carbon's electronegativity but frequently misjudge their relative acid strength. This leads to errors in predicting whether a specific base will deprotonate the alkyne, either overestimating its acidity (e.g., expecting reaction with weak bases like NaOH) or underestimating it in certain contexts.
💭 Why This Happens:
This mistake stems from a lack of precise comparative understanding of pKa values for various functional groups. Students may understand the concept of sp-hybridization and acidity but fail to relate it to the actual strength required for a reaction with a given base, approximating 'acidic' to mean 'reacts with any base' or 'reacts with common bases'.
✅ Correct Approach:
Understand the typical pKa range of terminal alkynes is approximately 25. Compare this to the pKa of the conjugate acid of the base being used. For a deprotonation reaction to be favorable, the pKa of the alkyne must be significantly lower than the pKa of the conjugate acid of the base.
- Terminal alkynes are stronger acids than water (pKa ~15.7), alcohols (pKa ~16-18), and ammonia (pKa ~38).
- Terminal alkynes are weaker acids than carboxylic acids (pKa ~4-5) and phenols (pKa ~10).
- They require very strong bases like sodium amide (NaNH2, conjugate acid NH3 pKa ~38), Grignard reagents, or alkyllithiums for significant deprotonation.
- They generally do NOT react significantly with weaker bases like NaOH (conjugate acid H2O pKa ~15.7) or NaHCO3.
📝 Examples:
❌ Wrong:
Assuming 1-butyne will react with sodium hydroxide (NaOH) to form sodium butynide, due to a general understanding of its acidity.
CH3CH2C≡CH + NaOH ⟶ CH3CH2C≡CNa + H2OThis reaction is not favorable because water (conjugate acid of NaOH) is a stronger acid (pKa ~15.7) than the terminal alkyne (pKa ~25). The equilibrium lies far to the left.
✅ Correct:
Using a sufficiently strong base, such as sodium amide (NaNH2), to deprotonate 1-butyne and form the corresponding alkynide.
CH3CH2C≡CH + NaNH2 → CH3CH2C≡CNa + NH3This reaction is highly favorable because ammonia (NH3, conjugate acid of NaNH2) is a much weaker acid (pKa ~38) than the terminal alkyne (pKa ~25). The equilibrium lies far to the right.💡 Prevention Tips:
- Memorize Key pKa Values: Have a rough idea of the pKa values for common acidic functional groups (e.g., terminal alkynes ~25, water ~15.7, alcohols ~16-18, carboxylic acids ~4-5, phenols ~10, ammonia ~38).
- Apply Acid-Base Principles: Always compare the acidity of the reactant (alkyne) with the basicity of the reagent. A reaction proceeds significantly only if the conjugate acid formed is weaker than the original acid (i.e., the pKa of the original acid is lower than the pKa of the conjugate acid of the base).
- Practice Reaction Prediction: Work through problems involving various bases to predict the feasibility of alkyne deprotonation.
JEE_Advanced
Minor
Sign Error
❌ Incorrect Base Selection for Terminal Alkyne Deprotonation
A common 'sign error' students make is incorrectly assuming that readily available strong bases like NaOH or KOH are potent enough to deprotonate a terminal alkyne and form an acetylide ion. This leads to predicting an incorrect reaction outcome.
💭 Why This Happens:
This error stems from a lack of precise understanding of relative acid-base strengths and pKa values. Students often categorize NaOH as a 'strong base' without considering its specific strength relative to a terminal alkyne's acidity (pKa ~25). They might incorrectly apply the general concept of 'strong base' without a quantitative comparison.
✅ Correct Approach:
To deprotonate a terminal alkyne, a base much stronger than hydroxide is required. The conjugate acid of the base must have a pKa significantly higher than 25. Examples of appropriate bases include sodium amide (NaNH₂, pKa of NH₃ ≈ 38), sodium hydride (NaH), or organolithium reagents like n-butyllithium (n-BuLi, pKa of butane ≈ 50).
📝 Examples:
❌ Wrong:
CH≡CH + NaOH → CH≡C⁻Na⁺ + H₂O (Incorrect reaction)
✅ Correct:
This reaction proceeds because the acetylide ion (pKa of conjugate acid ~25) is a weaker base than the amide ion (pKa of conjugate acid ~38), driving the equilibrium to the right.
CH≡CH + NaNH₂ → CH≡C⁻Na⁺ + NH₃ (Correct reaction)
This reaction proceeds because the acetylide ion (pKa of conjugate acid ~25) is a weaker base than the amide ion (pKa of conjugate acid ~38), driving the equilibrium to the right.
💡 Prevention Tips:
- Quantitative Comparison: Always compare the pKa of the terminal alkyne (~25) with the pKa of the conjugate acid of the base being used.
- Memorize Key Bases: Remember that for alkynes, bases like NaNH₂, NaH, and organolithium reagents are essential. NaOH/KOH are insufficient.
- Acidity Order: Keep in mind the general acidity order: Carboxylic acids > Water/Alcohols > Terminal Alkynes > Ammonia > Alkanes. This helps in understanding conjugate base strengths.
- JEE Focus: For JEE Advanced, precise understanding of which reagents achieve specific transformations is crucial.
JEE_Advanced
Minor
Unit Conversion
❌ <strong>Incorrectly Converting Mass to Moles in Stoichiometric Calculations</strong>
A common oversight for students is failing to convert the given mass (grams, milligrams, etc.) of an alkyne or a reacting agent into its corresponding molar quantity before applying stoichiometric ratios from the balanced chemical equation. This directly leads to erroneous calculations for product yields, limiting reagents, or the amount of reactant required for a specific reaction involving alkynes.
💭 Why This Happens:
This mistake often arises from an over-eagerness to apply stoichiometric coefficients directly to mass values, or a simple oversight in unit tracking. Students may rush calculations without ensuring all quantities are in moles, which is the fundamental unit for stoichiometry. It can also stem from a weak foundation in basic mole concepts.
✅ Correct Approach:
Always convert all given masses of reactants or products into moles using their respective molar masses. Similarly, for solutions, convert volume and concentration into moles (Moles = Molarity × Volume in Liters). Only once all quantities are expressed in moles should stoichiometric ratios from the balanced equation be applied. For JEE Advanced, this step is critical as multi-step reactions or complex limiting reagent problems heavily rely on accurate molar calculations.
📝 Examples:
❌ Wrong:
When reacting 10 g of Propyne (CH3C≡CH) with 10 g of Sodamide (NaNH2) for deprotonation, a student might incorrectly assume they are present in roughly equal amounts, or compare their masses directly to determine the limiting reagent. This ignores their different molar masses.
✅ Correct:
Consider the deprotonation of Propyne (CH3C≡CH, Molar Mass = 40 g/mol) with Sodamide (NaNH2, Molar Mass = 39 g/mol). If 10 g of each reactant is used:
- Moles of Propyne = 10 g / 40 g/mol = 0.25 mol
- Moles of Sodamide = 10 g / 39 g/mol = 0.256 mol
💡 Prevention Tips:
- Write a Balanced Equation: Always start by writing and balancing the chemical equation.
- Convert to Moles First: Make it a habit to convert all given masses or volumes/concentrations to moles immediately.
- Track Units: Pay meticulous attention to units throughout your calculations to ensure consistency (e.g., g/mol, mol/L).
- Practice Stoichiometry: Regularly solve numerical problems involving mass-to-mole conversions in various organic reactions, particularly those related to alkynes.
JEE_Advanced
Minor
Formula
❌ Misidentifying Products of Oxidative Cleavage of Terminal Alkynes
Students often correctly identify strong oxidizing agents (like hot KMnO4 or ozonolysis with oxidative workup) for alkynes but fail to correctly write the structural formula of all products, especially when dealing with terminal alkynes. The common error is not fully oxidizing the terminal carbon atom.
💭 Why This Happens:
This mistake stems from a lack of precise understanding of how the triple bond is cleaved and the subsequent oxidation states of the resulting carbon fragments under strong conditions. Students might recall that alkynes generally yield carboxylic acids but forget the specific outcome for the terminal -CH group, which is highly susceptible to further oxidation to CO2.
✅ Correct Approach:
For strong oxidative cleavage of alkynes (e.g., hot alkaline KMnO4 or O3 followed by H2O2):
- Internal Alkynes: The triple bond cleaves, and each carbon atom involved in the triple bond is oxidized to a carboxylic acid (R-COOH).
- Terminal Alkynes: The internal carbon of the triple bond is oxidized to a carboxylic acid (R-COOH), while the terminal -CH group is completely oxidized to carbon dioxide (CO2) and water.
📝 Examples:
❌ Wrong:
CH3-CH2-C≡CH (1-Butyne)
+ Hot KMnO4/H+
→ CH3-CH2-COOH (Propanoic Acid) + HCOOH (Formic Acid)
Why this is wrong: Formic acid (HCOOH) is an aldehyde derivative and is readily oxidized further to CO2 and H2O under strong oxidative conditions.
✅ Correct:
CH3-CH2-C≡CH (1-Butyne)
+ Hot KMnO4/H+
→ CH3-CH2-COOH (Propanoic Acid) + CO2 + H2O
This is correct: The terminal carbon fully oxidizes to carbon dioxide.
💡 Prevention Tips:
- JEE Advanced Tip: Always remember the complete oxidation of the terminal carbon of an alkyne to CO2 under strong oxidative conditions.
- Clearly distinguish between terminal and internal alkynes before predicting products.
- Practice writing the full structural formulas for all products, ensuring carbon atom balance and correct functional groups.
- CBSE vs. JEE: While CBSE might accept HCOOH as an intermediate product, JEE Advanced expects the final, most stable oxidation product under given strong conditions.
JEE_Advanced
Minor
Conceptual
❌ <strong><span style='color: #FF0000;'>Confusing Acidity of Terminal vs. Internal Alkynes and Misjudging Base Strength for Deprotonation</span></strong>
Students often correctly identify terminal alkynes as acidic due to the sp-hybridized carbon. However, they might incorrectly assume internal alkynes also exhibit similar acidity or may not correctly judge the strength of the base required to deprotonate a terminal alkyne, leading to incorrect reaction predictions, especially in multi-step syntheses for JEE Advanced.
💭 Why This Happens:
- Lack of Structural Specificity: Students understand 'alkynes are acidic' but overlook that only C-H bonds directly attached to an sp-hybridized carbon are acidic. Internal alkynes lack such a proton.
- Insufficient pKa Comparison: An incomplete understanding of the relative pKa values of terminal alkynes (~25) compared to common bases and other acids (e.g., water ~15.7, ethanol ~16, ammonia ~38) leads to using inappropriate bases.
- Over-generalization: Applying the acidity concept broadly without considering the specific structural requirements and base strength needed.
✅ Correct Approach:
- Understand the Basis of Acidity: The high s-character (50%) of the sp-hybridized carbon in terminal alkynes makes it more electronegative, polarizing the C-H bond and stabilizing the resulting acetylide anion. This specific C-H bond is absent in internal alkynes.
- Compare pKa Values: Terminal alkynes are stronger acids than ammonia (pKa ~38), alkenes (pKa ~44), and alkanes (pKa ~50). Thus, strong bases like NaNH2, NaH, or Grignard reagents are effective for deprotonation. They are, however, weaker acids than water or alcohols, so bases like NaOH (in aqueous solution) or RO- are generally not strong enough to deprotonate them significantly (CBSE vs. JEE: JEE Advanced often tests the subtle pKa differences and implications for synthetic routes).
- Identify only Terminal Alkynes for Acidic Reactions: Always check for the presence of a terminal proton before considering deprotonation reactions.
📝 Examples:
❌ Wrong:
1. Incorrect acidity for internal alkyne:
2. Insufficient base strength for terminal alkyne:
CH3-C≡C-CH3 + NaNH2 → No reaction (as 2-butyne has no acidic proton)2. Insufficient base strength for terminal alkyne:
CH3-CH2-C≡CH + NaOH (aq) → No significant reaction (NaOH is not strong enough to deprotonate 1-butyne effectively)
✅ Correct:
1. Correct deprotonation of terminal alkyne:
(Sodium butynide is formed, which can then act as a nucleophile in further reactions, e.g., SN2).
2. Subsequent Reaction using Acetylide Anion:
(Formation of an internal alkyne via SN2 reaction with an alkyl halide).
CH3-CH2-C≡CH + NaNH2 → CH3-CH2-C≡C-Na+ + NH3(Sodium butynide is formed, which can then act as a nucleophile in further reactions, e.g., SN2).
2. Subsequent Reaction using Acetylide Anion:
CH3-CH2-C≡C-Na+ + CH3I → CH3-CH2-C≡C-CH3 + NaI(Formation of an internal alkyne via SN2 reaction with an alkyl halide).
💡 Prevention Tips:
- Visualize the Proton: Always draw the structure and identify if a C-H bond is directly attached to an sp-hybridized carbon.
- Memorize Key pKa Values: Keep in mind the approximate pKa values for terminal alkynes (~25), water (~15.7), alcohols (~16-18), and ammonia (~38) to quickly assess feasible acid-base reactions.
- Practice Reaction Conditions: Focus on recognizing which bases (e.g., NaNH2, NaH) are strong enough to deprotonate terminal alkynes and differentiate them from weaker bases (e.g., NaOH).
JEE_Advanced
Minor
Calculation
❌ Incorrectly Determining Stoichiometry for Deprotonation of Polyalkynes or Full Addition Reactions
Students frequently miscalculate the exact number of moles of base required to deprotonate a polyalkyne (a molecule with multiple terminal alkyne groups) or the moles of reagent needed for complete addition across all pi bonds in an alkyne. This leads to errors in predicting the final product or understanding the reaction's extent, especially in JEE Advanced where precise quantitative understanding is critical.
💭 Why This Happens:
- Overlooking Multiple Reactive Sites: Students often fail to identify all terminal alkyne hydrogens in a complex molecule.
- Incomplete Understanding of Reaction Extent: Assuming partial addition when complete addition is implied by the reagent quantity, or vice-versa, without careful consideration of the stoichiometry provided.
- Rote Memorization without Conceptual Grasp: Knowing 'alkynes react with 2 moles of H2 for full saturation' but not applying it rigorously when specific mole ratios are given or when multiple alkyne groups are present.
✅ Correct Approach:
- Count Acidic Hydrogens: For acidity reactions, precisely count the number of H-atoms directly attached to sp-hybridized carbons (terminal alkynes). Each such hydrogen requires one mole of strong base for deprotonation.
- Count Pi Bonds: For addition reactions (hydrogenation, hydrohalogenation, halogenation), remember each triple bond consists of two pi bonds, each capable of undergoing an addition reaction. Therefore, two moles of a reagent (like H2, HX, X2) are typically required for complete saturation/addition across one triple bond.
- Match Stoichiometry: Carefully match the provided moles of reagent to the moles of the reactant, predicting the product based on the limiting reagent or the specified extent of reaction.
📝 Examples:
❌ Wrong:
Consider 1,3-butadiyne (HC≡C-C≡CH). Reacting 1 mole of 1,3-butadiyne with 1 mole of NaNH2 and expecting complete deprotonation to form Na+ -C≡C-C≡C- Na+. This is incorrect as 1,3-butadiyne has two acidic hydrogens.
✅ Correct:
To completely deprotonate 1 mole of 1,3-butadiyne (HC≡C-C≡CH), 2 moles of NaNH2 are required, forming Na+ -C≡C-C≡C- Na+.
Similarly, complete hydrogenation of 1 mole of 1-butyne (CH3CH2C≡CH) to butane (CH3CH2CH2CH3) requires 2 moles of H2 over a catalyst like Pd/C.
Similarly, complete hydrogenation of 1 mole of 1-butyne (CH3CH2C≡CH) to butane (CH3CH2CH2CH3) requires 2 moles of H2 over a catalyst like Pd/C.
💡 Prevention Tips:
- Systematic Counting: Before attempting any reaction, identify all acidic hydrogens or pi bonds in the reactant molecule.
- Balanced Equations: Always try to visualize or mentally balance the reaction equation, especially for multi-step or multi-site reactions.
- Practice Stoichiometric Problems: Regularly solve problems where specific mole ratios of reactants are given, and the product's structure depends on these ratios.
- Review Reaction Mechanisms: Understanding why a certain number of moles is consumed (e.g., each pi bond reacts separately) reinforces the stoichiometric understanding.
JEE_Advanced
Important
Approximation
❌ <span style='color: #FF0000;'>Misjudging the Relative Acidity of Terminal Alkynes</span>
Students often correctly identify terminal alkynes as acidic due to the s-character of the sp-hybridized carbon atom, which stabilizes the resulting carbanion. However, a common 'approximation understanding' error is to overestimate their acidity or incorrectly place them in the general acidity order. They frequently approximate terminal alkynes to be stronger acids than water or alcohols, which is generally incorrect, leading to errors in predicting acid-base reactions or selecting appropriate reagents for deprotonation.
💭 Why This Happens:
- Oversimplification: Remembering 'terminal alkynes are acidic' without understanding the extent of that acidity relative to other common compounds.
- Lack of Comparative pKa Knowledge: Not knowing the approximate pKa values for common functional groups for relative comparison.
- Focus on Isolated Facts: Failing to integrate the acidity of alkynes into a broader understanding of organic acid strengths and the factors affecting them.
✅ Correct Approach:
Understand that while terminal alkynes are indeed more acidic than alkanes and alkenes, and can react with strong bases like NaNH₂, they are weaker acids than water (pKa ~15.7), alcohols (pKa ~16-18), phenols (pKa ~10), and carboxylic acids (pKa ~5). Their pKa value is approximately 25. Always compare the pKa of the acid and its conjugate base with the pKa of the reactant base and its conjugate acid to predict the feasibility of a reaction. The equilibrium favors the formation of the weaker acid and weaker base.
📝 Examples:
❌ Wrong:
Scenario: Will propyne react with NaOH to form sodium propynide?
CH₃-C≡CH + NaOH → ?
Students often incorrectly assume NaOH is strong enough, leading to the product: CH₃-C≡C⁻Na⁺ + H₂O. This is wrong because water (conjugate acid of NaOH) is a stronger acid than propyne.
✅ Correct:
Correct Prediction for NaOH:
CH₃-C≡CH (pKa ~25) + NaOH (conjugate acid H₂O, pKa ~15.7) → No significant reaction
Since H₂O is a stronger acid than propyne, the equilibrium for deprotonation by NaOH lies overwhelmingly to the left. The reverse reaction (proton transfer from water to propynide) is favored.
Correct Base for Deprotonation:
CH₃-C≡CH (pKa ~25) + NaNH₂ (conjugate acid NH₃, pKa ~38) → CH₃-C≡C⁻Na⁺ + NH₃
Here, ammonia (NH₃) is a weaker acid than propyne. Thus, the amide ion (NH₂⁻) is a strong enough base to deprotonate propyne, and the equilibrium lies to the right, forming sodium propynide.
💡 Prevention Tips:
- Memorize Approximate pKa Values: Be familiar with the approximate pKa values for key functional groups (e.g., Carboxylic Acids ~5, Phenols ~10, Water ~15.7, Alcohols ~16-18, Terminal Alkynes ~25, Ammonia ~38, Alkenes ~44, Alkanes ~50).
- Compare Strengths Systematically: For any acid-base reaction, identify the acid and base on both sides. The reaction will proceed significantly in the direction that forms the weaker acid and weaker base.
- JEE Tip (CBSE and JEE Main): Questions often test this comparative acidity to determine if a specific acid-base reaction will proceed or to identify suitable reagents for selective deprotonation of terminal alkynes. Always verify the relative acidities.
JEE_Main
Important
Other
❌ <span style='color: #FF0000;'>Confusing Acidity of Terminal vs. Internal Alkynes</span>
Students often fail to distinguish the acidic nature of terminal alkynes (having a hydrogen directly bonded to an sp-hybridized carbon) from internal alkynes (which lack such a hydrogen). This leads to errors in predicting reactions with strong bases or using incorrect reagents for deprotonation.
💭 Why This Happens:
- Lack of understanding of the effect of hybridization on carbon's electronegativity and the stability of the conjugate base (acetylide anion).
- Generalizing that all hydrocarbons are non-acidic, overlooking the unique sp-hybridized C-H bond in terminal alkynes.
- Inadequate practice with reactions requiring deprotonation of terminal alkynes.
✅ Correct Approach:
- Identify Terminal Alkynes: A terminal alkyne has a hydrogen atom attached to an sp-hybridized carbon (R-C≡C-H).
- Acidity: This C-H bond is acidic (pKa ≈ 25) due to the high s-character (50%) of the sp-hybridized carbon, making it more electronegative and stabilizing the resulting carbanion (acetylide anion).
- Bases: Strong bases like NaNH₂ (sodamide), Na, RMgX (Grignard reagents), or n-BuLi are required to deprotonate terminal alkynes. Weaker bases like NaOH or NaHCO₃ are generally ineffective.
- Internal Alkynes: Internal alkynes (R-C≡C-R') do not have an acidic hydrogen and will not react with these strong bases to form acetylide anions.
📝 Examples:
❌ Wrong:
Students might incorrectly attempt to deprotonate an internal alkyne or use a weak base for a terminal alkyne.
CH₃-C≡C-CH₃ + NaNH₂ → No reaction (or incorrect product)CH₃-C≡C-H + NaOH → No reaction (NaOH is too weak a base for terminal alkyne)
✅ Correct:
The acidity of terminal alkynes is exploited in C-C bond formation (chain lengthening).
1. CH₃-C≡C-H + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃
2. CH₃-C≡C⁻Na⁺ + CH₃-Br → CH₃-C≡C-CH₃ + NaBr💡 Prevention Tips:
- Understand Hybridization: Revisit the concept of hybridization and its effect on bond character and acidity.
- Memorize pKa: Be aware that terminal alkynes have a pKa significantly lower than alkanes/alkenes, placing them in the range where strong bases can deprotonate them.
- Reagent Specificity: Know which bases are strong enough to deprotonate terminal alkynes (e.g., NaNH₂, NaH, Grignard reagents).
- Practice Synthesis Problems: Regularly solve multi-step synthesis problems involving alkynes, especially those requiring C-C bond formation.
JEE_Main
Important
Sign Error
❌ <span style='color: #ff0000;'>Sign Error: Incorrectly Identifying Acidic Hydrogens and Resultant Charges in Alkynes</span>
Students frequently make a fundamental 'sign error' by incorrectly identifying which hydrogens in alkynes are acidic or by misunderstanding the charge distribution after deprotonation. This critical error leads to:
- Assuming all alkynes possess acidic hydrogens, resulting in attempts to deprotonate internal alkynes.
- Failing to recognize the acidic nature of terminal alkynes, thereby missing crucial acid-base reactions.
- Incorrectly drawing the charge on the carbon after deprotonation, or misidentifying the leaving group during proton transfer.
💭 Why This Happens:
- Insufficient grasp of the relationship between hybridization (sp) and the electronegativity of carbon, which is pivotal for stabilizing the negative charge of the acetylide ion.
- Confusion between terminal (R-C≡C-H) and internal (R-C≡C-R') alkyne structures.
- Overlooking the relative pKa values, where terminal alkynes (pKa ~25) are significantly more acidic than alkenes (pKa ~44) or alkanes (pKa ~50), but still weaker acids than water or alcohols.
✅ Correct Approach:
The correct approach involves a systematic application of acid-base principles and an understanding of structural effects on acidity:
- Identify Terminal Alkynes: Remember that only the hydrogen directly attached to the sp-hybridized carbon of a terminal alkyne (R-C≡C-H) is acidic. Internal alkynes lack such a hydrogen.
- Understand Stabilization: The high s-character (50%) in an sp-hybridized orbital makes the carbon more electronegative. This enhanced electronegativity effectively stabilizes the resulting acetylide anion (R-C≡C⁻), making the proton relatively easy to remove by a sufficiently strong base.
- Assign Charges Correctly: Upon deprotonation, the carbon atom formally acquires a negative charge, forming a carbanion (acetylide ion), which is a powerful nucleophile or base in subsequent reactions.
📝 Examples:
❌ Wrong:
A common mistake is attempting to deprotonate an internal alkyne, for example:
Or, for a terminal alkyne, drawing the product without the negative charge on the carbon:
CH₃-C≡C-CH₃ + NaNH₂ → No reaction (or incorrect deprotonation of methyl groups)Or, for a terminal alkyne, drawing the product without the negative charge on the carbon:
R-C≡CH + Base → R-C≡C + BaseH (missing the negative charge).
✅ Correct:
For 1-butyne (CH₃CH₂C≡CH) reacting with sodium amide:
Here, the acidic hydrogen is correctly removed, and the negative charge is accurately placed on the sp-hybridized carbon atom, forming the sodium butynide (an acetylide ion). This acetylide ion can then participate in further reactions like alkylation.
CH₃CH₂C≡CH + NaNH₂ → CH₃CH₂C≡C⁻Na⁺ + NH₃Here, the acidic hydrogen is correctly removed, and the negative charge is accurately placed on the sp-hybridized carbon atom, forming the sodium butynide (an acetylide ion). This acetylide ion can then participate in further reactions like alkylation.
💡 Prevention Tips:
- Structure First: Always draw the full structure to unequivocally distinguish between terminal and internal alkynes.
- Hybridization Check: Mentally (or physically) check the hybridization of the carbon atom to which the hydrogen is attached. Only sp-hybridized C-H bonds are significantly acidic in alkynes.
- JEE Specific: Pay close attention to the reagents. The presence of strong bases such as NaNH₂, NaH, LDA, or organometallic reagents (e.g., Grignard reagents) is a strong indicator of an acid-base reaction involving a terminal alkyne. Internal alkynes will generally not react with these bases in an acid-base manner. For CBSE, understanding the fundamental concept of terminal alkyne acidity is key.
JEE_Main
Important
Conceptual
❌ Incorrectly Identifying Acidic Hydrogens and Choosing Inappropriate Bases for Alkyne Reactions
Students frequently fail to recognize that only terminal alkynes (R-C≡C-H) possess sufficiently acidic hydrogens to react with strong bases to form acetylide ions. They often attempt to deprotonate internal alkynes (R-C≡C-R') which lack acidic protons, or they use bases that are not strong enough for the deprotonation of terminal alkynes.
💭 Why This Happens:
This conceptual error stems from a lack of understanding of the acidity order of C-H bonds (sp > sp2 > sp3) and the relative strengths of various bases. Students often overgeneralize 'strong base' without considering the pKa values. For a successful acid-base reaction, the base must be stronger than the conjugate base of the acid being deprotonated. For example, OH- is not strong enough to deprotonate a terminal alkyne, but NH2- is.
✅ Correct Approach:
To correctly approach alkyne acidity:
- Identify if the alkyne is terminal (R-C≡C-H). Only these have acidic hydrogens.
- Recall the approximate pKa of terminal alkynes (around 25).
- Choose a base whose conjugate acid has a pKa significantly higher than 25. For instance, NaNH2 (conjugate acid NH3, pKa ~38) is suitable. NaOH (conjugate acid H2O, pKa ~15.7) is not strong enough.
- Internal alkynes do not participate in such acid-base reactions to form acetylides.
📝 Examples:
❌ Wrong:
Reaction: CH3-C≡C-CH3 + NaNH2 → No reaction
Explanation: 2-butyne is an internal alkyne; it has no acidic hydrogens. Thus, it cannot react with NaNH2 to form an acetylide.
✅ Correct:
Reaction: CH3-CH2-C≡C-H + NaNH2 (in liquid NH3) → CH3-CH2-C≡C-Na+ + NH3
Explanation: 1-butyne is a terminal alkyne with an acidic hydrogen. NaNH2 is a strong enough base to deprotonate it, forming sodium 1-butynide.
💡 Prevention Tips:
- Memorize Key pKas: Terminal alkynes (~25), Water (~15.7), Ammonia (~38). This helps predict reaction feasibility.
- Always Check for Terminal H: Before assuming an alkyne can react as an acid, confirm the presence of a hydrogen directly attached to the sp-hybridized carbon.
- JEE Advanced Context: Questions often test the selectivity of reagents. If a reagent requires an acidic proton (e.g., for acetylide formation and subsequent SN2 reactions), an internal alkyne will not react.
JEE_Advanced
Important
Other
❌ Overlooking the Acidity of Terminal Alkynes in Reactions
Students frequently fail to recognize the significant acidity of terminal alkynes (RC≡CH) compared to other hydrocarbons. This leads to errors in predicting the outcome of reactions involving strong bases or in designing multi-step synthesis where acetylide anions are key intermediates for C-C bond formation.
💭 Why This Happens:
- Generalization Error: Students incorrectly generalize that all hydrocarbons are non-acidic, ignoring the unique electronic properties of sp-hybridized carbons.
- Lack of Conceptual Depth: Insufficient understanding of how the increased s-character in sp-hybridized orbitals leads to greater electronegativity and polarization of the C-H bond, making the hydrogen more acidic.
- Confusing Base Strength: Misconception about which bases are strong enough to deprotonate terminal alkynes (e.g., attempting with NaOH instead of NaNH₂).
✅ Correct Approach:
The proton attached to the sp-hybridized carbon in a terminal alkyne is acidic (pKa ≈ 25). Strong bases like sodium amide (NaNH₂), organolithium reagents (e.g., n-BuLi), or Grignard reagents (RMgX) can deprotonate these alkynes to form acetylide anions (RC≡C⁻). These anions are strong nucleophiles and can participate in reactions such as SN2 alkylation with primary alkyl halides or additions to carbonyl compounds, extending the carbon chain.
📝 Examples:
❌ Wrong:
Predicting no reaction or an incorrect product for the following:
(Incorrect, as NaOH is generally not strong enough to deprotonate terminal alkynes effectively in typical lab conditions; students might mistakenly think it will react like an acid-base reaction with any base)
CH₃-C≡C-H + NaOH → ?(Incorrect, as NaOH is generally not strong enough to deprotonate terminal alkynes effectively in typical lab conditions; students might mistakenly think it will react like an acid-base reaction with any base)
✅ Correct:
Understanding the two-step process:
Step 1 (Deprotonation): Formation of the acetylide anion
Step 2 (Alkylation/C-C Bond Formation): Nucleophilic attack by the acetylide anion
(This demonstrates the acidity allowing for new C-C bond formation, crucial for synthesis in JEE Advanced.)
Step 1 (Deprotonation): Formation of the acetylide anion
CH₃-C≡C-H + NaNH₂ → CH₃-C≡C⁻ Na⁺ + NH₃Step 2 (Alkylation/C-C Bond Formation): Nucleophilic attack by the acetylide anion
CH₃-C≡C⁻ Na⁺ + CH₃CH₂Br → CH₃-C≡C-CH₂CH₃ + NaBr(This demonstrates the acidity allowing for new C-C bond formation, crucial for synthesis in JEE Advanced.)
💡 Prevention Tips:
- Hybridization and Acidity: Always correlate sp-hybridization with increased s-character and enhanced acidity for terminal C-H bonds.
- Base Strength Hierarchy: Memorize and understand the relative strengths of common bases (e.g., NaNH₂ > NaOH). Only very strong bases effectively deprotonate terminal alkynes.
- Identify Terminal Alkynes: In any reaction involving alkynes, immediately check if it's a terminal alkyne (has a C-H bond on the triple bond) as its acidity will dictate potential reaction pathways.
- Practice Synthetic Problems: Work through multi-step synthesis problems where forming an acetylide is an essential intermediate for chain extension.
JEE_Advanced
Important
Approximation
❌ Incorrect Assessment of Terminal Alkyne Acidity and Base Selection
Students frequently misjudge the relative acidity of terminal alkynes compared to other C-H bonds (alkenes, alkanes) or even water/alcohols. This 'approximation' often leads to the selection of an inappropriate base for deprotonation, incorrectly assuming a reaction will occur with a base that is not sufficiently strong.
💭 Why This Happens:
- Lack of Quantitative pKa Understanding: An imprecise grasp of pKa values for various organic acids prevents accurate comparison.
- Over-generalization of 'Strong Base': Students might consider any common strong base (like NaOH) universally effective without evaluating its conjugate acid's pKa relative to the alkyne's pKa.
- Confusion with Other Acidic Protons: Terminal alkyne acidity is often confused with alcohols or carboxylic acids, which are significantly more acidic.
✅ Correct Approach:
To successfully deprotonate a terminal alkyne, the chosen base must be stronger than the conjugate base of the alkyne. This means the conjugate acid of the base must have a pKa significantly higher than that of a terminal alkyne (approx. 25).
Key pKa Values to Remember for JEE Advanced:
Key pKa Values to Remember for JEE Advanced:
- Terminal Alkynes: ~25
- Water: ~15.7
- Alcohols: ~16-18
- Ammonia (NH3): ~38
- Alkanes: ~50
📝 Examples:
❌ Wrong:
Attempting to synthesize sodium phenylacetylide by reacting phenylacetylene with NaOH.
Reason: H₂O (pKa ~15.7) is a much stronger acid than phenylacetylene (pKa ~25), so the equilibrium lies to the left.
C₆H₅-C≡C-H + NaOH → No significant reactionReason: H₂O (pKa ~15.7) is a much stronger acid than phenylacetylene (pKa ~25), so the equilibrium lies to the left.
✅ Correct:
Synthesis of sodium phenylacetylide by reacting phenylacetylene with NaNH₂.
Reason: NH₃ (pKa ~38) is a much weaker acid than phenylacetylene (pKa ~25), so the equilibrium strongly favors product formation.
C₆H₅-C≡C-H + NaNH₂ → C₆H₅-C≡C⁻Na⁺ + NH₃Reason: NH₃ (pKa ~38) is a much weaker acid than phenylacetylene (pKa ~25), so the equilibrium strongly favors product formation.
💡 Prevention Tips:
- Prioritize pKa Values: For JEE Advanced, a precise understanding of relative pKa values for different functional groups, especially terminal alkynes, is critical.
- Master Acid-Base Principles: Always assess the acid-base equilibrium; a reaction will proceed in the direction of forming the weaker acid and weaker base.
- Practice Base Selection: For synthesis problems involving deprotonation, consciously select bases based on their strength relative to the acidic proton being removed.
JEE_Advanced
Important
Sign Error
❌ <span style='color: #FF0000;'>Incorrect Charge Assignment and Reaction Direction in Alkyne Chemistry</span>
Students frequently make 'sign errors' by misidentifying the most acidic proton, incorrectly assigning charges to intermediates, or misjudging the primary role of a reagent (e.g., base vs. nucleophile) in reactions involving alkynes. This leads to incorrect reaction pathways and products, particularly in acid-base reactions or nucleophilic/electrophilic additions. A common error is overlooking the highly acidic nature of terminal alkyne protons.
💭 Why This Happens:
This mistake often arises from:
- Lack of clear understanding of the relative acidities of different types of C-H bonds, especially the high acidity of sp-hybridized terminal alkyne protons (pKa ≈ 25) compared to sp2 (pKa ≈ 44) or sp3 (pKa ≈ 50) C-H bonds.
- Failure to recognize that strong bases (like NaNH2, Li, NaH) will preferentially deprotonate the most acidic proton present before any other reaction can occur.
- Confusing the electrophilic/nucleophilic nature of the triple bond itself versus the acidity of its terminal hydrogen.
- Inadequate practice in drawing reaction mechanisms, where each step, including charge movement and intermediate formation, must be correctly depicted.
✅ Correct Approach:
Always evaluate the acidity of all potential protons in a molecule when a strong base is present. For terminal alkynes, the C-H bond proton is significantly acidic due to the high s-character of the sp-hybridized carbon. Therefore, a strong base will deprotonate this proton first, forming an acetylide anion, which is a powerful nucleophile. Only after the acid-base reaction is complete can subsequent reactions (e.g., nucleophilic attack or further electrophilic reactions) proceed.
📝 Examples:
❌ Wrong:
Incorrect: Reaction of 1-butyne with NaNH2, showing nucleophilic attack of NaNH2 directly on the triple bond to form an addition product.
CH₃-CH₂-C≡C-H + NaNH₂ --X--> CH₃-CH₂-C(NH₂)=CHNa (Incorrect Addition)
✅ Correct:
Correct: Reaction of 1-butyne with NaNH2, showing deprotonation to form the acetylide anion.
CH₃-CH₂-C≡C-H + NaNH₂ ---> CH₃-CH₂-C≡C⁻ Na⁺ + NH₃ (Correct Acid-Base Reaction)The resulting acetylide anion (CH₃-CH₂-C≡C⁻) is then a potent nucleophile capable of reacting with electrophiles (e.g., alkyl halides in a substitution reaction).
💡 Prevention Tips:
- Prioritize Acid-Base Chemistry: In multi-step reactions, always consider and complete any possible acid-base reactions before proceeding to addition or substitution reactions.
- Know Your Reagents: Understand whether a reagent primarily acts as a base (e.g., NaNH2, LDA, Grignard reagents) or a nucleophile, or both, and its relative strength in each role.
- Practice pKa Values: Have a good conceptual understanding of the pKa values of common functional groups (terminal alkyne ≈ 25, water ≈ 15.7, ammonia ≈ 38) to predict proton transfer.
- Draw Mechanisms: Meticulously draw electron flow (curved arrows) and assign charges at each step of the reaction mechanism to avoid errors.
JEE_Advanced
Important
Unit Conversion
❌ <span style='color: #FF0000;'>Incorrect Conversion Between Moles, Mass, and Volume in Stoichiometric Calculations for Alkyne Reactions</span>
Students frequently err by failing to correctly convert between molar quantities (moles) required for a reaction and the practical units of mass (grams) for solid reagents or volume (mL/L) for liquid reagents/solutions. This leads to inaccurate amounts of reactants used or calculated, significantly impacting theoretical yield or the success of a reaction. This is particularly critical in reactions involving precise stoichiometry, such as the deprotonation of terminal alkynes with strong bases (e.g., NaNH₂, Grignard reagents) or the synthesis of longer chain alkynes where limiting reagents must be carefully determined.
💭 Why This Happens:
- Conceptual Confusion: Students may not fully grasp the concept of a 'mole' as a unit of amount and its central role in chemical stoichiometry.
- Ignoring Molar Mass: Overlooking the necessity of using molar mass (g/mol) as the primary conversion factor between mass and moles.
- Density and Concentration Oversights: Forgetting to use density (g/mL) for liquid reagents or molarity (mol/L) for solutions to convert between volume and moles.
- Unit Inconsistency: Not tracking units throughout calculations, leading to dimensional errors.
- Rushing Calculations: A common tendency to assume 'gram-for-gram' equivalence or 'volume-for-volume' equivalence without proper conversion.
✅ Correct Approach:
Always convert all given quantities (mass, volume, concentration) into moles first. Once all reactants are in moles, use the stoichiometric coefficients from the balanced chemical equation to determine the required or produced molar quantities. Finally, convert the calculated moles back to the desired practical units (grams, mL) using the appropriate conversion factors (molar mass, density, concentration). This systematic approach ensures accurate quantitative understanding.
📝 Examples:
❌ Wrong:
A student needs to deprotonate 0.5 g of 1-pentyne (Molar Mass ~68 g/mol) using a 2 M solution of n-Butyllithium (BuLi). The student incorrectly assumes that 0.5 mL of BuLi solution will be sufficient, without considering the molarity or the molar equivalents required. This is a common mistake in JEE Advanced practical applications, where direct 'gram-to-mL' or 'gram-to-gram' comparisons are made without stoichiometric conversions.
✅ Correct:
To deprotonate 0.5 g of 1-pentyne (C₅H₈, Molar Mass = 68.12 g/mol) with a 2 M solution of n-Butyllithium (BuLi):
1. Calculate moles of 1-pentyne:
Moles = Mass / Molar Mass = 0.5 g / 68.12 g/mol ≈ 0.00734 mol
2. Determine moles of BuLi required:
Since the reaction is 1:1, moles of BuLi = 0.00734 mol.
3. Calculate volume of 2 M BuLi solution:
Volume (L) = Moles / Molarity = 0.00734 mol / 2 mol/L = 0.00367 L
Volume (mL) = 0.00367 L * 1000 mL/L ≈ 3.67 mL
Thus, approximately 3.67 mL of 2 M BuLi solution is needed, not an arbitrary 0.5 mL.
1. Calculate moles of 1-pentyne:
Moles = Mass / Molar Mass = 0.5 g / 68.12 g/mol ≈ 0.00734 mol
2. Determine moles of BuLi required:
Since the reaction is 1:1, moles of BuLi = 0.00734 mol.
3. Calculate volume of 2 M BuLi solution:
Volume (L) = Moles / Molarity = 0.00734 mol / 2 mol/L = 0.00367 L
Volume (mL) = 0.00367 L * 1000 mL/L ≈ 3.67 mL
Thus, approximately 3.67 mL of 2 M BuLi solution is needed, not an arbitrary 0.5 mL.
💡 Prevention Tips:
- Master Molar Concepts: Ensure a strong understanding of moles, molar mass, molarity, and density.
- Use Dimensional Analysis: Always write down units for every quantity and cancel them systematically. This helps catch errors if units don't align.
- Create a Conversion Flowchart: Visualize the steps: Mass → Moles → Moles (of other substance) → Mass/Volume/Concentration.
- Practice Stoichiometry: Regularly solve quantitative problems, especially those involving limiting reagents and percentage yields in organic reactions.
- Double-Check: After calculation, ask if the answer makes logical sense (e.g., if you need grams of a light molecule, should the number be much larger than for a heavy molecule if moles are the same?).
JEE_Advanced
Important
Formula
❌ <span style='color: #FF0000;'>Confusing Acidity of Terminal vs. Non-Terminal Alkynes and Misapplying Related Reactions</span>
Students often make a critical error by failing to distinguish between terminal and internal alkynes regarding their acidity. They incorrectly assume that all alkynes possess acidic protons, or they overlook that only terminal alkynes have acidic hydrogen atoms. This fundamental misunderstanding leads to incorrect predictions of products for reactions involving strong bases (like NaNH₂) or heavy metal ions.
💭 Why This Happens:
This mistake primarily stems from a conceptual gap in understanding the electronic effects of sp-hybridized carbons. Students might not grasp why the C-H bond in terminal alkynes is significantly more acidic than in alkenes or alkanes. Overgeneralization, where a property observed in one variant of a functional group is extended to all variants without considering structural nuances, is also a common cause. Insufficient practice with diverse alkyne structures exacerbates this issue.
✅ Correct Approach:
The correct approach involves a clear understanding of alkyne structure and its correlation with acidity:
- Only terminal alkynes (R-C≡C-H) possess an acidic hydrogen atom directly attached to an sp-hybridized carbon.
- The high s-character (50%) of the sp-hybridized carbon makes it more electronegative, thereby stabilizing the resulting carbanion (acetylide anion) and making the C-H bond more acidic (pKa ≈ 25).
- Internal alkynes (R-C≡C-R') lack such a hydrogen and are therefore not acidic enough to react with strong bases like NaNH₂ to form acetylides.
- Reactions with strong bases (e.g., NaNH₂, R-Li) will selectively deprotonate terminal alkynes to form acetylide anions (R-C≡C⁻), which are powerful nucleophiles used in carbon-carbon bond formation.
- Heavy metal ions (e.g., Ag⁺, Cu⁺) also react specifically with terminal alkynes to form insoluble metal acetylides, which can be used for their detection or purification.
📝 Examples:
❌ Wrong:
Attempting to react but-2-yne with NaNH₂:
CH₃-C≡C-CH₃ + NaNH₂ → No ReactionThis is incorrect because but-2-yne is an internal alkyne and lacks acidic hydrogens.
✅ Correct:
Reacting but-1-yne with NaNH₂:
CH₃-CH₂-C≡C-H + NaNH₂ → CH₃-CH₂-C≡C⁻Na⁺ + NH₃
This is correct. But-1-yne is a terminal alkyne, and its acidic proton is removed by the strong base NaNH₂, forming a sodium but-1-ynide (acetylide) and ammonia.
💡 Prevention Tips:
- Visual Distinction: Always identify if the alkyne has a hydrogen directly bonded to the triple bond (terminal) or if it's flanked by carbons (internal).
- Pillar Concept: Reinforce the concept of sp-hybridization and its impact on electronegativity and acidity.
- Reagent Specificity: Understand that reagents like NaNH₂ are specific for terminal alkynes in terms of deprotonation.
- Practice Problems: Solve numerous problems involving both terminal and internal alkynes to solidify your understanding of their distinct reactivities.
JEE_Advanced
Important
Calculation
❌ Incorrect Stoichiometry of Base for Alkyne Deprotonation
Students frequently make errors in 'calculation understanding' by misjudging the number of acidic protons available in alkynes, leading to incorrect calculations for the equivalents of base required for deprotonation. A common blunder is attempting to deprotonate internal alkynes or using an insufficient/excessive amount of base for terminal alkynes.
💭 Why This Happens:
Lack of structural understanding: Students fail to recognize that only terminal alkynes (R-C≡C-H) possess an acidic proton directly attached to the sp-hybridized carbon. Internal alkynes (R-C≡C-R') lack such a proton.
Stoichiometric oversight: Forgetting the 1:1 molar ratio required between each terminal alkyne proton and a strong base (e.g., NaNH₂, BuLi) for complete deprotonation.
General confusion: Misapplying acidity concepts from other functional groups to alkynes without considering the specific structural requirements for alkyne acidity.
✅ Correct Approach:
To avoid errors, follow these steps:
- Identify Terminal Alkynes: Carefully examine the alkyne structure. An alkyne is terminal if one of the carbons of the triple bond is bonded to a hydrogen atom (e.g., R-C≡C-H).
- Count Acidic Protons: Each terminal alkyne group contributes exactly one acidic proton. Internal alkynes have no acidic protons for deprotonation by typical strong bases like NaNH₂.
- Determine Base Equivalents: For complete deprotonation, use 1 equivalent of a strong base for each terminal alkyne proton present in the molecule. If there are other functional groups with more acidic protons (e.g., -OH, -COOH), consider the relative pKa values and base strength.
📝 Examples:
❌ Wrong:
Scenario: Reacting 1 mole of 2-Butyne (CH₃-C≡C-CH₃) with 1 mole of NaNH₂.
Mistake: Assuming 2-Butyne, an internal alkyne, has an acidic proton to react with NaNH₂. 2-Butyne will not react with NaNH₂ for deprotonation, as it lacks a terminal hydrogen.
✅ Correct:
Scenario: Reacting 1 mole of 1-Butyne (CH₃-CH₂-C≡C-H) with 1 mole of NaNH₂.
Correct Approach: 1-Butyne is a terminal alkyne with one acidic proton. Therefore, 1 mole of NaNH₂ is correctly used to deprotonate 1 mole of 1-Butyne, forming sodium butynide (CH₃-CH₂-C≡C⁻Na⁺) and ammonia (NH₃).
💡 Prevention Tips:
- Visualize Structures: Always draw out the full structure of the alkyne to visually confirm if it's terminal or internal.
- Fundamental Principle: Ingrain the rule: "Only terminal alkynes are acidic; internal alkynes are not."
- Practice pKa Values: Understand that the pKa of acetylenic hydrogens (~25) allows deprotonation by very strong bases like NaNH₂ (conjugate acid NH₃, pKa ~38), but not by weaker bases like NaOH (conjugate acid H₂O, pKa ~15.7).
- Stoichiometry Check: Before 'calculating', confirm the presence and number of acidic protons.
JEE_Advanced
Important
Unit Conversion
❌ Incorrect Stoichiometry for Acidic Alkynes and Reagents
A common mistake is misinterpreting the quantitative requirement of reagents for reactions involving alkynes, especially concerning their acidity. Students often fail to recognize the number of acidic hydrogens in terminal alkynes or the correct molar equivalents needed for complete reaction, leading to incorrect product predictions or stoichiometric calculations.
💭 Why This Happens:
- Confusion between alkyne types: Not distinguishing between terminal (R-C≡C-H) and internal (R-C≡C-R') alkynes, where only terminal alkynes possess acidic hydrogens.
- Overlooking stoichiometry: Assuming a 1:1 reaction ratio without considering the specific reaction mechanism or the number of reactive sites.
- JEE Specific: In multi-step synthesis or quantitative analysis problems, students may incorrectly calculate the moles of reactants/products due to this fundamental misunderstanding.
✅ Correct Approach:
- Identify Terminal Alkynes: Always confirm if the alkyne is terminal (R-C≡C-H). Only these have an acidic hydrogen.
- Count Acidic Hydrogens: A terminal alkyne possesses exactly one acidic hydrogen atom attached to the sp-hybridized carbon.
- Match Reagent Equivalents: When a strong base (e.g., NaNH₂, LiNH₂, Grignard reagents) is used to deprotonate a terminal alkyne, ensure that the molar equivalents of the base match the number of terminal alkyne groups present. Similarly, for addition reactions, carefully consider how many equivalents of the adding reagent (e.g., Br₂, HBr, H₂) are required for mono- or di-addition.
📝 Examples:
❌ Wrong:
Scenario: A student reacts 1 mole of propyne (CH₃-C≡C-H) with 0.5 moles of NaNH₂ (sodium amide).
Incorrect Assumption: Thinking that 0.5 moles of base is sufficient to completely deprotonate 1 mole of propyne, or assuming partial deprotonation leads to the desired product in full yield.
Result: Only 0.5 moles of propyne would be deprotonated, leaving 0.5 moles of unreacted propyne. This leads to a mixture of products or an incorrect yield calculation if further reactions are expected from the alkyne anion.
✅ Correct:
Scenario: Reacting 1 mole of propyne (CH₃-C≡C-H) with 1 mole of NaNH₂ (a strong base).
Correct Approach: Since propyne is a terminal alkyne with one acidic hydrogen, 1 mole of NaNH₂ is required to completely deprotonate 1 mole of propyne.
Reaction: CH₃-C≡C-H + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃
Result: Complete formation of sodium propynide, which can then be used in subsequent reactions (e.g., alkylation).
💡 Prevention Tips:
- Structural Analysis: Always draw the full structure of the alkyne to visually confirm if it's terminal or internal.
- Reagent-to-Mole Mapping: Before solving, map the given moles/masses of reactants to their respective molar equivalents.
- CBSE vs JEE: While CBSE might focus on the general reaction, JEE often uses exact stoichiometry to test understanding of limiting reagents and yield, especially in multi-step problems.
- Practice Stoichiometry: Solve numerical problems that require calculating exact amounts of reactants or products based on molar ratios.
JEE_Main
Important
Other
❌ Confusing Acidity of Terminal vs. Internal Alkynes
Students frequently overlook or misapply the concept of acidity in alkynes. The most common error is failing to recognize that only terminal alkynes (R-C≡C-H) possess acidic hydrogen atoms due to the high electronegativity of the sp-hybridized carbon. Internal alkynes (R-C≡C-R') do not have such acidic hydrogens.
💭 Why This Happens:
This mistake often arises from:
- Lack of understanding of hybridization: Not fully grasping why sp-hybridized carbons make attached hydrogens acidic.
- Generalizing alkyne reactions: Treating all alkynes identically, without differentiating between terminal and internal structures.
- Confusion with other hydrocarbons: Assuming similar reactivity to alkenes or alkanes, which lack acidic protons.
✅ Correct Approach:
Always identify the type of alkyne. If it's a terminal alkyne, its terminal hydrogen is weakly acidic and can be deprotonated by strong bases (like NaNH2) or react with active metals (like Na) to form acetylide anions. These acetylide anions are strong nucleophiles and can be used in C-C bond formation reactions (e.g., alkylation). Internal alkynes do not exhibit this acidity and will not react under these conditions to form acetylides.
📝 Examples:
❌ Wrong:
Predicting the reaction of 2-butyne (an internal alkyne) with sodium amide (NaNH2):
CH3-C≡C-CH3 + NaNH2 → CH3-C≡C-Na+CH3 + NH3 (Incorrect)
✅ Correct:
Predicting the reaction of propyne (a terminal alkyne) with sodium amide (NaNH2):
The 2-butyne in the wrong example would simply not react with NaNH2 in this manner, as it has no acidic proton.
CH3-C≡C-H + NaNH2 → CH3-C≡C-Na+ + NH3 (Correct, formation of sodium propynide)The 2-butyne in the wrong example would simply not react with NaNH2 in this manner, as it has no acidic proton.
💡 Prevention Tips:
- Structural Analysis: Always draw out the structure and locate the triple bond. If there's a hydrogen directly attached to an sp-hybridized carbon of the triple bond, it's a terminal alkyne.
- Reagent Recognition: Recognize that reagents like NaNH2, Na, or Grignard reagents are specifically used to exploit the acidic nature of terminal alkynes.
- Mechanism Focus: Understand that these reactions are acid-base reactions, not electrophilic additions to the triple bond itself.
CBSE_12th
Important
Approximation
❌ Confusing Terminal and Internal Alkyne Acidity
A common mistake is incorrectly assuming that all alkynes possess acidic hydrogen atoms, leading to erroneous predictions in reactions with strong bases. Students often fail to distinguish between terminal and internal alkynes regarding their acidity, treating the hydrogens of an internal alkyne as replaceable.
💭 Why This Happens:
This error stems from an overgeneralization of the 'acidity of alkynes' without fully grasping the underlying reason – the sp-hybridization of the carbon atom in the C≡C bond. Only the hydrogen directly attached to an sp-hybridized carbon (i.e., in a terminal alkyne, R-C≡C-H) is acidic. Internal alkynes (R-C≡C-R') have no such hydrogen.
✅ Correct Approach:
Always identify if an alkyne is terminal (has a C≡C-H group) or internal (has no C≡C-H group). Only terminal alkynes are acidic due to the higher electronegativity of the sp-hybridized carbon, which withdraws electron density and weakens the C-H bond, allowing the hydrogen to be removed by a strong base. Internal alkynes do not react with bases like NaNH2 or Grignard reagents to form acetylides.
📝 Examples:
❌ Wrong:
Students might incorrectly predict a reaction between but-2-yne and sodium amide (NaNH2):
CH3-C≡C-CH3 + NaNH2 → No reaction (incorrectly shown as salt formation)
✅ Correct:
For but-1-yne (a terminal alkyne), the reaction with sodium amide is:
CH3-CH2-C≡C-H + NaNH2 → CH3-CH2-C≡C-Na+ + NH3
(Sodium but-1-ynide)
💡 Prevention Tips:
- Identify 'H' on sp-carbon: Before predicting acidity or reactions with bases, verify the presence of a hydrogen directly attached to the triple bond (R-C≡C-H).
- Understand sp-hybridization: Remember that sp-hybridized carbons are ~50% s-character, making them more electronegative than sp2 or sp3 carbons. This electron-withdrawing effect makes the C-H bond weakly acidic.
- Practice: Draw structures of various alkynes and explicitly mark or identify the acidic hydrogens, if any. This concept is fundamental for both CBSE Board and JEE Main/Advanced examinations.
CBSE_12th
Important
Sign Error
❌ <span style='color: red;'>Incorrect Placement of Negative Charge (Carbanion) in Alkyne Deprotonation</span>
Students frequently make a 'sign error' by incorrectly placing the negative charge when a terminal alkyne is deprotonated, or by attempting to deprotonate an internal alkyne, which lacks an acidic hydrogen. This leads to an incorrect representation of the acetylide ion and subsequent reaction mechanisms.
💭 Why This Happens:
- Lack of a strong understanding of sp-hybridization and its electron-withdrawing effect, which makes the terminal C-H bond slightly acidic.
- Confusing the acidity of terminal alkynes with internal alkynes, which do not have acidic hydrogens.
- Insufficient practice in drawing reaction mechanisms and identifying the precise site of deprotonation.
✅ Correct Approach:
Only terminal alkynes (R-C≡C-H) have an acidic hydrogen, which can be removed by a strong base (e.g., NaNH₂, NaH) to form an acetylide ion (carbanion). The negative charge must be correctly placed on the sp-hybridized carbon atom that originally bore the acidic hydrogen. This carbanion is relatively stable due to the high s-character (50%) of the sp orbital, which holds the lone pair closer to the more electronegative carbon nucleus.
📝 Examples:
❌ Wrong:
1. Attempting to deprotonate an internal alkyne:
CH₃-C≡C-CH₃ + NaNH₂ ⟶ No reaction (Internal alkynes are not acidic)2. Incorrect placement of negative charge in a terminal alkyne:
CH₃-C≡C-H + NaNH₂ ⟶ CH₃-CH₂-C⁻≡CH + NH₃ (Incorrect charge placement on CH₂)
✅ Correct:
Correct deprotonation of a terminal alkyne:
CH₃-C≡C-H + NaNH₂ ⟶ CH₃-C≡C⁻Na⁺ + NH₃(Prop-1-yne) + (Sodium Amide) ⟶ (Sodium Prop-1-ynide) + (Ammonia)
Here, the negative charge is correctly placed on the sp-hybridized carbon, forming the stable acetylide ion.
💡 Prevention Tips:
- Distinguish carefully: Always identify if an alkyne is terminal (R-C≡C-H) or internal (R-C≡C-R') before considering its acidity. Only terminal alkynes are acidic.
- Understand sp-hybridization: Recall that sp-hybridized carbon is more electronegative than sp² or sp³ carbon due to higher s-character, making the C-H bond more polarized and the resulting carbanion more stable.
- Practice drawing mechanisms: For deprotonation reactions, consistently show the base abstracting the terminal hydrogen and the electron pair forming a lone pair on the sp-hybridized carbon, thus creating the negative charge.
- JEE vs. CBSE: For JEE, understanding the relative acidity order (terminal alkynes > alkanes) and the stability of the conjugate base is crucial. For CBSE, correctly identifying acidic hydrogens and drawing the products of deprotonation with strong bases like NaNH₂ is key.
CBSE_12th
Important
Unit Conversion
❌ Confusing Terminal and Internal Alkynes Regarding Acidity and Specific Reactions
A common mistake students make is indiscriminately applying the concept of acidity to all alkynes. They often assume that all alkynes, whether terminal or internal, possess an acidic hydrogen and will therefore react with strong bases like NaNH₂ or specific reagents such as Tollens' reagent or ammoniacal cuprous chloride. This leads to incorrect predictions about reaction products and non-existent reactions.
💭 Why This Happens:
This mistake stems from an incomplete understanding of what makes a terminal alkyne acidic. Students often generalize the acidity of the C≡C-H bond without realizing that the acidic proton must be directly attached to an sp-hybridized carbon. Internal alkynes (R-C≡C-R) lack such a hydrogen atom, and thus do not exhibit acidic properties or react with reagents specific to acidic alkynes. This conceptual gap can be due to memorizing reactions without understanding the underlying structural requirements.
✅ Correct Approach:
It is crucial to differentiate between terminal alkynes (R-C≡C-H) and internal alkynes (R-C≡C-R). Only terminal alkynes are acidic because the sp-hybridized carbon atom is more electronegative than sp² or sp³ carbons, making the C-H bond polar and the hydrogen easily removable as a proton. Internal alkynes, lacking such a hydrogen, do not show acidic behavior. For CBSE and JEE, understanding this distinction is fundamental for predicting reactions, especially those involving the formation of acetylides.
📝 Examples:
❌ Wrong:
A student might incorrectly predict that 2-butyne (CH₃-C≡C-CH₃) will react with Tollens' reagent (ammoniacal silver nitrate) to form a silver mirror or precipitate. This is incorrect because 2-butyne is an internal alkyne and lacks an acidic hydrogen.
✅ Correct:
The correct understanding is that 1-butyne (CH₃-CH₂-C≡C-H), being a terminal alkyne, will react with Tollens' reagent to form a white precipitate of silver acetylide (CH₃-CH₂-C≡C⁻Ag⁺). Similarly, 1-butyne will react with NaNH₂ to form sodium acetylide, whereas 2-butyne will not.
💡 Prevention Tips:
- Structural Distinction: Always verify if the alkyne is terminal (has C≡C-H) or internal (has no C≡C-H).
- Acidity Principle: Reinforce the concept that acidity in alkynes is due to the s-character of the sp-hybridized carbon and its ability to stabilize the conjugate base. This requires a hydrogen directly attached to the triple bond.
- Reagent Specificity: Learn which reagents (e.g., NaNH₂, Tollens' reagent, ammoniacal Cu₂Cl₂) are specific for terminal alkynes and why.
- Practice: Solve problems involving distinguishing between terminal and internal alkynes based on their reactions.
CBSE_12th
Important
Formula
❌ Confusing Acidity of Terminal vs. Internal Alkynes
Students frequently misunderstand the acidity of alkynes, incorrectly assuming that all alkynes are acidic or trying to apply acid-base reactions to internal alkynes. This leads to errors in predicting reaction outcomes and formulating product structures, particularly when strong bases are involved.
💭 Why This Happens:
This common mistake arises from an incomplete understanding of the structural requirements for alkyne acidity. Students might know that alkynes are acidic but fail to differentiate between terminal (R-C≡C-H) and internal (R-C≡C-R') alkynes, generalizing the acidity property without considering the specific position of the triple bond and the presence of an acidic hydrogen.
✅ Correct Approach:
It is crucial to remember that only terminal alkynes are acidic. This acidity is attributed to the high s-character (50%) of the sp-hybridized carbon atom, which makes it more electronegative. This increased electronegativity pulls electron density away from the C-H bond, making the hydrogen atom sufficiently acidic to be abstracted by strong bases (e.g., NaNH₂, Na, RMgX) to form acetylide ions. Internal alkynes lack such a hydrogen and are therefore not acidic.
📝 Examples:
❌ Wrong:
Trying to deprotonate an internal alkyne:
CH₃-C≡C-CH₃ + NaNH₂ → No Reaction / Incorrect Product: CH₃-C≡C-CH₂⁻Na⁺Attempting this reaction demonstrates a misunderstanding of alkyne acidity, as internal alkynes do not possess an acidic hydrogen.
✅ Correct:
Correct acid-base reaction of a terminal alkyne:
CH₃-C≡C-H + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃Explanation: Prop-1-yne (a terminal alkyne) reacts with sodium amide (a strong base) to form sodium propynide and ammonia. This is a correct acid-base reaction because prop-1-yne has an acidic hydrogen.
💡 Prevention Tips:
- Focus on Structure: Always identify if an alkyne is terminal (has C≡C-H) or internal (has C≡C-C) before predicting its acidity or reactions with bases.
- Understand the 'Why': Grasp that the sp-hybridization of terminal alkyne carbons makes the C-H bond weak enough to be deprotonated, a property absent in internal alkynes.
- Practice Reactant-Product Mapping: Solve problems specifically involving terminal alkyne reactions with strong bases to reinforce the formation of acetylides. For CBSE 12th, questions often test this fundamental distinction.
CBSE_12th
Important
Calculation
❌ Misjudging Stoichiometry in Alkyne Addition Reactions
Students frequently predict incorrect products for alkyne addition reactions by miscalculating the required stoichiometry of reagents. They fail to differentiate between partial (one π bond reaction) and complete (both π bonds react) saturation, leading to errors when specific equivalents of reactants are provided.
💭 Why This Happens:
This mistake largely stems from overlooking the specified amounts of reactants or reaction conditions ('1 equivalent', 'excess', or specific catalysts like Lindlar's). Students often forget that an alkyne has two π bonds, each capable of addition, thus requiring two equivalents for full saturation.
✅ Correct Approach:
Always scrutinize the reagent amount.
- One equivalent of reagent (H₂, X₂, HX) reacts with one π bond, yielding an alkene derivative.
- Two equivalents or 'excess' react with both π bonds, leading to a fully saturated product.
- Specific catalysts (e.g., Lindlar's for H₂, Na/Li in liquid NH₃) explicitly indicate partial hydrogenation to an alkene.
📝 Examples:
❌ Wrong:
Question: Propyne + 1 equivalent Br₂ → ?
Incorrect Answer: 1,1,2,2-Tetrabromopropane
Reasoning: Assumed complete bromination, which requires two equivalents of Br₂, despite only one being specified.
✅ Correct:
Question: Propyne + 1 equivalent Br₂ → ?
Correct Answer: 1,2-Dibromopropene
Explanation: With one equivalent of Br₂, addition occurs across only one π bond. Complete saturation to 1,1,2,2-Tetrabromopropane would require two equivalents of Br₂.
💡 Prevention Tips:
- ☑ Check reagent quantities: Note '1 equivalent', '2 equivalents', or 'excess'.
- ☑ Identify specific catalysts: Learn catalysts for partial reactions (e.g., Lindlar's).
- ☑ Count π bonds: Alkynes have two π bonds, allowing two additions.
- ☑ Practice: Solve problems with varying stoichiometries.
CBSE_12th
Important
Conceptual
❌ Misunderstanding the Acidity of Terminal Alkynes
Students often fail to recognize that terminal alkynes possess an acidic hydrogen atom, unlike internal alkynes, alkenes, or alkanes. This leads to incorrect predictions in reactions involving strong bases or specific metal salts (e.g., Tollens' reagent, ammoniacal cuprous chloride) in both CBSE and JEE examinations.
💭 Why This Happens:
This conceptual error stems from an incomplete understanding of hybridization and its effect on electronegativity. The sp-hybridized carbon atom in terminal alkynes has a higher s-character (50%) than sp2 (33%) or sp3 (25%) carbons. This makes the sp-hybridized carbon more electronegative, causing it to hold the shared electron pair of the C-H bond more tightly, thus making the hydrogen more prone to removal as a proton.
✅ Correct Approach:
Always remember that the hydrogen attached to the triple bond in a terminal alkyne is acidic. This allows terminal alkynes to react with strong bases (like NaNH2, Na, Grignard reagents) to form alkynide anions, and to give characteristic tests with Tollens' reagent (forming silver acetylides) and ammoniacal cuprous chloride (forming red cuprous acetylides). Internal alkynes do not exhibit this acidity as they lack such a hydrogen atom.
📝 Examples:
❌ Wrong:
Wrong Prediction:
Predicting the reaction of but-2-yne with Tollens' Reagent:
CH3-C≡C-CH3 + [Ag(NH3)2]OH → CH3-C≡C-CH3 (Silver mirror)Reasoning Error: But-2-yne is an internal alkyne and does not have an acidic hydrogen. Therefore, it will not react with Tollens' reagent to form a silver acetylide. The student wrongly assumes all alkynes react with Tollens'.
✅ Correct:
Correct Prediction:
Predicting the reaction of but-1-yne with Tollens' Reagent:
CH3-CH2-C≡C-H + [Ag(NH3)2]OH → CH3-CH2-C≡C-Ag↓ + H2O + 2NH3Correct Approach: But-1-yne is a terminal alkyne with an acidic hydrogen. It reacts with Tollens' reagent to form a white precipitate of silver acetylide.
💡 Prevention Tips:
- Conceptual Clarity: Understand why sp-hybridization leads to increased acidity (higher s-character, greater electronegativity) compared to sp2 and sp3 carbons.
- Distinguish: Clearly differentiate between terminal (R-C≡C-H) and internal (R-C≡C-R') alkynes. Only terminal alkynes are acidic.
- Key Reactions: Memorize reactions specifically testing alkyne acidity: reaction with Na, NaNH2, Grignard reagents, Tollens' reagent (ammoniacal silver nitrate), and ammoniacal cuprous chloride.
- JEE vs CBSE: Both examinations emphasize this concept. For JEE, understanding the alkynide ion as a strong nucleophile in C-C bond formation reactions (like alkylation) is also crucial.
CBSE_12th
Important
Conceptual
❌ <span style='color: #FF0000;'>Confusing Acidity of Terminal Alkynes and Appropriate Bases</span>
Students often incorrectly assume that any strong base can deprotonate a terminal alkyne, or they fail to recognize the unique acidity of terminal alkynes compared to other hydrocarbons. They might also confuse the conditions needed to form an acetylide ion.
💭 Why This Happens:
- Lack of deep understanding of s-character and its impact on acidity in organic molecules.
- Misconception that all strong bases (e.g., NaOH, KOH) are sufficient for completely deprotonating terminal alkynes to form acetylides.
- Not comparing the pKa values of the alkyne and the conjugate acid of the base, which is crucial for predicting acid-base reaction feasibility.
✅ Correct Approach:
To deprotonate a terminal alkyne, the base used must be stronger than the acetylide ion (conjugate base of the alkyne). This means the conjugate acid of the base must have a pKa higher than that of the terminal alkyne (pKa ≈ 25). Appropriate very strong bases include:
- Sodium Amide (NaNH₂)
- Lithium Amide (LiNH₂)
- Sodium Hydride (NaH)
- Grignard Reagents (RMgX)
- Alkyllithiums (RLi)
📝 Examples:
❌ Wrong:
CH₃-C≡C-H + NaOH --> No significant reaction to form CH₃-C≡C⁻Na⁺
✅ Correct:
CH₃-C≡C-H + NaNH₂ --> CH₃-C≡C⁻Na⁺ + NH₃
💡 Prevention Tips:
- Understand pKa values: Familiarize yourself with approximate pKa values for different functional groups (Alkynes ~25, Alkenes ~44, Alkanes ~50, Water ~15.7, Alcohols ~16-18, Ammonia ~35-40).
- Compare acid-base strengths: A reaction proceeds significantly when a stronger acid reacts with a stronger base to form a weaker acid and a weaker base.
- JEE Specific: This concept is foundational for understanding reactions like alkylation of alkynes (forming new C-C bonds) and is frequently tested in multi-step synthesis problems. Recognize that typical strong inorganic bases like NaOH/KOH are insufficient for this purpose.
JEE_Main
Important
Calculation
❌ Incorrect Stoichiometry for Terminal Alkyne Deprotonation
Students frequently make errors in calculating the stoichiometric amount of base required to deprotonate terminal alkynes. This often stems from failing to recognize all acidic hydrogen atoms in a molecule, especially in compounds with multiple terminal alkyne groups, or misjudging the extent of reaction with a given amount of base.
💭 Why This Happens:
This mistake occurs due to a lack of careful structural analysis to identify every C≡CH unit, underestimating the basicity of reagents like NaNH2 (which can deprotonate all available terminal alkyne hydrogens), or simply rushing the mole-to-mole calculation during problem-solving. It's a blend of conceptual oversight and calculation error.
✅ Correct Approach:
The correct approach involves a two-step process:
- Identify all terminal alkyne units: Carefully scan the given alkyne structure for every -C≡CH group. Each such group possesses an acidic hydrogen.
- Apply 1:1 Stoichiometry for each acidic hydrogen: For every -C≡CH group identified, one equivalent of a strong base (like NaNH2, Grignard reagent, or organolithium reagent) will react to deprotonate it. Calculate the total moles of base required by summing up the moles for each acidic hydrogen.
📝 Examples:
❌ Wrong:
Consider the reaction:
This is incorrect if the aim is complete deprotonation, as it only accounts for one acidic hydrogen when two are present.
HC≡C-CH2-C≡CH + 1 equiv. NaNH2 → HC≡C-CH2-C≡C-Na+This is incorrect if the aim is complete deprotonation, as it only accounts for one acidic hydrogen when two are present.
✅ Correct:
For the complete deprotonation of 1,4-pentadiyne:
Here, two equivalents of NaNH2 are correctly used because there are two terminal alkyne groups, each contributing an acidic hydrogen.
HC≡C-CH2-C≡CH + 2 equiv. NaNH2 → Na+-C≡C-CH2-C≡C-Na+ + 2 NH3Here, two equivalents of NaNH2 are correctly used because there are two terminal alkyne groups, each contributing an acidic hydrogen.
💡 Prevention Tips:
- Systematic Structural Analysis: Always highlight or circle all terminal alkyne (-C≡CH) groups in the molecule first.
- Know Your Bases: Understand that strong bases like NaNH2 are capable of deprotonating all acidic terminal alkyne hydrogens present.
- Double-Check Moles: Before concluding, verify if the number of equivalents of base matches the number of acidic hydrogens identified.
- JEE Focus: Questions often test this exact concept to differentiate students who apply careful stoichiometry.
JEE_Main
Important
Formula
❌ Confusing Terminal and Internal Alkyne Acidity
Students often make the critical error of treating all alkynes as acidic, failing to differentiate between terminal and internal alkynes. Only terminal alkynes (R-C≡C-H) possess an acidic hydrogen directly attached to the sp-hybridized carbon, making them capable of reacting with strong bases.
💭 Why This Happens:
This mistake stems from a weak understanding of the
- Effect of Hybridization on Acidity: The high s-character (50%) of the sp-hybridized carbon in terminal alkynes makes the C-H bond more polarized, and the conjugate base (acetylide anion) more stable.
- Memorization without Conceptual Clarity: Students might memorize reactions without grasping the fundamental acid-base principles governing them.
- Overgeneralization: Assuming that since 'alkynes' have special properties, all alkynes share the same acidity profile.
✅ Correct Approach:
Always analyze the structure of the alkyne. Identify if there is a hydrogen atom directly bonded to the sp-hybridized carbon of the triple bond. Only if R-C≡C-H is present, the alkyne is acidic. Internal alkynes (R-C≡C-R') lack such an acidic hydrogen and thus do not react with strong bases like NaNH₂, Na, or n-BuLi to form acetylide ions.
📝 Examples:
❌ Wrong:
Incorrect: Reaction of 2-butyne with sodium amide.
CH₃-C≡C-CH₃ + NaNH₂ → No Reaction / Incorrect product (assuming acetylide formation)Explanation: 2-butyne is an internal alkyne and has no acidic hydrogen. NaNH₂ is a strong base but cannot deprotonate 2-butyne.
✅ Correct:
Correct: Reaction of propyne with sodium amide.
CH₃-C≡C-H + NaNH₂ → CH₃-C≡C⁻Na⁺ (Sodium propynide) + NH₃Explanation: Propyne is a terminal alkyne with an acidic hydrogen. NaNH₂ (a very strong base) can deprotonate it to form the stable acetylide ion.
💡 Prevention Tips:
- Visualize the Structure: Before attempting any reaction, draw the structure and explicitly identify the C-H bonds adjacent to the triple bond.
- Understand Hybridization: Reinforce your understanding of sp, sp², and sp³ hybridizations and their impact on bond polarity and acidity.
- Practice with Bases: Solve problems involving various strong bases (NaNH₂, Na, Grignard reagents, n-BuLi) and identify which alkynes will react.
- JEE Main Specific: This concept is fundamental for synthesis problems where acetylide anions act as nucleophiles for C-C bond formation.
JEE_Main
Critical
Sign Error
❌ Misinterpreting Acidity: Incorrectly Identifying Acidic Hydrogens in Alkynes
Students frequently make a 'sign error' by incorrectly identifying which hydrogens in an alkyne are acidic. This leads to erroneous predictions about acid-base reactions and subsequent synthetic pathways. The critical mistake is often assuming all hydrogens are acidic or that non-terminal alkynes react with strong bases to form acetylides.
💭 Why This Happens:
This error stems from a lack of clear understanding of the unique electronic properties of sp-hybridized carbons. Students often:
- Fail to grasp that the increased electronegativity of sp-hybridized carbon makes the adjacent C-H bond polar and the hydrogen acidic.
- Do not fully appreciate the stability of the resulting acetylide anion (RC≡C-) due to the higher s-character of the sp-hybridized carbon accommodating the negative charge.
- Overlook the requirement for a terminal alkyne (R-C≡C-H) to possess an acidic proton. Internal alkynes (R-C≡C-R') lack such hydrogens.
- Confuse the strength of bases required. For CBSE and JEE, remember strong bases like NaNH2 are needed, not weak bases like NaOH for deprotonation.
✅ Correct Approach:
The key is to recognize that only hydrogens directly attached to an sp-hybridized carbon at the end of an alkyne chain (terminal alkynes) are acidic. This acidity, while weak compared to carboxylic acids, is strong enough for reactions with very strong bases.
- CBSE/JEE Focus: Understand that the sp-hybridized carbon has 50% s-character, making it more electronegative than sp2 (33% s) or sp3 (25% s) carbons. This pulls electron density from the C-H bond, making the hydrogen easier to remove.
- The resulting carbanion (acetylide ion) is more stable because the negative charge resides on the more electronegative sp carbon, which can better accommodate it.
📝 Examples:
❌ Wrong:
Reaction: CH3-C≡C-CH3 (2-Butyne) + NaNH2 → CH3-C≡C-Na+-CH3 (Sodium Butynide)
Reason for Error: 2-Butyne is an internal alkyne and has no terminal acidic hydrogen. NaNH2 will not react to form an acetylide.
✅ Correct:
Reaction: CH3-CH2-C≡C-H (1-Butyne) + NaNH2 → CH3-CH2-C≡C-Na+ + NH3
Explanation: 1-Butyne is a terminal alkyne with an acidic hydrogen. NaNH2 is a sufficiently strong base to deprotonate it, forming sodium butynide and ammonia.
💡 Prevention Tips:
- Identify Terminal vs. Internal: Always check if there's a hydrogen directly attached to a triple-bonded carbon (terminal). If not, it's not acidic.
- Hybridization Matters: Recall that increased s-character (sp > sp2 > sp3) leads to higher electronegativity and thus more acidic C-H bonds.
- Base Strength: Remember that only very strong bases like NaNH2 (sodium amide), NaH (sodium hydride), or organometallic reagents (e.g., Grignard reagents) are strong enough to deprotonate terminal alkynes. NaOH is generally not sufficient.
- Practice: Work through examples of identifying acidic hydrogens and predicting the outcome of alkyne reactions with various bases.
CBSE_12th
Critical
Approximation
❌ Misconception: All Alkynes Are Acidic
Students frequently make the critical mistake of assuming that all alkynes possess acidic protons. This 'approximation understanding' leads to incorrect predictions, where internal alkynes (R-C≡C-R') are treated as if they can react with bases, or where terminal alkynes are expected to react with weak bases. This fundamentally misinterprets the structural requirement for alkyne acidity.
💭 Why This Happens:
This error stems from an overgeneralization of the acidity concept. Students often associate 'alkyne' directly with 'acidity' without fully appreciating that only the hydrogen directly attached to an sp-hybridized carbon (i.e., in a terminal alkyne) is acidic. They may also confuse the strength of bases required, treating all bases equally effective.
✅ Correct Approach:
Understand that acidity in alkynes is strictly limited to terminal alkynes (R-C≡C-H) due to the high s-character (50%) of the sp-hybridized carbon atom. This makes the C-H bond more polar and the proton more easily removed. Furthermore, only strong bases like sodamide (NaNH2), organolithium reagents (e.g., n-BuLi), or Grignard reagents (e.g., CH3MgBr) are strong enough to deprotonate terminal alkynes. Weaker bases like NaOH or NaHCO3 are ineffective.
📝 Examples:
❌ Wrong:
Wrong Reaction Prediction:
Predict the product of CH3-C≡C-CH3 (But-2-yne) + NaNH2.
Incorrect Answer: CH3-C≡C-CH2-Na+ + NH3
Reasoning: But-2-yne is an internal alkyne; it has no hydrogen directly attached to the sp-hybridized carbons. Therefore, it has no acidic proton to react with NaNH2.
✅ Correct:
Correct Reaction Prediction:
Predict the product of CH≡C-CH2-CH3 (But-1-yne) + NaNH2.
Correct Answer: CH≡C-CH2-CH3 + NaNH2 → Na+-C≡C-CH2-CH3 + NH3
Explanation: But-1-yne is a terminal alkyne. The hydrogen attached to the sp-hybridized carbon is acidic and can be abstracted by a strong base like NaNH2 to form sodium but-1-ynide.
💡 Prevention Tips:
- Structural Analysis: Always first identify if an alkyne is terminal (R-C≡C-H) or internal (R-C≡C-R'). Only terminal alkynes are acidic.
- Base Strength: Recognize that only very strong bases (e.g., NaNH2, LiNH2, organolithiums, Grignard reagents) are capable of deprotonating terminal alkynes.
- Practice: Work through problems distinguishing between terminal and internal alkynes and predicting their reactions with various bases (both strong and weak).
CBSE_12th
Critical
Other
❌ Confusing the Acidity of Terminal Alkynes and their Specific Reactions
Students frequently fail to recognize that only terminal alkynes (R-C≡C-H) possess acidic hydrogen atoms. This misunderstanding leads to incorrect application of unique reactions, such as the formation of alkynides with strong bases (e.g., NaNH2) or heavy metal ions (e.g., Tollens' reagent, ammoniacal cuprous chloride). They might incorrectly assume all alkynes are acidic or attempt to apply these specific reactions to internal alkynes (R-C≡C-R').
💭 Why This Happens:
- Lack of a deep conceptual understanding of hybridization (specifically sp-hybridization in terminal alkynes) and its direct impact on the electronegativity of carbon.
- Failure to connect the increased s-character of sp-hybridized carbon to the acidity of the attached hydrogen atom.
- Overlooking the crucial distinction between 'terminal' and 'internal' alkynes and generalizing properties to the entire alkyne class.
- Insufficient practice with reaction mechanisms and distinguishing tests for different types of hydrocarbons.
✅ Correct Approach:
1. Understand that in a terminal alkyne, the carbon atom bonded to the hydrogen is sp-hybridized.
2. An sp-hybridized carbon has 50% 's' character. This high 's' character makes the carbon more electronegative than sp2 (33% 's') or sp3 (25% 's') carbons.
3. Due to increased electronegativity, the sp-hybridized carbon pulls the shared electrons in the C-H bond closer, making the bond slightly polar and the hydrogen atom weakly acidic.
4. Consequently, only terminal alkynes can donate this acidic hydrogen to strong bases or react with specific metal ion reagents to form alkynides. Internal alkynes, lacking such a hydrogen, do not exhibit this property.
📝 Examples:
❌ Wrong:
CH3-CH2-C≡C-CH3 + [Ag(NH3)2]+OH- (Tollens' reagent) → White precipitate (Incorrect)
Explanation: Pent-2-yne is an internal alkyne. It does not have an acidic hydrogen and therefore will not react with Tollens' reagent to form a silver alkynide precipitate. This is a common mistake in CBSE 12th distinguishing tests.
✅ Correct:
CH3-CH2-C≡C-H + [Ag(NH3)2]+OH- → CH3-CH2-C≡C-Ag↓ (White precipitate) + NH3 + H2O
Explanation: Pent-1-yne is a terminal alkyne, possessing an acidic hydrogen. It readily reacts with Tollens' reagent (ammoniacal silver nitrate) to form a white precipitate of silver pentynide. This reaction is a key way to distinguish terminal from internal alkynes in CBSE and JEE exams.
💡 Prevention Tips:
- Master Hybridization: Ensure a solid understanding of sp-hybridization and its implications for bond properties and acidity.
- Differentiate Clearly: Always identify if an alkyne is terminal (R-C≡C-H) or internal (R-C≡C-R') before predicting its reactions. This is crucial for both CBSE and JEE objective questions.
- Practice Distinguishing Tests: Regularly review and practice the characteristic reactions of terminal alkynes (e.g., with NaNH2, Tollens' reagent, ammoniacal CuCl) as these are frequently asked as distinguishing tests.
- Conceptual Map: Create a mental or written flowchart to categorize hydrocarbons and their unique reactions based on their structural features (e.g., presence of terminal C≡C-H bond).
CBSE_12th
Critical
Unit Conversion
❌ <span style='color: #FF0000;'>Incorrect Molar Volume or Quantity Conversion in Alkyne Reaction Stoichiometry</span>
Students frequently make critical errors in stoichiometric calculations involving alkynes, specifically when converting between mass, moles, and volumes of gaseous reactants or products. This often stems from an incorrect understanding or application of molar mass and molar volume, leading to quantitatively wrong answers for questions related to reaction yields or required reagent amounts.
💭 Why This Happens:
- Confusion of Molar Quantities: Students often mix up molar mass (g/mol) with molar volume (L/mol or mL/mol).
- Incorrect Molar Volume at STP: Forgetting the standard molar volume (22.4 L/mol) or misapplying its milliliter equivalent (22400 mL/mol).
- Unit Inconsistency: Failing to convert all units to a consistent system before calculation (e.g., mixing grams with kilograms without conversion).
- Ignoring Final Unit Requirement: Overlooking the specific units requested for the final answer (e.g., calculating in liters but the question asks for milliliters).
- JEE Specific: In advanced problems, neglecting to use the Ideal Gas Law (PV=nRT) for non-STP conditions or improper conversion of pressure/temperature units.
✅ Correct Approach:
To avoid these critical errors, follow a systematic approach:
- Balance the Chemical Equation: Ensure the reaction is correctly balanced to establish the correct stoichiometric ratios.
- Convert Mass to Moles: Use the molar mass (g/mol) to convert any given mass (in grams) into moles.
- Apply Stoichiometry: Use the mole ratio from the balanced equation to find the moles of the desired reactant or product.
- Convert Moles to Volume (for gases at STP): Remember that 1 mole of any gas at STP (Standard Temperature and Pressure: 0°C/273.15 K and 1 atm pressure) occupies 22.4 L or 22400 mL.
- JEE Specific: For non-STP conditions, use the Ideal Gas Law (PV=nRT), ensuring consistent units for P (pressure), V (volume), n (moles), R (gas constant), and T (temperature in Kelvin).
- Final Unit Check: Always convert your final answer to the units specified in the question.
📝 Examples:
❌ Wrong:
Problem: Calculate the volume of NH3 gas (in mL) evolved at STP when 0.52 g of propyne (CH3C≡CH) reacts completely with excess NaNH2.
Reaction: CH3C≡CH + NaNH2 → CH3C≡C-Na+ + NH3(g)
Wrong Calculation Attempt:
- Molar mass of propyne (C3H4) ≈ 40 g/mol.
- Moles of propyne = 0.52 g / 40 g/mol = 0.013 mol.
- From stoichiometry, moles of NH3 = 0.013 mol.
- Student's Mistake: Volume of NH3 = 0.013 mol × 22.4 mL/mol (Incorrectly assuming 22.4 is in mL/mol directly).
- Result: Volume = 0.2912 mL. This answer is significantly incorrect due to a unit conversion error.
✅ Correct:
Problem: Calculate the volume of NH3 gas (in mL) evolved at STP when 0.52 g of propyne (CH3C≡CH) reacts completely with excess NaNH2.
Reaction: CH3C≡CH + NaNH2 → CH3C≡C-Na+ + NH3(g)
Correct Calculation:
- Molar mass of propyne (C3H4) ≈ 40 g/mol.
- Moles of propyne = 0.52 g / 40 g/mol = 0.013 mol.
- From the balanced reaction, 1 mole of propyne yields 1 mole of NH3. Therefore, moles of NH3 = 0.013 mol.
- At STP, 1 mole of any gas occupies 22.4 L or 22400 mL.
- Volume of NH3 = 0.013 mol × 22400 mL/mol = 291.2 mL.
💡 Prevention Tips:
- Always Show Units: Write down units at every step of your calculation. This helps in tracking and identifying inconsistencies.
- Unit Reference Card: Keep a quick reference of common conversion factors (e.g., L to mL, STP molar volume).
- Practice Stoichiometry: Regularly practice quantitative problems that involve gas volumes and mass-mole conversions.
- Read Carefully: Pay close attention to the units mentioned in the question and the units required for the final answer. Circle or highlight them.
- Dimensional Analysis: Use dimensional analysis to ensure that units cancel out correctly, leading to the desired final unit.
CBSE_12th
Critical
Formula
❌ Confusing Acidity of Terminal vs. Internal Alkynes
Students frequently make the critical error of failing to differentiate between terminal alkynes (R-C≡C-H) and internal alkynes (R-C≡C-R') when considering their acidic nature and subsequent reactions. This leads to incorrect application of reaction formulas, particularly for reactions involving the acidic hydrogen, such as the formation of metal acetylides. They might incorrectly predict a reaction with an internal alkyne or omit a reaction with a terminal alkyne.
💭 Why This Happens:
This mistake stems from an incomplete understanding of the factors governing alkyne acidity. Students often generalize 'alkyne' without focusing on the crucial presence of a terminal hydrogen. The concept of sp-hybridization leading to increased electronegativity of the carbon atom and thus making the C-H bond acidic is often overlooked or poorly applied. Forgetting that internal alkynes lack this acidic hydrogen is a common oversight.
✅ Correct Approach:
Understand that only terminal alkynes possess an acidic hydrogen directly attached to the sp-hybridized carbon of the triple bond. This hydrogen can be abstracted by strong bases (e.g., Na, NaNH₂, Grignard reagents) to form metal acetylides. Internal alkynes, having alkyl or aryl groups on both sides of the triple bond, do not possess such an acidic hydrogen and therefore do not undergo these reactions.
📝 Examples:
❌ Wrong:
Reaction of 2-Butyne (an internal alkyne) with Sodium Amide:
CH₃-C≡C-CH₃ + NaNH₂ → CH₃-C≡C⁻Na⁺-CH₃ + NH₃
(Incorrect: Internal alkynes do not react with NaNH₂ to form acetylides.)
✅ Correct:
Reaction of 1-Butyne (a terminal alkyne) with Sodium Amide:
CH₃-CH₂-C≡C-H + NaNH₂ → CH₃-CH₂-C≡C⁻Na⁺ + NH₃
(Correct: Sodium butynide is formed due to the acidic terminal hydrogen.)
💡 Prevention Tips:
- Tip 1 (CBSE & JEE): Always identify the structure of the alkyne given in a reaction. Is it a terminal alkyne (e.g., 1-butyne) or an internal alkyne (e.g., 2-butyne)?
- Tip 2 (CBSE & JEE): Remember the specific reagents that react with acidic terminal hydrogens (e.g., Na, NaNH₂, RMgX). If an internal alkyne is paired with these, the answer is generally 'No reaction'.
- Tip 3 (CBSE): For 'Name the product' questions, ensure you apply the acidic reaction only when a terminal alkyne is the reactant.
CBSE_12th
Critical
Conceptual
❌ Confusing Acidity of Terminal vs. Internal Alkynes and Misapplying Related Reactions
Students frequently fail to recognize that only terminal alkynes (those with a hydrogen atom directly attached to an sp-hybridized carbon, i.e., C≡C-H) are acidic. They might incorrectly assume internal alkynes also possess acidic hydrogens or apply reactions requiring acetylide formation to internal alkynes, leading to incorrect products or 'no reaction' scenarios when a reaction should occur.
💭 Why This Happens:
- Lack of Conceptual Clarity: Students often overlook the fundamental difference in hybridization (sp vs. sp2 vs. sp3) and its direct impact on C-H bond acidity. The higher s-character (50%) in sp-hybridized carbon makes it more electronegative, stabilizing the carbanion formed after proton removal.
- Overgeneralization: Applying general 'alkyne reactions' without considering the specific structural requirement (terminal vs. internal) for acidity-dependent reactions.
- Insufficient Practice: Limited practice with multi-step synthesis problems involving acetylide formation and its subsequent use as a nucleophile.
✅ Correct Approach:
To avoid this mistake, follow these steps:
- Identify Terminal Alkynes: Always look for the presence of a C≡C-H bond. If this hydrogen is absent, the alkyne is internal and not acidic in the same way.
- Understand sp-Hybridization: Recall that the sp-hybridized carbon's high electronegativity (due to 50% s-character) makes the terminal C-H bond polarized, allowing strong bases to abstract the proton.
- Form Acetylides: Terminal alkynes react with strong bases (e.g., NaNH₂, Na metal, Grignard reagents) to form highly nucleophilic acetylide anions (R-C≡C⁻).
- Utilize Acetylides: These acetylide anions are potent nucleophiles and crucial for C-C bond formation reactions, such as SN2 reactions with primary alkyl halides or addition to aldehydes/ketones.
📝 Examples:
❌ Wrong:
Reaction of 2-Butyne (Internal Alkyne):
Expected Product: Incorrectly, some students might predict acetylide formation, which is wrong as 2-butyne has no acidic hydrogen.
CH₃-C≡C-CH₃ + NaNH₂ (Incorrectly Assumed Reaction)Expected Product: Incorrectly, some students might predict acetylide formation, which is wrong as 2-butyne has no acidic hydrogen.
✅ Correct:
Reaction of 1-Butyne (Terminal Alkyne):
Explanation: The terminal hydrogen of 1-butyne is acidic and reacts with the strong base NaNH₂ to form the acetylide, which then acts as a nucleophile in an SN2 reaction to extend the carbon chain.
CH₃CH₂-C≡C-H + NaNH₂ → CH₃CH₂-C≡C⁻Na⁺ (Sodium butynide) + NH₃CH₃CH₂-C≡C⁻Na⁺ + CH₃CH₂Br → CH₃CH₂-C≡C-CH₂CH₃ (3-Hexyne) + NaBrExplanation: The terminal hydrogen of 1-butyne is acidic and reacts with the strong base NaNH₂ to form the acetylide, which then acts as a nucleophile in an SN2 reaction to extend the carbon chain.
💡 Prevention Tips:
- Visual Inspection: Always visually check the alkyne structure for the presence of a terminal C≡C-H bond.
- Comparative Acidity: Remember the general order of acidity: Terminal Alkyne > Alkene > Alkane.
- Reagent Awareness: Pay close attention to the reagents used. Reagents like NaNH₂, Na, or BuLi are indicators that an acidic hydrogen reaction is likely.
- JEE Focus: For JEE, this concept is fundamental for designing synthetic routes involving C-C bond formation, especially in multi-step problems. Practicing these types of synthesis questions is crucial.
CBSE_12th
Critical
Calculation
❌ Incorrect Stoichiometry and Product Prediction in Acidity Reactions of Alkynes
Students frequently misunderstand the number of acidic hydrogens present in a given terminal alkyne, leading to critical errors in predicting the stoichiometry of its reaction with strong bases (like NaNH2) and the structure of the resulting salt. A common mistake is treating alkynes like propyne (which has only one acidic hydrogen) as if they have two, or conversely, reacting acetylene (which has two acidic hydrogens) with only one equivalent of base.
✅ Correct Approach:
To avoid this critical mistake, adopt a systematic approach:
- Identify Terminal Alkynes: Always confirm if the alkyne is terminal (R-C≡C-H), as internal alkynes (R-C≡C-R') are non-acidic.
- Count Acidic Hydrogens: Precisely count the number of hydrogens directly bonded to the sp-hybridized carbon(s) of the triple bond. This number dictates the moles of strong base required for complete reaction and the valency of the resulting acetylide ion.
- Apply Stoichiometry: For each acidic hydrogen, one mole of a strong base (e.g., NaNH2, Na metal) will react to remove the proton, forming a corresponding alkyne salt and the conjugate acid of the base.
📝 Examples:
❌ Wrong:
Explanation: Propyne (CH3-C≡C-H) only has one acidic hydrogen. Reacting it with two moles of NaNH2 is stoichiometrically incorrect for complete reaction, and the product shown is structurally impossible if only one acidic hydrogen exists.
Wrong Reaction: Treating propyne with 2 equivalents of strong base.
CH3-C≡C-H + 2 NaNH2 → Na+ -C≡C-CH3 (Incorrectly implying two acidic H's reacted)
Explanation: Propyne (CH3-C≡C-H) only has one acidic hydrogen. Reacting it with two moles of NaNH2 is stoichiometrically incorrect for complete reaction, and the product shown is structurally impossible if only one acidic hydrogen exists.
✅ Correct:
Explanation: Propyne has only one acidic hydrogen, so it reacts with one mole of NaNH2 to form a monosodium acetylide.
Explanation: Acetylene has two acidic hydrogens, thus it reacts with two moles of NaNH2 to form a disodium acetylide.
Correct Reactions:
CH3-C≡C-H + 1 NaNH2 → CH3-C≡C-Na+ + NH3
Explanation: Propyne has only one acidic hydrogen, so it reacts with one mole of NaNH2 to form a monosodium acetylide.
H-C≡C-H + 2 NaNH2 → Na+ -C≡C-Na+ + 2 NH3
Explanation: Acetylene has two acidic hydrogens, thus it reacts with two moles of NaNH2 to form a disodium acetylide.
💡 Prevention Tips:
- Visualize the Structure: Always draw the complete structural formula of the alkyne to clearly see all C≡C-H bonds.
- Count sp-H Atoms: Explicitly count the number of hydrogen atoms directly attached to sp-hybridized carbons.
- Memorize Key Cases: Remember that R-C≡C-H (any alkyl group) has one acidic H, while H-C≡C-H (acetylene) has two acidic H's.
- Practice Stoichiometry: Solve several problems involving acid-base reactions of different terminal alkynes, focusing on balancing the equation and predicting the correct product based on the number of acidic sites.
- CBSE & JEE Insight: This concept is fundamental for both CBSE board exams and JEE, as questions often involve predicting products or identifying reagents based on the acidity of alkynes.
CBSE_12th
Critical
Conceptual
❌ <h3 style='color: #FF5733;'>Confusing Acidity of Terminal vs. Internal Alkynes</h3>
Students frequently fail to distinguish between the acidic nature of terminal alkynes (alk-1-ynes, having a C≡C-H bond) and the non-acidic nature of internal alkynes (alk-2-ynes, etc., R-C≡C-R'). This leads to incorrect product prediction in reactions involving strong bases (like Na or NaNH₂) or heavy metal ions found in Tollens' reagent (AgNO₃/NH₃) or ammoniacal cuprous chloride (CuCl/NH₃). They often apply terminal alkyne reactions universally.
💭 Why This Happens:
- Lack of clear conceptual understanding about the role of sp-hybridization and increased s-character, which enhances the electronegativity of carbon, making the C-H bond polar and the hydrogen acidic.
- Overlooking the critical structural requirement: the presence of a hydrogen atom directly attached to the sp-hybridized carbon (i.e., being a terminal alkyne).
- Rote memorization of reactions without understanding the underlying mechanism or structural prerequisites.
✅ Correct Approach:
Always begin by identifying if an alkyne is terminal (R-C≡C-H) or internal (R-C≡C-R'). Only terminal alkynes possess an acidic hydrogen atom. This acidic hydrogen can be abstracted by strong bases (e.g., Na, NaNH₂) to form acetylide anions, or replaced by certain metal ions (e.g., Ag⁺, Cu⁺) to form metal acetylides (often precipitates). Internal alkynes, lacking this acidic proton, do not undergo these specific acid-base or metal-replacement reactions.
📝 Examples:
❌ Wrong:
Reason: But-2-yne is an internal alkyne and has no acidic hydrogen to react with Tollens' reagent.
Reacting but-2-yne with Tollens' reagent:
CH₃-C≡C-CH₃ + [Ag(NH₃)₂]OH → White precipitate (Incorrect)Reason: But-2-yne is an internal alkyne and has no acidic hydrogen to react with Tollens' reagent.
✅ Correct:
Reacting but-1-yne with Tollens' reagent:
CH₃-CH₂-C≡C-H + [Ag(NH₃)₂]OH → CH₃-CH₂-C≡C-Ag ↓ + NH₃ + H₂O
Explanation: But-1-yne is a terminal alkyne; its acidic hydrogen is replaced by Ag⁺, forming a silver acetylide precipitate.Reacting but-2-yne with Tollens' reagent:
CH₃-C≡C-CH₃ + [Ag(NH₃)₂]OH → No reaction
Explanation: But-2-yne is an internal alkyne and does not have an acidic hydrogen, hence no reaction with Tollens' reagent.
💡 Prevention Tips:
- Structural Analysis: Before attempting any alkyne reaction, always draw its structure and explicitly identify if it has a C≡C-H bond. This is crucial for distinguishing between terminal and internal alkynes.
- Conceptual Clarity: Reinforce your understanding of why terminal alkynes are acidic (high s-character in sp-hybridized carbon) and why internal alkynes are not (no hydrogen on the sp-hybridized carbons).
- Reagent-Function Mapping: Create a mental or physical map linking specific reagents (e.g., NaNH₂, Tollens' reagent, ammoniacal CuCl) directly to their ability to react *only* with acidic terminal alkynes.
- Practice Diagnostic Tests: Solve numerous problems that involve using chemical tests to differentiate between terminal and internal alkynes.
JEE_Main
Critical
Other
❌ Confusing Reactivity of Terminal vs. Internal Alkynes in Acetylide Formation and Alkylation
Students frequently fail to differentiate between terminal alkynes (R-C≡C-H) and internal alkynes (R-C≡C-R') when considering reactions that require the formation of an acetylide anion. They mistakenly assume that internal alkynes can also undergo deprotonation with strong bases like NaNH2, leading to incorrect predictions for subsequent reactions such as alkylation (reaction with RX).
💭 Why This Happens:
This critical misunderstanding stems from an incomplete grasp of the fundamental principle of acidity in organic chemistry, specifically how hybridization affects the acidity of C-H bonds. While students may know that sp-hybridized carbons are more electronegative, they often don't fully connect this to the unique acidity of the terminal alkyne proton. This leads to an overgeneralization of alkyne reactivity, ignoring the specific structural requirement (a terminal hydrogen) for acetylide formation.
✅ Correct Approach:
Always begin by identifying the type of alkyne. Only terminal alkynes possess an acidic hydrogen directly attached to an sp-hybridized carbon. This proton can be removed by a sufficiently strong base (e.g., NaNH2) to form a nucleophilic acetylide anion (R-C≡C:-). This acetylide can then participate in SN2 reactions, like alkylation with primary alkyl halides (RX), to form new carbon-carbon bonds and extend the alkyne chain. Internal alkynes lack such an acidic proton and thus cannot form acetylides or undergo these deprotonation-alkylation reactions.
📝 Examples:
❌ Wrong:
CH3-C≡C-CH3 + NaNH2 followed by CH3I → CH3-C≡C-CH2CH3 (Incorrect)Explanation: But-2-yne is an internal alkyne. It does not have an acidic hydrogen, so NaNH2 cannot deprotonate it to form an acetylide. Therefore, alkylation with CH3I will not occur at the alkyne carbon.
✅ Correct:
CH3-C≡C-H + NaNH2 → CH3-C≡C:- Na+ (Propyne acetylide)Explanation: Propyne is a terminal alkyne. Its acidic proton is deprotonated by NaNH2 to form an acetylide anion. This anion then acts as a nucleophile to attack methyl iodide, forming But-2-yne via an SN2 mechanism.
CH3-C≡C:- Na+ + CH3I → CH3-C≡C-CH3 + NaI (Correct)
💡 Prevention Tips:
- Deep Dive into Acidity: Ensure a thorough understanding of how the high s-character of sp-hybridized orbitals increases carbon's electronegativity, making the attached hydrogen in terminal alkynes significantly acidic (pKa ≈ 25).
- Categorize First: As a routine, before attempting any reaction involving alkynes, mentally (or physically) classify it as terminal or internal. This simple step can prevent many errors.
- Mechanism-Based Reasoning: For JEE Advanced, always think mechanistically. Recognize that the formation of an acetylide anion is an essential prerequisite for nucleophilic attack in reactions like alkylation. If the acetylide cannot form, the reaction pathway is blocked.
- Practice with Varied Examples: Solve problems that explicitly test the distinction between terminal and internal alkyne reactivity to solidify this understanding.
JEE_Advanced
Critical
Approximation
❌ Critical Misjudgment of Terminal Alkyne Acidity and Base Specificity
Students often correctly identify terminal alkynes as acidic due to the sp-hybridized carbon's high electronegativity. However, a critical mistake arises from an oversimplified approximation of their relative acidity. They might incorrectly assume that any strong base, such as NaOH or NaOR (alkoxide), is sufficient for complete deprotonation. This oversight leads to incorrect predictions about reaction feasibility, product formation, and the design of synthetic routes, especially in competitive scenarios or when dealing with multi-functional molecules. This is a common trap in JEE Advanced synthesis problems.
💭 Why This Happens:
- Oversimplification of 'Strong Base': Students often group all strong bases together without considering their relative strengths or the pKa of their conjugate acids.
- Lack of pKa Comparison: Insufficient practice in rigorously comparing the pKa values of the terminal alkyne (approx. 25) with the pKa of the conjugate acid of the chosen base. For effective deprotonation, the base's conjugate acid must be significantly weaker (i.e., have a higher pKa) than the alkyne.
- Focus on Presence, Not Extent: Knowing that it's acidic is not enough; understanding the degree of acidity and the equilibrium involved is crucial for JEE Advanced.
✅ Correct Approach:
To avoid this critical error, students must adopt a rigorous approach to acid-base chemistry for terminal alkynes:
- Quantitative pKa Analysis: Always compare the pKa of the terminal alkyne (approx. 25) with the pKa of the conjugate acid of the base. For a reaction to proceed to completion, the base's conjugate acid must have a pKa significantly higher than 25 (e.g., NH3, pKa ~38; CH4, pKa ~50).
- Specific Base Selection: Understand that only very strong bases like Sodium Amide (NaNH2), Alkyl Lithium (RLi), or Grignard reagents (RMgX) are capable of effectively deprotonating terminal alkynes. Bases like NaOH (conjugate acid H2O, pKa ~15.7) or NaOR (conjugate acid ROH, pKa ~16-18) are generally not strong enough for complete deprotonation.
- Contextual Application: In multi-functional molecules, identify all acidic protons and their relative pKa values to predict which proton will be preferentially abstracted by a given base.
📝 Examples:
❌ Wrong:
CH≡C-CH2CH3 + NaOH → No reaction / Poor yield of acetylide saltExplanation: Water (conjugate acid of NaOH) has a pKa of ~15.7. Since 1-butyne has a pKa of ~25, water is a stronger acid than 1-butyne. Therefore, the equilibrium lies far to the left, and NaOH cannot effectively deprotonate the terminal alkyne to a significant extent.
✅ Correct:
CH≡C-CH2CH3 + NaNH2 → Na+[CH≡C-CH2CH3]- + NH3Explanation: Ammonia (conjugate acid of NaNH2) has a pKa of ~38. Since 1-butyne (pKa ~25) is a much stronger acid than ammonia, NaNH2 can completely deprotonate the terminal alkyne, forming the acetylide anion effectively.
💡 Prevention Tips:
- Memorize Key pKa Values: Know approximate pKa values for common functional groups (terminal alkynes, alcohols, water, amines) and the conjugate acids of strong bases used in organic chemistry. This is essential for JEE Advanced.
- Practice Acid-Base Equilibria: Consistently practice problems involving acid-base reactions to predict the direction of equilibrium and the extent of deprotonation.
- Understand Reagent Specificity: Learn which specific reagents are used for which reactions and why. For terminal alkynes, NaNH2, RLi, and RMgX are key.
- JEE Advanced Specific: Be particularly cautious in multi-step synthesis or reaction mechanism problems where the choice of base can dictate the entire reaction pathway. Overlooking this detail is a common trap leading to critical errors.
JEE_Advanced
Critical
Sign Error
❌ <span style='color: #FF0000;'>Misinterpreting Relative Acidity and Base Strength for Terminal Alkynes</span>
Students frequently correctly identify terminal alkynes as acidic but then make critical errors in comparing their acidity to other common compounds (like water, alcohols, or other hydrocarbons) or in selecting an appropriate strong base for deprotonation. This leads to an incorrect prediction of whether a deprotonation reaction will occur significantly, essentially a 'sign error' in the reaction's favorability.
💭 Why This Happens:
- Overgeneralization: Students assume that because terminal alkynes are 'acidic', any strong base can deprotonate them effectively.
- Lack of Comparative pKa Knowledge: Insufficient understanding of the relative pKa values for different classes of compounds (e.g., alkanes, alkenes, alkynes, water, alcohols, amines).
- Confusion with Brønsted-Lowry Principles: Forgetting that for a deprotonation to be favorable, the base used must be significantly stronger than the conjugate base of the alkyne (or, the conjugate acid of the base must be weaker than the alkyne).
✅ Correct Approach:
For effective deprotonation of terminal alkynes to form alkynide anions, follow these principles:
- pKa Trends: Understand that terminal alkynes (pKa ≈ 25) are more acidic than alkanes (pKa ≈ 50) and alkenes (pKa ≈ 45) due to sp-hybridization.
- Relative Acidity: However, terminal alkynes are weaker acids than water (pKa ≈ 15.7), alcohols (pKa ≈ 16-18), and carboxylic acids (pKa ≈ 3-5).
- Appropriate Bases: Therefore, weaker bases like hydroxide (OH⁻) or alkoxide (RO⁻) are generally not strong enough to completely deprotonate terminal alkynes.
- Strong Bases Required: For complete deprotonation, a base whose conjugate acid has a pKa value significantly higher than 25 is needed. Common examples include:
- Amide bases: NaNH₂ (Sodium Amide), conjugate acid NH₃ (pKa ≈ 38)
- Alkyl lithium reagents: n-BuLi (n-Butyllithium), conjugate acid Butane (pKa ≈ 50)
- Grignard reagents: CH₃MgBr (Methylmagnesium bromide), conjugate acid Methane (pKa ≈ 50)
📝 Examples:
❌ Wrong:
Predicting the product when propyne reacts with sodium hydroxide (NaOH):
CH₃-C≡C-H + NaOH → CH₃-C≡C⁻Na⁺ + H₂OExplanation: This reaction is highly unfavorable and would not proceed significantly. Water (pKa ≈ 15.7) is a much stronger acid than propyne (pKa ≈ 25). Therefore, the equilibrium lies far to the left, meaning hydroxide is not a strong enough base to deprotonate propyne.
✅ Correct:
Predicting the product when propyne reacts with sodium amide (NaNH₂):
CH₃-C≡C-H (pKa ≈ 25) + NaNH₂ (conjugate acid NH₃, pKa ≈ 38) → CH₃-C≡C⁻Na⁺ + NH₃Explanation: This reaction proceeds efficiently to the right. Ammonia (NH₃, pKa ≈ 38) is a much weaker acid than propyne (pKa ≈ 25). This means its conjugate base (NH₂⁻) is significantly stronger than the alkynide anion, driving the deprotonation to completion.
💡 Prevention Tips:
- Memorize Key pKa Values and Trends: Understand the relative order of acidity for common functional groups (e.g., Alkanes < Alkenes < Terminal Alkynes < NH₃ < Alcohols ≈ H₂O < Carboxylic Acids).
- Apply the Rule of Equilibrium: For an acid-base reaction (A-H + B⁻ → A⁻ + B-H) to favor product formation, the acid on the left (A-H) must be stronger than the conjugate acid on the right (B-H).
- Contextualize for JEE Advanced: These concepts are crucial for multi-step synthesis problems where selecting the correct base for alkyne deprotonation is often a critical initial step. An error here can lead to an entirely wrong reaction pathway and products.
JEE_Advanced
Critical
Unit Conversion
❌ <span style='color: red;'>Incorrect Volume Conversion in Stoichiometric Calculations for Alkyne Reactions</span>
A critical and frequent error observed in JEE Advanced problems, particularly those involving quantitative analysis of alkyne reactions, is the failure to correctly convert volume units. Students often use volume in milliliters (mL) directly with molarity (mol/L), leading to an incorrect calculation of moles. This fundamental unit conversion mistake results in errors of several orders of magnitude, making the entire stoichiometric analysis or yield calculation fundamentally flawed.
💭 Why This Happens:
This mistake commonly arises from:
- Lack of attention to units: Rushing through calculations without explicitly writing and cancelling units.
- Incomplete understanding of molarity: Forgetting that molarity is defined as moles per liter, thus requiring volume to be in liters.
- Over-reliance on calculators: Inputting numbers without critical thinking about the magnitude or units of the result.
✅ Correct Approach:
Always ensure that all volume units are consistent with the concentration units before performing calculations. When using molarity (mol/L), the volume must always be in liters (L). A simple rule is to convert all given quantities to their base SI units (or standard chemical units like L for volume) at the beginning of a problem.
📝 Examples:
❌ Wrong:
Problem: A solution containing a terminal alkyne is reacted with 100 mL of 0.1 M ammoniacal silver nitrate solution (Tollen's reagent). Calculate the moles of Ag+ ions supplied for the reaction.
Wrong Approach:
Moles of Ag+ = Molarity × Volume = 0.1 mol/L × 100 mL = 10 moles.
Explanation: This calculation incorrectly uses mL directly with mol/L, leading to an answer that is 1000 times larger than the correct value. 10 moles from 100 mL of 0.1 M solution is chemically impossible.
✅ Correct:
Problem: A solution containing a terminal alkyne is reacted with 100 mL of 0.1 M ammoniacal silver nitrate solution (Tollen's reagent). Calculate the moles of Ag+ ions supplied for the reaction.
Correct Approach:
- Convert volume to Liters:
Volume in Liters = 100 mL × (1 L / 1000 mL) = 0.1 L - Calculate moles of Ag+:
Moles of Ag+ = Molarity × Volume in Liters = 0.1 mol/L × 0.1 L = 0.01 moles.
💡 Prevention Tips:
- Explicitly write units: Always include units with every numerical value in your calculations and ensure they cancel out correctly. This is called dimensional analysis.
- Standardize units: Convert all quantities to a consistent set of units (e.g., grams, liters, moles) at the very start of the problem.
- Perform a sanity check: After calculating, quickly assess if the magnitude of your answer makes logical sense in a chemical context.
- Practice unit conversions: Regularly practice problems that require unit conversions to build proficiency.
JEE_Advanced
Critical
Formula
❌ Misconception of Terminal Alkyne Acidity and Required Bases
Students frequently misunderstand the acidity of alkynes, incorrectly assuming that all alkynes are acidic or that milder bases are sufficient to deprotonate terminal alkynes. This leads to critical errors in predicting reaction outcomes, especially in reactions involving acetylide ion formation for C-C bond formation.
💭 Why This Happens:
- Lack of Structural Understanding: Failure to connect the sp-hybridization of terminal carbon in alkynes with the increased s-character, which makes the C-H bond slightly acidic. Internal alkynes (R-C≡C-R) lack this acidic proton.
- Insufficient Knowledge of pKa Values: Not knowing that terminal alkynes have a pKa of approximately 25, which is significantly weaker than carboxylic acids or phenols, and thus requires much stronger bases for deprotonation.
- Confusing Base Strength: Misconception that common strong bases like NaOH or KOH are strong enough to deprotonate terminal alkynes.
✅ Correct Approach:
Only terminal alkynes (R-C≡C-H) possess an acidic hydrogen atom. Due to the high s-character (50%) of the sp-hybridized carbon, the C-H bond is more polar, allowing for deprotonation. To form an acetylide ion, a very strong base is required, whose conjugate acid has a pKa significantly higher than 25. Key strong bases include NaNH₂ (sodamide), organometallic reagents (e.g., R-Li, R-MgX), and NaH.
📝 Examples:
❌ Wrong:
| Incorrect Reaction | Reason for Error |
|---|---|
CH₃-C≡C-CH₃ + NaNH₂ → No reaction (wrongly assumed acidic) | Internal alkyne; no acidic proton. |
HC≡CH + NaOH → Na⁺⁻C≡C⁻Na⁺ + 2 H₂O (wrongly assumed base strength) | NaOH is too weak a base (pKa of H₂O ≈ 15.7) to deprotonate alkyne (pKa ≈ 25). |
✅ Correct:
| Correct Reaction | Explanation |
|---|---|
CH₃-C≡C-H + NaNH₂ → CH₃-C≡C⁻Na⁺ + NH₃ | NaNH₂ (sodamide) is a strong enough base (conjugate acid NH₃, pKa ≈ 38) to deprotonate terminal alkyne. |
HC≡CH + 2 BuLi → Li⁺⁻C≡C⁻Li⁺ + 2 BuH | Butyllithium (BuLi) is an extremely strong base, capable of deprotonating both acidic hydrogens of acetylene. |
💡 Prevention Tips:
- JEE Advanced Focus: Always check the alkyne's structure. Only terminal alkynes (with a hydrogen directly attached to the sp-hybridized carbon) are acidic.
- Memorize the common strong bases used for alkyne deprotonation: NaNH₂, RLi, RMgX, and NaH. These are crucial for subsequent reactions like alkylation.
- Understand the pKa scale: A base can deprotonate an acid effectively if its conjugate acid is weaker (has a higher pKa) than the acid being deprotonated.
- Regularly practice problems involving acetylide formation and subsequent reactions (e.g., SN2 reactions with primary alkyl halides).
JEE_Advanced
Critical
Calculation
❌ Misjudging Stoichiometry and Relative Acidity in Alkyne Deprotonation
Students frequently make critical errors by miscalculating the equivalents of base required to fully deprotonate a terminal alkyne, or by incorrectly determining which acidic proton will react preferentially when multiple acidic sites (e.g., alcohol, amine, terminal alkyne) are present in the same molecule. This leads to an incorrect intermediate structure, drastically affecting subsequent reaction outcomes in multi-step syntheses.
💭 Why This Happens:
- Ignoring Relative Acidity: Failing to compare the pKa values of different acidic protons (e.g., terminal alkyne H vs. alcohol H) when a strong base is used.
- Overlooking Stoichiometry: Not paying close attention to the number of equivalents of base specified (e.g., '1 eq.', '2 eq.', or 'excess').
- Assuming Single Deprotonation: Believing only one acidic proton will react, even if the molecule has multiple terminal alkyne groups or other more acidic functional groups, and sufficient base is provided.
- Lack of Conceptual Clarity: Weak understanding of strong bases like NaNH2 which can deprotonate not only terminal alkynes but also alcohols and primary/secondary amines.
✅ Correct Approach:
- Identify All Acidic Protons: Thoroughly scan the molecule for all potentially acidic protons (terminal alkyne, -OH, -NH, -COOH).
- Compare Relative Acidities: Recall or estimate the relative pKa values. Remember, a lower pKa means higher acidity. Generally, RCOOH < ROH < RNH2 < RC≡CH.
- Assess Base Strength and Stoichiometry: Determine if the base is strong enough to deprotonate the identified acidic protons. Then, carefully consider the molar equivalents of the base provided. 'Excess' implies all protons stronger than the conjugate acid of the base will be removed.
- Step-by-Step Deprotonation: If limited equivalents of base are given, deprotonate the most acidic proton first, then the next most acidic, and so on, until the base is consumed.
📝 Examples:
❌ Wrong:
A common mistake involves propargyl alcohol (HO-CH2-C≡CH):
Wrong Reaction: If a student reacts HO-CH2-C≡CH with 1 equivalent of NaNH2 and shows deprotonation of the terminal alkyne proton, forming HO-CH2-C≡C-Na+.
Wrong Reaction: If a student reacts HO-CH2-C≡CH with 1 equivalent of NaNH2 and shows deprotonation of the terminal alkyne proton, forming HO-CH2-C≡C-Na+.
✅ Correct:
For propargyl alcohol (HO-CH2-C≡CH):
Recall that the pKa of an alcohol (~16) is lower (more acidic) than that of a terminal alkyne (~25).
Correct Reaction (Case 1): HO-CH2-C≡CH + 1 eq. NaNH2
The more acidic hydroxyl proton is preferentially removed:
HO-CH2-C≡CH + NaNH2 → Na+-O-CH2-C≡CH + NH3
Correct Reaction (Case 2): HO-CH2-C≡CH + 2 eq. NaNH2 (or NaNH2 in excess)
Both acidic protons are removed:
HO-CH2-C≡CH + 2 NaNH2 → Na+-O-CH2-C≡C-Na+ + 2 NH3
Recall that the pKa of an alcohol (~16) is lower (more acidic) than that of a terminal alkyne (~25).
Correct Reaction (Case 1): HO-CH2-C≡CH + 1 eq. NaNH2
The more acidic hydroxyl proton is preferentially removed:
HO-CH2-C≡CH + NaNH2 → Na+-O-CH2-C≡CH + NH3
Correct Reaction (Case 2): HO-CH2-C≡CH + 2 eq. NaNH2 (or NaNH2 in excess)
Both acidic protons are removed:
HO-CH2-C≡CH + 2 NaNH2 → Na+-O-CH2-C≡C-Na+ + 2 NH3
💡 Prevention Tips:
- Memorize Key pKa Values: Have a mental hierarchy of common functional group acidities.
- Read Carefully: Always scrutinize the amount of reagent (equivalents, 'excess') specified in the problem statement.
- Draw All Possible Products: In complex molecules, draw out the structure clearly and mark all acidic protons to aid in identifying reaction sites.
- Practice Multi-Step Synthesis: Work through problems involving multiple acidic sites and varying stoichiometries of bases to solidify understanding.
- JEE Advanced Focus: JEE Advanced often tests these nuanced understandings of stoichiometry and relative reactivity, so a thorough conceptual grasp is essential.
JEE_Advanced
Critical
Conceptual
❌ Misinterpreting Acidity of Terminal Alkynes and its Synthetic Consequences
Students frequently misunderstand the acidic nature of the C-H bond in terminal alkynes (R-C≡C-H). This leads to incorrect predictions of their reactivity with various bases and subsequent errors in multi-step synthesis involving acetylide formation and alkylation. This is a critical conceptual gap for JEE Advanced.
💭 Why This Happens:
- Lack of clarity on hybridization and s-character: The sp-hybridized carbon in a terminal alkyne is more electronegative than sp2 or sp3 carbons, making the C-H bond more polarized and the proton more acidic. Students often compare its acidity incorrectly with alcohols or carboxylic acids.
- Not recognizing the importance of strong bases: Students might attempt to deprotonate terminal alkynes with weak bases (e.g., NaOH, NaHCO3), which are insufficient to form the acetylide anion.
- Failing to link acidity to nucleophilicity: The acetylide anion (R-C≡C-) formed after deprotonation is a strong nucleophile, crucial for C-C bond forming reactions like alkylation. If the acetylide isn't formed, the subsequent reaction fails.
✅ Correct Approach:
Understand that:
- Terminal alkynes are significantly more acidic than alkenes or alkanes due to the high s-character (50%) of the sp-hybridized carbon. This stabilizes the conjugate base (acetylide anion) by effectively accommodating the negative charge closer to the nucleus.
- They are, however, less acidic than water or alcohols.
- Only very strong bases like sodium amide (NaNH2), n-butyllithium (n-BuLi), or sodium metal (Na) are capable of deprotonating terminal alkynes to form acetylide anions.
- The resulting acetylide anions are powerful nucleophiles and excellent reagents for SN2 reactions with primary alkyl halides (or tosylates) to form new C-C bonds, thereby extending the carbon chain.
JEE Advanced Tip: Always assess the relative acidity and basicity of reactants when planning a synthesis.
📝 Examples:
❌ Wrong:
CH3-C≡C-H + NaOH --> No Reaction / Incorrectly forms CH3-C≡C-Na+Explanation: NaOH is not a strong enough base to quantitatively deprotonate the terminal alkyne. The pKa of terminal alkynes is ~25, while water (conjugate acid of NaOH) has a pKa of ~15.7. The equilibrium does not favor acetylide formation significantly.
✅ Correct:
1. CH3-C≡C-H + NaNH2 --> CH3-C≡C-Na+ + NH3
2. CH3-C≡C-Na+ + CH3CH2Br --> CH3-C≡C-CH2CH3 + NaBrExplanation: Sodium amide (NaNH2) is a sufficiently strong base (conjugate acid NH3, pKa ~38) to quantitatively deprotonate the terminal alkyne, forming the nucleophilic acetylide. This acetylide then undergoes an SN2 reaction with the primary alkyl bromide to form a new alkyne with a longer carbon chain.
💡 Prevention Tips:
- Review Acidity Order: Understand and internalize the relative acidities: Carboxylic Acids > Phenols > Water ~ Alcohols > Terminal Alkynes > Alkenes > Alkanes.
- Match Base Strength: Always choose a base whose conjugate acid is weaker (higher pKa) than the acid being deprotonated. For terminal alkynes, think NaNH2, NaH, RLi, or Grignard reagents.
- Practice Multi-step Synthesis: Work through problems that specifically require deprotonation and subsequent alkylation or addition reactions to solidify the conceptual link.
- Understand Hybridization: Revisit the concept of sp-hybridization and its impact on electronegativity and C-H bond acidity.
JEE_Advanced
Critical
Calculation
❌ Stoichiometric Errors in Alkyne Addition Reactions
Students frequently make 'calculation understanding' errors by misjudging the number of equivalents of reagents (e.g., HBr, Br2, H2) that can react with an alkyne. This leads to incorrect product prediction or inaccurate determination of reactant quantities, which is critical in JEE Main problems involving multi-step synthesis or quantitative analysis.
💭 Why This Happens:
- Ignoring Two Pi Bonds: Forgetting that alkynes possess two π bonds, allowing for two equivalents of addition in many cases.
- Misinterpretation of 'Equivalents': Failing to correctly interpret phrases like '1 equivalent' vs. 'excess' reagent.
- Overlooking Regioselectivity/Stereoselectivity: While more conceptual, sometimes incorrect product structure leads to a 'calculation' error in counting atoms or bonds.
- Confusing Partial vs. Complete Addition: Not distinguishing conditions for partial hydrogenation (e.g., Lindlar's catalyst) from complete hydrogenation.
✅ Correct Approach:
To avoid these errors, follow these steps:
- Identify the Alkyne Type: Determine if it's a terminal or internal alkyne, as this affects certain reactions (e.g., acidity of terminal alkynes).
- Count Reactive Sites: Remember an alkyne has two π bonds, meaning it can add up to two equivalents of reagents like HX or X2 for complete saturation.
- Analyze Reagent Stoichiometry: Always pay close attention to the number of equivalents specified in the question (e.g., '1 equivalent HBr', 'excess Br2', '1 mol H2 with Lindlar's catalyst').
- Understand Reaction Conditions: Specific catalysts or conditions dictate the extent of reaction (e.g., Lindlar's catalyst for cis-alkene, Na/NH3 for trans-alkene, both for one addition).
📝 Examples:
❌ Wrong:
A student might react 1 mole of 1-butyne with 1 mole of HBr and predict the product to be 2,2-dibromobutane, mistakenly performing a di-addition with only one equivalent of reagent. They calculate as if one equivalent reacts to saturate both pi bonds.
✅ Correct:
Consider the reaction of 1-butyne (a terminal alkyne):
Case 1: 1-butyne + 1 equivalent HBr → Predominantly 2-bromobut-1-ene (Markovnikov addition to one π bond).
Case 2: 1-butyne + Excess HBr → 2,2-dibromobutane (Markovnikov addition to both π bonds, leading to geminal dihalide).
The key 'calculation understanding' here is that excess HBr implies reaction with both π bonds, requiring two moles of HBr per mole of alkyne for full saturation.
Case 1: 1-butyne + 1 equivalent HBr → Predominantly 2-bromobut-1-ene (Markovnikov addition to one π bond).
Case 2: 1-butyne + Excess HBr → 2,2-dibromobutane (Markovnikov addition to both π bonds, leading to geminal dihalide).
The key 'calculation understanding' here is that excess HBr implies reaction with both π bonds, requiring two moles of HBr per mole of alkyne for full saturation.
💡 Prevention Tips:
- Practice Stoichiometric Calculations: Work through problems specifically focusing on mole ratios in alkyne reactions.
- Mechanism-Based Thinking: Understand the step-by-step addition process to visualize how many bonds are formed/broken and how many equivalents are consumed.
- Flashcards for Reagents & Conditions: Memorize specific reagents and their corresponding stoichiometric implications (e.g., Lindlar's vs. H2/Pd/C).
- Cross-Check Products: After drawing a product, quickly count carbons and verify the degree of saturation to ensure it matches the reactant and reagent equivalents.
JEE_Main
Critical
Formula
❌ Confusing Stereochemistry in Partial Hydrogenation of Alkynes
Students frequently misidentify the stereochemistry (cis or trans) of the alkene produced during partial hydrogenation of alkynes, confusing Lindlar's catalyst with Sodium (or Lithium) in liquid ammonia.
💭 Why This Happens:
This critical error stems from a lack of precise understanding of how specific reagents dictate stereochemistry. Lindlar's catalyst facilitates syn-addition (cis-alkene), while Na/Li in liquid NH3 promotes anti-addition (trans-alkene) via a radical mechanism. Failing to distinguish between these mechanisms leads to incorrect product prediction.
✅ Correct Approach:
Always correctly pair the reagent with its stereochemical specificity:
This distinction is crucial for JEE Main.
- Lindlar's Catalyst (Pd/CaCO3/PbO/Quinoline) → cis-alkene formation (syn-addition)
- Sodium (Na) or Lithium (Li) in liquid Ammonia (NH3) → trans-alkene formation (anti-addition)
This distinction is crucial for JEE Main.
📝 Examples:
❌ Wrong:
Consider the reaction of 2-butyne with Na/Li in liquid NH3:
CH3-C≡C-CH3 + Na/Li, NH3(l) → cis-CH3CH=CHCH3 (Incorrect)
✅ Correct:
The correct outcomes for 2-butyne's partial hydrogenation are:
| Reagent | Product | Stereochemistry |
|---|---|---|
| Lindlar's Catalyst | CH3-CH=CH-CH3 | cis-2-butene |
| Na/Li in liquid NH3 | CH3-CH=CH-CH3 | trans-2-butene |
CH3-C≡C-CH3 + Na/Li, NH3(l) → trans-CH3CH=CHCH3 (Correct)💡 Prevention Tips:
- Flashcards: Create dedicated flashcards for each reagent, explicitly stating the product's stereochemistry.
- Comparative Practice: Solve problems that require both types of partial hydrogenation side-by-side.
- Conceptual Link: Briefly review the underlying mechanisms to understand *why* each stereochemistry is preferred.
- JEE Focus: This is a frequently tested concept in JEE Main; master it for easy marks.
JEE_Main
Critical
Unit Conversion
❌ Misinterpreting pKa Values for Acidity Comparison
Students frequently make the critical error of associating a higher pKa value with stronger acidity, leading to incorrect conclusions when comparing the acid strengths of various compounds, including terminal alkynes.
💭 Why This Happens:
This mistake stems from a fundamental misunderstanding of the pKa scale. Since pKa is defined as the negative logarithm of the acid dissociation constant (Ka), there is an inverse relationship. A larger Ka (indicating a stronger acid, meaning more dissociation) results in a smaller pKa value due to the negative logarithmic function. Students often apply a direct proportionality, thinking 'larger number means more' without considering the negative logarithm.
✅ Correct Approach:
Always remember that a lower pKa value signifies a stronger acid. The pKa scale is logarithmic, where each unit difference represents a tenfold difference in acidity. For terminal alkynes, their pKa (~25) indicates they are weak acids, significantly weaker than carboxylic acids (pKa ~4-5) or even water (pKa ~15.7), but stronger than alkanes or alkenes.
📝 Examples:
❌ Wrong:
Question: Compare the acidity of Acetylene (pKa ≈ 25) and Water (pKa ≈ 15.7).
Student's Incorrect Reasoning: 'Since 25 is greater than 15.7, Acetylene is a stronger acid than Water.'
Student's Incorrect Reasoning: 'Since 25 is greater than 15.7, Acetylene is a stronger acid than Water.'
✅ Correct:
Question: Compare the acidity of Acetylene (pKa ≈ 25) and Water (pKa ≈ 15.7).
Correct Reasoning: 'Water has a pKa of approximately 15.7, while Acetylene has a pKa of approximately 25. Since a lower pKa indicates a stronger acid, Water (pKa 15.7) is significantly more acidic than Acetylene (pKa 25).'
Correct Reasoning: 'Water has a pKa of approximately 15.7, while Acetylene has a pKa of approximately 25. Since a lower pKa indicates a stronger acid, Water (pKa 15.7) is significantly more acidic than Acetylene (pKa 25).'
💡 Prevention Tips:
- Invert the Thinking: Train yourself to immediately associate 'lower pKa' with 'stronger acid' and 'higher pKa' with 'weaker acid'.
- Understand the Logarithm: Recall that pKa = -log(Ka). A large Ka (strong acid) results in a small, often negative, value for log(Ka), making -log(Ka) a small positive number.
- JEE Specific pKa Range Awareness: Memorize approximate pKa values for key functional groups: Carboxylic Acids (4-5), Phenols (10), Alcohols (16-18), Water (15.7), Terminal Alkynes (25), Amines (35-40). This provides a quick reference for comparison.
- Practice Regularly: Solve numerous problems involving acidity comparison based on pKa values to solidify this concept.
JEE_Main
Critical
Sign Error
❌ Critical Sign Error: Overlooking Terminal Alkyne Acidity
Students frequently make a critical 'sign error' by failing to recognize the highly acidic nature of the hydrogen atom attached to a terminal alkyne (R-C≡C-H). This oversight leads to incorrect predictions in reactions involving strong bases, where they might miss the formation of the crucial acetylide anion (R-C≡C:⁻) or incorrectly depict its charge distribution. This fundamental error can completely alter the predicted reaction products and pathways, especially in multi-step synthesis problems common in JEE.
💭 Why This Happens:
- Lack of Conceptual Understanding: Many students do not fully grasp how the high s-character of sp-hybridized carbon stabilizes the resulting carbanion, making the C-H bond more acidic compared to sp2 (alkenes) or sp3 (alkanes) hybridized carbons.
- Confusing Reactivity: They might incorrectly assume the terminal carbon as an electrophilic site for direct nucleophilic attack or attempt addition reactions instead of the primary acid-base reaction.
- Inadequate Practice: Insufficient practice in writing mechanisms and predicting products for reactions involving strong bases with terminal alkynes.
✅ Correct Approach:
Always identify if an alkyne is terminal. If so, its C-H bond is significantly acidic (pKa ≈ 25). In the presence of strong bases (e.g., NaNH₂, RLi, Grignard reagents), the primary reaction will be deprotonation to form the corresponding acetylide anion (R-C≡C:⁻), which is a powerful nucleophile. Correctly assigning the negative charge to the terminal carbon and understanding the ionic nature of the metal acetylide formed is crucial for subsequent reactions.
📝 Examples:
❌ Wrong:
Consider the reaction of Propyne with Sodium Amide (NaNH₂):
CH₃-C≡C-H + NaNH₂ → CH₃-C≡C-Na + NH₃
(Incorrectly showing a covalent bond between C and Na, or missing the ionic nature and charge separation, which is a sign error in representing the product.)
✅ Correct:
The correct representation of the same reaction is:
CH₃-C≡C-H + NaNH₂ → CH₃-C≡C:⁻ Na⁺ + NH₃
(The terminal hydrogen is removed as H⁺ by the strong base, forming the propynyl anion with a negative charge on the sp-hybridized carbon and a spectator Na⁺ ion.)
💡 Prevention Tips:
- Prioritize Acidity: For terminal alkynes, always consider acid-base reactions first when a strong base is present.
- Understand Hybridization: Reinforce your understanding of how sp-hybridization (50% s-character) makes the C-H bond more acidic.
- Learn pKa Values: Memorize relative pKa values: Alkynes (~25) < Alkenes (~44) < Alkanes (~50).
- Practice Anion Formation: Consistently draw the acetylide anion (R-C≡C:⁻) with the correct charge distribution. This understanding is particularly vital for JEE Main where subsequent reactions (e.g., with alkyl halides for chain lengthening) depend entirely on the correct formation and representation of this anion.
JEE_Main
Critical
Approximation
❌ Misjudging the Relative Acidity of Terminal Alkynes
Students frequently understand that terminal alkynes are acidic due to the s-character of the sp-hybridized carbon. However, a critical mistake is the incorrect approximation of their relative acidity compared to other common organic compounds or even water. This leads to erroneous predictions in acid-base reactions, especially regarding which bases can effectively deprotonate a terminal alkyne.
💭 Why This Happens:
- Lack of a well-internalized comparative acidity scale (pKa values) for common functional groups.
- Overemphasis on the 'acidic' nature without quantifying 'how acidic' it truly is.
- Confusing terminal alkynes with much stronger organic acids like carboxylic acids or phenols.
- Insufficient practice with predicting the feasibility of acid-base reactions based on relative strengths.
✅ Correct Approach:
Always place the terminal alkyne (pKa ≈ 25) on an approximate acidity scale relative to other compounds.
- Terminal alkynes are significantly stronger acids than alkanes (pKa ≈ 50), alkenes (pKa ≈ 44), and ammonia (pKa ≈ 38).
- However, they are significantly weaker acids than water (pKa ≈ 15.7), alcohols (pKa ≈ 16-18), phenols (pKa ≈ 10), and carboxylic acids (pKa ≈ 3-5).
📝 Examples:
❌ Wrong:
Predicting the reaction of a terminal alkyne with NaOH:
(This reaction typically does not proceed significantly because H₂O (pKa ~15.7) is a much stronger acid than the terminal alkyne (pKa ~25), meaning NaOH is not a strong enough base to deprotonate the alkyne.)
R-C≡C-H + NaOH → R-C≡C⁻Na⁺ + H₂O(This reaction typically does not proceed significantly because H₂O (pKa ~15.7) is a much stronger acid than the terminal alkyne (pKa ~25), meaning NaOH is not a strong enough base to deprotonate the alkyne.)
✅ Correct:
Predicting the reaction of a terminal alkyne with NaNH₂:
(This reaction proceeds because NH₃ (pKa ~38) is a significantly weaker acid than the terminal alkyne (pKa ~25), making NaNH₂ a strong enough base to deprotonate the alkyne.)
R-C≡C-H + NaNH₂ → R-C≡C⁻Na⁺ + NH₃(This reaction proceeds because NH₃ (pKa ~38) is a significantly weaker acid than the terminal alkyne (pKa ~25), making NaNH₂ a strong enough base to deprotonate the alkyne.)
💡 Prevention Tips:
- Memorize Key pKa Values: Keep a mental table of approximate pKa values for important functional groups (e.g., terminal alkyne, H₂O, alcohol, NH₃, phenol, carboxylic acid).
- Compare pKa Values: For any acid-base reaction, compare the pKa of the acid with the pKa of the conjugate acid of the base. The reaction favors the formation of the weaker acid (higher pKa).
- Practice Acid-Base Equilibrium: Work through problems specifically designed to test relative acidity and basicity.
- JEE Main Specific: Often, questions will involve multiple acidic protons or potential bases. Correctly identifying the strongest acid/base and its pKa is crucial for predicting the major product.
JEE_Main
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Alkynes: acidity and important reactions
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📝 CBSE Problems: 12
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📐 Formulas: 6
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⚠️ Mistakes: 62
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