📖Topic Explanations

🌐 Overview
Hello students! Welcome to Alkanes: free radical substitution! Get ready to unlock the secrets behind how seemingly unreactive molecules can undergo fascinating transformations.

Imagine you have a compound, an alkane, that's often considered the "lazy" molecule of organic chemistry – generally unreactive, stable, and content. These are the saturated hydrocarbons, abundant in fuels like petrol and natural gas. Their strong C-C and C-H bonds make them quite resistant to many common reagents. So, how do chemists get them to react and transform into other useful compounds? This is where the intriguing world of free radical substitution comes into play!

This topic is all about understanding the specific conditions and unique mechanism through which alkanes, typically with halogens like chlorine or bromine, can undergo a reaction where a hydrogen atom is replaced by a halogen atom. But it's not just any reaction; it's a chain reaction, driven by highly reactive species called free radicals. Think of free radicals as the "wild cards" of chemistry – they are atoms or molecules with an unpaired electron, making them extremely eager to react and achieve stability.

Understanding free radical substitution is not just crucial for alkanes; it's a fundamental concept that underpins many reactions across organic chemistry. From the industrial synthesis of various chemicals to atmospheric reactions that influence our environment, free radical chemistry plays a pivotal role. For your JEE and board exams, mastering this topic will not only equip you to predict the products of alkane halogenation but also build a strong foundation for understanding other complex reaction mechanisms, including those involved in polymer formation and anti-knock agents in fuels. It teaches you how to think about reaction pathways, energy changes, and the role of reactive intermediates.

In this journey, you'll uncover the fascinating three-step process of free radical substitution:

  • Initiation: How the first free radicals are formed, kickstarting the reaction.

  • Propagation: The "chain reaction" phase where free radicals react with stable molecules, generating new free radicals and keeping the reaction going.

  • Termination: How the free radicals eventually combine with each other, bringing the chain reaction to a halt.


You'll also explore factors like selectivity, reactivity, and how different conditions influence the outcome of these reactions, leading to various products.

Get ready to demystify the reactivity of alkanes and grasp the power of free radicals. Let's dive in and unlock the intricate chemistry that drives these essential transformations!
📚 Fundamentals
Hello, my dear students! Welcome to a fascinating journey into the world of hydrocarbons. Today, we're going to unravel the mystery behind how one of the most 'boring' and unreactive families of organic compounds – the Alkanes – can actually undergo a very important and common reaction called Free Radical Substitution.

Now, before we jump into the 'how', let's quickly refresh our memory about alkanes.

### What are Alkanes and Why are They "Sleepy"?

Remember alkanes? They are hydrocarbons that contain only single bonds between carbon atoms and between carbon and hydrogen atoms. Think of them as the 'saturated' members of the hydrocarbon family, meaning they have the maximum possible number of hydrogen atoms for their carbon skeleton. Examples include methane (CH₄), ethane (C₂H₆), propane (C₃H₈), and so on.

Chemically, alkanes are often described as unreactive or inert. Why is that?
Well, it's because:

  1. Their C-C and C-H bonds are very strong. It takes a lot of energy to break them.

  2. These bonds are also non-polar or very weakly polar. This means there isn't much positive or negative charge concentration for other charged species (like ions) to attack.


So, alkanes usually don't react with acids, bases, or common oxidizing and reducing agents under normal conditions. They just sit there, minding their own business. Because of this, they were sometimes called "paraffins," which comes from the Latin "parum affinis," meaning "little affinity" or "little reactivity."

But wait! Nature always finds a way, right? Even these 'sleepy' molecules can be woken up to react, but it requires a special kind of 'kick' and a special kind of reacting species. And that's where Free Radicals come into play!

### Introducing Our Special Guests: Free Radicals!

Imagine you have a pair of shoes. Both are perfectly fine. Now, imagine you lose one shoe. What happens? You're incomplete, right? You'd be scrambling to find the other shoe or at least find a replacement. That's a bit like a free radical!

A free radical is an atom or a group of atoms that has an unpaired electron. Yes, just one electron that isn't paired up with another electron in an orbital. We usually represent a free radical with a single dot (•) next to the atom or group. For example, a chlorine radical is Cl•, and a methyl radical is CH₃•.


Cl – Cl ⟶ Cl• + Cl•

(Chlorine molecule) (Chlorine free radicals)



Now, why is this unpaired electron a big deal? Because electrons naturally prefer to be paired up. So, an atom or molecule with an unpaired electron is in a very unstable and high-energy state. This makes free radicals extremely reactive! They are desperate to find another electron to pair up with, and they'll go to great lengths to grab one from anywhere they can. They are like hungry 'electron snatchers'.

#### How are Free Radicals Formed? Homolytic Cleavage

Free radicals are typically formed when a covalent bond breaks in a special way called homolytic cleavage (or homolysis).
In homolytic cleavage, the bond breaks symmetrically, and each atom gets one of the two bonding electrons.


A — B ⟶ A• + B•



This usually requires a significant amount of energy, typically supplied by:

  • Heat (high temperatures)

  • Light (especially ultraviolet or UV light)

  • Presence of certain peroxides (compounds with O-O bonds, which are weak)



This is different from heterolytic cleavage, where one atom gets both bonding electrons, forming ions (e.g., A—B → A⁺ + :B⁻). But for alkanes and halogens, we are mostly interested in homolytic cleavage.

### The Dance Begins: Why Alkanes React with Free Radicals (Halogenation)

So, we have our 'sleepy' alkane with its strong C-H bonds, and we have our 'super-energetic' and 'electron-hungry' free radical. When these two meet, the radical, with its intense desire for an electron, can actually abstract (pull off) a hydrogen atom from the alkane. In doing so, it forms a new bond and leaves behind *another* free radical – this time, an alkyl radical!

This type of reaction, where one atom or group is replaced by another, is called a substitution reaction. Since it's initiated and propagated by free radicals, we call it Free Radical Substitution.

A classic example of free radical substitution in alkanes is their reaction with halogens (like chlorine, Cl₂, or bromine, Br₂). This reaction is often called halogenation.

Let's take the simplest alkane, methane (CH₄), and react it with chlorine (Cl₂).


CH₄ + Cl₂ CHCl₃ + HCl



The UV light or heat is crucial here! Without it, the reaction simply won't happen at any noticeable rate because the chlorine molecule (Cl₂) won't break apart into its highly reactive free radicals.

### The Story Unfolds: The Three Stages of Free Radical Substitution

The free radical substitution reaction proceeds via a chain mechanism, which means it happens in a series of steps where the products of one step become reactants for the next. It's like a domino effect! This chain reaction can be divided into three main stages:

#### 1. Initiation Step: Getting the Party Started!

This is where the first free radicals are formed. For halogenation, the halogen molecule (like Cl₂) absorbs energy from UV light or heat. This energy is enough to break the weak Cl-Cl bond homolytically, creating two highly reactive chlorine free radicals.














Reaction Explanation

Cl — Cl 2 Cl•
 The chlorine molecule breaks into two chlorine free radicals, each with an unpaired electron. These are our 'electron snatchers'!


This step consumes energy and doesn't produce any stable products, but it's vital because it generates the radicals needed to start the entire process.

#### 2. Propagation Steps: The Chain Reaction!

This is the main part of the reaction where the 'domino effect' really takes off. In these steps, a radical reacts with a stable molecule to form a new stable molecule and *another* radical. This new radical then goes on to react with another stable molecule, continuing the chain!

Let's see how our chlorine radical (Cl•) reacts with methane (CH₄):


















Reaction Explanation

Step 1: Cl• + CH₄ ⟶ HCl + CH₃•
 A chlorine radical is so reactive that it can abstract a hydrogen atom from methane. This forms a stable hydrogen chloride (HCl) molecule and, importantly, a new free radical called a methyl radical (CH₃•).

Step 2: CH₃• + Cl₂ ⟶ CH₃Cl + Cl•
 The newly formed methyl radical (CH₃•) is also highly reactive. It attacks another chlorine molecule (Cl₂) and abstracts a chlorine atom. This forms a stable chloromethane (CH₃Cl, our desired product!) and regenerates a chlorine radical (Cl•).


Notice something crucial here: a chlorine radical (Cl•) is consumed in the first propagation step, but *another* chlorine radical (Cl•) is produced in the second propagation step. This means the reaction can continue over and over again, as long as there are reactants (methane and chlorine molecules) available. This is why it's called a chain reaction!

#### 3. Termination Steps: The Party's Over!

Eventually, the radicals don't always find a stable molecule to react with. Instead, they might bump into each other. When two free radicals combine, their unpaired electrons pair up, forming a stable covalent bond. This stops the chain reaction because no new radicals are generated.

These steps remove radicals from the system and bring the reaction to a halt.






















Reaction Explanation

Cl• + Cl• ⟶ Cl₂
 Two chlorine radicals combine to form a stable chlorine molecule.

CH₃• + CH₃• ⟶ CH₃ — CH₃ (Ethane)
 Two methyl radicals combine to form a stable ethane molecule. This is often a minor side product.

CH₃• + Cl• ⟶ CH₃Cl (Chloromethane)
 A methyl radical and a chlorine radical combine to form the stable product, chloromethane.


These termination steps essentially 'clean up' the radicals, preventing the chain reaction from continuing indefinitely.

### Key Requirements for Free Radical Substitution:

To make an alkane undergo free radical substitution, you typically need three things:

  • An Alkane (the starting material)

  • A Halogen (like Cl₂ or Br₂)

  • An energy source: UV light or heat to initiate the reaction by forming the first radicals.



### Analogy: The Gossip Chain

Think of the initiation step as the moment someone overhears a juicy piece of gossip. That person (the initial radical) is now 'activated'.
The propagation steps are when this activated person tells one other person, who then tells another, and so on. The gossip (the 'radical' nature) keeps spreading, causing reactions! Each person tells one, but then *they* become the 'activated' one to tell someone else.
The termination steps are when two people who already know the gossip meet and have no one new to tell, or when someone just stops spreading it. The gossip chain effectively ends.

### CBSE vs. JEE Focus (Fundamentals)

For CBSE/MP Board/ICSE at the fundamentals level, understanding what free radicals are, why they are reactive, and the three main steps (initiation, propagation, termination) of the mechanism with a simple example like methane chlorination is absolutely crucial. You should be able to write down these steps clearly.

For JEE Main/Advanced, this fundamental understanding is the *starting point*. JEE will build upon this by asking about:

  • The stability of different alkyl radicals (primary, secondary, tertiary).

  • The relative reactivity and selectivity of different halogens (e.g., Cl₂ vs. Br₂).

  • The formation of multiple products from larger alkanes and how to predict their proportions.

  • The energy profile diagram of the reaction.


But for today, the basics are what we've covered, and they are essential for everything that comes next!

So, the next time you see an alkane reacting, remember the little 'electron snatchers' – the free radicals – and their high-energy dance under UV light!
🔬 Deep Dive
Welcome to this deep dive into the fascinating world of Free Radical Substitution reactions in Alkanes! This is a cornerstone topic for understanding organic reactions, especially for your JEE preparations. We'll start from the very basics and build up to the intricate details that are crucial for predicting products and understanding reaction mechanisms.

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### Understanding Free Radical Substitution: The Basics

Alkanes are known for their relative inertness. They are saturated hydrocarbons, meaning they contain only single bonds, and their C-C and C-H bonds are quite strong. They don't readily undergo addition reactions like alkenes or alkynes, nor do they typically react with strong acids, bases, or oxidizing/reducing agents under normal conditions.

However, under specific conditions – typically exposure to ultraviolet (UV) light or high temperatures – alkanes can undergo reactions where one or more hydrogen atoms are replaced by another atom or group. This type of reaction is called a substitution reaction. When this substitution involves highly reactive species known as free radicals, it's specifically termed a Free Radical Substitution reaction. The most common and studied example is the halogenation of alkanes, where hydrogen atoms are replaced by halogen atoms (F, Cl, Br, I).

#### What are Free Radicals?

Before we dive into the mechanism, let's understand our key player: the free radical.
A free radical is an atom or group of atoms possessing an unpaired electron. This unpaired electron makes them extremely reactive and short-lived.

How are they formed? Free radicals are usually formed by homolytic cleavage (or homolysis) of a covalent bond. In homolytic cleavage, each atom involved in the bond retains one electron from the shared pair when the bond breaks. This often requires energy input, typically in the form of UV light (represented as hν) or heat (Δ).



A — B  --heat or hν-->  A•  +  B•


Here, A• and B• are free radicals. The dot represents the unpaired electron.

Stability of Alkyl Free Radicals:
Just like carbocations, the stability of alkyl free radicals follows a specific order:
Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (CH₃•)
This stability order is primarily due to hyperconjugation and the inductive effect. Alkyl groups (electron-donating groups) can stabilize the radical center by delocalizing the unpaired electron density through hyperconjugation with adjacent C-H bonds, and also by pushing electron density towards the electron-deficient radical center (though to a lesser extent than for carbocations).

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### The Mechanism of Free Radical Halogenation: A Chain Reaction

The free radical halogenation of alkanes, such as the chlorination of methane, proceeds via a chain reaction mechanism. This mechanism can be broken down into three distinct steps:

1. Initiation Step
2. Propagation Steps
3. Termination Steps

Let's illustrate this using the chlorination of methane (CH₄ + Cl₂ → CH₃Cl + HCl) as our prime example.

#### 1. Initiation Step: Creating the Radicals

This step requires energy (UV light or heat) to create the initial free radicals. It's the "kick-off" for the entire reaction.
Only the halogen molecule (e.g., Cl₂) undergoes homolytic cleavage. The C-H bond in methane is much stronger and does not break at this stage.



Cl — Cl  --hν or Δ-->  Cl•  +  Cl•

*Explanation:* A chlorine molecule absorbs a photon of UV light (hν) or enough thermal energy (Δ), causing the weak Cl-Cl bond (bond dissociation energy ~242 kJ/mol) to break homolytically, forming two highly reactive chlorine free radicals (Cl•).

#### 2. Propagation Steps: The Chain Continues

These are the "heart" of the chain reaction. In these steps, a radical reacts with a stable molecule to form a new stable molecule and a new radical. This allows the reaction to continue without requiring further initiation.

There are two propagation steps:

a) Hydrogen Abstraction: A chlorine radical (Cl•) is highly reactive and seeks to pair its unpaired electron. It does so by abstracting a hydrogen atom from a stable methane molecule (CH₄). This breaks a C-H bond homolytically.



Cl•  +  CH₃—H  -->  HCl  +  CH₃•

*Explanation:* The Cl• attacks a methane molecule, breaking a C-H bond. A new H-Cl bond is formed, releasing hydrochloric acid (HCl), and generating a methyl free radical (CH₃•). The C-H bond (bond dissociation energy ~439 kJ/mol) is much stronger than the Cl-Cl bond, but the overall reaction is exothermic because the H-Cl bond formed is very strong.

b) Halogen Abstraction: The newly formed methyl radical (CH₃•) is also highly reactive. It reacts with another stable chlorine molecule (Cl₂) to abstract a chlorine atom.



CH₃•  +  Cl—Cl  -->  CH₃Cl  +  Cl•

*Explanation:* The CH₃• attacks a chlorine molecule, breaking the Cl-Cl bond. A new C-Cl bond is formed, producing chloromethane (CH₃Cl) (our product), and regenerating a chlorine free radical (Cl•). The regeneration of Cl• is crucial because it can now go back and react with another methane molecule, continuing the chain reaction. This is why it's called a *chain* reaction.

NET PROPAGATION REACTION:
If you sum the two propagation steps, you'll see the radicals cancel out, giving the overall substitution reaction:
Cl• + CH₄ → HCl + CH₃•
CH₃• + Cl₂ → CH₃Cl + Cl•
------------------------------------
CH₄ + Cl₂ → CH₃Cl + HCl

#### 3. Termination Steps: Stopping the Chain

Eventually, the chain reaction needs to stop. This happens when two free radicals combine to form a stable, non-radical molecule. These steps consume radicals without generating new ones. Since radical concentrations are usually very low, termination steps are less frequent than propagation steps.

Possible termination steps:

* Radical-Radical Combination 1 (Cl• + Cl•):


    Cl•  +  Cl•  -->  Cl₂

*Explanation:* Two chlorine radicals combine to reform a chlorine molecule.

* Radical-Radical Combination 2 (CH₃• + CH₃•):


    CH₃•  +  CH₃•  -->  CH₃—CH₃  (Ethane)

*Explanation:* Two methyl radicals combine to form ethane. This is a common side product in free radical reactions.

* Radical-Radical Combination 3 (CH₃• + Cl•):


    CH₃•  +  Cl•  -->  CH₃Cl  (Chloromethane)

*Explanation:* A methyl radical and a chlorine radical combine to form the desired product, chloromethane. While this forms a product, it also removes two radicals, thus terminating the chain.

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### Reactivity and Selectivity in Halogenation

The choice of halogen and the structure of the alkane significantly influence the outcome of the reaction, particularly the reactivity and selectivity.

#### 1. Reactivity of Halogens

The reactivity of halogens in free radical substitution follows the order:
F₂ > Cl₂ > Br₂ > I₂

* Fluorination (F₂): Extremely exothermic and often explosive. It's difficult to control and typically leads to extensive fragmentation of the alkane. Not practically useful.
* Chlorination (Cl₂): Exothermic, but controllable. It can lead to multiple substitutions (e.g., CH₄ can form CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄). It is generally less selective, meaning it reacts almost equally with primary, secondary, and tertiary hydrogens.
* Bromination (Br₂): Less exothermic and much slower than chlorination. It is significantly more selective, preferring to substitute tertiary hydrogens over secondary, and secondary over primary. This makes it more useful for preparing specific monobrominated products.
* Iodination (I₂): Endothermic and reversible. The HI produced is a strong reducing agent and can reduce the alkyl iodide back to the alkane. Thus, it usually doesn't occur without an oxidizing agent (like HIO₃ or HNO₃) to remove HI and drive the reaction forward. It's generally not a practical method for synthesizing alkyl iodides.




































Halogen Relative Reactivity Characteristics Selectivity
F₂ Very High Explosive, uncontrollable, forms mixtures. Very Low
Cl₂ High Exothermic, manageable, multiple substitutions possible. Low (but somewhat selective)
Br₂ Moderate Less exothermic, slower, requires heat/light. High
I₂ Very Low Endothermic, reversible, requires oxidizing agent. N/A (Impractical)


#### 2. Reactivity of Hydrogen Atoms (and Radical Stability)

The relative ease with which a hydrogen atom is abstracted by a halogen radical (Cl• or Br•) is critical for determining the product distribution. This directly correlates with the stability of the intermediate alkyl radical formed.
The order of abstraction for hydrogen atoms is:
Tertiary (3°) > Secondary (2°) > Primary (1°)

This is because the stability of the free radical formed in the propagation step (R•) follows the same order:
3° R• > 2° R• > 1° R•

Example: Chlorination of Propane (CH₃-CH₂-CH₃)

Propane has two types of hydrogen atoms:
* Six primary (1°) hydrogens (on the CH₃ groups)
* Two secondary (2°) hydrogens (on the CH₂ group)

A chlorine radical can abstract either a 1° or a 2° hydrogen.
1. Abstraction of 1° H:


    Cl• + CH₃-CH₂-CH₃  -->  HCl + CH₃-CH₂-CH₂• (1° radical)

    CH₃-CH₂-CH₂• + Cl₂  -->  CH₃-CH₂-CH₂Cl (1-chloropropane) + Cl•


2. Abstraction of 2° H:


    Cl• + CH₃-CH₂-CH₃  -->  HCl + CH₃-CH•-CH₃ (2° radical)

    CH₃-CH•-CH₃ + Cl₂  -->  CH₃-CHCl-CH₃ (2-chloropropane) + Cl•


Relative Rates of Abstraction:
Experimental data shows that the relative rates of abstraction for different types of hydrogens at 25°C are approximately:

* For Chlorination (Cl₂):
* 1° H : 2° H : 3° H ≈ 1.0 : 3.8 : 5.0
This means a 2° H is 3.8 times more likely to be abstracted than a 1° H, and a 3° H is 5 times more likely. While there is a preference for more substituted hydrogens, the differences are not huge, leading to mixtures of products.

* For Bromination (Br₂):
* 1° H : 2° H : 3° H ≈ 1.0 : 82 : 1600
This stark difference indicates that bromination is highly selective. A 3° H is 1600 times more likely to be abstracted than a 1° H! This makes bromination very useful for synthesizing specific products, especially when a tertiary hydrogen is available.

#### Why the Difference in Selectivity?

The difference in selectivity between chlorination and bromination is explained by the Hammond Postulate.

* Chlorination: The hydrogen abstraction step (R-H + Cl• → R• + HCl) is highly exothermic. According to the Hammond Postulate, the transition state for an exothermic reaction resembles the reactants (early transition state). In an early transition state, the C-H bond is only slightly broken, and the radical character on carbon is not fully developed. Therefore, the stability of the incipient alkyl radical (R•) has less influence on the activation energy, leading to lower selectivity. The reaction is fast and less discriminating.

* Bromination: The hydrogen abstraction step (R-H + Br• → R• + HBr) is much less exothermic, almost thermoneutral, or even slightly endothermic for 1° H. The transition state for an endothermic reaction resembles the products (late transition state). In a late transition state, the C-H bond is significantly broken, and the radical character on carbon is well developed. Thus, the stability of the forming alkyl radical (R•) strongly influences the activation energy. A more stable radical (3° > 2° > 1°) will have a significantly lower activation energy, leading to a much faster reaction and high selectivity. The reaction is slower but highly discriminating.

Think of it this way:
* Chlorine is like a greedy, fast attacker. It's so reactive that it will attack almost any hydrogen it encounters without much thought, just getting the job done quickly. So, you get a mix of products.
* Bromine is like a picky, strategic attacker. It's less reactive, so it takes its time, looks for the easiest target (the most stable radical intermediate), and only then attacks. This leads to a highly selective outcome.

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### Predicting Products and Solving Problems (JEE Focus)

For JEE, you must be able to:
1. Identify the type of hydrogens (1°, 2°, 3°) in a given alkane.
2. Understand the relative reactivity of halogens (Cl vs. Br).
3. Apply the relative rates of abstraction to calculate the approximate percentage of different monohalogenated products.

Example 1: Monochlorination of Isobutane (2-Methylpropane)



        CH₃

        |

CH₃ — CH — CH₃


* Types of Hydrogens:
* Nine primary (1°) hydrogens (on the three CH₃ groups).
* One tertiary (3°) hydrogen (on the central CH group).

* Possible products:
1. Substitution at a 1° carbon: 1-chloro-2-methylpropane (isobutyl chloride)
2. Substitution at the 3° carbon: 2-chloro-2-methylpropane (tert-butyl chloride)

* Calculation of Percentage Yields (using Cl₂ relative rates 1:3.8:5):
* Number of 1° H: 9
* Number of 3° H: 1
* Relative rate for 1° H = 1.0
* Relative rate for 3° H = 5.0

* "Reactivity Index" for 1° H substitution: 9 (hydrogens) × 1.0 (rate) = 9
* "Reactivity Index" for 3° H substitution: 1 (hydrogen) × 5.0 (rate) = 5

* Total Reactivity Index: 9 + 5 = 14

* % 1-chloro-2-methylpropane: (9 / 14) × 100 ≈ 64.3%
* % 2-chloro-2-methylpropane: (5 / 14) × 100 ≈ 35.7%

Conclusion: Even though the 3° position is more reactive, there are many more 1° hydrogens, leading to a significant amount of the 1° substituted product. Chlorination is not very selective.

Example 2: Monobromination of Isobutane

Using the same alkane, let's see the effect of bromination (Br₂ relative rates 1:82:1600):

* "Reactivity Index" for 1° H substitution: 9 (hydrogens) × 1.0 (rate) = 9
* "Reactivity Index" for 3° H substitution: 1 (hydrogen) × 1600 (rate) = 1600

* Total Reactivity Index: 9 + 1600 = 1609

* % 1-bromo-2-methylpropane: (9 / 1609) × 100 ≈ 0.56%
* % 2-bromo-2-methylpropane: (1600 / 1609) × 100 ≈ 99.44%

Conclusion: Bromination is highly selective. The tertiary substituted product is overwhelmingly favored, making it a very useful method for targeted synthesis if a tertiary hydrogen is available.

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### Advanced Considerations for JEE

* Multiple Substitutions: In excess halogen, especially with chlorine, multiple hydrogen atoms can be substituted, leading to di-, tri-, and tetra-halogenated products. For example, in the chlorination of methane, you can get CH₃Cl, CH₂Cl₂, CHCl₃, and CCl₄. Controlling stoichiometry (using excess alkane) helps minimize polysubstitution.
* Stereochemistry: If the substitution creates a chiral center, a racemic mixture of enantiomers will be formed because the planar free radical intermediate can be attacked from either face with equal probability.
* Allylic/Benzylic Halogenation: Free radical mechanisms are also involved in allylic and benzylic halogenation (e.g., using NBS for bromination at allylic/benzylic positions), which are much more reactive due to the resonance stabilization of the resulting allylic or benzylic radical. While this falls under "alkenes" or "aromatics," the core free radical mechanism is the same.

By mastering the initiation, propagation, and termination steps, and understanding the nuances of reactivity and selectivity, you'll be well-equipped to tackle any free radical substitution problem in your JEE exams!
🎯 Shortcuts

Free radical substitution is a fundamental reaction for alkanes, typically involving halogens in the presence of UV light or heat. Remembering the key steps, conditions, and selectivity aspects is crucial for success in exams. Here are some mnemonics and shortcuts to help you.



1. Mechanism Steps: "I P T"


The free radical substitution mechanism always proceeds in three distinct stages:



  • Initiation

  • Propagation

  • Termination


Mnemonic: Think of it as a logical sequence: "I Pass the Test" or "I Proceed Through".




  • I - Initiation: This is where radicals are formed. Typically, the halogen molecule (e.g., Cl-Cl) undergoes homolytic cleavage due to UV light or heat.

    Think: "The fight starts here, radicals are born!"

  • P - Propagation: Radicals react with stable molecules to form new radicals, carrying the chain forward. This involves two steps:

    1. Hydrogen abstraction from alkane by a halogen radical.

    2. Alkyl radical reacts with a halogen molecule to form the product and a new halogen radical.


    Think: "The fight continues, radicals keep creating more radicals."

  • T - Termination: Radicals combine with each other to form stable molecules, ending the chain reaction.

    Think: "The fight ends, radicals combine to make stable molecules."




2. Reaction Conditions: "HULL" for Halogens


To initiate free radical substitution, you need:



  • Halogen (Cl2, Br2)

  • UV Light or Light (hν)

  • Light (or Low temperature is not suitable, high heat is) - let's focus on light as the primary initiator.


Mnemonic: Think "HULL" – a strong, foundational part of a ship, representing the strong conditions needed to break bonds. Or, simply Halogen + Light/Heat.



3. Radical Stability: "3,2,1 GO!"


The stability of alkyl free radicals follows the order:


Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl


Mnemonic: Think of a race countdown: "3, 2, 1 GO!" – the higher the number, the more stable the radical and the more preferred the site for hydrogen abstraction. This is due to hyperconjugation and inductive effects.



  • 3° Radical: Most stable (e.g., (CH3)3C•)

  • 2° Radical: More stable than primary (e.g., (CH3)2CH•)

  • 1° Radical: Least stable among alkyl radicals (e.g., CH3CH2•)

  • Methyl Radical: Least stable (CH3•)


JEE Tip: This stability order dictates the regioselectivity of the reaction. Abstraction of a 3° hydrogen is easiest because it leads to the most stable radical intermediate.



4. Selectivity of Chlorination vs. Bromination: "Chlorine is Crazy, Bromine is Brainy"


This is a critical distinction for JEE and board exams:






































Feature Chlorination (Cl2) Bromination (Br2)
Mnemonic "Chlorine is Crazy" (less selective) "Bromine is Brainy" (more selective)
Selectivity Low: Reacts almost indiscriminately with all types of H (1°, 2°, 3°). Product distribution is mainly statistical. High: Highly selective for abstracting 3° H > 2° H > 1° H. Major product is almost always the one from the most stable radical.
Reactivity/Speed Very Fast: The Cl• radical is very reactive. Slower: The Br• radical is less reactive.
Controlling Factor Mainly kinetic control (speed of reaction). Mainly thermodynamic control (stability of the radical intermediate).
Product Example Isobutane + Cl2 → Mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane (significant amounts of both). Isobutane + Br2 → Primarily 2-bromo-2-methylpropane (very high yield).


Mnemonic Breakdown:



  • Chlorine is Crazy: It reacts wildly and quickly, attacking almost any hydrogen in its path. This means you get a *mixture* of products, reflecting the number of each type of hydrogen present. It doesn't "think" much about stability.

  • Bromine is Brainy: It's more discerning and selective. It "thinks" about forming the most stable radical intermediate, so it will preferentially abstract a 3° H over a 2° H, and a 2° H over a 1° H. This leads to a *single major product*.


By using these mnemonics and understanding the underlying chemical principles, you can confidently approach questions on free radical substitution of alkanes in your exams.

💡 Quick Tips

📝 Quick Tips: Alkanes - Free Radical Substitution 📝



Free radical substitution is a cornerstone reaction for alkanes. Mastering its nuances is vital for both board exams and competitive tests like JEE. Here are some quick, exam-focused tips to ace this topic:




  • Mechanism Overview: Remember the three essential steps:

    • Initiation: Homolytic cleavage of the halogen molecule (e.g., Cl–Cl) by UV light or heat to form two highly reactive halogen free radicals (e.g., Cl•). This is the energy-demanding step.

    • Propagation: This is the chain-carrying step.

      1. A halogen radical abstracts a hydrogen from the alkane, forming an alkyl radical (R•) and HX.

      2. The alkyl radical then reacts with another halogen molecule (X2) to form the alkyl halide (RX) and a new halogen radical (X•), which continues the chain.



    • Termination: Any two free radicals combine to form a stable molecule. This stops the chain reaction. Examples: X• + X• → X2; R• + X• → RX; R• + R• → R–R (dimerization, a minor byproduct).



  • Radical Stability & Selectivity: This is arguably the most critical concept for predicting products.

    • Stability Order: Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl radical. A more stable radical forms faster and is preferred.

    • Chlorination (Cl2/hν): Highly reactive, less selective. It attacks almost all types of hydrogens. Product distribution is a mix, influenced by both the number of equivalent hydrogens and radical stability (though stability has a lesser effect compared to bromination). You'll get significant amounts of all possible monochlorinated products.

    • Bromination (Br2/hν): Less reactive, but highly selective. It preferentially abstracts hydrogen atoms that lead to the most stable radical. Therefore, the major product is usually the one formed by substituting a 3° H (if available), followed by 2° H, and then 1° H.


    💡 JEE Tip: For bromination, expect the product from the most stable radical to be the overwhelming major product. For chlorination, consider all possibilities and their relative amounts based on both statistical factor and radical stability ratios (e.g., 1°:2°:3° reactivity ratios for Cl often ~1:3.8:5.0 at 25°C).



  • Polyhalogenation: A common issue. Since the product alkyl halide (RX) also has C-H bonds, it can undergo further substitution. This leads to di-, tri-, and even tetra-substituted products. This makes it difficult to obtain a single desired product in high yield.

  • Reaction Conditions: Always remember that UV light (hν) or high temperature is required to initiate the reaction by breaking the halogen-halogen bond. In its absence, the reaction will not proceed.

  • Chirality: If the substitution creates a chiral center, a racemic mixture of enantiomers will be formed because the radical intermediate is planar, allowing attack from either face.



By keeping these quick tips in mind, you'll be well-prepared to tackle questions on free radical substitution of alkanes.

🧠 Intuitive Understanding

Welcome to the intuitive understanding of Free Radical Substitution in Alkanes! This concept is fundamental to understanding alkane reactivity and a frequent topic in competitive exams like JEE.



Alkanes, often called paraffins (from Latin 'parum affinis' meaning 'little affinity'), are known for their relative unreactivity. This inertness stems from the strong, non-polar C-C and C-H sigma bonds, which are difficult to break under normal conditions. So, how do we get them to react, especially with halogens?



The Core Idea: Breaking Inertness with Energy and Radicals


Imagine you have a very stable, tightly bound system (the alkane). To make it react, you need to introduce something highly reactive or provide significant energy. Free radical substitution employs both:



  • High Energy Input: The reaction typically requires UV light (hν) or high temperatures (e.g., 250-400°C) to initiate. This energy is crucial for breaking the relatively strong halogen-halogen bond (like Cl-Cl or Br-Br) homolytically, generating highly reactive free radicals.

  • Free Radicals: A free radical is an atom or group of atoms possessing an unpaired electron, making it extremely reactive. These 'electron-hungry' species are the driving force behind the substitution.



The "Domino Effect" – A Chain Reaction


Once formed, these highly reactive halogen radicals don't just sit there; they attack the alkane. The entire process unfolds as a chain reaction, typically divided into three intuitive stages:



  1. Initiation: Starting the Chain

    • Energy (light/heat) breaks the halogen molecule (e.g., Cl-Cl) into two highly reactive halogen free radicals (e.g., 2Cl•). This is the 'spark' that starts the reaction.



  2. Propagation: Continuing the Chain

    • A halogen radical (Cl•) attacks an alkane, abstracts a hydrogen atom, and forms HCl and a new alkyl radical (R•). This is the substitution step, where a hydrogen is replaced.

    • This alkyl radical (R•) is also highly reactive. It then reacts with another halogen molecule (e.g., Cl-Cl), forming the desired haloalkane (R-Cl) and regenerating a new halogen radical (Cl•).

    • This regenerated halogen radical can then attack another alkane molecule, continuing the cycle. This is why it's a "chain" – one radical produces another, propagating the reaction.



  3. Termination: Ending the Chain

    • Eventually, two free radicals might collide with each other (e.g., Cl• + Cl• → Cl₂; R• + R• → R-R; R• + Cl• → R-Cl). These collisions remove radicals from the reaction mixture, breaking the chain. This is how the reaction 'stops'.





Key Insights for JEE/CBSE:



  • Multiple Substitution: Because the halogen radical is regenerated, it can keep reacting. This often leads to a mixture of products where more than one hydrogen atom on the alkane is substituted (e.g., CH₄ can give CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄). This is a major limitation for selective synthesis.

  • Regioselectivity: The abstraction of hydrogen by a halogen radical is not entirely random. The stability of the intermediate alkyl radical dictates the preference. More substituted alkyl radicals are more stable (tertiary > secondary > primary). Therefore, a tertiary hydrogen is abstracted more readily than a secondary, which is more readily abstracted than a primary hydrogen. This results in the formation of more substituted haloalkanes as major products.

  • Reactivity of Halogens: Fluorine reacts explosively, chlorine and bromine react controllably, and iodine reacts reversibly (requiring an oxidizing agent to remove HI). The reactivity order is F₂ > Cl₂ > Br₂ > I₂.



In essence, free radical substitution is like a 'radical party' where a little energy starts a chain of highly reactive species attacking alkanes, replacing hydrogens, and only stopping when the radicals run out of partners to react with!


Keep these intuitive ideas in mind as you delve deeper into the specific mechanisms!

🌍 Real World Applications

Real World Applications: Alkanes - Free Radical Substitution


Free radical substitution reactions, though often studied in a theoretical context, underpin several crucial industrial processes and natural phenomena. Understanding these mechanisms helps us appreciate the chemistry behind everyday materials and energy sources.



1. Industrial Production of Halogenated Organic Compounds


The most direct application of free radical substitution in alkanes is their halogenation. This process is used to synthesize a variety of industrially important halogenated alkanes, which serve as:




  • Solvents: Compounds like dichloromethane (CH2Cl2) and trichloromethane (CHCl3, chloroform) are excellent solvents for fats, waxes, resins, and other organic materials. Tetrachloromethane (CCl4) was widely used as a cleaning agent, though its use is now restricted due to toxicity and environmental concerns.


  • Refrigerants & Propellants: Chlorofluorocarbons (CFCs), such as CCl2F2 (Freon-12), were historically produced via free radical halogenation and used extensively in refrigeration and aerosol propellants. However, their role in stratospheric ozone depletion led to their phasing out under the Montreal Protocol. Newer, more environmentally friendly alternatives (e.g., HFCs) are now used, some of which still involve halogenation steps.


  • Chemical Intermediates: Halogenated alkanes are valuable intermediates in the synthesis of many other organic compounds, including pharmaceuticals, pesticides, and polymers, through subsequent substitution or elimination reactions.



2. Cracking of Petroleum Hydrocarbons


One of the most significant industrial applications related to alkanes and free radical chemistry is the cracking of crude oil. Crude oil is a mixture of various hydrocarbons, primarily alkanes. Demand for shorter-chain hydrocarbons (like those found in gasoline) is often higher than their natural abundance in crude oil, while longer-chain alkanes are less valuable.




  • Process: Cracking involves breaking down larger alkane molecules into smaller, more useful ones (e.g., gasoline components like C5-C12 alkanes) and unsaturated hydrocarbons (alkenes like ethene and propene).


  • Mechanism: While catalytic cracking exists, thermal cracking, particularly at high temperatures, proceeds through a complex free radical mechanism, involving the homolytic cleavage of C-C bonds to initiate radical chains.


  • Importance: This process is fundamental to the petrochemical industry, increasing the yield of valuable fuels and providing alkenes, which are crucial monomers for the production of plastics (e.g., polyethylene, polypropylene).



3. Combustion of Fuels


The burning of alkanes, our primary energy source for vehicles, heating, and power generation, is a complex process that involves intricate free radical chain reactions. While it's an oxidation reaction rather than a substitution in the typical sense, free radicals (like hydroxyl radicals, •OH, and hydroperoxyl radicals, •OOH) play a central role in initiating and propagating the combustion process.




  • Energy Release: This is the most widespread and economically important application of alkanes, enabling the release of substantial energy.


  • Environmental Impact: Understanding the free radical mechanisms in combustion helps in designing more efficient engines and catalytic converters to minimize the formation of harmful byproducts (e.g., NOx, soot).



4. Atmospheric Chemistry


Free radical reactions of alkanes also occur naturally in the atmosphere. For example, the reaction of methane (CH4), a potent greenhouse gas, with hydroxyl radicals (•OH) is a key process in its removal from the atmosphere. This reaction is a free radical abstraction (a type of substitution):


CH4 + •OH → •CH3 + H2O


This initiates a series of reactions that ultimately lead to the oxidation of methane into CO2 and H2O.




Key Takeaway for JEE/CBSE: While detailed mechanisms of industrial processes are beyond the scope, recognizing that free radical substitution is crucial for producing valuable chemicals (like halogenated compounds) and for energy production (combustion, cracking) is important for understanding the broader relevance of organic chemistry.


🔄 Common Analogies

Common Analogies for Free Radical Substitution


Understanding complex chemical mechanisms can often be simplified by relating them to everyday experiences. For free radical substitution in alkanes, analogies help grasp the concept of initiation, propagation, and termination, especially the chain reaction nature.



The "Tag" Game (or "Hot Potato") Analogy


Imagine a game of "Tag" where one person is "it" and tries to tag others. This game beautifully illustrates the mechanism of free radical substitution:





  1. The "Tagger" (or "It") / Hot Potato = The Free Radical:

    • In the game, the person who is "it" is unstable and wants to get rid of this status by tagging someone else.

    • In chemistry, a free radical (e.g., Cl•) is highly reactive and unstable due to its unpaired electron. It actively seeks to react and achieve stability.




  2. Initiation Phase = Starting the Game:

    • In the "Tag" game, someone (like a referee) has to pick the first person to be "it" and start the game.

    • In free radical substitution, a stable molecule (like Cl₂) is provided with energy (e.g., UV light or heat). This energy cleaves the bond homolytically, creating two highly reactive free radicals (e.g., Cl₂ → 2Cl•). This is the initial "tagger" entering the field.




  3. Propagation Phase = The Game in Progress (Chain Reaction):

    • This is the main part of the game where the "it" status is passed around, keeping the game going.

    • Step 1 (Radical attacks Alkane): The initial "it" player (Cl• radical) runs and tags a stable, unaware player (an alkane molecule, R-H). When tagged, the stable player becomes the new "it" (forms an alkyl radical, R•), and the original "it" player is now free (forms a stable HCl molecule). The "it" status has been successfully transferred.

    • Step 2 (New Radical attacks Reagent): The newly created "it" player (R• radical) then quickly tags another bystander (another Cl₂ molecule). This bystander splits: one part becomes the new "it" (a new Cl• radical is formed), and the other part becomes a stable player who is "out" (forms R-Cl, the desired alkyl halide product).

    • This cycle continues, as one "it" player (radical) always creates another "it" player, perpetuating the chain reaction.




  4. Termination Phase = Ending the Game:

    • The game ends when two "it" players meet and decide to stop, or they are somehow removed from play (e.g., by high-fiving and both becoming "not it").

    • In chemistry, the chain reaction ends when two free radicals combine to form a stable, non-radical molecule. This removes the reactive species from the system. Examples include:

      • Cl• + Cl• → Cl₂

      • R• + R• → R-R (e.g., butane from two ethyl radicals)

      • R• + Cl• → R-Cl







This analogy highlights how a small input of energy (to start the game) can lead to a long chain of reactions, and how the "reactive nature" is transferred until two such reactive species eventually neutralize each other.

📋 Prerequisites

Prerequisites for Free Radical Substitution of Alkanes


To effectively understand and master the free radical substitution reactions of alkanes, a strong foundation in the following core concepts is essential. These prerequisites ensure that you can grasp the mechanism, product formation, and selectivity aspects without difficulty.



1. Basic Organic Chemistry Fundamentals



  • Hydrocarbons & Alkanes: A clear understanding of what hydrocarbons are, specifically the definition and general formula of alkanes (CnH2n+2). Knowledge of their saturated nature is key.

  • Nomenclature: Ability to name simple alkanes using IUPAC rules and identify primary, secondary, and tertiary carbon atoms and hydrogen atoms. This is crucial for predicting reaction sites and product identification.

  • Isomerism: Familiarity with structural isomerism, particularly chain and position isomerism, as free radical substitution often leads to the formation of multiple isomeric products.



2. Chemical Bonding and Structure



  • Covalent Bonding: Understanding the nature of C-C and C-H single covalent bonds (sigma bonds).

  • Hybridization & Geometry: Knowledge of sp3 hybridization in carbon atoms of alkanes, leading to a tetrahedral geometry around each carbon. This impacts bond angles and steric hindrance.

  • Bond Dissociation Energy (BDE): An awareness of BDE, especially for C-H bonds, is important to understand the relative ease of hydrogen abstraction. Understanding the concept of homolytic cleavage, where a covalent bond breaks such that each fragment retains one electron.



3. Introduction to Reaction Mechanisms



  • Types of Bond Cleavage: Distinguish between homolytic and heterolytic bond cleavage. Free radical reactions exclusively involve homolytic cleavage.

  • Free Radicals: Crucial: You must know what a free radical is (a species with an unpaired electron), how it is formed (typically by homolytic cleavage of a non-polar bond using heat or light), and understand the relative stability of alkyl free radicals (tertiary > secondary > primary > methyl).

  • Arrow Pushing: Basic understanding of using single-headed (half-hooked) arrows to depict the movement of single electrons in radical reactions.



⚠️ Common Exam Traps

Free radical substitution of alkanes, while seemingly straightforward, harbors several conceptual and application-based traps that students frequently fall into during exams. Understanding these common pitfalls can significantly improve your score.





⚠ Common Exam Traps in Free Radical Substitution ⚠





  • Trap 1: Ignoring Regioselectivity and Statistical Factors (especially for Chlorination)


    • Mistake: Assuming all hydrogens in an alkane are equally reactive or only considering the most stable radical. Forgetting to account for the number of equivalent hydrogens.


    • Correction: While the order of hydrogen abstraction reactivity is 3° > 2° > 1° (due to stability of the intermediate free radical), the actual product distribution is a combination of this relative reactivity and the statistical factor (number of each type of hydrogen).

      • JEE Tip: For chlorination, the difference in reactivity between 1°, 2°, and 3° H atoms is small (approx. 1:3.8:5 for Cl). Therefore, statistical factors play a significant role, and you often get a complex mixture where the product from the most numerous hydrogens might be major, even if they are primary.

      • JEE Tip: For bromination, the difference in reactivity is much larger (approx. 1:82:1600 for Br). This makes bromination highly selective for the most stable (3°) radical, virtually ignoring statistical factors.


      Example: In the chlorination of isobutane (2-methylpropane), despite the tertiary H being more reactive, the product derived from primary H (1-chloro-2-methylpropane) often forms in a higher amount due to there being nine primary H's compared to only one tertiary H. However, in bromination, 2-bromo-2-methylpropane (from 3° H) would be the almost exclusive product.





  • Trap 2: Forgetting about Polyhalogenation


    • Mistake: Only considering mono-substituted products in your answer.


    • Correction: Free radical halogenation is generally a non-selective process. Once an alkane is monohalogenated, the resulting haloalkane can also undergo further substitution, leading to di-, tri-, and even poly-halogenated products. To favor mono-substitution, alkanes must be used in a large excess relative to the halogen. If not specified, assume a mixture, especially for chlorination.





  • Trap 3: Incorrectly Identifying All Possible Isomeric Products


    • Mistake: Missing out on some structural isomers that can be formed from the initial alkane.


    • Correction: Systematically identify all chemically non-equivalent hydrogen atoms in the given alkane. Each set of non-equivalent hydrogens can potentially lead to a distinct monohalogenated product. For example, n-butane has two types of non-equivalent hydrogens, leading to 1-halobutane and 2-halobutane.





  • Trap 4: Overlooking Reaction Conditions (UV light/Heat)


    • Mistake: Assuming the reaction occurs spontaneously or at room temperature without specific initiation conditions.


    • Correction: Free radical substitution requires initiation by high energy, typically UV light (hν) or high temperatures (>250°C), to homolytically cleave the halogen molecule and generate the initial radicals. Without these conditions, the reaction will not proceed. Always include hν or Δ (heat) over the reaction arrow.





  • Trap 5: Misunderstanding Halogen Reactivity Extremes


    • Mistake: Believing fluorination and iodination are as practical as chlorination and bromination.


    • Correction: Fluorination is extremely vigorous and often explosive, making it difficult to control. Iodination is very slow and reversible; the HI formed is a strong reducing agent that can reduce the iodoalkane back to the alkane. Therefore, free radical halogenation is predominantly used for chlorine and bromine.





By being mindful of these common traps, you can approach questions on free radical substitution with greater accuracy and confidence.

⭐ Key Takeaways

This section summarizes the most crucial concepts regarding the Free Radical Substitution of Alkanes. Mastering these points is essential for both conceptual understanding and exam performance.





Key Takeaways: Alkanes - Free Radical Substitution





  • Characteristic Reaction: Free radical substitution is the most characteristic reaction of alkanes, occurring in the presence of UV light (hν) or high temperature. It involves the replacement of a hydrogen atom by a halogen atom.


  • Mechanism - Three Steps:


    1. Initiation: Homolytic cleavage of the halogen molecule (e.g., Cl2 → 2Cl•) by UV light or heat, generating free radicals. This step requires energy.


    2. Propagation: A chain reaction involving two steps:

      • A halogen radical (X•) abstracts a hydrogen from the alkane, forming an alkyl radical (R•) and HX.

      • The alkyl radical (R•) reacts with another halogen molecule (X2), forming the alkyl halide (RX) and regenerating a halogen radical (X•). This step sustains the chain.




    3. Termination: Combination of any two free radicals (e.g., X• + X•, R• + R•, R• + X•) to form stable, non-radical products, ending the chain reaction.




  • Free Radical Stability & Regioselectivity:

    • The stability of alkyl free radicals follows the order: Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl.

    • This stability dictates the ease of hydrogen abstraction. A more stable alkyl radical is formed more readily. Therefore, the rate of abstraction of hydrogen atoms by a halogen radical is: 3° H > 2° H > 1° H.

    • This leads to the formation of multiple isomeric products, with the product formed from the most stable radical being the major product (JEE Focus).




  • Reactivity vs. Selectivity:

    • Chlorination: Less selective, but highly reactive. Produces a mixture of all possible monochloroalkanes in significant amounts. The ratio of products is influenced by both the number of available hydrogens and their relative reactivity (e.g., 3°:2°:1° reactivity ratio approx. 5:4:1).

    • Bromination: More selective, but less reactive. It predominantly forms the product resulting from the abstraction of the most stable hydrogen (e.g., 3° H). The 3°:2°:1° reactivity ratio is significantly higher (e.g., 1600:82:1). Bromination is often preferred for preparing a single major product due to its higher selectivity.

    • JEE Insight: Understand that the more selective a reaction, the less reactive it generally is. This is due to a later transition state resembling the stable radical.




  • Limitations:

    • Lack of control over the reaction, often leading to polysubstitution (e.g., CH4 can form CH3Cl, CH2Cl2, CHCl3, CCl4).

    • Formation of a mixture of isomeric products, which can be difficult to separate.

    • Fluorination is too violent; iodination is too slow and reversible.





Exam Tip: For CBSE, focus on writing the mechanism correctly and identifying the type of reaction. For JEE, pay close attention to predicting major products based on free radical stability and understanding the differences in selectivity between chlorination and bromination.

🧩 Problem Solving Approach

Problem Solving Approach for Free Radical Substitution in Alkanes


Solving problems related to free radical substitution (FRS) reactions in alkanes requires a systematic approach, focusing on understanding the mechanism, predicting product regioselectivity, and distinguishing between different halogens.



General Problem-Solving Strategy:



  1. Identify Reactants and Conditions:

    • Look for an alkane and a halogen (Cl2 or Br2).

    • Crucially, identify the presence of UV light or heat ($Delta$), which signifies free radical conditions. Absence of these means FRS won't occur, or other reactions (e.g., electrophilic addition for alkenes) might be implied.



  2. Understand the Mechanism (Conceptual):

    • Recall the three steps: Initiation, Propagation, and Termination. While you might not need to draw the full mechanism for every problem, understanding which bonds break homolytically and the formation of radical intermediates is key.

    • Initiation: Homolytic cleavage of the weakest bond (usually X-X, e.g., Cl-Cl or Br-Br).

    • Propagation:

      1. Hydrogen abstraction: A halogen radical (X•) abstracts a hydrogen from the alkane, forming HX and an alkyl radical (R•).

      2. Halogenation: The alkyl radical (R•) reacts with a halogen molecule (X2), forming the alkyl halide (RX) and a new halogen radical (X•).





  3. Predict Possible Alkyl Radicals:

    • Identify all unique types of hydrogen atoms in the given alkane (primary, secondary, tertiary).

    • Abstaction of each type of hydrogen will lead to a different alkyl radical.

    • Remember the stability order of alkyl radicals: 3° > 2° > 1° > methyl. This stability order directly correlates with the ease of hydrogen abstraction.



  4. Determine Product Regioselectivity:

    • The halogen radical will preferentially abstract a hydrogen atom that leads to the most stable alkyl radical.

    • For JEE Main: This is a critical distinction between chlorination and bromination.

      • Chlorination (Cl2, hν): Highly reactive and less selective. While it prefers 3° > 2° > 1° abstraction, the differences in activation energies are small. Therefore, chlorination often yields a mixture of all possible monochlorinated products, with ratios influenced by both radical stability and statistical factors (number of available hydrogens). Predicting a single major product can be challenging without specific reactivity ratios.

      • Bromination (Br2, hν): Less reactive but highly selective. It strongly favors the abstraction of hydrogen that forms the most stable alkyl radical (3° >> 2° > 1°). Thus, bromination usually yields predominantly the product resulting from substitution at the most substituted carbon.





  5. Consider Stereochemistry (if applicable):

    • If the substitution occurs at a carbon that becomes a chiral center, the planar nature of the radical intermediate means that the incoming halogen can attack from either face. This leads to a racemic mixture of enantiomers (if a single chiral center is formed) or diastereomers.





Illustrative Example: Monohalogenation of 2-Methylpropane (Isobutane)


Let's consider the monohalogenation of 2-methylpropane:




   CH3

   |

CH3-CH-CH3


  • Identify H-atoms:

    • Nine primary (1°) hydrogens on the three methyl groups. All are equivalent.

    • One tertiary (3°) hydrogen on the central carbon.



  • Possible Radical Intermediates:

    • Abstracing a 1° H leads to a 1° radical: (CH3)2CH-CH2•

    • Abstracing a 3° H leads to a 3° radical: (CH3)3C•


    The 3° radical is significantly more stable than the 1° radical.



  • Predict Products:

    • Chlorination (Cl2, hν): Both 1-chloro-2-methylpropane (major, due to statistical factor) and 2-chloro-2-methylpropane (significant, due to radical stability) will be formed in substantial amounts. While the 3° C-H bond is more reactive, there are 9 primary hydrogens vs. 1 tertiary hydrogen.

    • Bromination (Br2, hν): Due to high selectivity, the reaction will overwhelmingly favor the formation of the product from the more stable 3° radical. Thus, 2-bromo-2-methylpropane will be the major product (almost exclusively).





Mastering this distinction between chlorination and bromination is crucial for acing related questions in JEE Main and advanced exams. Always consider the stability of the intermediate free radical and the relative reactivity/selectivity of the halogen.

📝 CBSE Focus Areas

CBSE Focus Areas: Alkanes - Free Radical Substitution



For CBSE Board examinations, the topic of free radical substitution in alkanes is fundamental. Students are primarily expected to understand the mechanism, the factors influencing the reaction, and its practical implications, particularly concerning product formation.

1. Understanding the Mechanism


The most critical aspect for CBSE is a clear understanding and ability to describe the three distinct steps of the free radical substitution mechanism. This is often asked as a direct question.


  • Initiation Step: Focus on the homolytic cleavage of the halogen molecule (e.g., Cl-Cl bond) by UV light or heat, leading to the formation of highly reactive free radicals.

    Cl-Cl --(UV light/heat)--> Cl• + Cl•

  • Propagation Steps: These are chain-carrying steps.

    • The halogen radical abstracts a hydrogen atom from the alkane, forming an alkyl radical and HX.

      R-H + Cl• --> R• + HCl

    • The alkyl radical then reacts with another halogen molecule to form the halogenated alkane product and regenerates a halogen radical, which continues the chain.

      R• + Cl-Cl --> R-Cl + Cl•


    CBSE Tip: Be prepared to write out the propagation steps clearly, illustrating the regeneration of the radical.

  • Termination Steps: These steps lead to the consumption of free radicals and halt the chain reaction. Understand that any two radicals can combine.

    • Cl• + Cl• --> Cl2

    • R• + R• --> R-R

    • R• + Cl• --> R-Cl





2. Reactivity and Selectivity


CBSE expects students to know the relative reactivity of hydrogen atoms (primary, secondary, tertiary) and how this influences product distribution.


  • Relative Reactivity of H-atoms: The order of ease of abstraction of hydrogen atoms by free radicals is 3° > 2° > 1°. This is due to the greater stability of the corresponding tertiary alkyl radical.

    Key Concept: Stability of alkyl radicals: 3° > 2° > 1° > methyl.

  • Product Prediction: For alkanes with different types of hydrogen atoms, substitution will preferentially occur at the more substituted carbon. However, a mixture of products is always formed because the reaction is not highly selective, especially with chlorine.

    Example: Chlorination of propane yields both 1-chloropropane and 2-chloropropane, with 2-chloropropane being the major product due to the abstraction of a secondary hydrogen.



3. Factors Affecting the Reaction


Knowledge of basic reaction conditions is essential.


  • Light/Heat: UV light or high temperatures are necessary to initiate the reaction by causing homolytic cleavage of the halogen bond.

  • Nature of Halogen:

    • Chlorination: Highly reactive and less selective. It often leads to a mixture of products.

    • Bromination: Less reactive but much more selective. It preferentially substitutes at the most hindered (3° > 2° > 1°) position, making it more useful for synthesizing specific products.


    CBSE Warning: While JEE delves deep into selectivity ratios, for CBSE, just understanding that bromination is more selective than chlorination is generally sufficient.



4. Limitations of Free Radical Halogenation


Recognize that this method is not ideal for synthesizing a single, pure haloalkane, especially for longer alkanes.


  • Formation of a mixture of mono-halogenated products.

  • Further substitution can occur, leading to di-, tri-, or poly-halogenated products if the reaction is not controlled.



Mastering these points will ensure a strong performance in CBSE examinations regarding free radical substitution in alkanes.

🎓 JEE Focus Areas

JEE Focus Areas: Alkanes - Free Radical Substitution


Free radical substitution reactions of alkanes, particularly halogenation, are fundamental in organic chemistry and a recurring topic in JEE Main. Understanding the underlying mechanism and factors influencing product distribution is crucial for success.



1. Mechanism: Focus on Propagation Steps



  • While all three steps (Initiation, Propagation, Termination) are important, JEE questions often test understanding related to the Propagation steps.

  • Chain Propagation Steps:

    1. Hydrogen Abstraction: R-H + X• → R• + H-X (Rate-determining step, determines regioselectivity)

    2. Halogen Abstraction: R• + X-X → R-X + X• (Regenerates halogen radical, continues chain)



  • The relative stability of the alkyl radical (R•) formed in the first propagation step dictates the ease of hydrogen abstraction and thus, the regioselectivity of the reaction.



2. Regioselectivity and Stability of Alkyl Radicals



  • The order of stability for alkyl radicals is 3° > 2° > 1° > methyl. This is due to hyperconjugation, similar to carbocations.

  • Consequently, the ease of hydrogen abstraction follows the same order: 3° H > 2° H > 1° H.

  • This means a tertiary hydrogen is most easily removed, leading to the formation of a more stable tertiary radical, which then gives the major product.



3. Reactivity vs. Selectivity of Halogens


This is a critical distinction for JEE:
































Feature Chlorination (Cl₂) Bromination (Br₂)
Reactivity Highly reactive (less selective) Less reactive (highly selective)
Relative Rate of H abstraction (3°:2°:1°) ~5:4:1 (Cl: slightly favors 3°/2° H, but statistical factor is significant) ~1600:82:1 (Br: strongly favors 3° H)
Product Distribution Mixture of isomers (significant amounts of all possible mono-halogenated products, often statistical) Predominantly the product from 3° H abstraction (major product due to high selectivity)
Nature of RDS Exothermic, less sensitive to Ea differences Endothermic, very sensitive to Ea differences (Hammond's Postulate applies)

JEE Tip: For chlorination, always consider both reactivity and the number of each type of hydrogen. For bromination, the most substituted hydrogen (3° > 2°) will almost always give the major product.



4. Predicting Major Products and Stereochemistry



  • In unsymmetrical alkanes, identify all unique types of hydrogens (1°, 2°, 3°).

  • For bromination, the product from abstracting the 3° H (if present) will be the major product. If only 2° and 1° H are present, the 2° product will be major.

  • Stereochemistry: If a chiral center is formed or destroyed during the reaction, consider the stereochemical outcomes.

    • Free radical intermediates are planar. If a new chiral center is formed, a racemic mixture (equal amounts of R and S enantiomers) will be produced.

    • Example: Chlorination of (R)-2-methylbutane at C1 or C4 maintains configuration (no chiral center formed/affected). Chlorination at C2 or C3 will lead to racemic products if a new chiral center is formed.





5. Polyhalogenation



  • Free radical halogenation often leads to polyhalogenated products (e.g., CH₄ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄).

  • To minimize polyhalogenation and favor monohalogenation, use a large excess of the alkane relative to the halogen.




Master these nuances, and you'll confidently tackle any JEE question on alkane halogenation!


📊 AI Generation Metadata

AI Model: Gemini Full Parallel API Client
Status: AI Generated
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🌐 Overview
Alkane halogenation (Cl2/Br2) proceeds via free radical chain: initiation (radical formation), propagation (abstraction and halogenation), termination (radical coupling). Reactivity/selectivity: Cl2 is more reactive, Br2 more selective (Hammond).
📚 Fundamentals
• Initiation often via homolysis of X2 under light/heat.
• Propagation: H abstraction and radical substitution steps.
• Br2 is more selective due to later transition state (Hammond).
🔬 Deep Dive
Energetics of abstraction vs addition; chain length and inhibition; NBS allylic bromination; solvent and temperature effects on selectivity.
🎯 Shortcuts
“Clo2se is Crazy, Bro2 is Balanced” — Cl2 more reactive (crazy), Br2 more selective (balanced).
💡 Quick Tips
• Use hv/peroxides to initiate radicals.
• Beware of over-halogenation; control equivalents.
• Consider allylic/benzylic positions for selective bromination.
🧠 Intuitive Understanding
Once a few radicals form (by light/heat), they “steal” H from alkanes, creating new radicals and continuing the chain—until radicals find each other and stop.
🌍 Real World Applications
Industrial halogenation (controlled), synthesis planning (selective bromination), and understanding side reactions under photochemical conditions.
🔄 Common Analogies
A spark in dry grass: initiation ignites a chain reaction that spreads (propagation) until it runs out of fuel (termination).
📋 Prerequisites
Bond dissociation energies; radical stability (3° > 2° > 1° > methyl); photochemical initiation; Hammond postulate.
🔗 Related Topics
Radical polymerization; allylic/benzylic bromination (NBS); radical inhibitors; selectivity vs reactivity trends.
⚠️ Common Exam Traps
• Ignoring tertiary/allylic/benzylic preference.
• Forgetting termination pathways.
• Overlooking rearrangements when radicals rearrange (rare but possible).
⭐ Key Takeaways
• Radical chain mechanisms can be long-lived.
• Radical stability guides product distribution.
• Bromination is selective; chlorination tends to be less selective/more reactive.
🧩 Problem Solving Approach
Identify possible radical sites; assess stability; write propagation steps; predict major product(s) and common side products; consider stereochemistry if chiral centers form.
📝 CBSE Focus Areas
Basic mechanism steps; selectivity trend; simple product prediction for methane/propane/butane chlorination/bromination.
🎓 JEE Focus Areas
Quantitative product ratio reasoning (radical stability), stereochemical consequences, competing reactions under photochemical conditions.

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📐Important Formulas (2)

Product Yield Calculation in Free Radical Substitution
Moles propto (N_{H} imes R_{rel})
Text: Mole Percentage (X) = frac{ ext{(Number of H at position X)} imes ext{(Relative Reactivity at X)}}{sum [( ext{Number of H} imes ext{Relative Reactivity})]_{ ext{all positions}}} imes 100
This fundamental proportionality governs the outcome of free radical halogenation (Chlorination or Bromination). The quantity of a specific product formed is directly proportional to the product of two factors: 1) The statistical factor (number of available equivalent hydrogens), and 2) The chemical factor (the relative reactivity of those hydrogens, which depends on the stability of the resulting free radical). <br/><b>JEE Tip:</b> If the relative reactivity factor ($R_{rel}$) is large (e.g., in Bromination), the statistical factor becomes less important, leading to higher selectivity.
Variables: To calculate the percentage yield or molar ratio of different isomeric haloalkane products resulting from the monohalogenation of an alkane (e.g., chlorination of propane or butane).
Relative Reactivity Ratios (R_rel) by Halogen
R_{3^{circ}}: R_{2^{circ}}: R_{1^{circ}}
Text: Relative reactivity ratios for primary (1°), secondary (2°), and tertiary (3°) hydrogens.
These empirical constants quantify the selectivity of the halogen radical (X•) used. You must use these values (provided in the question or standard tables) to perform product ratio calculations.<br/><table><thead><tr><th>Halogen</th><th>R(3° H)</th><th>R(2° H)</th><th>R(1° H)</th></tr></thead><tbody><tr><td><b>Chlorination (Cl•)</b></td><td>5.0</td><td>3.8</td><td>1.0</td></tr><tr><td><b>Bromination (Br•)</b></td><td>1600</td><td>82</td><td>1.0</td></tr></tbody></table><br/>The large disparity in Bromination values shows why Br• is highly selective (Reaction Control) compared to Cl•.
Variables: Required as input constants for the Product Yield Calculation formula to account for the stability differences in $3^{circ} > 2^{circ} > 1^{circ}$ free radicals.

📚References & Further Reading (10)

Book
Organic Chemistry
By: Paula Yurkanis Bruice
Offers a clear, step-by-step presentation of the free radical halogenation mechanism (initiation, propagation, termination) suitable for Board exams, while also providing practical examples of product distribution relevant for JEE Main.
Note: Excellent pedagogical structure, making the radical mechanism highly accessible for foundational understanding and rapid review.
Book
By:
Website
Free Radical Halogenation (Selectivity and Reactivity)
By: LibreTexts Chemistry Library
Covers the comparative reactivity and selectivity factors between F₂, Cl₂, Br₂, and I₂, providing the quantitative reactivity ratios (e.g., 5:3.8:1 for bromination at 3°, 2°, 1° positions) crucial for advanced numerical problems.
Note: Crucial for solving advanced multi-product percentage problems frequently found in JEE Advanced based on relative rates of substitution.
Website
By:
PDF
Organic Chemistry Class 11 & 12 Study Material (Alkanes)
By: Aakash Educational Services Ltd. / FIITJEE Materials
Compiled summary sheets providing condensed facts, common pitfalls, and specific reaction conditions (e.g., presence of UV light or heat) tailored strictly for competitive exam environments.
Note: Highly practical for rapid revision and ensuring all critical conditions and exception cases are memorized for objective-type questions.
PDF
By:
Article
Free Radical Halogenation: Exploring Selectivity in Undergraduate Laboratories
By: L. M. D. L. V. G. A. Wimalasena et al.
Focuses on the experimental verification of selectivity (e.g., using 2-methylpropane) and product analysis via spectroscopic methods, connecting theory (radical stability) to practical observation.
Note: Helps advanced students bridge the gap between theoretical radical stability rules and real-world results, improving conceptual depth for advanced problems.
Article
By:
Research_Paper
Relative Rates of Free-Radical Chlorination of Alkanes
By: G. A. Russell
Original research that established the quantitative reactivity ratios (e.g., 3°, 2°, 1° hydrogen abstraction) used widely in competitive exams for chlorination calculations. Provides the raw data underpinning selectivity rules.
Note: Provides the foundational quantitative data for highly selective numerical problems, essential for rigorous JEE preparation.
Research_Paper
By:

⚠️Common Mistakes to Avoid (63)

Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th
Important Other

❌ Ignoring the Statistical Factor in Product Ratio Calculation (Minor Products)

Students often determine the ratio of monohalogenated products based solely on the known relative reactivities of the radical intermediates (3° > 2° > 1°), forgetting to account for the statistical probability, which is the total number of equivalent hydrogen atoms available for abstraction at each position.
💭 Why This Happens:
Over-emphasis on the relative stability of the radicals (e.g., for chlorination, the 1°:2°:3° reactivity ratio is 1:3.8:5) and neglecting the fact that if there are significantly more 1° hydrogens than 2° hydrogens, the minor product yield can be surprisingly high.
✅ Correct Approach:

For calculating the relative amount of any isomer (I):

Relative amount of I = (Number of equivalent H atoms at that position) × (Reactivity Factor)

For JEE Advanced, the statistical factor is mandatory for accurately estimating the yield of minor products or for problems involving fractional distillation yields.

📝 Examples:
❌ Wrong:
ReactantMistake
Chlorination of Propane (CH₃CH₂CH₃)Assuming that since 2° radical is much more stable than 1° radical, the yield of 2-chloropropane is overwhelmingly high (>80%), neglecting the six 1° hydrogens.
✅ Correct:

Example: Chlorination of Propane (Reactivity Ratio 1°:2° = 1:3.8)

  • 1-Chloropropane (1° H): 6 H atoms × 1 (Reactivity) = 6 units
  • 2-Chloropropane (2° H): 2 H atoms × 3.8 (Reactivity) = 7.6 units

Total Units = 13.6. Relative Yield of 1-chloropropane ≈ 44% and 2-chloropropane ≈ 56%.

The yield of the primary product (44%) is significant, which is often missed if the statistical factor is ignored.

💡 Prevention Tips:
Always start by drawing the molecule and identifying the total number of chemically equivalent 1°, 2°, and 3° hydrogens.
Use the relative reactivity factors only after multiplying by the statistical availability (number of H atoms).
Crucial Tip for Br₂: Since Bromination is highly selective (1:82:1600), the statistical factor rarely changes the outcome (the 3° product dominates completely). This calculation is critical mainly for Chlorination.
CBSE_12th

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Alkanes: free radical substitution

Subject: Chemistry
Unit: 20 - HYDROCARBONS
Sub-unit: 20.1 - Alkanes & Alkenes
Complexity: Mid
Syllabus: JEE_Main

Content Completeness: 33.3%

33.3%
📚 Explanations: 0
📝 CBSE Problems: 0
🎯 JEE Problems: 0
🎥 Videos: 0
🖼️ Images: 0
📐 Formulas: 2
📚 References: 10
⚠️ Mistakes: 63
🤖 AI Explanation: No