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Alkyl Halide Peroxide Effect Addition Reaction Question JEE Main 2026

Quick Summary

Question Type: Peroxide effect, radical addition of HBr, and correct statement identification
Chapter: Alkyl Halide – Peroxide Effect in HBr Addition
Difficulty: ⭐⭐⭐ Medium
Time to Solve: 3-4 minutes
Key Formula: In presence of peroxide, HBr adds to alkene by free radical mechanism and gives anti-Markovnikov product; in absence of peroxide, addition proceeds through carbocation and gives Markovnikov product
Correct Answer: (C) A, C & E Only
Why: The reaction with peroxide proceeds through the more stable benzylic radical intermediate, benzene is formed during initiation as a byproduct, and in absence of peroxide the reaction follows carbocation mechanism. Statement B is false because peroxide does not generate H•, and D is false because 1-Bromo-2-phenylethane is the major, not minor, product.

The Question

JEE Main 2026 (Online) 28 January Morning Shift – Alkyl Halide

Ph-CH=CH₂   –(PhCOO)₂ / HBr–>   Product

Consider the above reaction.

• A. The reaction proceeds through a more stable radical intermediate.
• B. The role of peroxide is to generate H• (Hydrogen radical).
• C. During this reaction, benzene is formed as a byproduct.
• D. 1-Bromo-2-phenylethane is formed as the minor product.
• E. The same reaction in absence of peroxide proceeds via carbocation intermediate.

Identify the correct statements. Choose the correct answer from the options given below:

(A) A & E Only
(B) A, B & D Only
(C) A, C & E Only
(D) C, D & E Only

Quick Answer

Correct Option: (C) A, C & E Only

Reasoning: In presence of peroxide, HBr adds to styrene through a free radical mechanism. Bromine radical adds in such a way that a benzylic radical is formed, and this radical is highly stable due to resonance. Therefore statement A is correct.

• A is correct: a more stable benzylic radical intermediate is formed
• B is false: peroxide helps generate radicals, but not H• as the chain-carrying species
• C is correct: benzene can form as a byproduct during radical initiation
• D is false: 1-Bromo-2-phenylethane is the major anti-Markovnikov product, not the minor one
• E is correct: in absence of peroxide, the reaction proceeds through carbocation intermediate

Therefore, the correct set of statements is A, C and E.

Video Solution

If you want the full explanation in a clear step-by-step teaching format, watch the video solution below:

Watch Full Video Solution on YouTube

Understanding the Concept

Why Peroxide Changes the Product

Normally, HBr adds to an alkene through an electrophilic addition mechanism, where the reaction proceeds through a carbocation intermediate and gives the Markovnikov product.

But in presence of peroxide, the reaction follows a free radical chain mechanism. In this case, bromine radical adds first and the pathway that forms the more stable radical is preferred.

For styrene, bromine radical adds in a way that produces a benzylic radical, which is resonance-stabilized. This makes the anti-Markovnikov product the major product.

The Key Principle

1. In presence of peroxide, HBr follows radical addition
2. The more stable radical intermediate is favored
3. Benzylic radical is especially stable due to resonance
4. Without peroxide, HBr addition proceeds via carbocation

Detailed Step-by-Step Solution

Step 1: Identify the Nature of the Reaction

• The alkene is styrene, Ph-CH=CH₂
• Reagent is HBr in presence of peroxide
• This means the reaction proceeds by free radical mechanism

Step 2: Check Statement A

• Bromine radical adds to the terminal carbon of the double bond
• This produces a benzylic radical at the carbon attached to phenyl ring
• Benzylic radical is resonance-stabilized and therefore more stable

So statement A is correct.

Step 3: Check Statement B

• Peroxide does not directly generate H• radical as its main role
• Peroxide first forms radicals such as benzoyloxy radical
• These radicals help generate Br•, which continues the chain reaction

So statement B is false.

Step 4: Check Statement C

Download Handwritten Solution

• Benzoyl peroxide forms benzoyloxy radicals
• These can decarboxylate to phenyl radicals
• Phenyl radical can abstract hydrogen from HBr to form benzene

So statement C is correct.

Step 5: Check Statement D

• With peroxide, the reaction gives the anti-Markovnikov product
• The major product is 1-Bromo-2-phenylethane
• So calling it the minor product is incorrect

So statement D is false.

Step 6: Check Statement E

• In absence of peroxide, HBr adds through the normal electrophilic addition pathway
• That mechanism involves formation of a benzylic carbocation
• Hence the addition proceeds via carbocation intermediate

So statement E is correct.

Step 7: Write Final Conclusion

• Correct statements are A, C and E

Therefore, the correct option is (C).

Final Answer

Option (C): A, C & E Only ✓

The correct statements are A, C and E.

Essential Formulas for This Topic

Primary Rules

1. Peroxide Effect Rule:
   • In presence of peroxide, HBr adds by radical mechanism
   • The product is generally anti-Markovnikov
   • This effect is important for HBr addition to alkenes
2. Radical Stability Rule:
   • More stable radical intermediate is favored
   • Benzylic radical is highly stable because of resonance
   • This controls the direction of radical addition
3. No Peroxide Rule:
   • In absence of peroxide, HBr adds by electrophilic addition
   • The reaction proceeds through carbocation intermediate
   • Markovnikov product is formed

Important Constants

• Benzylic radical is resonance-stabilized
• Br• is the important propagating radical in this reaction
• Benzene may form as a byproduct during initiation
• 1-Bromo-2-phenylethane is the major product in presence of peroxide

Common Mistakes to Avoid

❌ Mistake 1: Thinking Peroxide Generates H•

Wrong Thinking: “Peroxide directly produces hydrogen radical.”

Correct Approach: Peroxide initiates radical formation, but the important radical in this chain process is bromine radical.

❌ Mistake 2: Ignoring Radical Stability

Wrong Approach: Not checking which radical intermediate is more stable

Correct Approach: The pathway forming the resonance-stabilized benzylic radical is preferred.

❌ Mistake 3: Taking Anti-Markovnikov Product as Minor

Common Error:

• Confusing peroxide-effect product with a side product
• Assuming normal Markovnikov product is still major

Correct Approach: In presence of peroxide, anti-Markovnikov product becomes the major product.

❌ Mistake 4: Forgetting the No-Peroxide Mechanism

Wrong Thinking: “The mechanism stays radical even without peroxide.”

Correct Understanding:

• Without peroxide, the reaction proceeds by carbocation formation
• That gives the normal Markovnikov addition product
• This is why statement E is true

Key Concept Summary

What You Must Remember

1. Peroxide changes the mechanism: HBr adds through radicals
2. Stable radical decides the path: Benzylic radical is favored
3. Benzene can form as byproduct: Through phenyl radical pathway
4. Anti-Markovnikov product is major: In presence of peroxide
5. Without peroxide, carbocation forms: So normal electrophilic addition occurs

The Golden Rule for Peroxide Effect Questions

“In presence of peroxide, think radical mechanism; in absence of peroxide, think carbocation mechanism.”

Frequently Asked Questions

Q1: Why is statement A correct?

A: Because the reaction forms a resonance-stabilized benzylic radical intermediate, which is more stable.

Q2: Why is statement B false?

A: Because peroxide initiates radical formation, but it does not serve mainly to generate H• radical.

Q3: Why is statement C correct?

A: Because phenyl radical formed during initiation can abstract hydrogen from HBr to give benzene.

Q4: Why is statement D false?

A: Because 1-Bromo-2-phenylethane is the major anti-Markovnikov product in presence of peroxide.

Q5: What happens in absence of peroxide?

A: The reaction proceeds through benzylic carbocation intermediate and gives Markovnikov addition product.

Prerequisites to Solve This Question

Before attempting this problem, you should understand:

1. Free radical mechanism: Initiation, propagation and termination steps
2. Peroxide effect: Anti-Markovnikov addition of HBr
3. Benzylic stability: Resonance stabilization of radicals and carbocations
4. Electrophilic addition: Normal HBr addition without peroxide
5. Major-minor product logic: How mechanism controls final product

After Solving This, You Can:

✅ Identify peroxide effect questions confidently
✅ Predict anti-Markovnikov product of HBr addition
✅ Compare radical and carbocation pathways clearly
✅ Recognize benzylic radical stability in reactions
✅ Avoid confusion between major and minor products
✅ Solve JEE Main organic mechanism questions faster

Study Tips for This Topic

For JEE Main:

1. Memorize the peroxide effect: HBr shows anti-Markovnikov addition in presence of peroxide
2. Always check intermediate stability: Radical or carbocation stability decides the pathway
3. Remember benzylic stabilization: Styrene-based intermediates are often resonance-stabilized
4. Compare both conditions: With peroxide and without peroxide

Common JEE Variants:

• Identify the major product in presence of peroxide
• Choose correct statements about radical mechanism
• Compare peroxide and no-peroxide pathways
• Predict benzylic radical or carbocation stability

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium
JEE Main Frequency: High – Peroxide effect and HBr addition are common organic chemistry concepts
JEE Advanced Frequency: Medium – Often tested with mechanism and intermediate stability
Topic Importance: Very High – Radical addition and peroxide effect are core reaction concepts

Written by PadhoLikhoJEE Team
Chemistry Educators for JEE Aspirants

Last Updated: March 2026
Question Source: JEE Main 2026 PYQ
Topic: Alkyl Halide – Peroxide Effect in HBr Addition
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