Chemistry · JEE Advanced only

Allylic and Benzylic Halogenation (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 2 to 3 hours for concepts and mechanism, plus 1 hour of practice problems focusing on selectivity and product prediction.

Last updated: 26 October 2026

Direct answer

Allylic and benzylic halogenation are selective free radical substitution reactions where a halogen (usually bromine) replaces a hydrogen atom at the allylic position (next to a C=C double bond) or benzylic position (next to a benzene ring).

Allylic and benzylic halogenation are selective free radical substitution reactions where a halogen (usually bromine) replaces a hydrogen atom at the allylic position (next to a C=C double bond) or benzylic position (next to a benzene ring). These positions are special because the resulting radicals are stabilized by resonance, making the reaction highly selective compared to random halogenation of alkanes.

Official syllabus

What the JEE Advanced syllabus says

“halogenation (including allylic and benzylic halogenation)”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: mechanism of halogenation of alkanes.

JEE Main covers basic halogenation of alkanes, but Advanced expects you to predict which hydrogen gets replaced when multiple types are present. You need to understand radical stability and resonance to solve selectivity problems that Main does not ask.

Key ideas

What you must understand

Allylic carbon is the carbon directly attached to a C=C double bond; benzylic carbon is directly attached to a benzene ring
Both allylic and benzylic radicals are resonance-stabilized, making them more stable than primary, secondary, or tertiary alkyl radicals
NBS (N-bromosuccinimide) in CCl₄ with light or peroxide is the classic reagent for allylic and benzylic bromination
Radical stability order: benzylic ≈ allylic > 3° > 2° > 1° > methyl
The reaction proceeds via a free radical chain mechanism: initiation, propagation, and termination steps
At low Br₂ concentration (maintained by NBS), substitution dominates over addition to the double bond

Formulas

Formulas and key facts

Radical stability: C₆H₅CH₂• ≈ CH₂=CH-CH₂• > (CH₃)₃C• > (CH₃)₂CH• > CH₃CH₂• > CH₃•
NBS + hν → Br• (bromine radical source)
R-H + Br• → R• + HBr (hydrogen abstraction)
R• + Br₂ → R-Br + Br• (chain propagation)
Bond dissociation energy: allylic C-H ≈ 361 kJ/mol, benzylic C-H ≈ 356 kJ/mol, 3° C-H ≈ 381 kJ/mol
HBr + NBS → Br₂ + succinimide (regenerates low concentration of Br₂)
Selectivity factor for benzylic H with Br₂ at 127°C ≈ 64000 times methyl H

Avoid

Common mistakes

Confusing allylic position with vinylic position: vinylic is ON the double bond (C=C-H), allylic is NEXT to it (C=C-C-H)
Forgetting that NBS is used specifically to keep Br₂ concentration low, preventing addition across the double bond
Ignoring resonance when predicting products: allylic bromination can give a mixture of products if the allylic radical has non-equivalent resonance structures
Assuming chlorination is equally selective: Cl• is more reactive and less selective than Br•, so allylic selectivity is much lower with Cl₂

Worked example

A JEE Advanced-level example

Question. When 1-butene (CH₂=CH-CH₂-CH₃) is treated with NBS in CCl₄ under UV light, what is the major monobrominated product? Draw the mechanism and explain your answer.

Show solution

Identify the allylic positions in 1-butene: C3 (the CH₂ between the double bond and CH₃) has allylic hydrogens. C1 is vinylic, C2 is vinylic, C4 is not allylic.
NBS generates a low concentration of Br₂, which under UV light forms Br• radicals.
Br• abstracts the allylic hydrogen from C3, forming the allylic radical CH₂=CH-CH•-CH₃.
This radical is resonance-stabilized: CH₂=CH-CH•-CH₃ ↔ •CH₂-CH=CH-CH₃ (two resonance forms).
The radical reacts with Br₂. Due to resonance, bromine can attach at C1 or C3, giving two products: CH₂Br-CH=CH-CH₃ (1-bromo-2-butene) and CH₂=CH-CHBr-CH₃ (3-bromo-1-butene).
The major product is 1-bromo-2-butene (CH₃-CH=CH-CH₂Br in its more stable trans form) because the internal double bond is more substituted and thermodynamically more stable.

Answer: Major product: 1-bromo-2-butene (trans-CH₃CH=CHCH₂Br), formed due to resonance in the allylic radical and preference for the more stable internal alkene.

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Allylic and Benzylic Halogenation in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is mechanism of halogenation of alkanes.

How long does Allylic and Benzylic Halogenation take to prepare for JEE Advanced?

About 2 to 3 hours for concepts and mechanism, plus 1 hour of practice problems focusing on selectivity and product prediction, if you already know the related JEE Main topic.