JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Hydrocarbons

The product B formed in the following reaction sequence is :

(A)(see structure)
(B)(see structure)
(C)(see structure)
(D)(see structure)
✓ Correct

Answer: (D)

Solution

4-Methylstyrene reacts with HCl by Markovnikov addition: the proton adds to the terminal

= C H_{2}

carbon, generating the more stable benzylic carbocation, which

C l^{-}

then captures to give the benzylic chloride,

p - C H_{3} C_{6} H_{4} C H (C l) C H_{3}

, as the major product [A]. This benzylic halide reacts readily with

A g C N

: the silver ion assists ionisation of the good benzylic leaving group, and the resulting carbocation is attacked by the softer, ambident cyanide nucleophile through nitrogen, giving the isocyanide (carbylamine),

p - C H_{3} C_{6} H_{4} C H (N C) C H_{3}

, as the major product B.

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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