Haloalkanes and Haloarenes · Question 25

The product B formed in the following reaction sequence is: [figure]

1. [figure]
2. [figure]
3. [figure]
4. [figure]

Answer

(4) [figure]

Solution

Step 1 (HCl, electrophilic addition): H+ adds to the side-chain double bond so that the more stable carbocation is formed. Addition of H+ to the terminal-side carbon gives a secondary benzylic carbocation, stabilised by resonance with the ring (and by hyperconjugation/+I of the p-methyl group):

Ar-CH=CH-CH3 H+ Ar- +CH-CH2-CH3 (Ar = 4-methylphenyl)

Cl- attacks the benzylic carbon: (A) = Ar-CHCl-CH2-CH3 (major).

Step 2 (AgCN): AgCN is mainly covalent, so only the nitrogen lone pair of the cyanide group is free to attack; the C–N bond is formed and an isocyanide results.

Ar-CHCl-CH2CH3 + AgCN Ar-CH(NC)-CH2CH3 + AgCl

(B) = 1-(1-isocyanopropyl)-4-methylbenzene: the –NC group is on the benzylic carbon and the side chain is three carbons long, as drawn in option (4).

Options (1) and (3) place –NC on the wrong carbon (anti-Markovnikov position) and option (2) is a nitrile with a two-carbon chain.

Answer: (4)