Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with $\mathrm{Br}_{2}$ to form product B. $A$ and $B$ are respectively :

JEE Main · 2018 · 08 Apr 2018 · Offline · Alcohols Phenols and Ethers

Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with

Br_{2}

to form product B.

A

and

B

are respectively :

Answer: (D)

Phenol first loses its acidic proton to

NaOH

, and the phenoxide then attacks the carbonyl carbon of methyl chloroformate, displacing chloride; this esterification gives phenyl methyl carbonate,

C X_{6} H X_{5} - O - CO - O - CH X_{3}

, as product A. The carbonate group still releases some electron density into the ring but is only weakly activating, so when A meets

Br X_{2}

the electrophilic substitution occurs at the less hindered para position, giving the para-bromo carbonate as product B.

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option