Question

Phenol reacts with methyl chloroformate in the presence of $\mathrm{NaOH}$ to form product A. A reacts with ${\mathrm{Br}}_2$ to form product B. A and B are respectively

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The correct option is B

The explanation for the correct option:

Option (B):

1. Sodium hydroxide, a strong base, removes the proton from phenol to create the phenoxide ion. The carbonyl carbon of methyl chloroformate, which is an electron-deficient carbon, is attacked by phenoxide ion, an electron-rich species.
2. As a result, methyl chloroformate's chlorine ion was removed. Compound A is produced as a result of this.
3. After compound A is created, it interacts with ${\mathrm{Br}}_2$ to create compound B. The electron density is pushed into the para position by compound A's functional group, a para-directing group. As a result, the combination forms compound B, which attacks ${\mathrm{Br}}_2$ via its para location.
4. Hence, it is the correct option.

Therefore, A and B are