Question

Phenol reacts with methyl chloroformate in the presence of $NaOH$ to form product A. A reacts with $B{r_2}$ to form product B. A and B are respectively:
A.
and
B.
and

C.
and
D.
and

Answer 1

Here, in this reaction it is given that phenol reacts with methyl chloroformate which is a methyl ester of chloroformic acid in the presence of a strong base $NaOH$ and it produces a product named A. The product A then reacts with $B{r_2}$ to produce product B.

Complete step by step solution:
Sodium hydroxide, being a strong base, abstracts the proton from phenol to form phenoxide ion. Phenoxide ion, which is an electron rich species attacks the carbonyl carbon of methyl chloroformate which is an electron deficient carbon. This led to the removal of $C{l^ - }$ ion from methyl chloroformate. This leads to the formation of compound A.

After the formation of compound A, it reacts with $B{r_2}$ to form compound B. The functional group of compound A, which is a para-directing group, pushes the electron density to para position. This leads the compound to attack $B{r_2}$ through its para position to form compound B.

Hence, the option (a) is the correct answer.

Additional Information:
Methyl chloroformate is a common reagent used in organic synthesis whenever there is a need to introduce a methoxycarbonyl functional group to attack a nucleophile in a chemical reaction (called carboxymethylation). Methyl chloroformate is also generally used for the derivatization of functional groups such as carboxylic acids, , phenols and amines.
Methyl chloroformate is synthesised using methanol and phosgene.
$COC{l_2} + MeOH \to ClCOOC{H_3} + HCl$

Note:
It is very important to note the nature of each of the compounds involved in the reaction. One should keep in mind that, in a chemical reaction, whenever there is possibility of an acid-base reaction to take place, we must make sure that it happens at first.