First reaction: phenol with methyl chloroformate in presence of $NaOH$

Phenol in presence of $NaOH$ forms phenoxide ion: $C_6H_5OH + NaOH \rightarrow C_6H_5O^-Na^+ + H_2O$

Methyl chloroformate has the structure: $ClCOOCH_3$

Phenoxide attacks the carbonyl carbon and replaces $Cl^-$, giving the carbonate ester: $C_6H_5O^- + ClCOOCH_3 \rightarrow C_6H_5OCOOCH_3$

So, product $A$ is methyl phenyl carbonate: $A = C_6H_5OCOOCH_3$

Nature of substituent on benzene ring in $A$

In $A$, the group attached to benzene through oxygen is: $-OCOOCH_3$

Since the ring is bonded through oxygen, the oxygen can donate electron density by resonance to the ring. Therefore this substituent is ortho/para directing (though less activating than $-OCH_3$).

Second reaction: bromination of $A$ with $Br_2$

Because the substituent $-OCOOCH_3$ is ortho/para directing, bromination occurs at the ortho and para positions.

Thus product $B$ is 2,4,6-tribromo derivative only if the ring is strongly activated like phenol/aniline, but here the carbonate group reduces activation compared to phenol. Hence ordinary bromination gives mainly para-bromo and some ortho product depending on conditions. In standard JEE-type interpretation for aryl carbonate, bromination is directed to ortho/para, with para major.

Therefore the expected identified product is p-bromo phenyl methyl carbonate.

Conclusion

$A = C_6H_5OCOOCH_3 \quad \text{(methyl phenyl carbonate)}$ B = p\text{-}BrC_6H_4OCOOCH_3 \quad \text{(para-bromo derivative)}}

Comparison with stored answer

The stored correct answer is D. Based on the above chemistry, the option corresponding to:

• $A =$ methyl phenyl carbonate
• $B =$ para-bromo methyl phenyl carbonate

should be correct. Hence I agree with the stored answer D.