Reaction of \( C_6H_5Br + CH_3Br \) with \( \text{NaOH (alc.)} \) to Form B:
In the first step, the reaction of bromobenzene (\( C_6H_5Br \)) with methyl bromide (\( CH_3Br \)) in the presence of alcoholic NaOH undergoes a nucleophilic substitution to form B, which is ortho-bromophenol.
Reaction of B with \( \text{HBr (ether)} \) to Form C:
In the second step, treating ortho-bromophenol with \( \text{HBr} \) in ether leads to the formation of C, which is para-bromophenol.
Identifying the Relationship Between A and C:
Compound A (bromobenzene) and compound C (para-bromophenol) are position isomers because they differ in the position of the bromine and hydroxyl groups on the benzene ring.
Conclusion:
The structures of B and C are as shown in Option (3), and A and C are position isomers.
