Question

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is C

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.

Answer 2

Step 1: Understanding the question.
We are given a compound (A) — a bromobenzyl bromide derivative. It undergoes a two-step reaction sequence:
(1) Reaction with alcoholic NaOH to form compound (B).
(2) Reaction of (B) with HBr in ether to form compound (C).
We need to identify (B) and (C) and explain how (A) and (C) are related.

Step 2: Step 1 — Reaction with NaOH (alc.).
The starting compound (A) has the structure:

p–bromobenzyl bromide

When treated with alcoholic NaOH, a dehydrohalogenation (elimination) reaction takes place, forming a double bond between the α–carbon and the benzene ring.
Thus, compound (B) is:

p–bromostyrene

(Structure: a benzene ring with Br at the para position and a CH=CH₂ side group).

Step 3: Step 2 — Reaction with HBr (in ether).
In the presence of HBr and ether (a non-polar solvent), the addition of HBr to the double bond follows the Markovnikov rule. Hence, Br attaches to the more substituted carbon atom of the alkene.
Thus, compound (C) is:

1–bromo–2–(p–bromophenyl)ethane

(Structure: a benzene ring with Br at para position and CHBr–CH₃ side chain).

Step 4: Relationship between (A) and (C).
Both (A) and (C) are position isomers — they have the same molecular formula but differ in the position of the bromine atoms attached to the side chain and the benzene ring.

Step 5: Final Answer.

(B): p–bromostyrene

(C): 1–bromo–2–(p–bromophenyl)ethane

Relationship: (A) and (C) are position isomers.