1. Find the molecular formula

Given percentage composition:

• $\%C = 52.2$
• $\%H = 4.9$
• $\%Br = 42.9$

Take $100\,g$ compound:

• C $= 52.2\,g \Rightarrow \dfrac{52.2}{12} = 4.35$ mol
• H $= 4.9\,g \Rightarrow \dfrac{4.9}{1} = 4.9$ mol
• Br $= 42.9\,g \Rightarrow \dfrac{42.9}{80} \approx 0.536$ mol

Divide by the smallest:

• C: $\dfrac{4.35}{0.536} \approx 8.1$
• H: $\dfrac{4.9}{0.536} \approx 9.1$
• Br: $1$

So the formula is approximately $\mathrm{C_8H_9Br}$.

Check molar mass: $8(12)+9(1)+80=96+9+80=185$ Using bromine as $79$: $96+9+79=184$ Hence molecular formula is: $\boxed{\mathrm{C_8H_9Br}}$

2. Use oxidation products

Both isomers on oxidation with $KMnO_4$ give:

• (A) $\to$ benzoic acid
• (B) $\to$ $p$-bromobenzoic acid

Oxidation of alkyl side chains on benzene converts the entire side chain to $-COOH$, provided there is at least one benzylic H.

• If oxidation gives benzoic acid, then bromine is in the side chain, not on the ring.
• If oxidation gives $p$-bromobenzoic acid, then bromine is on the ring and the oxidizable alkyl group is para to it.

So:

• (A) must be a brominated side-chain derivative of ethylbenzene type.
• (B) must be a bromine-substituted ring compound with an ethyl group para to Br.

3. Identify isomer A

Formula: $\mathrm{C_8H_9Br}$

A gives benzoic acid on oxidation, so likely structure is: $\mathrm{C_6H_5-CHBr-CH_3}$ This is 1-bromo-1-phenylethane.

Check properties:

• It has a chiral carbon attached to:
  • $\mathrm{Br}$
  • $\mathrm{H}$
  • $\mathrm{CH_3}$
  • $\mathrm{C_6H_5}$

Hence it is optically active.

• With alcoholic $AgNO_3$, benzylic bromides react readily and give $AgBr$, a pale yellow precipitate.

So (A) is: $\boxed{\mathrm{C_6H_5-CHBr-CH_3}}$

4. Identify isomer B

B gives $p$-bromobenzoic acid on oxidation, so bromine must be on the benzene ring and the side chain must oxidize to $-COOH$.

With formula $\mathrm{C_8H_9Br}$, the suitable structure is: $\mathrm{p{-}Br-C_6H_4-CH_2CH_3}$ This is $p$-bromoethylbenzene.

On oxidation: $\mathrm{p{-}Br-C_6H_4-CH_2CH_3 \xrightarrow[heat]{KMnO_4} p{-}Br-C_6H_4-COOH}$ which is $p$-bromobenzoic acid.

5. Final identification

Thus,

• $A = \mathrm{C_6H_5-CHBr-CH_3}$
• $B = \mathrm{p{-}Br-C_6H_4-CH_2CH_3}$

So the correct option is the one containing: $\boxed{A = \text{1-bromo-1-phenylethane, } B = p\text{-bromoethylbenzene}}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C.