Two isomers (A) and (B) with Molar mass 184 g/mol and elemental composition C, 52.2%; H, 4.9% and Br 42.9% gave benzoic acid and p-bromobenzoic acid, respectively on oxidation with KMnO₄. Isomer 'A' is optically active and gives a pale yellow precipitate when warmed with alcoholic AgNO₃. Isomer 'A' and 'B' are, respectively

Solution

Chemistry ma’am · 3 steps

Get the formula first, then let each clue decide where the Br sits: side chain or ring.

1. Molecular formula. Divide each percentage by the atomic mass: C 52.2/12 = 4.35, H 4.9/1 = 4.9, Br 42.9/80 = 0.536. Dividing by the smallest gives about 8 : 9 : 1, so the formula is C₈H₉Br, with a molar mass of about 185, close to the given 184.
2. Read the oxidation. KMnO₄ cuts any side chain that has a benzylic H down to one -COOH, but it leaves a ring Br untouched. Benzoic acid means A has one side chain and no Br on the ring. p-Bromobenzoic acid means B has Br on the ring and one side chain para to it, so B is p-BrC₆H₄CH₂CH₃.

   Clue	What it tells you	Fits C₆H₅CHBrCH₃?
   Gives benzoic acid	One side chain, Br not on the ring	Yes, the -CHBrCH₃ chain becomes -COOH
   Optically active	Needs a chiral carbon	Yes, the CHBr carbon carries H, Br, CH₃ and C₆H₅
   Pale yellow ppt with alcoholic AgNO₃	Br must leave easily as Br⁻ to form AgBr	Yes, it is a benzylic bromide and gives a stable benzylic carbocation

3. So. A is H₃C-CHBr-C₆H₅ (1-bromo-1-phenylethane) and B is 1-bromo-4-ethylbenzene, which is option (c).

The idea to keep. KMnO₄ turns every side chain with a benzylic H into one -COOH and leaves a ring halogen untouched. A ring halogen gives no precipitate with AgNO₃. A benzylic halogen gives one easily.

Answer from the NTA final answer key. Solution by PrepWiser.