Q.A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO2. (B) on heating with liquid ammonia followed by treating with Br2/KOH gives (C), which on treating with NaNO2 and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.

Step 1. Working backward from D: a monobasic acid of molecular mass 74 is propanoic acid, CH3CH2COOH (C3H6O2: 3x12 + 6x1 + 2x16 = 36+6+32 = 74) -- fixing the final product and, from there, every earlier intermediate's carbon count.

Step 2. D is reached by oxidising an alcohol, which itself came from C (a primary amine) reacting with NaNO2/HCl at low temperature -- exactly the primary-aliphatic-amine-plus-nitrous-acid outcome (loss of N2, giving the alcohol with the SAME carbon skeleton). So C's alcohol, oxidised to propanoic acid, must itself be propan-1-ol, meaning C = propan-1-amine, CH3CH2CH2-NH2.

Step 3. C is the Hofmann-degradation product of an amide with ONE MORE carbon: that amide is therefore butanamide, CH3CH2CH2-CONH2 (4 carbons, one more than propan-1-amine's 3) -- and this amide was itself formed by heating B (a monobasic/monocarboxylic acid) with liquid ammonia, so B = butanoic acid, CH3CH2CH2-COOH (matching the amide's own 4 carbons exactly). …

Work the whole chain BACKWARD from the one hard numeric clue (D, molecular mass 74, a monobasic acid = propanoic acid), then use each named reaction's own carbon-count rule (Hofmann degradation removes one carbon; malonic-acid-type decarboxylation removes …

• Trying to solve the sequence forward from an assumed A without using the molecular-mass-74 clue on D -- with a gem-dihalide's exact chain length otherwise unconstrained, the problem only pins down to ONE specific answer b …