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Propanoic acid undergoes the following sequence of reactions: Step 1: Treatment with red phosphorus and Br2, followed by hydrolysis (Hell-Volhard-Zelinsky reaction) gives a compound more acidic than propanoic acid. Step 2: The product from Step 1 is treated with NH2OH, then acidified with H+ to give compound Q. Step 3: Q is treated with NaNO2 / HCl. Identify compound Q in this reaction sequence.
- CH3CH(NH2)COOH
- CH3CH(OH)COOH
- CH3CH2CH(NH2)COOH
- CH3CH2CH(OH)COOH
Correct answer: CH3CH(NH2)COOH
Solution
The HVZ reaction on propanoic acid gives 2-bromopropanoic acid; substitution of Br by NH2 (via NH3 or related nucleophile) yields 2-aminopropanoic acid (alanine, Q). The NaNO2/HCl treatment (diazotization) confirms Q has a primary amine group.
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