JEE Advanced 2022 Paper 1 · Q13 · Carboxylic Acids & Derivatives
Considering the reaction sequence given below, the correct statement(s) is(are)
CH$_3$CH$_2$COOH $\xrightarrow[\text{2. H}_2\text{O}]{\text{1. Br}_2\text{, red phosphorous}}$ P $\xrightarrow[\text{3. H}_3\text{O}^+]{\text{1. potassium phthalimide / 2. NaOH}}$ Q + phthalic acid (HOOC-C$_6$H$_4$-COOH)
Reveal answer + step-by-step solution
Correct answer:B, C, D
Solution
Step 1 — Hell–Volhard–Zelinsky reaction: CH$_3$CH$_2$COOH (propionic acid) + Br$_2$/red P, then H$_2$O, gives the $\alpha$-bromo acid P = CH$_3$CHBrCOOH ($\alpha$-bromopropionic acid / 2-bromopropanoic acid).
Step 2 — Gabriel-type substitution with potassium phthalimide: the $\alpha$-Br of P is displaced by the phthalimide nitrogen; subsequent NaOH/H$_3$O$^+$ hydrolysis liberates the primary amine and phthalic acid. The amine on the carbon $\alpha$ to -COOH is the $\alpha$-amino acid: Q = CH$_3$CH(NH$_2$)COOH (alanine).
(A) FALSE. NaBH$_4$ does not reduce carboxylic acids (it reduces only aldehydes/ketones effectively). P contains -COOH that cannot be reduced by NaBH$_4$. (LiAlH$_4$ would be needed.)
(B) TRUE. Treating P (an $\alpha$-bromo acid) with conc. NH$_4$OH (an ammonia source) effects nucleophilic substitution of Br by NH$_2$; acidification gives the $\alpha$-amino acid Q (alanine).
(C) TRUE. Q is a primary aliphatic amine (R-CH(NH$_2$)-COOH). With NaNO$_2$ in aq. HCl, primary aliphatic amines form an unstable diazonium salt that decomposes immediately to evolve N$_2$ gas, giving an $\alpha$-hydroxy acid (lactic acid).
(D) TRUE. The electron-withdra … [truncated]
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