Identify X in the following reactions: R-COOH ->[(i) X][(ii) H2O/HCl] R-CH2OH R-CH=CH2 ->[(i) X][(ii) H2O, NaOH, H2O2] R-CH2-CH2-OH

NEET 2023 Phase 2 · ChemistryCarbonyl reductionPrevious Year Question

Identify $X$ in the following reactions: $R - COOH (i) X (ii) H2O/HCl R - CH X_{2} OH$ $R - CH = CH X_{2} (i) X (ii) H2O, NaOH, H2O2 R - CH X_{2} - CH X_{2} - OH$

Answer: (C) (C) $B X_{2} H X_{6}$ . \textbf{Answer:} (C) $B X_{2} H X_{6}$ \textbf{Solution:} Diborane $B X_{2} H X_{6}$ reduces a carboxylic acid $- COOH$ to a primary alcohol $R - CH X_{2} OH$ after aqueous acidic work-up, whereas $NaBH X_{4}$ cannot reduce $- COOH$ .

A.(A) $NaBH X_{4}$
B.(B) $H X_{2} / Pd$
C.(C) $B X_{2} H X_{6}$ ✓
D.(D) $LiAlH X_{4}$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $B X_{2} H X_{6}$ \textbf{Solution:} Diborane $B X_{2} H X_{6}$ reduces a carboxylic acid $- COOH$ to a primary alcohol $R - CH X_{2} OH$ after aqueous acidic work-up, whereas $NaBH X_{4}$ cannot reduce $- COOH$ . The same reagent also carries out hydroboration of the alkene, which on oxidation with $H X_{2} O X_{2}$ /NaOH gives the anti-Markovnikov alcohol $R - CH X_{2} CH X_{2} OH$ . A single reagent fits both transformations, so $X$ = $B X_{2} H X_{6}$ .

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Identify X in the following reactions: R−COOH(i) X(ii) H2O/HCl​R−CHX2​OH R−CH=CHX2​(i) X(ii) H2O, NaOH, H2O2​R−CHX2​−CHX2​−OH

Solution & Explanation

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Identify $X$ in the following reactions: $R - COOH (i) X (ii) H2O/HCl R - CH X_{2} OH$ $R - CH = CH X_{2} (i) X (ii) H2O, NaOH, H2O2 R - CH X_{2} - CH X_{2} - OH$