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

Correct answer: (C) (C) B2H6. Answer: (C) B2H6 Solution: Diborane B2H6 reduces a carboxylic acid R-COOH to the primary alcohol R-CH2OH, and it also performs hydroboration of an alkene which, on oxidation with H2O2/OH-, gives the anti-Markovnikov primary alcohol R-CH2-CH2-OH. NaBH4 does not reduce carboxylic acids, H2/Pd does not effect these conversions, and although LiAlH4 reduces -COOH to alcohol it does not do the alkene hydroboration step. Only B2H6 accounts for both transformations.

NEET 2023 Phase 2 · ChemistryPrevious Year Question

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

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 $R - COOH$ to the primary alcohol $R - CH X_{2} OH$ , and it also performs hydroboration of an alkene which, on oxidation with $H X_{2} O X_{2} / OH X^{-}$ , gives the anti-Markovnikov primary alcohol $R - CH X_{2} - CH X_{2} - OH$ .

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 $R - COOH$ to the primary alcohol $R - CH X_{2} OH$ , and it also performs hydroboration of an alkene which, on oxidation with $H X_{2} O X_{2} / OH X^{-}$ , gives the anti-Markovnikov primary alcohol $R - CH X_{2} - CH X_{2} - OH$ . $NaBH X_{4}$ does not reduce carboxylic acids, $H X_{2} / Pd$ does not effect these conversions, and although $LiAlH X_{4}$ reduces $- COOH$ to alcohol it does not do the alkene hydroboration step. Only $B X_{2} H X_{6}$ accounts for both transformations.

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Identify 'X' in the following reactions: R−COOH(i) X(ii) HX2​O/HCl​R−CHX2​OH R−CH=CHX2​(i) X(ii) HX2​O, NaOH, HX2​OX2​​R−CHX2​−CHX2​−OH

Solution & Explanation

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