$\begin{aligned} R - COOH \frac{(i) " X "}{(ii) H_{2} O / HCl} \to R - {CH}_{2} OH \\ R - CH = {CH}_{2} \underset{(ii) H_{2} O, NaOH, H_{2} O_{2}}{⟶} R - {CH}_{2} - {CH}_{2} - OH \end{aligned}$
Identify ' X ' in above reactions

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$B_{2} H_{6}$

${NaBH}_{4}$

${LiAlH}_{4}$

$H_{2} / Pd$

answer is A.

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Detailed Solution

$R - COOH \frac{(i) B_{2} H_{6}}{(ii) H_{2} O / HCl} \to R - {CH}_{2} OH$
$R - CH = {CH}_{2} \frac{\underset{(i) B_{2} H_{6}}{}}{(ii) H_{2} O, NaOH, H_{2} O_{2}} R - {CH}_{2} - {CH}_{2} - OH$
(2) LiAlH4 reduces –COOH to alcohol but not alkene
(3) NaBH4, does not reduce, esters and alkene
(4) H2/Pd reduces alkenes, alkynes to corresponding alkanes but not to alcohol.