$\begin{array}{r}\mathrm{R}-\mathrm{COOH}\frac{\text{ (i) " }\mathrm{X}"}{\text{ (ii) }{\mathrm{H}}_2\mathrm{O}/\mathrm{HCl}}\to \mathrm{R}-{\mathrm{CH}}_2\mathrm{OH}\\ \mathrm{R}-\mathrm{CH}={\mathrm{CH}}_2\underset{\text{ (ii) }{\mathrm{H}}_2\mathrm{O},\mathrm{NaOH},{\mathrm{H}}_2{\mathrm{O}}_2}{⟶}\mathrm{R}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{OH}\end{array}$
Identify ' X ' in above reactions

$\mathrm{R}-\mathrm{COOH}\frac{\text{ (i) }{\mathrm{B}}_2{\mathrm{H}}_6}{\text{ (ii) }{\mathrm{H}}_2\mathrm{O}/\mathrm{HCl}}\to R-{\mathrm{CH}}_2\mathrm{OH}$
$\mathrm{R}-\mathrm{CH}={\mathrm{CH}}_2\frac{{\displaystyle \underset{\text{ (i) }{\mathrm{B}}_2{\mathrm{H}}_6}{ }}}{\text{ (ii) }{\mathrm{H}}_2\mathrm{O},\mathrm{NaOH},{\mathrm{H}}_2{\mathrm{O}}_2}\mathrm{R}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{OH}$
(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.