Chemistry - ALCOHOL & ETHER Question with Solution | TestHub

$\mathrm{R}-\mathrm{COOH}\frac{\text{ (i) }{\mathrm{B}}_2{\mathrm{H}}_6}{\text{ (ii) }{\mathrm{H}}_2\mathrm{O}/\mathrm{HCl}}\mathrm{R}-{\mathrm{CH}}_2\mathrm{OH}$ $\mathrm{R}-\mathrm{CH}={\mathrm{CH}}_2\frac{\text{ (i) }{\mathrm{B}}_2{\mathrm{H}}_6}{\text{ (ii) }{\mathrm{H}}_2\mathrm{O},\mathrm{NaOH},{\mathrm{H}}_2{\mathrm{O}}_2}\mathrm{R}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{OH}$ - ${\mathrm{LiAlH}}_4$ and ${\mathrm{NaBH}}_4$ cannot reduce $\mathrm{C}=\mathrm{C}$ - ${\mathrm{H}}_2/\mathrm{Pd}$ reduce alkene to alkane