Question

In the above reaction, 'A' is carbide of $3^{rd}$ period element which forms amphoteric oxide.

D can be converted into B by using :

• A. $Ni$
• B. $Na/ether$
• C. $Zn-Hg/HCl$
• D. ${LiAlH}_4$

Answer 1

The correct option is C $Zn-Hg/HCl$

$A{l}_4{C}_3+H_{2}O\to C{H}_4\stackrel{O_2/M{o}_2{O}_3/{500}^o}{\to }HCHO$
A B D

Clemmensen reduction:

Aldehyde and ketone are reduced to alkanes in presence of $Zn-Hg$ and conc. $HCl$
$HCHO\stackrel{Zn-Hg/HCl}{\to }C{H}_4$

The hydrolysis of metal carbide A gives metal hydroxide and methane B.

Methane B on oxidation with $M{o}_2{O}_3$ at ${500}^{0}C$ gives Formaldehyde D.

The carbonyl group in formaldehyde D can be reduced to methylene group to obtain methane B by using zinc amalgam in conc HCl. This reaction is known as Clemmnsen's reduction.