Question

$CH\equiv CH\underset{Excess}{\overset{HCl}{\to }}A\underset{\Delta }{\overset{Hydrolysis}{\to }}B\underset{Na-Hg/H_{2}O}{\overset{Reduction}{\to }}C$

Compound $C$ is:

• A. ethanal
• B. propanal
• C. ethanol
• D. 2-methyl butan-1-ol

Answer 1

The correct option is C ethanol

Acetylene on treatment with excess HCl gives 1,1-dichloroethane. Two molecules of HCl add to carbon carbon triple bond. Both Cl atom is added to same C atom. 1,1-dichloroethane on hydrolysis gives ethane-1,1-diol. On heating, ethane-1,1-diol loses a molecule of water to form acetaldehyde. Acetaldehyde on reduction with sodium amalgam gives ethanol which is compound C.