Looking for classes? Ksquare Career Institute, Bengaluru →
In a set of reactions propionic acid yielded a compound D.
$CH_3CH_2COOH \xrightarrow{SOCl_2} B \xrightarrow{NH_3} C \xrightarrow[Br_2]{KOH} D$
The structure of D would be :-
$CH_3CH_2COOH \xrightarrow{SOCl_2} B \xrightarrow{NH_3} C \xrightarrow[Br_2]{KOH} D$
The structure of D would be :-
A
$CH_3CH_2CH_2NH_2$
B
$CH_3CH_2CONH_2$
C
$CH_3CH_2NHCH_3$
D
$CH_3CH_2NH_2$
Detailed Solution
Step 1: thionyl chloride converts the carboxylic acid into the acid chloride.
$CH_3CH_2COOH \xrightarrow{SOCl_2} CH_3CH_2COCl$ (B, propanoyl chloride) $+ SO_2 + HCl$
Step 2: ammonia converts the acid chloride into the amide.
$CH_3CH_2COCl \xrightarrow{NH_3} CH_3CH_2CONH_2$ (C, propanamide)
Step 3: the amide on treatment with $Br_2$ and KOH undergoes Hoffmann bromamide degradation, giving a primary amine with one carbon atom less than the amide.
$CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$
So D is ethylamine, $CH_3CH_2NH_2$.
$CH_3CH_2COOH \xrightarrow{SOCl_2} CH_3CH_2COCl$ (B, propanoyl chloride) $+ SO_2 + HCl$
Step 2: ammonia converts the acid chloride into the amide.
$CH_3CH_2COCl \xrightarrow{NH_3} CH_3CH_2CONH_2$ (C, propanamide)
Step 3: the amide on treatment with $Br_2$ and KOH undergoes Hoffmann bromamide degradation, giving a primary amine with one carbon atom less than the amide.
$CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$
So D is ethylamine, $CH_3CH_2NH_2$.