Looking for classes? Ksquare Career Institute, Bengaluru →
Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?
A
$PCl_5$
B
$NaOH-Br_2$
C
Sodalime
D
Hot conc. $H_2SO_4$
Detailed Solution
An amide treated with bromine and aqueous or ethanolic NaOH gives a primary amine with one carbon atom less (Hoffmann bromamide degradation).
$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Acetamide has two carbons; the carbonyl carbon is lost as carbonate and the product is methyl amine.
$PCl_5$ dehydrates acetamide to acetonitrile ($CH_3CN$); soda lime and hot conc. $H_2SO_4$ do not give methyl amine (the latter hydrolyses it to acetic acid).
Hence the reagent is $NaOH-Br_2$.
$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Acetamide has two carbons; the carbonyl carbon is lost as carbonate and the product is methyl amine.
$PCl_5$ dehydrates acetamide to acetonitrile ($CH_3CN$); soda lime and hot conc. $H_2SO_4$ do not give methyl amine (the latter hydrolyses it to acetic acid).
Hence the reagent is $NaOH-Br_2$.