Question

(a) in the above reaction is:
$HCONHR\underset{Pyridine}{\overset{POC{l}_3}{\to }}\left(a\right)+H_{2}O$
(1) $\mathrm{R}\mathrm{C}\mathrm{H}=\mathrm{N}\mathrm{O}\mathrm{H}$
(2) $\mathrm{R}-\mathrm{N}=\mathrm{C}=\mathrm{O}$
(3) $\mathrm{R}-\mathrm{C}\equiv \mathrm{N}$
(4) $R-\stackrel{+}{\mathrm{N}}\equiv \stackrel{-}{\mathrm{C}}$

Answer 1

Hint: We know that amide is a functional group which contains an acyl group (R-C=O) linked to a nitrogen atom. Simply we can say that amide is formed by replacing one hydrogen atom of ammonia by an acyl group.

Complete step by step solution:
Let’s understand the types of amide in detail. There are three types of amide, that is, primary amide, secondary amide and tertiary amide.

Primary amide is the amide in which N atom is bonded to only one carbon atom.

Secondary amide is the amide in which N atom is bonded to two carbon atoms.

Tertiary amide is the amide in which N atom is bonded to three carbon atoms.

Let’s discuss the reagent POCl3. A primary amide can be converted to a nitrile using POCl3 . POCl3 is a dehydrating agent that removes water from the amide.

Now, come to the question. Here, a secondary amide is given which reacts with POCl3 in presence of pyridine. One product formed in the reaction is water and we have to find the other product.

Let’s discuss the mechanism of the reaction.

$\mathrm{P}\mathrm{O}\mathrm{C}{\mathrm{l}}_3\to \mathrm{P}\mathrm{C}{\mathrm{l}}_3+\left[\mathrm{O}\right]$

Now, oxygen abstracts hydrogen from amide.

Therefore, (a) of the reaction is $\mathrm{R}-\mathrm{N}=\mathrm{C}=\mathrm{O}$.

So, the correct answer is Option 2.

Note: Primary amine can be synthesized by reacting amide with bromine in presence of aqueous ethanolic solution of sodium hydroxide. This reaction is known as Hoffmann bromamide reaction.

In this reaction, an acyl or aryl group migrates from carbonyl carbon of the amide to the nitrogen atom. The amine formed in the reaction possesses one less carbon atom than the amide.