Direction for the question 41 to 45: Question is based on following passage:

Methyl nitrile  is hydrolysed to carboxylic acid in the presence of H+ or OH- as catalyst. An intermediate compound [A] is formed during this reaction.

The intermediate compound [Al is:

The correct option is Ethan amide.

Step-by-step Explanation:
1. Identification of the Starting Material:
The reaction scheme in the image starts with Methyl nitrile (also known as acetonitrile or ethanenitrile), which has the chemical formula:
$${\mathrm{CH}}_3-C\equiv N$$

2. Partial Hydrolysis (Formation of Intermediate [A]):
As shown in the reaction flowchart, treating methyl nitrile with water in the presence of an acid ($H^{+}$) or base (${\mathrm{OH}}^{-}$) catalyst results in partial hydrolysis. This converts the nitrile group ($-C\equiv N$) into an amide group ($-\mathrm{CO}-{\mathrm{NH}}_2$). The intermediate compound [A] formed is ethanamide (commonly called acetamide):
$${\mathrm{CH}}_3-C\equiv N+H_{2}O\to {\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{NH}}_2$$

3. Verification via Complete Hydrolysis to Carboxylic Acid [B]:
According to the scheme, heating ($\Delta$) Intermediate [A] in acidic or basic conditions completes the hydrolysis to form Carboxylic Acid [B], which is ethanoic acid (acetic acid):
$${\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{NH}}_2+H_{2}O\stackrel{H^{+}/{\mathrm{OH}}^{-},\Delta }{\to }{\mathrm{CH}}_3-\mathrm{COOH}+{\mathrm{NH}}_3$$

4. Verification via Hoffmann Bromamide Reaction (Formation of Amine [C]):
Intermediate [A] reacts with bromine and sodium hydroxide (${\mathrm{Br}}_2/\mathrm{NaOH}$). This is the Hoffmann bromamide degradation, which degrades an amide to a primary amine with one less carbon atom. Ethanamide degrades to methylamine, which corresponds to Amine [C]:
$${\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{NH}}_2+{\mathrm{Br}}_2+4\mathrm{NaOH}\to {\mathrm{CH}}_3-{\mathrm{NH}}_2+{\mathrm{Na}}_2{\mathrm{CO}}_3+2\mathrm{NaBr}+2H_{2}O$$

5. Esterification (Formation of Sweet-Smelling Compound [D]):
Carboxylic Acid [B] (ethanoic acid) reacts with methanol (${\mathrm{CH}}_3\mathrm{OH}$) under acidic catalyst conditions ($H^{+}$) to undergo esterification, producing the sweet-smelling ester methyl ethanoate (methyl acetate) as Compound [D]:
$${\mathrm{CH}}_3-\mathrm{COOH}+{\mathrm{CH}}_3\mathrm{OH}\stackrel{H^{+}}{\to }{\mathrm{CH}}_3-{\mathrm{COOCH}}_3+H_{2}O$$
All reactions in the scheme confirm that the intermediate compound [A] is ethanamide.