Question

Identify the major product C formed in the following reaction sequence: 

• A. Propylamine
• B. Butylamine
• C. butanamide
• D. α-bromobutane acid

Answer 1

The Correct Option is A

Let's break down this reaction sequence step by step to figure out the major product C.

Step 1: Reaction of 1-iodopropane with NaCN
The starting material is 1-iodopropane ($ \mathrm{CH_3 - CH_2 - CH_2 - I} $). Sodium cyanide ($ \mathrm{NaCN} $) is a nucleophile. This reaction is a nucleophilic substitution reaction ($ S_N2 $ mechanism is favored for primary alkyl halides). The cyanide ion ($ \mathrm{CN}^- $) will displace the iodide ion ($ \mathrm{I}^- $).

Therefore, product A is propanenitrile (also known as butyronitrile):

$ \mathrm{CH_3 - CH_2 - CH_2 - I + NaCN \longrightarrow CH_3 - CH_2 - CH_2 - CN + NaI} $

Step 2: Partial hydrolysis of propanenitrile (A) with $ \mathrm{OH}^- $
Nitriles can be hydrolyzed under basic or acidic conditions. Partial hydrolysis of a nitrile yields an amide.

So, product B is butanamide:

$ \mathrm{CH_3 - CH_2 - CH_2 - CN + H_2O \xrightarrow{OH^-} CH_3 - CH_2 - CH_2 - CONH_2} $

Step 3: Reaction of butanamide (B) with $ \mathrm{Br_2} $ and $ \mathrm{NaOH} $
This reaction is the Hofmann bromamide degradation (also known as the Hofmann rearrangement). In this reaction, an amide reacts with bromine in an aqueous solution of sodium hydroxide to give a primary amine with one carbon atom less than the original amide. The carbonyl group is eliminated as carbonate.

So, butanamide ($ \mathrm{CH_3 - CH_2 - CH_2 - CONH_2} $) will react with $ \mathrm{Br_2} $ and $ \mathrm{NaOH} $ to give propylamine:

$ \mathrm{CH_3 - CH_2 - CH_2 - CONH_2 + Br_2 + 4NaOH \longrightarrow CH_3 - CH_2 - CH_2 - NH_2 + Na_2CO_3 + 2NaBr + 2H_2O} $

Final Product:
Therefore, the major product C is propylamine.

The correct answer is (1) propylamine.

Final Answer:
The final answer is: $ \boxed{(1)} $