Problem 88 - A Complete Resource Book in Chemistry for JEE Main

Question

When propanamide reacts with \(\mathrm{Br}_{2}\) and \(\mathrm{NaOH}\) then which of the following compounds is formed? (a) propyl alcohol (b) ethyl alcohol (c) ethylamine (d) propylamine

Step-by-Step Solution

Answer

The compound formed is ethylamine, which corresponds to option (c).

Step 1: Understanding the Reaction Type

The reaction of an amide with bromine (\( \mathrm{Br}_2 \)) and sodium hydroxide (\( \mathrm{NaOH} \)) is known as the Hofmann bromamide degradation reaction. This reaction is used to reduce an amide to an amine with one less carbon atom.

Step 2: Identify the Starting Material

Our starting material is propanamide, which has the molecular formula \( \mathrm{CH_3CH_2CONH_2} \). This amide will be the substrate in the Hofmann bromamide degradation reaction.

Step 3: Perform the Degradation Step

In the Hofmann bromamide degradation, the amide (propanamide) will lose its carbonyl carbon atom, leading to the formation of a primary amine. Since the carbonyl carbon is removed, the chain is shortened by one carbon.

Step 4: Determine the Amine Product

Propanamide (\( \mathrm{CH_3CH_2CONH_2} \)) turns to an amine with one less carbon. So, the propanamide transforms into ethylamine (\( \mathrm{CH_3CH_2NH_2} \)) given the removal of the carbonyl carbon.

Step 5: Choose the Correct Option

The compound formed is ethylamine, which matches option (c) in the list provided.

Key Concepts

Propanamide

Propanamide is an organic compound that serves as the starting material for a reaction known as the Hofmann bromamide degradation. This compound belongs to the family of amides, which are characterized by having a carbonyl group (a carbon double-bonded to an oxygen) attached to a nitrogen atom. In the case of propanamide, its molecular structure is represented by the formula \( \mathrm{CH_3CH_2CONH_2} \), where a two-carbon ethyl group (\( \mathrm{CH_3CH_2} \)) is attached to the carboxamide group \( \mathrm{CONH_2} \). Understanding the structure of propanamide is crucial because it forms the basis for the transformations that occur in the Hofmann bromamide degradation.

During this reaction, something interesting happens: the carbon atom double-bonded to the oxygen in the carbonyl group is actually removed. This results in a reduction in the length of the carbon chain by one carbon atom. This is a defining trait of the Hofmann degradation, as it allows for the conversion of amides to amines that possess one carbon atom less than the original amide.

Ethylamine

Ethylamine is a primary amine that is the resultant product when propanamide undergoes the Hofmann bromamide degradation reaction. When propanamide with three carbon atoms, represented by \( \mathrm{CH_3CH_2CONH_2} \), loses its carbonyl group in this degradation, it turns into ethylamine, represented by \( \mathrm{CH_3CH_2NH_2} \). Ethylamine contains two carbon atoms in its structure, derived from its ethyl part. What makes ethylamine significant in this reaction is its showcasing of how amides are transformed into their corresponding amines with a reduced carbon chain length. In this instance, the removal of the carbonyl group's carbon atom from propanamide effectively converts it into an amine group by forming an NH₂ group in place of the \( \mathrm{CONH_2} \) group.

Ethylamine itself is a colorless gas at room temperature and has a strong, ammonia-like odor. It's a common building block in organic chemistry, used in the creation of dyes, pharmaceuticals, and other complex chemical compounds.

Reaction Mechanism

The reaction mechanism of the Hofmann bromamide degradation is particularly fascinating and a good example of how chemical transformations can yield dramatic structural changes. It involves several key steps that lead to the conversion of propanamide into ethylamine with one less carbon atom.

Firstly, the amide, which in our case is propanamide \( \mathrm{CH_3CH_2CONH_2} \), reacts with bromine \( \mathrm{Br_2} \) and sodium hydroxide \( \mathrm{NaOH} \). The reaction starts with the halogenation of the nitrogen atom. During this process, the nitrogen atom in the amide picks up a bromine atom, creating an N-bromoamine intermediate.

After halogenation, a base-mediated deprotonation takes place. The presence of the base \( \mathrm{NaOH} \) in the reaction facilitates the removal of a proton, aiding in the formation of a nitrene intermediate. This nitrene is a crucial intermediate as it undergoes rearrangement, losing the carbonyl carbon, which was initially part of the original propanamide. As this carbon is lost, an isocyanate intermediate is formed.

The isocyanate reacts further with water present in the reaction mixture, eventually yielding the primary amine, ethylamine \( \mathrm{CH_3CH_2NH_2} \). Thus, through a series of well-orchestrated steps including halogenation, deprotonation, and rearrangement, the Hofmann bromamide degradation efficiently reshapes the original structure of the amide to form an amine.