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 product is ethylamine (option c).

Step 1: Understanding the Reaction

The reaction of propanamide with bromine (\(\mathrm{Br}_{2}\)) and sodium hydroxide (\(\mathrm{NaOH}\)) is known as Hofmann's bromamide reaction. This reaction converts an amide to an amine with one fewer carbon atom than the starting amide.

Step 2: Identify the Products

In the reaction, propanamide (\(\text{CH}_3 ext{CH}_2 ext{CONH}_2\)) will lose the \(\text{-CONH}_2\) group, resulting in the formation of an amine with one fewer carbon. This means that the product will be ethylamine (\(\text{CH}_3 ext{CH}_2 ext{NH}_2\)).

Step 3: Verify the Reaction Type

Hofmann's bromamide reaction is characterized by an amide losing its \(\text{-CO}\) group to form an amine. Since propanamide contains three carbon atoms, the resultant amine will contain two carbon atoms, confirming the product as ethylamine.

Key Concepts

Propanamide

Propanamide, also known as propionamide, is an organic compound with a distinct amide group. Its chemical formula is \( ext{CH}_3 ext{CH}_2 ext{CONH}_2\). The amide group consists of a carbonyl carbon (C=O) bonded to a nitrogen atom, this is an important functional group in organic chemistry.
Propanamide is often used as an example in organic reactions due to its relatively simple structure. It primarily becomes the starting material in various transformation reactions, one being Hofmann's bromamide reaction.
When propanamide reacts in this specific way, it undergoes a transformation that removes its carbonyl group. This leaves the compound with a smaller structure.

Ethylamine

Ethylamine is the product of reacting propanamide with bromine and sodium hydroxide in Hofmann's bromamide reaction. Its chemical formula is \( ext{CH}_3 ext{CH}_2 ext{NH}_2\). This compound is an amine containing two carbon atoms along with an amino group \( ext{NH}_2\).
Amines like ethylamine are similar to ammonia but have one or more alkyl groups replacing hydrogen atoms. In ethylamine's case, the ethyl group (\( ext{C}_2 ext{H}_5\)) is attached to the nitrogen.
Ethylamine is used in various chemical syntheses and as a building block in the manufacturing of dyes, drugs, and pesticides. Its formation from propanamide highlights a significant carbon reduction process in organic chemistry.

Carbon Reduction

In organic chemistry, carbon reduction refers to reducing the number of carbon atoms in a molecule. Hofmann's bromamide reaction is a perfect example of this principle. By acting on propanamide, this reaction reduces the three-carbon structure to a two-carbon amine, ethylamine.
The process:

• The reaction starts with an amide group found in propanamide.
• In the presence of bromine and sodium hydroxide, the amide group undergoes transformation.
• The main transformation is the removal of one carbon atom from the structure.

The overall outcome is the conversion of a higher carbon structure into a simpler one. This is crucial in creating specific carbon-chains in chemical synthesis.

Amines from Amides

The transformation of amides to amines is central in many chemical processes. Hofmann's bromamide reaction showcases this concept by converting an amide (propanamide) into an amine (ethylamine) with one less carbon atom.
This process involves:

• The amide group \( ext{CONH}_2\) losing its carbonyl group \( ext{CO}\).
• The remaining organic structure forming a bond with the nitrogen atom, resulting in an amine.
• The resultant amine having a simpler molecular formula compared to the starting amide.

Amines are valuable in many industrial applications, ranging from pharmaceuticals to agrochemicals. Their synthesis from amides allows for the creation of numerous compounds and is a fundamental step in advanced organic transformations.