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

Ethylamine is formed.

Step 1: Recognize the Reaction Type

The reaction involves propanamide reacting with bromine (\(\mathrm{Br}_2\)) and sodium hydroxide (\(\mathrm{NaOH}\)), which is a classical example of the Hofmann Bromamide Reaction. This reaction typically converts a primary amide into a primary amine with one less carbon atom.

Step 2: Identify the Structure of Propanamide

Propanamide has the chemical structure \(\mathrm{CH_3CH_2CONH_2}\), containing three carbon atoms.

Step 3: Apply Hofmann Bromamide Reaction

In the Hofmann Bromamide Reaction, the amide group (\(\mathrm{CONH_2}\)) loses a carbon atom to form an amine group with one less carbon. Hence, propanamide (\(\mathrm{CH_3CH_2CONH_2}\)) transforms into ethylamine (\(\mathrm{CH_3CH_2NH_2}\)).

Step 4: Determine the Final Product

After the reaction, the product formed is ethylamine, which results from the removal of the carbonyl group and reduction in carbon count by one.

Key Concepts

Understanding Propanamide

Propanamide is an organic compound that serves as the starting point in the Hofmann Bromamide Reaction. Chemically, it can be represented by the formula \(\mathrm{CH_3CH_2CONH_2}\). This molecule comprises three carbon atoms, making it a primary amide. Amides are characterized by a carbonyl group \(\mathrm{(C=O)}\) bonded to a nitrogen atom \((\mathrm{NH_2})\).

In propanamide, the carbonyl group is sandwiched between two alkyl groups: a methyl group \(\mathrm{(CH_3)}\) and an ethyl group \(\mathrm{(CH_2)}\). Therefore, understanding propanamide’s structure is crucial as it provides the basis for applying the Hofmann Bromamide reaction to form ethylamine.

When propanamide undergoes the Hofmann Bromamide reaction, it will lose a carbon atom. This carbon atom is part of the amide group \(\mathrm{(CONH_2)}\), which is converted to an amine group \(\mathrm{(NH_2)}\) with one less carbon atom in its structure.

Formation of Ethylamine from Propanamide

Ethylamine is the resulting compound from the Hofmann Bromamide Reaction involving propanamide. It is an organic compound with the chemical formula \(\mathrm{CH_3CH_2NH_2}\), consisting of two carbon atoms. In this reaction, ethylamine is produced when the propanamide loses its carbonyl group and a decarbonylation process occurs.

• The removed carbon atom is part of the carboxamide group \(\mathrm{CONH_2}\), which is essential for transforming the molecule into a primary amine.
• Ethylamine exhibits the properties of a primary amine, which includes possessing a distinct fishy odor, being soluble in water, and having both nucleophilic and basic characteristics.
• The simplicity of ethylamine's structure also makes it a versatile building block in organic synthesis.

Understanding the transformation from propanamide to ethylamine in the Hofmann Bromamide Reaction aids in grasping the broader capabilities of this method in reducing carbon atoms in organic compounds.

The Reaction Mechanism Explored

The Hofmann Bromamide Reaction mechanism is a fascinating process that demonstrates how a molecular structure can be transformed by strategic chemical manipulation. When propanamide is treated with \(\mathrm{Br}_2\) and \(\mathrm{NaOH}\), a sequence of steps occurs.

Initially, an \(\mathrm{N-bromoamide}\) intermediate is formed. Here are the steps involved in this complex reaction:

• **Bromination**: The amide reacts with bromine, leading to the formation of an \(\mathrm{N-bromoamide}\).
• **Deprotonation**: The strong base (\(\mathrm{NaOH}\)) removes a proton from the \(\mathrm{N-bromoamide}\), facilitating the migration of the \(\mathrm{R-NH}\) segment and the release of the carbonyl group.
• **Rearrangement**: This migration effectively shortens the carbon chain, culminating in the loss of carbon dioxide, and transforming the molecule into the corresponding amine.

The reaction ultimately reduces the number of carbon atoms in the molecule by one. This mechanism is crucial for understanding how primary amides are converted into primary amines with one less carbon atom, a feature central to the usefulness of the Hofmann Bromamide Reaction in synthetic organic chemistry.