Question

A compound having molecular formula $ C_4H_{11}N $, reacts with $ \text{CHCl}_3 $ in alcoholic KOH, on heating, to form a compound with a foul smell. Identify the optically active isomer of the compound which also shows the above reaction.

• A. 2-Methylpropan-1-amine
• B. Butan-1-amine
• C. Butan-2-amine
• D. N-Methylpropan-1-amine

Hint:

When identifying optically active compounds, look for chiral centers that do not have identical substituents. The Hofmann degradation typically removes a carbon atom, which is useful for identifying the products.

Answer 1

The Correct Option is C

The given compound is \( C_4H_{11}N \), which represents an amine group. The reaction described involves the formation of a compound with a foul smell upon treatment with chloroform (\( \text{CHCl}_3 \)) in the presence of alcoholic KOH, known as the Hofmann degradation reaction. 
This reaction typically involves the removal of one carbon atom from the amine chain, leading to the formation of an amine with one less carbon atom. 
The Hofmann reaction usually occurs with primary amines, where a foul-smelling product such as an amine with fewer carbon atoms (like a lower alkyl amine) is formed after the removal of one carbon from the parent chain. 
Now, let’s evaluate the given options: 
- (A) 2-Methylpropan-1-amine: This compound has a branched structure and would likely produce a different set of products, thus it's not the correct answer. 
- (B) Butan-1-amine: This is a primary amine, but it will undergo the Hofmann reaction to produce ethylamine. However, it is not optically active, so it cannot be the correct answer. 
- (C) Butan-2-amine: This compound is a secondary amine and will show the reaction with chloroform and alcoholic KOH, resulting in a foul-smelling product. This compound is also optically active due to the presence of a chiral center at carbon 2. 
- (D) N-Methylpropan-1-amine: This is a substituted amine and will not undergo the Hofmann degradation reaction. 
Thus, the correct answer is  (C) Butan-2-amine, as it is the optically active isomer and undergoes the described reaction.