Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify (a) in the following sequence of reactions. \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{OCl}(\mathrm{a}) \stackrel{\mathrm{NH}_{3}}{\longrightarrow} \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{ON} \stackrel{\mathrm{Br}_{2}+\mathrm{NaOH}}{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}}\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCl}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{Cl}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{Cl}\)

Step-by-Step Solution

Answer

The compound (a) is \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCl} \) (option a).

Step 1: Analyze the Reaction Pathway

The given sequence starts with a compound (a) having the formula \( \mathrm{C}_4 \mathrm{H}_7 \mathrm{OCl} \). This compound undergoes reaction with \( \mathrm{NH}_3 \) to form \( \mathrm{C}_4 \mathrm{H}_9 \mathrm{ON} \), which is an amide. Since amides are typically formed from acyl chlorides and amines, the starting compound should be an acyl chloride.

Step 2: Determine the Structure of (a)

The probable acyl chloride structure having four carbon atoms is \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCl} \). This compound can react with \( \mathrm{NH}_3 \) to form butyramide, \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CONH}_2 \).

Step 3: Confirm the Reaction Process

Next, the product (butyramide) undergoes a Hoffmann bromamide reaction with \( \mathrm{Br}_2 + \mathrm{NaOH} \) which results in a primary amine with one less carbon atom compared to the starting molecule. This produces \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2 \), which matches the terminal product given.

Step 4: Choose the Correct Option

The sequence analysis confirms that compound (a) is \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCl} \) because it successfully converts into butyramide and ultimately into propylamine as per the given reactions.

Key Concepts

Understanding Acyl Chloride

Acyl chlorides are a fascinating group of organic compounds crucial in synthetic chemistry. They feature the functional group \( \text{RCOCl} \), where \( \text{R} \) represents a radical or alkyl group attached to the carbonyl carbon. These compounds are known for their reactivity, particularly in forming amides. Each acyl chloride includes a carbon-oxygen double bond (\( \text{C=O} \)) and a chlorine atom. This dual presence of a strong electrophile (carbon in \( \text{C=O} \)) and a good leaving group (chloride) makes acyl chlorides highly reactive intermediates in various organic transformations.

• They readily react with nucleophiles such as ammonia, water, and alcohols.
• One of their most significant reactions is with ammonia to form amides, which we see in this problem.

Understanding these reactions can be essential in many synthetic processes where modification of compounds is required. Thus, whenever you encounter an organic synthesis problem involving an unknown compound possibly leading to an amide, consider acyl chlorides first.

Amide Formation from Acyl Chloride

Amide formation from acyl chlorides is a classic example of a nucleophilic acyl substitution reaction. In such a reaction, the nucleophile (like ammonia) attacks the electrophilic carbon atom in the carbonyl group, replacing the chloride ion. This reaction is favored due to:

• The carbonyl group's highly polarized nature, making it a prime target for nucleophiles.
• The excellent leaving group ability of chlorine, which departs as chloride upon attack.

This reaction typically proceeds under mild conditions and is quite efficient. When applying this reaction, ensure you have the correct molar ratios for optimal yields. The simplicity and efficiency of this conversion make it a staple in organic synthesis, particularly in industrial applications for creating polymers and pharmaceuticals. Understanding the mechanism - nucleophile addition followed by loss of the leaving group - is crucial for grasping this transformation.

Exploring the Hoffmann Bromamide Reaction

The Hoffmann bromamide reaction is an intriguing route to transform amides into amines. This reaction leads to primary amines with one less carbon atom than the original amide. The process involves the use of bromine and a strong base like NaOH. Here's how it works:

• The initial step involves bromine reacting with the amide to form an N-bromoamide.
• A subsequent rearrangement driven by the base (NaOH) leads to the migration of an alkyl group from carbon to nitrogen.
• Ultimately, this rearrangement results in the formation of an amine and liberates \( \text{CO}_2 \).

This reaction is unique due to its ability to achieve such a transformation by seemingly removing a carbon atom (decarbonylation). In the context of the problem, applying this reaction to butyramide results in propylamine, which has one less carbon than butyramide. The Hoffmann bromamide reaction is highly valued in synthetic chemistry for creating amines from amides in a single step, demonstrating the conversion of functional groups in a new light.