Product (P) and (Q) are respectively

Question

Complete the following:

1. Reactant: RCOCl
Reagent: H2 / Pd-BaSO4
Major Product: RCHO
Name of Reaction: Rosenmund Reduction

2. Reactant: CH3COOH
Reagent: Cl2 / Red P
Major Product: Cl-CH2COOH
Name of Reaction: HVZ Reaction

3. Reactant: CH3CHO
Reagent: Zn-Hg / conc. HCl
Major Product: CH3CH3
Name of Reaction: Clemmensen Reduction

Answer 1

In the given reaction sequence, we are tasked with identifying the products (P) and (Q) formed from the initial compound. The structure given is an amide: acetanilide.

Let's analyze each reaction step-by-step:

Reduction with LiAlH₄:
- LiAlH₄ is a strong reducing agent, typically used to reduce carboxylic acids, esters, aldehydes, and amides.
- In this reaction, the amide group (CONH) is reduced to an amine.
- Thus, the reduction of acetanilide by LiAlH₄ results in the formation of aniline, where the carbonyl group (C=O) is removed.
Product (P) is aniline.
Hofmann Bromamide Reaction:
- This portion involves Br₂ and CH₃COOH, which are reagents for the Hofmann Bromamide reaction.
- The reaction involves converting an amide into an amine with one fewer carbon atom (amide to amine decarbonylation).
- Therefore, acetanilide undergoes this reaction to form phenylamine (aniline), just like in the previous step.
Product (Q) is also aniline.

Both reactions yield aniline as the final product:

Conclusion: The answer is that Product (P) and (Q) are both aniline, derived from the two distinct pathways leading to the same product.

Answer 2

In the given reaction sequence, we are tasked with identifying the products (P) and (Q) formed from the initial compound. The structure given is an amide: acetanilide.

Let's analyze each reaction step-by-step:

Reduction with LiAlH₄:
- LiAlH₄ is a strong reducing agent, typically used to reduce carboxylic acids, esters, aldehydes, and amides.
- In this reaction, the amide group (CONH) is reduced to an amine.
- Thus, the reduction of acetanilide by LiAlH₄ results in the formation of aniline, where the carbonyl group (C=O) is removed.
Product (P) is aniline.
Hofmann Bromamide Reaction:
- This portion involves Br₂ and CH₃COOH, which are reagents for the Hofmann Bromamide reaction.
- The reaction involves converting an amide into an amine with one fewer carbon atom (amide to amine decarbonylation).
- Therefore, acetanilide undergoes this reaction to form phenylamine (aniline), just like in the previous step.
Product (Q) is also aniline.

Both reactions yield aniline as the final product:

Conclusion: The answer is that Product (P) and (Q) are both aniline, derived from the two distinct pathways leading to the same product.

Product (P) and (Q) are respectively

The Correct Option is C

Solution and Explanation

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