Question

What are the products N and Q in the following reaction sequences?

• A. A
• B. B
• C. C
• D. D

Hint:

Stephen reduction is a very selective method to convert nitrilies (\(-\text{CN}\)) to aldehydes (\(-\text{CHO}\)).
Similarly, the Gattermann-Koch reaction is a standard method to formylate benzene to benzaldehyde.
Recognizing these named reactions helps solve multi-step synthesis questions instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to identify the final organic products \(\text{N}\) and \(\text{Q}\) in two different chemical reaction pathways starting from benzene derivatives.

Step 2: Key Formula or Approach:
Let's analyze the steps in both reaction sequences:
1. Sequence 1: Treatment of benzenediazonium chloride with mild reducing agents (\(\text{H}_3\text{PO}_2\)), followed by formulation of the ring using \(\text{CO}\) and \(\text{HCl}\) in the presence of anhydrous \(\text{AlCl}_3\) (Gattermann-Koch reaction).
2. Sequence 2: Partial reduction of benzonitrile with tin(II) chloride and hydrochloric acid, followed by acidic hydrolysis (Stephen reduction).

Step 3: Detailed Explanation:
- Analysis of Reaction Sequence 1:
- Starting material: Benzenediazonium chloride (\(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\)).
-
Step 1: Reaction with hypophosphorous acid (\(\text{H}_3\text{PO}_2\)) and water reduces the diazonium group to a hydrogen atom, yielding benzene (\(\text{M}\)).
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_3\text{PO}_2, \text{H}_2\text{O}} \text{C}_6\text{H}_6\text{ (M)} \]
-
Step 2: Benzene (\(\text{M}\)) is reacted with carbon monoxide (\(\text{CO}\)) and hydrogen chloride (\(\text{HCl}\)) gas in the presence of anhydrous aluminum chloride (\(\text{AlCl}_3\)). This is the Gattermann-Koch reaction, which introduces a formyl group (\(-\text{CHO}\)) onto the benzene ring to yield benzaldehyde (\(\text{N}\)).
\[ \text{C}_6\text{H}_6\text{ (M)} \xrightarrow{\text{CO, HCl / anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{CHO (N)} \]
- Analysis of Reaction Sequence 2:
- Starting material: Benzonitrile (\(\text{C}_6\text{H}_5\text{CN}\)).
-
Step 1: Benzonitrile is treated with tin(II) chloride (\(\text{SnCl}_2\)) in the presence of \(\text{HCl}\). This partially reduces the nitrile group to an iminium chloride intermediate (\(\text{P}\)).
\[ \text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{SnCl}_2, \text{HCl}} \text{C}_6\text{H}_5\text{CH}=\text{NH}\cdot\text{HCl (P)} \]
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Step 2: Acidic hydrolysis (\(\text{H}_3\text{O}^+\)) of the imine intermediate (\(\text{P}\)) converts it into benzaldehyde (\(\text{Q}\)). This classic reaction is known as the Stephen reduction.
\[ \text{C}_6\text{H}_5\text{CH}=\text{NH}\cdot\text{HCl (P)} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{CHO (Q)} \]
Comparing both pathways, we find that both \(\text{N}\) and \(\text{Q}\) are benzaldehyde (\(\text{C}_6\text{H}_5\text{CHO}\)).

Step 4: Final Answer:
Both reaction sequences yield benzaldehyde as the final product (\(\text{N} = \text{Q} = \text{Benzaldehyde}\)), which corresponds to option (A).