Question:
What are X and Y respectively in the following reaction sequence? (\(anhy = \text{anhydrous}\))
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- \xrightarrow{\text{C}_2\text{H}_5\text{OH}} \text{X} \xrightarrow{\text{CO, HCl, anhy. AlCl}_3} \text{Y} \]
% The question implies the first X is an anion, the second X is an organic product. Let's call the organic product P1 to avoid confusion.
% So, C6H5N2+Anion- → P1 → Y
What are X and Y respectively in the following reaction sequence? (\(anhy = \text{anhydrous}\))
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- \xrightarrow{\text{C}_2\text{H}_5\text{OH}} \text{X} \xrightarrow{\text{CO, HCl, anhy. AlCl}_3} \text{Y} \]
% The question implies the first X is an anion, the second X is an organic product. Let's call the organic product P1 to avoid confusion.
% So, C6H5N2+Anion- → P1 → Y
Show Hint
Reactions of Diazonium Salts (\(ArN_2^+X^-\)):
- Reduction to Arene (Ar-H): With H\(_3\)PO\(_2\) (hypophosphorous acid) or CH\(_3\)CH\(_2\)OH (ethanol). Ethanol is oxidized to ethanal.
- Replacement by -OH (Phenol): Warming with water or dilute acid.
- Sandmeyer reactions (replacement by -Cl, -Br, -CN using Cu(I) salts).
- Replacement by -I (using KI).
- Replacement by -F (Balz-Schiemann reaction, using HBF\(_4\) then heat).
Gattermann-Koch Reaction:
- Formylation of benzene or its derivatives (introducing -CHO group).
- Reagents: CO, HCl, anhydrous AlCl\(_3\) (Lewis acid), often CuCl (co-catalyst).
- Forms aryl aldehydes. Example: Benzene \( \rightarrow \) Benzaldehyde.
Updated On: Jun 5, 2025
- Benzene for P1 (X in Q), Benzaldehyde for Y
- Benzene for P1, Benzoic acid for Y
- Phenol for P1, Salicylic acid (o-Hydroxybenzoic acid) for Y
- Phenol for P1, Salicylaldehyde (o-Hydroxybenzaldehyde) for Y
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The Correct Option is A
Solution and Explanation
The starting material is a benzenediazonium salt, \( \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- \).
Let's assume X\(^-\) is Cl\(^-\) or HSO\(_4^-\) typically.
Step 1: \( \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- \xrightarrow{\text{C}_2\text{H}_5\text{OH}} \) Product (referred to as X in options, let's call it P1).
Reaction of diazonium salts with ethanol (C\(_2\)H\(_5\)OH) can lead to reduction of the diazonium group to -H (forming benzene) or substitution by -OC\(_2\)H\(_5\) (forming phenetole, an ether).
The reduction to benzene is favored, especially if a reducing agent like H\(_3\)PO\(_2\) is also present or if the ethanol acts as the reducing agent (it gets oxidized to ethanal).
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 + \text{CH}_3\text{CHO} + \text{HX} \] So, P1 (X in the options) is Benzene (\( \text{C}_6\text{H}_6 \)).
Step 2: P1 (\( \text{C}_6\text{H}_6 \)) \( \xrightarrow{\text{CO, HCl, anhy.
AlCl}_3} \) Y This reaction of benzene with carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of anhydrous AlCl\(_3\) (often with CuCl as a co-catalyst) is the Gattermann-Koch reaction.
It is used to introduce an aldehyde group (-CHO) onto the benzene ring.
\[ \text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{Anhy.
AlCl}_3/\text{CuCl}} \text{C}_6\text{H}_5\text{CHO} + \text{HCl} \] (The HCl is formally regenerated or used in forming the electrophile [HCO]\(^+\)).
Product Y is Benzaldehyde (\( \text{C}_6\text{H}_5\text{CHO} \)).
So, X (P1) is Benzene and Y is Benzaldehyde.
This corresponds to the structures shown in option (1).
Let's check other possibilities if P1 were Phenol: If \( \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- \) reacts with warm water or dilute acid, it forms phenol (\( \text{C}_6\text{H}_5\text{OH} \)).
Ethanol can sometimes lead to ether formation too.
If P1 was Phenol: Phenol \( \xrightarrow{\text{CO, HCl, anhy.
AlCl}_3} \) Y This reaction (Gattermann-Koch) on phenol is not standard for simple formylation.
Phenols are highly activated.
Formylation of phenol usually occurs via Reimer-Tiemann (CHCl\(_3\)/NaOH) or Gattermann aldehyde synthesis (using HCN/HCl then hydrolysis, not CO/HCl).
If Gattermann-Koch were applied to phenol, it would likely give p-hydroxybenzaldehyde or a mixture.
Options (3) and (4) suggest P1 is Phenol.
Option (3) makes Y Salicylic acid (o-Hydroxybenzoic acid).
Option (4) makes Y Salicylaldehyde (o-Hydroxybenzaldehyde).
Salicylic acid is COOH, Gattermann-Koch gives CHO.
Salicylaldehyde is formylation at ortho.
The most standard interpretations are: - Diazonium salt + Ethanol \( \rightarrow \) Benzene (reduction).
- Benzene + CO/HCl/AlCl\(_3\) (Gattermann-Koch) \( \rightarrow \) Benzaldehyde.
This makes X=Benzene, Y=Benzaldehyde.
Let's assume X\(^-\) is Cl\(^-\) or HSO\(_4^-\) typically.
Step 1: \( \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- \xrightarrow{\text{C}_2\text{H}_5\text{OH}} \) Product (referred to as X in options, let's call it P1).
Reaction of diazonium salts with ethanol (C\(_2\)H\(_5\)OH) can lead to reduction of the diazonium group to -H (forming benzene) or substitution by -OC\(_2\)H\(_5\) (forming phenetole, an ether).
The reduction to benzene is favored, especially if a reducing agent like H\(_3\)PO\(_2\) is also present or if the ethanol acts as the reducing agent (it gets oxidized to ethanal).
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 + \text{CH}_3\text{CHO} + \text{HX} \] So, P1 (X in the options) is Benzene (\( \text{C}_6\text{H}_6 \)).
Step 2: P1 (\( \text{C}_6\text{H}_6 \)) \( \xrightarrow{\text{CO, HCl, anhy.
AlCl}_3} \) Y This reaction of benzene with carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of anhydrous AlCl\(_3\) (often with CuCl as a co-catalyst) is the Gattermann-Koch reaction.
It is used to introduce an aldehyde group (-CHO) onto the benzene ring.
\[ \text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{Anhy.
AlCl}_3/\text{CuCl}} \text{C}_6\text{H}_5\text{CHO} + \text{HCl} \] (The HCl is formally regenerated or used in forming the electrophile [HCO]\(^+\)).
Product Y is Benzaldehyde (\( \text{C}_6\text{H}_5\text{CHO} \)).
So, X (P1) is Benzene and Y is Benzaldehyde.
This corresponds to the structures shown in option (1).
Let's check other possibilities if P1 were Phenol: If \( \text{C}_6\text{H}_5\text{N}_2^+\text{X}^- \) reacts with warm water or dilute acid, it forms phenol (\( \text{C}_6\text{H}_5\text{OH} \)).
Ethanol can sometimes lead to ether formation too.
If P1 was Phenol: Phenol \( \xrightarrow{\text{CO, HCl, anhy.
AlCl}_3} \) Y This reaction (Gattermann-Koch) on phenol is not standard for simple formylation.
Phenols are highly activated.
Formylation of phenol usually occurs via Reimer-Tiemann (CHCl\(_3\)/NaOH) or Gattermann aldehyde synthesis (using HCN/HCl then hydrolysis, not CO/HCl).
If Gattermann-Koch were applied to phenol, it would likely give p-hydroxybenzaldehyde or a mixture.
Options (3) and (4) suggest P1 is Phenol.
Option (3) makes Y Salicylic acid (o-Hydroxybenzoic acid).
Option (4) makes Y Salicylaldehyde (o-Hydroxybenzaldehyde).
Salicylic acid is COOH, Gattermann-Koch gives CHO.
Salicylaldehyde is formylation at ortho.
The most standard interpretations are: - Diazonium salt + Ethanol \( \rightarrow \) Benzene (reduction).
- Benzene + CO/HCl/AlCl\(_3\) (Gattermann-Koch) \( \rightarrow \) Benzaldehyde.
This makes X=Benzene, Y=Benzaldehyde.
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