Question

The Gattermann-Koch reaction is used in the industrial preparation of benzaldehyde. The electrophile involved in this reaction is

• A. HCO+
• B. HCl + CO$_2$ + anhydrous AlCl$_3$
• C. CO+
• D. CO + anhydrous AlCl$_3$

Answer 1

The Correct Option is A

The Gattermann-Koch reaction is an electrophilic aromatic substitution used for the synthesis of benzaldehyde from benzene. It involves the generation of an electrophile which reacts with the benzene ring. To identify the correct electrophile, we need to evaluate the options given:

• HCO⁺
• HCl + CO₂ + anhydrous AlCl₃
• CO⁺
• CO + anhydrous AlCl₃

In the Gattermann-Koch reaction, the electrophile is formed when carbon monoxide (CO) reacts with hydrochloric acid (HCl) in the presence of a Lewis acid catalyst, usually anhydrous aluminum chloride (AlCl₃). This reaction generates the formyl cation (HCO⁺), which then acts as the electrophile in the reaction with benzene to form benzaldehyde.

Therefore, the correct electrophile involved in the Gattermann-Koch reaction is: HCO⁺.

Answer 2

The Gattermann-Koch reaction is a method used to introduce an aldehyde group (-CHO) directly into an aromatic ring. It involves reacting benzene or a substituted benzene with carbon monoxide (CO) and hydrochloric acid (HCl) in the presence of anhydrous aluminum chloride (AlCl₃) as a catalyst.

The electrophile generated in this reaction is the formyl cation (HCO⁺), which is formed through the interaction of CO and HCl with AlCl₃.

Here's a breakdown of the electrophile formation:

CO + HCl + AlCl₃ → HCO⁺[AlCl₄]^-

The formyl cation (HCO⁺) is the electrophile that attacks the aromatic ring.

Let's analyze the given options:

• Option 1: CO⁺: This is not the correct electrophile.
• Option 2: HCl + CO₂ + anhydrous AlCl₃: CO₂ is not involved in the Gattermann-Koch reaction.
• Option 3: HCO⁺: This is the correct electrophile, the formyl cation.
• Option 4: CO + anhydrous AlCl₃: This combination does not directly form the electrophile; HCl is also required.

Therefore, the correct answer is:

Option 1: HCO⁺