Question

Identify the product 'B' in the following reaction:

\(CH_3MgBr \xrightarrow{\text{Dry ice, Dry ether}} A \xrightarrow{\text{dil. HCl}, H_2O} B\)

• A. Methanoic acid
• B. Ethanoic acid
• C. Methanol
• D. Ethanol

Hint:

Logic Tip: Reaction with $CO_2$ is a classic chain-lengthening reaction. A Grignard reagent with $n$ carbons will produce a carboxylic acid with $n+1$ carbons. Here, methyl ($1$ carbon) becomes ethanoic acid ($2$ carbons).

Answer 1

The Correct Option is B

Grignard Reagent Reaction 

Concept:

Grignard reagents (\(R-MgX\)) react with solid carbon dioxide (dry ice) to form an intermediate magnesium carboxylate complex. Upon subsequent acid hydrolysis, this complex yields a carboxylic acid with one more carbon atom than the original alkyl group in the Grignard reagent.

Step 1: 
Identify the first step (Nucleophilic addition). 
Methyl magnesium bromide (\(CH_3MgBr\)) acts as a nucleophile and attacks the electrophilic carbon of carbon dioxide (\(CO_2\), dry ice) in the presence of dry ether.

\[ CH_3MgBr + O=C=O \xrightarrow{\text{Dry ether}} CH_3-C(=O)-O^-MgBr^+ \]

This forms the intermediate product A, which is a magnesium acetate complex. Step 2: 
Identify the second step (Acid Hydrolysis). 
The intermediate A is then subjected to acid hydrolysis using dilute HCl and water. This protonates the carboxylate oxygen, yielding a carboxylic acid and a basic magnesium halide by-product.

\[ CH_3-COOMgBr + H_2O \xrightarrow{H^+} CH_3-COOH + Mg(OH)Br \]

The final organic product B is \(CH_3COOH\), which is ethanoic acid (also known as acetic acid).