Question

Identify the product obtained when isopropyl magnesium chloride in dry ether reacts with dry ice forming a complex, hydrolysed further.

• A. Propanoic acid
• B. Propanal
• C. 2 - Methylpropanoic acid
• D. 2 - Methylpropanol

Hint:

A Grignard reagent adds to CO2 to give a carboxylic acid with one more carbon.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A Grignard reagent behaves as a carbanion source. It attacks the carbon of \(\text{CO}_2\) (dry ice), giving a magnesium carboxylate complex. Acid hydrolysis then liberates the carboxylic acid.

Step 2: Key Formula or Approach:
\[ \text{R-MgX} + \text{CO}_2 \to \text{R-COOMgX} \xrightarrow{\text{H}_3\text{O}^+} \text{R-COOH} \]
The acid has one more carbon than the Grignard's alkyl group.

Step 3: Detailed Explanation:
Isopropyl magnesium chloride is \((\text{CH}_3)_2\text{CH-MgCl}\), with R = isopropyl (3 carbons).
Adding \(\text{CO}_2\) gives \((\text{CH}_3)_2\text{CH-COOMgCl}\). Hydrolysis gives \((\text{CH}_3)_2\text{CH-COOH}\).
This is a four-carbon acid with a methyl branch on C2, named 2-methylpropanoic acid (isobutyric acid).
Propanoic acid in (A) would come from ethyl magnesium halide. Aldehyde (B) and alcohol (D) are not made from \(\text{CO}_2\).

Final Answer:
The product is 2-methylpropanoic acid, option (C). \[ \boxed{\text{2-Methylpropanoic acid (C)}} \]