Question:medium

In a reaction,
$(\text{CH}_3)_2\text{CHMgBr} + \text{CO}_2 \xrightarrow[\text{ether}]{\text{dry}} \text{A} \xrightarrow[\text{dilHCl}]{\text{H.OH}} \text{B}$.
Find the product ' B ' of above reaction.

Show Hint

$R\text{MgX} + \text{CO}_2 \to R\text{COOH}$. The acid always has one more carbon than the alkyl group of the Grignard reagent.
Updated On: May 7, 2026
  • Propanoic acid
  • 2-Methyl propanoic acid
  • Butanoic acid
  • 2,2-Dimethyl ethanoic acid
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Grignard reagents (\(\text{RMgX}\)) react with carbon dioxide (\(\text{CO}_2\)) to form a magnesium salt of a carboxylic acid, which upon hydrolysis yields the free carboxylic acid.
Step 2: Key Formula or Approach:
\[ \text{R-MgX} + \text{O=C=O} \to \text{R-COOMgX} \xrightarrow{\text{H}_3\text{O}^+} \text{R-COOH} \] Step 3: Detailed Explanation:
The starting material is Isopropyl magnesium bromide, \((\text{CH}_3)_2\text{CHMgBr}\).
1. Addition step: The nucleophilic isopropyl group attacks the electrophilic carbon of \(\text{CO}_2\) in dry ether to form the complex 'A', which is \((\text{CH}_3)_2\text{CH-COOMgBr}\).
2. Hydrolysis step: Addition of dilute HCl results in the formation of the product 'B'.
\[ (\text{CH}_3)_2\text{CH-COOMgBr} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} (\text{CH}_3)_2\text{CH-COOH} + \text{Mg(OH)Br} \] The compound \((\text{CH}_3)_2\text{CH-COOH}\) is 2-Methyl propanoic acid (also known as isobutyric acid).
Step 4: Final Answer:
The product 'B' is 2-Methyl propanoic acid.
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