Question

Identify the product formed in the following reaction: $CH_{3}CH_{2}MgBr \xrightarrow[ii)~dil.~HCl]{i)~Dry~ice/dry~ether} Product$

• A. Ethanoic acid
• B. Propanoic acid
• C. 2-Methylpropanoic acid
• D. Butanoic acid

Hint:

Reaction with $CO_2$ increases the carbon chain length by one.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a standard method to prepare carboxylic acids from Grignard reagents by carbonation.

Step 3: Detailed Explanation:
- Dry ice is solid Carbon dioxide ($\text{CO}_2$). - Reaction: \[ \text{CH}_3\text{CH}_2\text{MgBr} + \text{O=C=O} \xrightarrow{\text{dry ether}} \text{CH}_3\text{CH}_2\text{COOMgBr} \] The alkyl group ($\text{CH}_3\text{CH}_2^-$) attacks the carbon of $\text{CO}_2$. - Hydrolysis: \[ \text{CH}_3\text{CH}_2\text{COOMgBr} \xrightarrow{\text{dil. HCl}} \text{CH}_3\text{CH}_2\text{COOH} + \text{Mg(OH)Br} \] The starting Grignard reagent has 2 carbons (Ethyl), and $\text{CO}_2$ adds 1 more carbon. The total carbon count in the product is 3. - A 3-carbon carboxylic acid is Propanoic acid.

Step 4: Final Answer:
The product is Propanoic acid.