Step 1: Understanding the Question:
The question asks us to identify which of the given synthetic organic pathways results in the formation of butanoic acid ($\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$) as a product.
Step 2: Key Formula or Approach:
The key to solving this problem lies in analyzing the carbon skeleton changes for each reaction type:
1. Haloform Reaction: Converts a methyl ketone ($\text{R-CO-CH}_3$) to a carboxylic acid with one less carbon (R-COOH) and haloform ($\text{CHX}_3$).
2. Grignard Carbonation: Converts an alkyl halide (R-X) to a carboxylic acid with one additional carbon (R-COOH).
3. Ester Hydrolysis: Hydrolyzes an ester (R-COOR') to form R-COOH and R'-OH.
4. Nitrile Hydrolysis: Hydrolyzes a nitrile (R-CN) to a carboxylic acid with the same number of carbons (R-COOH).
Step 3: Detailed Explanation:
• Let us evaluate each option systematically to track the carbon count of the final products.
• In Option A, the starting material is pentan-2-one, which is a methyl ketone with five carbon atoms:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CO}-\text{CH}_3 \]
• Treatment of pentan-2-one with sodium hypoiodite (NaOI) triggers the haloform reaction.
• The active reagent iodinates the methyl group, which is then cleaved by the hydroxide ion to yield iodoform ($\text{CHI}_3$) and a carboxylate ion with four carbons:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{COO}^- \]
• In Option B, 1-bromobutane ($4\text{ carbons}$) reacts with Mg to form butylmagnesium bromide. Treatment with $\text{CO}_2$ followed by hydrolysis introduces a new carboxyl carbon, yielding pentanoic acid ($5\text{ carbons}$), not butanoic acid.
• In Option C, butyl acetate on acid hydrolysis yields acetic acid ($2\text{ carbons}$) and butanol ($4\text{ carbons}$), not butanoic acid.
• In Option D, pentanenitrile ($5\text{ carbons}$) on complete acid hydrolysis yields pentanoic acid ($5\text{ carbons}$).
Step 4: Final Answer:
Therefore, only the haloform reaction of pentan-2-one (Option A) yields butanoic acid.
