Question

Figure

Answer 1

2-methylcyclopentan-1-ol

Answer 2

The reaction sequence starts with 2-(cyanomethyl)cyclopentan-1-one.

Step 1: Acidic hydrolysis of the nitrile group ($\text{-CN}$) under heating ($\text{H}^+/\text{H}_2\text{O}, \Delta$).
Nitriles are hydrolyzed to carboxylic acids under acidic conditions with heating.
The $-\text{CH}_2\text{CN}$ group is converted to a $-\text{CH}_2\text{COOH}$ group.
The intermediate product is 2-(carboxymethyl)cyclopentan-1-one.
This is a $\beta$-keto carboxylic acid. The ketone carbonyl is at C1. The carboxymethyl group is attached to C2. The carbon of the $-\text{CH}_2-$ group is $\alpha$ to the carboxylic acid carbon. The carbon C2 of the ring is attached to this $-\text{CH}_2-$ group, so C2 is $\beta$ to the carboxylic acid carbon. C2 is also $\alpha$ to the ketone carbonyl. Thus, the ketone carbonyl is on the $\beta$-carbon relative to the carbon $\alpha$ to the carboxylic acid carbon. Therefore, this is a $\beta$-keto carboxylic acid.

Step 2: Heating ($\Delta$).
$\beta$-Keto carboxylic acids undergo decarboxylation upon heating, losing $\text{CO}_2$.
The decarboxylation of 2-(carboxymethyl)cyclopentan-1-one leads to the loss of the $-\text{COOH}$ group and the formation of a methyl group at the position where the $-\text{CH}_2\text{COOH}$ group was attached, but with the loss of the carbon from the $-\text{COOH}$ group. The carbon that was part of the $-\text{CH}_2-$ group becomes a methyl group.
So, the $-\text{CH}_2\text{COOH}$ group at C2 is replaced by a $-\text{CH}_3$ group at C2.
The product after decarboxylation is 2-methylcyclopentan-1-one.

Step 3: Reduction with $\text{LiAlH}_4$.
$\text{LiAlH}_4$ is a strong reducing agent that reduces ketones to secondary alcohols.
Reduction of 2-methylcyclopentan-1-one with $\text{LiAlH}_4$ reduces the ketone group at C1 to a secondary alcohol.
The product is 2-methylcyclopentan-1-ol.

The final product is 2-methylcyclopentan-1-ol.

Explanation of the solution: The starting nitrile undergoes acidic hydrolysis to form a $\beta$-keto carboxylic acid. The $\beta$-keto carboxylic acid then undergoes decarboxylation upon heating to form a ketone. Finally, the ketone is reduced to a secondary alcohol using $\text{LiAlH}_4$.

Let's reconfirm the decarboxylation of $\beta$-keto carboxylic acids. The mechanism involves a cyclic transition state where the enol form is formed, followed by tautomerization to the ketone. This process removes the carboxyl group as $\text{CO}_2$. In 2-(carboxymethyl)cyclopentan-1-one, the $-\text{CH}_2\text{COOH}$ group is attached to the $\alpha$-carbon of the ketone. The carbon of the $-\text{CH}_2-$ group is $\alpha$ to the $-\text{COOH}$ group. The carbon C2 of the ring is $\beta$ to the $-\text{COOH}$ group. The ketone is at C1, which is $\alpha$ to C2. So the ketone is on the $\beta$-carbon relative to the carbon $\alpha$ to the $-\text{COOH}$. This is indeed a $\beta$-keto carboxylic acid.