Identify the final product (Z) in the following sequence of reactions :
`Me_2C=O+HCNrarr(X) overset(H_3O^+)rarr(Y)overset(H_2SO_4)rarr`

Options

A. `(CH_3)_2C(OH) COOH`
B. `CH_2=C(CH_3)COOH`
C. `HOCH_2CH(CH_3)COOH`
D. `CH_3CH=CHCOOH`

Answer

To solve the problem step by step, we will analyze the sequence of reactions given in the question:

1. **Identify the starting compound**: The starting compound is acetone (Me2C=O), which is CH3COCH3.

2. **Reaction with HCN**:
- In the first reaction, acetone reacts with hydrogen cyanide (HCN). The nucleophilic cyanide ion (CN-) attacks the electrophilic carbon of the carbonyl group (C=O) in acetone.
- The reaction leads to the formation of a cyanohydrin. The structure of the cyanohydrin (X) will be:
\[ \text{CH}_3\text{C(OH)(CN)}\text{CH}_3 \]
- Here, the hydroxyl group (OH) and the cyanide group (CN) are added to the same carbon atom.

3. **Hydrolysis with H3O+**:
- The cyanohydrin (X) undergoes hydrolysis when treated with hydronium ion (H3O+). The CN group is converted into a carboxylic acid (COOH) group.
- The product of this reaction (Y) will be:
\[ \text{CH}_3\text{C(OH)(COOH)}\text{CH}_3 \]
- This compound has a hydroxyl group (OH) and a carboxylic acid group (COOH) on the same carbon.

4. **Reaction with H2SO4**:
- The next step involves treating the hydrolyzed product (Y) with sulfuric acid (H2SO4). This will lead to the protonation of the hydroxyl group, making it a better leaving group.
- The water molecule (H2O) will leave, forming a carbocation intermediate:
\[ \text{CH}_3\text{C}^+\text{(COOH)}\text{CH}_3 \]
- The carbocation is then stabilized by the adjacent methyl groups.

5. **Deprotonation**:
- Finally, one of the methyl groups will lose a proton (H+) to stabilize the carbocation, resulting in the formation of a double bond.
- The final product (Z) will be:
\[ \text{CH}_2=\text{C(CH}_3)\text{(COOH)} \]
- This compound is an α,β-unsaturated carboxylic acid.

6. **Final Product Identification**:
- The final product (Z) is:
\[ \text{CH}_2=\text{C(CH}_3)\text{COOH} \]
- This corresponds to option B in the question.

### Summary of Steps:
1. Start with acetone (Me2C=O).
2. React with HCN to form cyanohydrin (X).
3. Hydrolyze cyanohydrin with H3O+ to form carboxylic acid (Y).
4. Treat with H2SO4 to form carbocation.
5. Deprotonate to form the final product (Z).