Alcohol `(X)overset("aq. NaOH"+I_(2))rarrCHI_(3)+(Y)overset(H_(3)O^(+))rarrPhCH_(2)COOH`. The alcohol (X) is

Options

A. `CH_(3)underset(Ph)underset("| ")"CH"CH(OH)CH_(2)`
B. `PhCH_(2)CH(OH)CH_(3)`
C. `PhCH(OH)CH_(2)CH_(3)`
D. `PhCH(CH_(3))OH`

Answer

To solve the problem, we need to identify the alcohol (X) that undergoes a reaction to form iodoform (CHI3) and then, through acid hydrolysis, produces a carboxylic acid (PhCH2COOH).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction involves an alcohol (X) reacting with aqueous sodium hydroxide (NaOH) and iodine (I2) to produce iodoform (CHI3) and another compound (Y).
- The compound Y is then subjected to acid hydrolysis to yield PhCH2COOH.

2. **Identifying the Type of Reaction**:
- The reaction of alcohol with NaOH and iodine is characteristic of the iodoform reaction, which typically occurs with methyl alcohols or alcohols that can be oxidized to form a methyl ketone.

3. **Formulating Alcohol (X)**:
- For the iodoform reaction to occur, the alcohol must have the structure that allows for the formation of a methyl ketone.
- The general structure of alcohol (X) can be represented as R-CH(OH)-CH3, where R is a phenyl group (Ph).

4. **Choosing the Correct Alcohol**:
- Given that Y is PhCH2COOH, we can deduce that the alcohol (X) must be Ph-CH(OH)-CH3.
- This structure allows the formation of a methyl ketone (Ph-CO-CH3) upon oxidation, which can then react with NaOH and iodine to form iodoform.

5. **Confirming the Reaction Pathway**:
- When Ph-CH(OH)-CH3 reacts with NaOH and iodine, it is oxidized to form the corresponding methyl ketone (Ph-CO-CH3).
- This methyl ketone can then undergo the iodoform reaction to yield iodoform (CHI3) and the sodium salt of the carboxylic acid (Ph-CH2COONa).
- Upon acid hydrolysis of the sodium salt, we obtain the desired carboxylic acid, PhCH2COOH.

6. **Conclusion**:
- Therefore, the alcohol (X) is Ph-CH(OH)-CH3.

### Final Answer:
The alcohol (X) is **Ph-CH(OH)-CH3**.