An alcohol X on heating with concentrated `H_(2)SO_(4)` gives an alkene Y which can show geometrical isomerism . The alcohol X is

Options

A. `CH_(3)CH_(2)CH(OH)CH_(3)`
B. `(CH_(3))_(2)C(OH) CH(CH_(3))_(2)`
C. `(CH_(3))_(3) C(OH)`
D. `(CH_(3))_(2)C(OH) CH_(2)CH_(3)`

Explanation

To solve the question, we need to identify the alcohol (X) that, when heated with concentrated sulfuric acid (H₂SO₄), produces an alkene (Y) capable of exhibiting geometrical isomerism.

### Step-by-Step Solution:

1. **Understanding Geometrical Isomerism**:
Geometrical isomerism occurs in compounds with restricted rotation around a double bond, typically alkenes. For an alkene to show geometrical isomerism, it must have two different groups attached to each carbon of the double bond.

2. **Identifying the Alcohol**:
We need to consider the possible alcohols that can undergo dehydration (loss of water) when treated with concentrated H₂SO₄ to form an alkene.

3. **Analyzing the Options**:
- **Option A**: CH₃CH₂CHOHCH₃ (2-butanol)
- When heated with H₂SO₄, it can form CH₃CH=CHCH₃ (but-2-ene), which can show cis/trans isomerism.
- **Option B**: CH₃C(CH₃)CHOH (2-pentanol)
- This would also form an alkene but does not have the necessary structure to show geometrical isomerism.
- **Option C**: CH₃C(CH₃)₂CHOH (2-methyl-2-propanol)
- This would form an alkene that cannot show geometrical isomerism due to identical groups around the double bond.
- **Option D**: CH₃CH₂CHOHCH₂CH₃ (1-pentanol)
- This would form an alkene that cannot show geometrical isomerism either.

4. **Conclusion**:
From the analysis, the only alcohol that can form an alkene capable of geometrical isomerism is **Option A: CH₃CH₂CHOHCH₃ (2-butanol)**.

### Final Answer:
The alcohol X is **CH₃CH₂CHOHCH₃ (2-butanol)**.