An alcohol (A) on heating with concentrated `H_(2)SO_(4)` gives alkene (B) as major product and (B) can shown the geometrical isomerism . The compound (A) is `:`

Options

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

Explanation

To solve the question, we need to identify the alcohol (A) that, when heated with concentrated sulfuric acid (H₂SO₄), produces an alkene (B) that exhibits geometrical isomerism.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- When an alcohol is heated with concentrated H₂SO₄, it undergoes dehydration to form an alkene. The reaction involves the removal of a water molecule (H₂O) from the alcohol.

2. **Identifying the Alkene**:
- The problem states that the alkene (B) can show geometrical isomerism. For geometrical isomerism to occur, the alkene must have a carbon-carbon double bond (C=C) with different groups attached to each carbon of the double bond.

3. **Choosing the Alcohol**:
- A common alcohol that fits this description is 2-butanol. The structure of 2-butanol is:
\[ \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \]
- When 2-butanol is dehydrated, it forms 2-butene, which has the structure:
\[ \text{CH}_3\text{CH}=\text{CH}\text{CH}_3 \]

4. **Geometrical Isomerism**:
- 2-butene can exist in two forms: cis-2-butene and trans-2-butene. This is due to the different arrangements of the methyl (CH₃) groups around the double bond:
- **Cis-2-butene**: Both CH₃ groups are on the same side of the double bond.
- **Trans-2-butene**: The CH₃ groups are on opposite sides of the double bond.

5. **Conclusion**:
- Therefore, the alcohol (A) that meets all the criteria given in the question is 2-butanol.

### Final Answer:
The compound (A) is **2-butanol**.

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