Compound 'A' of molecula formula `C_(4)H_(10)O` on treatment with Lucas reagent at room temperature gives compound 'B'. When compound 'B' is heated with alcoholic KOH, it gives isobutene. Compound 'A' annd 'B' are respectively

Options

A. 2-methyl-2-propanol ad 2-chloro-2-methyl-propane
B. 2-methyl-1-propanol and 1-chloro-2-methyl-propane
C. 2-methyl-1-propanol and 2-chloro-2-methyl-propane
D. butan-2-ol and 2-chlorobutane

Explanation

To solve the problem, we need to identify the compounds A and B based on the given information about their reactions.

### Step-by-Step Solution:

1. **Identify the Molecular Formula**:
- Compound A has the molecular formula C₄H₁₀O. This indicates that it could be an alcohol since it contains an -OH group.

2. **Determine the Degree of Unsaturation**:
- The degree of unsaturation (DU) can be calculated using the formula:
\[ \text{DU} = C + 1 - \frac{H}{2} \]
- For C₄H₁₀O:
\[ \text{DU} = 4 + 1 - \frac{10}{2} = 4 + 1 - 5 = 0 \]
- A degree of unsaturation of 0 suggests that there are no rings or double bonds in compound A.

3. **Reaction with Lucas Reagent**:
- Lucas reagent is used to convert alcohols to alkyl halides. Since compound A reacts with Lucas reagent at room temperature, it indicates that compound A is a tertiary alcohol.
- The tertiary alcohol that fits the formula C₄H₁₀O is 2-methyl-2-propanol (tert-butanol).

4. **Formation of Compound B**:
- When 2-methyl-2-propanol (compound A) reacts with Lucas reagent, it forms 2-chloro-2-methylpropane (compound B).

5. **Heating Compound B with Alcoholic KOH**:
- When compound B (2-chloro-2-methylpropane) is heated with alcoholic KOH, it undergoes an elimination reaction (E2 mechanism) to form isobutene (2-methylpropene).

6. **Final Identification**:
- Thus, compound A is 2-methyl-2-propanol and compound B is 2-chloro-2-methylpropane.

### Summary:
- Compound A: 2-methyl-2-propanol
- Compound B: 2-chloro-2-methylpropane