Compound `P` Liberates `H_(2)` gas with `Na` metal. `P` gives the precipitate with tollen's reagent, there is no reponse towards Lucas reagent and compound `Q` gives instant turbidity with anhydrous `ZnCl_(2)//HCl_(1)` and with sodium metal `1` mole of compound `Q` liberates `11.2` litre `H_(2)` gas at `STP`. Find the structural formula of compound `P` and `Q`.

Options

A. `P` is `CH_(2)=CH-overset(O)overset(||)(C )-H`
B. `Q` is `CH_(3)-underset(OH)underset(|)overset(CH_(3))overset(|)(C )-CH_(2)-O-CH_(3)`
C. `P` is `CH_(3)-O-C-=C-H`
D. `Q` is `CH_(3)-underset(OH)underset(|)overset(CH_(3))overset(|)(C )-underset(OH)underset(|)(CH)-CH_(3)`

Answer

To solve the problem, we need to analyze the information given about compounds P and Q step by step.

### Step 1: Analyze Compound P
1. **Liberates H₂ gas with sodium metal**: This indicates that compound P must contain a functional group that can react with sodium to release hydrogen gas. Typically, this suggests the presence of an alcohol or an acidic hydrogen.

2. **Gives a precipitate with Tollen's reagent**: Tollen's reagent is used to test for aldehydes and some ketones. The formation of a precipitate indicates that compound P is likely an aldehyde or a compound that can be oxidized to an aldehyde.

3. **No response towards Lucas reagent**: The Lucas test is used to distinguish between primary, secondary, and tertiary alcohols. The fact that P does not respond suggests that it is not an alcohol or that it is a primary alcohol, which does not react quickly with Lucas reagent.

### Conclusion for Compound P
Given the above points, the only structure that fits all these criteria is an alkyne with a terminal hydrogen (which can react with sodium) and can be oxidized to an aldehyde. Thus, we can deduce that compound P is likely **propyne (CH₃C≡CH)**.

### Step 2: Analyze Compound Q
1. **Gives instant turbidity with anhydrous ZnCl₂/HCl**: This indicates that compound Q is likely a tertiary alcohol, as tertiary alcohols react quickly with Lucas reagent to form an alkyl chloride, leading to turbidity.

2. **With sodium metal, 1 mole of compound Q liberates 11.2 liters of H₂ gas at STP**: At STP, 1 mole of any gas occupies 22.4 liters. Therefore, if 1 mole of compound Q liberates 11.2 liters of H₂, it indicates that compound Q must contain two hydrogen atoms that can be liberated, suggesting it is a tertiary alcohol.

### Conclusion for Compound Q
Given the information, compound Q can be deduced to be **tert-butanol (C₄H₁₀O)**, which has the structure **(CH₃)₃COH**.

### Final Structural Formulas
- **Compound P**: Propyne (CH₃C≡CH)
- **Compound Q**: Tert-butanol ((CH₃)₃COH)