An optically active compound `H(C_(5)H_(6)O)` on treatment with `H_(2)` in the presence of Lindlar's catalyst gave a compound I `(C_(5)H_(8)O)`. Upon hydrogenation with `H_(2)` and Pd/C, Compound H gave J `(C_(5)H_(12)O)`. Both `I` and `J` were found to b optically inactive. Select the correct statements.

Options

A. J gives positive halloform test
B. H gives white precipitate with `[Ag(NH_(3))_(2)]^(+)`
C. Compound H liberates one mole of `H_(2)` per mole of H when treated with Na
D. J gives positives test with Jone's reagent

Explanation

To solve the problem, we need to analyze the reactions and compounds involved step by step.

### Step 1: Identify the Structure of Compound H
Given the molecular formula of compound H is \( C_5H_6O \), we can calculate the degree of unsaturation (DU) to understand its structure.

**Degree of Unsaturation Calculation:**
\[ DU = \frac{(2 \times \text{number of C}) + 2 - \text{number of H}}{2} \]
Substituting the values:
\[ DU = \frac{(2 \times 5) + 2 - 6}{2} = \frac{10 + 2 - 6}{2} = \frac{6}{2} = 3 \]
This indicates that compound H has 3 degrees of unsaturation, which could mean it contains either:
- One triple bond and one double bond, or
- One ring and one double bond.

Since compound H is optically active, it must have a chiral center.

### Step 2: Propose a Structure for Compound H
A possible structure for compound H that is optically active and fits the formula \( C_5H_6O \) could be:
\[ \text{H} - C \equiv C - C(OH) - CH_2 - CH_3 \]
This structure has a chiral center at the carbon with the hydroxyl group (OH), making it optically active.

### Step 3: Reaction with Hydrogen and Lindlar's Catalyst
When compound H reacts with \( H_2 \) in the presence of Lindlar's catalyst, it undergoes partial hydrogenation, leading to compound I \( (C_5H_8O) \).

The reaction can be represented as:
\[ \text{H} - C \equiv C - C(OH) - CH_2 - CH_3 + H_2 \xrightarrow{\text{Lindlar's catalyst}} \text{I} \]
The resulting compound I will have a double bond and will be optically inactive because it lacks a chiral center.

### Step 4: Further Hydrogenation with Pd/C
Next, compound H undergoes complete hydrogenation with \( H_2 \) and Pd/C to form compound J \( (C_5H_{12}O) \).

The reaction can be represented as:
\[ \text{I} + H_2 \xrightarrow{\text{Pd/C}} \text{J} \]
Compound J is a saturated compound and also optically inactive.

### Step 5: Analyze the Statements
Now, we need to evaluate the statements provided in the question:

1. **Statement A**: J gives a positive haloform test.
- **Incorrect**: Compound J does not have a methyl ketone group, so it will not give a positive haloform test.

2. **Statement B**: H gives a white precipitate with Ag(NH3)2^+ (Tollens' reagent).
- **Correct**: Compound H contains a terminal alkyne, which gives a positive Tollens' test.

3. **Statement C**: Compound H liberates 1 mole of H2 per mole of H when treated with Na.
- **Correct**: The acidic hydrogen in the terminal alkyne will react with sodium to release hydrogen gas.

4. **Statement D**: J gives a positive test with Jones reagent.
- **Correct**: Compound J is a secondary alcohol and will give a positive test with Jones reagent.

### Final Conclusion
The correct statements are B, C, and D.

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