An organic compound X with molecular formula `C_6H_10`, when treated with HBr, forms a gem dibromide. The compound X upon warming with `HgSO_4` and dil. `H_2SO_4`, produces a ketone which gives a positive iodoform test. The compound X is

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Answer

To solve the problem, we need to identify the organic compound X with the molecular formula C₆H₁₀ based on the given reactions and properties.

### Step-by-Step Solution:

1. **Understanding the Molecular Formula**:
- The molecular formula of the compound is C₆H₁₀.
- We can calculate the degree of unsaturation (DU) using the formula:
\[ \text{DU} = \frac{C - H + X}{2} + 1 \]
where C is the number of carbon atoms, H is the number of hydrogen atoms, and X is the number of halogen atoms.
- For C₆H₁₀, we have:
\[ \text{DU} = \frac{6 - 10 + 0}{2} + 1 = \frac{-4}{2} + 1 = -2 + 1 = -1 \]
- This indicates that there are 2 degrees of unsaturation, which could imply the presence of a triple bond or two double bonds.

2. **Reaction with HBr**:
- The compound X forms a geminal dibromide when treated with HBr.
- A geminal dibromide means that both bromine atoms are attached to the same carbon atom. This suggests that X likely has a double or triple bond.

3. **Reaction with HgSO₄ and Dilute H₂SO₄**:
- The compound X, when warmed with HgSO₄ and dilute H₂SO₄, produces a ketone that gives a positive iodoform test.
- The iodoform test is positive for methyl ketones (RCOCH₃), indicating that the ketone formed must have a methyl group adjacent to the carbonyl carbon.

4. **Identifying Possible Structures**:
- Given the information, we can deduce that X must be an alkyne (due to the presence of a triple bond) that can produce a methyl ketone upon hydration.
- The structure of X should be 3-heptyne (C₆H₁₀), which can be represented as:
\[ \text{CH}_3C \equiv CCH_2CH_3 \]

5. **Confirming the Structure**:
- When 3-heptyne reacts with HBr, it will add across the triple bond to form a geminal dibromide.
- When treated with HgSO₄ and dilute H₂SO₄, it will yield 2-pentanone (a methyl ketone), which gives a positive iodoform test.

### Conclusion:
The compound X is **3-heptyne**.