A hydrocarbon `C_6 H_12` decolourises bromine solution and yields `n-`hexane on hydrogenation. On oxidation with `KMnO_4` it affords two different monobasic acids of the type `R-COOH`. The compound is :

Options

A. Cyclohexene
B. 2-Hexene
C. 1-Hexene
D. 3-Hexene

Explanation

To solve the problem, we need to identify the hydrocarbon with the formula \( C_6H_{12} \) that meets the specified conditions. Let's break it down step by step.

### Step 1: Determine the Degree of Unsaturation
The degree of unsaturation can be calculated using the formula:
\[ \text{Degree of Unsaturation} = \frac{2C + 2 - H}{2} \]
For \( C_6H_{12} \):
\[ \text{Degree of Unsaturation} = \frac{2(6) + 2 - 12}{2} = \frac{12 - 12}{2} = 0 \]
This indicates that there is either a double bond or a ring structure. However, since the compound decolorizes bromine, it indicates the presence of a double bond.

**Hint:** The degree of unsaturation helps us determine if there are double bonds or rings in the structure.

### Step 2: Analyze the Reaction with Bromine
The hydrocarbon decolorizes bromine solution, which is characteristic of alkenes. This means our compound is likely an alkene.

**Hint:** Alkenes react with bromine to form dibromides, leading to decolorization.

### Step 3: Hydrogenation to Yield n-Hexane
The hydrocarbon yields n-hexane upon hydrogenation, indicating that it must be a straight-chain alkene without branching. This means that the alkene must have a double bond that, when hydrogenated, results in a straight-chain alkane.

**Hint:** The product of hydrogenation gives clues about the structure of the original compound.

### Step 4: Identify Possible Structures
Given that the compound is a straight-chain alkene with the formula \( C_6H_{12} \), the possible structures include:
1. Hex-1-ene
2. Hex-2-ene
3. Hex-3-ene

**Hint:** Consider the position of the double bond when identifying potential structures.

### Step 5: Oxidation with KMnO4
The compound must yield two different monobasic acids upon oxidation with KMnO4.
- **Hex-1-ene** would yield one type of acid.
- **Hex-2-ene** can yield two different acids: one from the terminal carbon and one from the internal carbon.
- **Hex-3-ene** would also yield one type of acid.

Thus, only hex-2-ene can produce two different acids upon oxidation.

**Hint:** The position of the double bond affects the products formed during oxidation.

### Conclusion
The hydrocarbon \( C_6H_{12} \) that decolorizes bromine, yields n-hexane on hydrogenation, and produces two different monobasic acids on oxidation with KMnO4 is **Hex-2-ene**.

**Final Answer:** The compound is **Hex-2-ene** (option B).