A hydrocarbon reacts with HI to give X which on reaction with aqueous KOH forms Y. Oxidation of Y gives 3-methyl, 2-butanone.
The hydrocarbon is

Options

A. `CH_(3)CH=overset(CH_(3))overset(|)(C)-CH_(3)`
B. `CH_(2)=CH-underset(CH_(3))underset(|)(CH)-CH_(3)`
C. `CH_(3)-CH_(2)-underset(CH_(3))underset(|)(CH)-CH_(3)`
D. `CH-=C-underset(CH_(3))underset(|)(CH)-CH_(3)`

Explanation

To solve the problem step by step, we will identify the hydrocarbon that reacts with HI to produce a compound that subsequently reacts with aqueous KOH and is oxidized to yield 3-methyl-2-butanone.

### Step 1: Understanding the Reaction with HI
The hydrocarbon reacts with hydrogen iodide (HI). This suggests that the hydrocarbon must have a double bond (alkene) because HI adds across the double bond.

### Step 2: Determining Compound X
When the hydrocarbon reacts with HI, it forms a compound X. The addition of HI across the double bond will result in the formation of an alkyl iodide. If we denote the hydrocarbon as an alkene with the structure \( C_4H_8 \) (since we will eventually form a ketone with 4 carbons), the addition of HI would yield an alkyl iodide.

### Step 3: Reaction with Aqueous KOH
Next, compound X (the alkyl iodide) reacts with aqueous KOH. Aqueous KOH typically leads to a substitution reaction where the iodide (I) is replaced by a hydroxyl group (OH), forming compound Y. This means Y is an alcohol.

### Step 4: Oxidation of Compound Y
The oxidation of compound Y leads to the formation of 3-methyl-2-butanone. The structure of 3-methyl-2-butanone indicates that Y must be a secondary alcohol, as secondary alcohols can be oxidized to ketones.

### Step 5: Identifying Compound Y
To form 3-methyl-2-butanone, Y must have the structure of a secondary alcohol with a methyl group on the third carbon. This suggests that Y could be 3-methyl-2-butanol.

### Step 6: Identifying the Original Hydrocarbon
Since Y is derived from an alkyl iodide, we can deduce that the original hydrocarbon must be an alkene that, when treated with HI, produces an alkyl iodide that can be converted to 3-methyl-2-butanol. The hydrocarbon that fits this description is 2-methylpropene (or isobutylene).

### Conclusion
Thus, the original hydrocarbon that reacts with HI to give compound X, which then reacts with aqueous KOH to form compound Y, and is oxidized to yield 3-methyl-2-butanone, is **2-methylpropene**.

### Final Answer
The hydrocarbon is **2-methylpropene**.

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