Compound X on treatment with HI give Y, Y on treatment with ethanolic KOH gives Z (an isomer of X). Ozonolysis of Z (with `H_2O_2` workup) gives a two -carbon carboxylic acid and four carbon ketone . Hence, X is :

Options

A. 2-methyl-2-pentene
B. 4-methyl-1-pentene
C. 2,3-dimethyl-2-butene
D. 3-methyl-1-pentene

Answer

To solve the problem step by step, we need to analyze the information given in the question and derive the structure of compound X.

### Step 1: Analyze the reaction with HI
- Compound X reacts with HI to give compound Y. This indicates that X is likely an alkene since HI adds across the double bond.
- The addition of HI follows Markovnikov's rule, meaning that the hydrogen atom from HI will attach to the carbon with more hydrogen substituents, while the iodine will attach to the other carbon.

### Step 2: Determine the structure of Y
- The product Y is formed from the addition of HI to X. Since we do not have the structure of X yet, we cannot directly draw Y. However, we know that Y will be a haloalkane.

### Step 3: Analyze the reaction with ethanolic KOH
- Y is treated with ethanolic KOH, leading to the formation of Z, which is an isomer of X. The reaction with KOH suggests that a dehydrohalogenation occurs, resulting in the formation of a double bond.
- Since Z is an isomer of X, it must have the same molecular formula but a different structural arrangement.

### Step 4: Ozonolysis of Z
- The ozonolysis of Z with a hydrogen peroxide workup yields a two-carbon carboxylic acid and a four-carbon ketone.
- The two-carbon carboxylic acid is likely acetic acid (CH₃COOH), and the four-carbon ketone could be butanone (CH₃COCH₂CH₃).

### Step 5: Deduce the structure of Z
- The ozonolysis products suggest that Z must have a structure that, when cleaved, produces these specific fragments.
- The presence of a two-carbon carboxylic acid and a four-carbon ketone indicates that Z could be a 5-carbon alkene.

### Step 6: Identify the structure of X
- Since Z is an isomer of X, we can deduce the structure of X based on the possible structures of Z.
- If Z is an alkene that can yield a two-carbon acid and a four-carbon ketone upon ozonolysis, we can hypothesize that Z could be 3-methyl-1-pentene (C₅H₁₀).
- Therefore, X must be 3-methyl-1-pentene, which upon treatment with HI gives a haloalkane (Y), and then dehydrohalogenation with KOH gives Z.

### Final Answer
Thus, the compound X is **3-methyl-1-pentene**.

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