In given sequence of reaction predict X
`X overset(KOH)underset(H_(2)O)rarr Y overset(Al_(2)O_(3))underset(633 K)rarr Z overset([O])rarr2` moles of `CH_(3)COOH`

Options

A. `CH_(3) CH_(2) CH (I) CH_(3)`
B. `CH_(3)CH_(2)CH_(2)CH_(2)I`
C. `CH_(3)CH(I) CH (I) CH_(3)`
D. `CH_(3)CH(I) CH_(2) CH_(2)I`

Explanation

To solve the problem step by step, we need to analyze the reactions and determine the structure of compound X that leads to the formation of 2 moles of acetic acid (CH₃COOH) after the series of reactions.

### Step 1: Identify the starting compound X
The first reaction involves X reacting with KOH in an aqueous medium. Since KOH is a strong base and is used in an aqueous solution, it suggests that X is likely an alkyl halide. When alkyl halides react with KOH (aq), they typically undergo hydrolysis to form alcohols.

**Hint:** Alkyl halides react with aqueous KOH to form alcohols.

### Step 2: Determine the product Y
From the reaction of X with KOH, we get Y, which is an alcohol. The structure of Y will depend on the structure of X. Since we are looking for a compound that eventually leads to the formation of 2 moles of acetic acid, we need to consider the possible structures of Y.

**Hint:** The structure of Y is derived from X by replacing the halogen with an -OH group.

### Step 3: Identify the reaction of Y with Al₂O₃
Next, Y reacts with Al₂O₃ at 633 K. Al₂O₃ is a dehydrating agent, which means it will remove water from Y to form an alkene Z. The dehydration of alcohols typically results in the formation of alkenes.

**Hint:** Dehydration of alcohols using a dehydrating agent like Al₂O₃ leads to the formation of alkenes.

### Step 4: Determine the structure of Z
We need to find an alkene Z that, upon oxidation, produces 2 moles of acetic acid. The oxidation of alkenes can lead to cleavage of the double bond, resulting in carboxylic acids.

The alkene that can produce 2 moles of acetic acid upon oxidation is 2-butene (CH₃-CH=CH-CH₃). When oxidized, it can yield 2 moles of acetic acid.

**Hint:** The alkene Z must be capable of yielding 2 moles of acetic acid upon oxidation.

### Step 5: Determine the structure of Y from Z
If Z is 2-butene, we can deduce the structure of Y (the alcohol) that would dehydrate to form this alkene. The alcohol corresponding to 2-butene would be 2-butanol (CH₃-CH(OH)-CH₂-CH₃).

**Hint:** The alcohol Y is formed by adding an -OH group to the alkene Z.

### Step 6: Determine the structure of X from Y
Finally, to find X, we need to convert 2-butanol back into an alkyl halide. This can be done by replacing the -OH group of 2-butanol with a halogen (in this case, iodine). The resulting compound would be 2-iodobutane (CH₃-CH(I)-CH₂-CH₃).

**Hint:** The alkyl halide X is formed by replacing the -OH group in Y with a halogen.

### Conclusion
Thus, the compound X that leads to the formation of 2 moles of acetic acid after the series of reactions is 2-iodobutane.

**Final Answer:** X = 2-iodobutane (option A).