The compound X on treatment with acidified `K_(2)Cr_(2)O_(7)` gives compound Y which reacts with `I_(2)` and `Na_(2)CO_(3)` to form tri-iodomethane. The compound X is :

Options

A. `CH_(3)OH`
B. `CH_(3)COCH`
C. `CH_(3)CHO`
D. `CH_(3)CHOHCH_(3)`

Explanation

To solve the problem, we need to identify the compound X based on the reactions described. Let's break down the steps:

### Step 1: Identify the reaction of compound X with acidified K2Cr2O7
- The compound X is treated with acidified potassium dichromate (K2Cr2O7), which is a strong oxidizing agent.
- This suggests that compound X is likely an alcohol, as alcohols are oxidized to carbonyl compounds (aldehydes or ketones) by K2Cr2O7.

### Step 2: Determine the structure of compound Y
- The reaction of compound X with K2Cr2O7 produces compound Y.
- Since Y is formed from an alcohol, it is likely a ketone or an aldehyde.
- The mention of the product being able to react with I2 and Na2CO3 to form triiodomethane (iodoform) indicates that Y must be a methyl ketone (RCOCH3).

### Step 3: Identify the structure of compound X
- The only alcohol that can be oxidized to a methyl ketone (which can then react to form iodoform) is 2-propanol (isopropyl alcohol), which has the structure CH3CHOHCH3.
- When 2-propanol is oxidized by K2Cr2O7, it forms acetone (CH3COCH3), which is a methyl ketone.

### Conclusion
- Therefore, the compound X is 2-propanol (isopropyl alcohol).

### Final Answer
- The compound X is: **2-propanol (CH3CHOHCH3)**.

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