An organic compound 'A' has the molecular formula `C_(3)H_(6)O`, it undergoes iodoform test. Whet satruated with HCl it gives 'B' of molecular foumula `C_(9)H_(14)O`. A and B, respectively are

Options

A. Propanal and mesitylene
B. Propanone and mesityl oxide
C. Propanone and 2, 6-dimethyl-2, 5-heptadien-4-one.
D. Propanone and mesitylene oxide.

Explanation

To solve the problem step by step, let’s analyze the information provided:

### Step 1: Identify Compound A
The molecular formula of compound A is \( C_3H_6O \). Since it undergoes the iodoform test, it must contain a methyl ketone group (a carbonyl group adjacent to a methyl group). The only structure that fits this requirement with three carbon atoms is propanone (acetone).

**Compound A:** Propanone (acetone) - \( CH_3COCH_3 \)

### Step 2: Determine the Reaction with HCl
When propanone (A) is saturated with HCl, it undergoes an addition reaction. Given that the molecular formula of the resulting compound B is \( C_9H_{14}O \), we can infer that propanone likely forms a trimer (a compound made up of three molecules of propanone).

### Step 3: Identify Compound B
The trimer of propanone can be represented as a cyclic compound. The molecular formula \( C_9H_{14}O \) suggests that the trimer is a compound with a ketone functional group. The structure of the cyclic trimer of propanone is known as 2,6-dimethyl-2,5-heptane-diene-4-one.

**Compound B:** 2,6-dimethyl-2,5-heptane-diene-4-one

### Conclusion
Thus, the compounds A and B are:
- **A:** Propanone (acetone)
- **B:** 2,6-dimethyl-2,5-heptane-diene-4-one