The vapour density of a compound is 29, which reacts with iodine and NaOH to form a yellow compound.The compound is-

Options

A. `CH_(3)`COOH
B. `CH_(3) "COCH"_(3)`
C. `CH_(3)"CHOHCH"_(3)`
D. `CH_(3)`OH

Answer

To solve the problem step-by-step, we need to determine the compound based on the given vapor density and the reaction with iodine and NaOH.

### Step 1: Calculate the Molar Mass from the Vapor Density
The vapor density (VD) of the compound is given as 29. The relationship between vapor density and molar mass (M) is given by the formula:

\[ \text{Vapor Density} = \frac{\text{Molar Mass}}{2} \]

Using this formula, we can rearrange it to find the molar mass:

\[ \text{Molar Mass} = \text{Vapor Density} \times 2 = 29 \times 2 = 58 \text{ g/mol} \]

### Step 2: Identify Possible Compounds
Now that we know the molar mass of the compound is 58 g/mol, we need to identify compounds that have this molar mass and can react with iodine and NaOH to form a yellow compound.

### Step 3: Check Common Organic Compounds
We can consider common organic compounds that might fit this description. Two likely candidates are:
1. Acetic acid (CH₃COOH)
2. Acetone (CH₃COCH₃)

### Step 4: Calculate the Molar Mass of Acetic Acid
For acetic acid (CH₃COOH):
- Carbon (C): 2 atoms × 12 g/mol = 24 g/mol
- Hydrogen (H): 4 atoms × 1 g/mol = 4 g/mol
- Oxygen (O): 2 atoms × 16 g/mol = 32 g/mol

Total molar mass = 24 + 4 + 32 = 60 g/mol

### Step 5: Calculate the Molar Mass of Acetone
For acetone (CH₃COCH₃):
- Carbon (C): 3 atoms × 12 g/mol = 36 g/mol
- Hydrogen (H): 6 atoms × 1 g/mol = 6 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol

Total molar mass = 36 + 6 + 16 = 58 g/mol

### Step 6: Determine the Correct Compound
Since the molar mass of acetone is 58 g/mol and it can react with iodine and NaOH to form a yellow compound (iodoform), we conclude that the compound is acetone.

### Final Answer
The compound is **acetone (CH₃COCH₃)**.

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