Consider the following transformations `CH_3COOHoverset(CaCO_3)toAoverset("Heat")toB underset("NaOH")overset(I_2)to C` .The molecular formula of C is

Options

A. (a) `CH_3-undersetunderset(I)(|)oversetoverset(OH)(|)C-CH_3`
B. (b) `ICH_2-COCH_3`
C. (c) `CHI_3`
D. (d) `CH_3I`

Explanation

To solve the question, we need to follow the transformations step by step:

### Step 1: Identify the initial compound
The initial compound is acetic acid, which has the molecular formula \( \text{CH}_3\text{COOH} \).

### Step 2: Reaction with Calcium Carbonate
When acetic acid reacts with calcium carbonate (\( \text{CaCO}_3 \)), it forms calcium acetate. The reaction can be represented as:

\[ 2 \text{CH}_3\text{COOH} + \text{CaCO}_3 \rightarrow \text{Ca(CH}_3\text{COO)}_2 + \text{H}_2\text{O} + \text{CO}_2 \]

The product A is calcium acetate, which has the formula \( \text{Ca(CH}_3\text{COO)}_2 \).

### Step 3: Heating the Calcium Acetate
On heating calcium acetate, it decomposes to form acetone and calcium carbonate:

\[ \text{Ca(CH}_3\text{COO)}_2 \xrightarrow{\text{Heat}} \text{CH}_3\text{C(O)CH}_3 + \text{CaCO}_3 \]

The product B is acetone, which has the molecular formula \( \text{CH}_3\text{C(O)CH}_3 \) or simply \( \text{C}_3\text{H}_6\text{O} \).

### Step 4: Reaction with Iodine and NaOH
Next, acetone (product B) is treated with iodine (\( \text{I}_2 \)) in the presence of sodium hydroxide (\( \text{NaOH} \)). This is known as the iodoform reaction, which typically occurs with methyl ketones. The reaction leads to the formation of iodoform (\( \text{CHI}_3 \)) and sodium acetate:

\[ \text{CH}_3\text{C(O)CH}_3 + \text{I}_2 + \text{NaOH} \rightarrow \text{CHI}_3 + \text{CH}_3\text{COONa} \]

The product C is iodoform, which has the molecular formula \( \text{CHI}_3 \).

### Conclusion
Thus, the molecular formula of compound C is \( \text{CHI}_3 \).

### Final Answer
The molecular formula of C is \( \text{CHI}_3 \).

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