` C_6 H_5- CO - CH_3 overset( NaOH//I_2) to ? + ? `

Options

A. `C_6H_5COOH+ CH_4`
B. ` C_6H_5COONa + CHl_3`
C. ` C_6H_6 + CH_3COONa+HI`
D. ` C_6H_5CH_2COOH`

Explanation

To solve the problem of the reaction of C₆H₅-CO-CH₃ with NaOH and I₂, we will follow these steps:

### Step 1: Identify the Reactant
The reactant is C₆H₅-CO-CH₃, which is known as acetophenone. It contains a carbonyl group (C=O) adjacent to a methyl group (CH₃).

**Hint:** Look for functional groups in the reactant that can participate in reactions with the given reagents.

### Step 2: Recognize the Reaction Type
The reagents NaOH and I₂ indicate that we are dealing with the Iodoform reaction. This reaction typically occurs with methyl ketones or compounds that can be converted into methyl ketones.

**Hint:** Remember that the Iodoform reaction is specific to methyl ketones and some alcohols.

### Step 3: Determine the Products
In the Iodoform reaction, the methyl ketone (acetophenone in this case) will react with NaOH and I₂ to produce two products:
1. Sodium benzoate (C₆H₅COONa)
2. Iodoform (CHI₃)

The methyl group (CH₃) from the ketone is converted into iodoform, while the carbonyl part (C₆H₅CO) is converted into its sodium salt.

**Hint:** Consider how the structure of the reactant changes during the reaction to identify the products.

### Step 4: Write the Balanced Reaction
The balanced reaction can be summarized as follows:

C₆H₅-CO-CH₃ + NaOH + I₂ → C₆H₅COONa + CHI₃

**Hint:** Ensure that both sides of the reaction are balanced in terms of atoms and charge.

### Final Answer
The products of the reaction are:
1. Sodium benzoate (C₆H₅COONa)
2. Iodoform (CHI₃)

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