In the reaction ,
`CH_(3)-underset(Br)underset(|)(CH)-CH_(3)overset(alc.KOH)(to)(A)overset(HBr)underset("Peroxide")(to)(B) overset(NaI)underset("Acetone")(to)(C)`
The compound (C) is :

Options

A. `CH_3CH_2CH_2I`
B. `CH_3CHICH_2I`
C. `CH_3CH_2CH_2CH_2CH_3`
D. `CH_3CH_2CHI_2`

Explanation

To solve the question step by step, let's analyze the reaction sequence provided:

1. **Identify the starting compound**: The starting compound is `CH3-CH(Br)-CH3`, which is 2-bromopropane.

2. **Step 1: Reaction with alcoholic KOH**:
- The reaction of 2-bromopropane with alcoholic KOH is a dehydrohalogenation reaction. In this reaction, a hydrogen atom and a bromine atom are eliminated to form an alkene.
- The elimination will occur between the bromine and a hydrogen atom from the adjacent carbon, leading to the formation of propene.
- **Product A**: `CH3-CH=CH2` (propene).

3. **Step 2: Reaction with HBr in the presence of peroxide**:
- The presence of peroxide leads to the anti-Markovnikov addition of HBr to the alkene. According to the peroxide effect, the bromine will add to the less substituted carbon, which in this case is the terminal carbon of propene.
- Thus, HBr adds across the double bond resulting in the formation of butane.
- **Product B**: `CH3-CH2-CH2-CH2Br` (butyl bromide).

4. **Step 3: Reaction with NaI in acetone**:
- The reaction of butyl bromide with sodium iodide in acetone is a nucleophilic substitution reaction (S_N2 mechanism). Here, iodide ion (I-) will replace the bromine atom.
- This leads to the formation of butyl iodide.
- **Product C**: `CH3-CH2-CH2-CH2I` (butyl iodide).

Thus, the final compound (C) formed in the reaction sequence is butyl iodide.

### Final Answer:
The compound (C) is `CH3-CH2-CH2-CH2I` (butyl iodide).