`underset(Br) underset(|) (CH_(3)CHCH_(3))overset(alc. KOH)rarr A underset("peroxide")overset(HBr)rarr B overset(CH_(3)ON a)rarrC`
In the above reaction sequence, the final product is :

Options

A. diethyl ether
B. 1-methoxypropane
C. isopropyl alcohol
D. propylene glycol

Explanation

To solve the given reaction sequence, let's break it down step by step:

### Step 1: Identify the starting compound
The starting compound is **CH₃CH=CH₃** (propene with a bromine substituent at the first carbon).

### Step 2: Reaction with alcoholic KOH
When propene reacts with **alcoholic KOH**, it undergoes an **E2 elimination reaction**. In this reaction, a β-hydrogen is abstracted, and bromine leaves, resulting in the formation of an alkene.

- **Elimination Reaction**:
- The β-hydrogen (from the second carbon) is removed by the base (KOH), and the bond between the first carbon and bromine breaks.
- This leads to the formation of **propene (CH₃-CH=CH₂)**.

### Step 3: Reaction with HBr in the presence of peroxide
Next, the alkene (propene) reacts with **HBr** in the presence of **peroxide**. According to the **anti-Markovnikov rule**, the hydrogen from HBr will add to the carbon with more hydrogen atoms (terminal carbon), and the bromine will add to the carbon with fewer hydrogen atoms.

- **Anti-Markovnikov Addition**:
- The bromine will add to the terminal carbon (CH₂), resulting in **1-bromopropane (CH₃-CH₂-CH₂Br)**.

### Step 4: Reaction with sodium methoxide (CH₃ONa)
The final step involves the reaction of **1-bromopropane** with **sodium methoxide (CH₃ONa)**. This reaction will lead to the formation of an ether through an **SN2 reaction**.

- **SN2 Reaction**:
- The methoxide ion (CH₃O⁻) attacks the carbon bonded to bromine, displacing bromine and forming **1-methoxypropane (CH₃-CH₂-CH₂-O-CH₃)**.

### Final Product
The final product of the reaction sequence is **1-methoxypropane**.

### Summary of the Steps
1. **Starting compound**: CH₃-CH=CH₂ (propene).
2. **E2 elimination with alcoholic KOH**: Forms CH₃-CH=CH₂ (propene).
3. **Anti-Markovnikov addition of HBr with peroxide**: Forms CH₃-CH₂-CH₂Br (1-bromopropane).
4. **SN2 reaction with sodium methoxide**: Forms CH₃-CH₂-CH₂-O-CH₃ (1-methoxypropane).

### Final Answer:
The final product is **1-methoxypropane**.