`CH_(3)CH_(2)CH_(2)Cl overset(alc. KOH)rarr B overset(HBr)rarr C overset("Na//ether")rarr D`
In the above reaction, the product D is

Options

A. Propane
B. 2,3-Dimethylbutane
C. Hexane
D. Allyl bromine

Explanation

To solve the given reaction sequence, we will go through each step one by one.

### Step 1: Starting Material
The starting material is **CH₃CH₂CH₂Cl** (1-chloropropane).

### Step 2: Reaction with Alcoholic KOH
When 1-chloropropane reacts with alcoholic KOH, a dehydrohalogenation reaction occurs. This means that a hydrogen atom and the chlorine atom are eliminated, leading to the formation of an alkene.

- **Mechanism**: The KOH acts as a base and abstracts a beta-hydrogen from the beta-carbon (the carbon adjacent to the one bonded to chlorine). The chlorine leaves as a chloride ion, resulting in the formation of a double bond.

- **Product**: The product formed is **CH₃CH=CH₂** (propene).

### Step 3: Reaction with HBr
Next, the alkene (propene) reacts with HBr. This is an addition reaction where HBr adds across the double bond.

- **Markovnikov's Rule**: According to Markovnikov's rule, the hydrogen atom from HBr will attach to the carbon with more hydrogen atoms (the less substituted carbon), while the bromine will attach to the carbon with fewer hydrogen atoms.

- **Product**: The product of this reaction will be **CH₃CHBrCH₃** (2-bromopropane).

### Step 4: Reaction with Sodium in Dry Ether
Finally, 2-bromopropane reacts with sodium in dry ether. This is a Wurtz reaction, which leads to the formation of an alkane.

- **Mechanism**: In the presence of sodium, the bromine atom is removed, and a new carbon-carbon bond is formed between the two alkyl groups.

- **Product**: The product formed is **C₆H₁₄** (hexane), specifically **n-hexane**.

### Final Product
Thus, the final product D is **n-hexane**.

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