n-propyl bromide on treating with alcoholic KOH produces

Options

A. Propyne
B. Propanol
C. Propane
D. Propene

Explanation

To solve the question regarding the reaction of n-propyl bromide with alcoholic KOH, we can follow these steps:

### Step 1: Identify the Reactants
The reactant in this case is n-propyl bromide, which has the chemical formula CH3-CH2-CH2-Br. It is a primary alkyl halide.

### Step 2: Understand the Reaction Conditions
The reaction is conducted in the presence of alcoholic KOH. KOH is a strong base, and when used in alcoholic conditions, it favors elimination reactions (specifically E2 mechanism) over substitution reactions.

### Step 3: Mechanism of the Reaction
In an E2 elimination reaction, the base (KOH) abstracts a hydrogen atom from the β-carbon (the carbon adjacent to the carbon bearing the bromine), while simultaneously the bromine atom leaves from the α-carbon. This results in the formation of a double bond between the α and β carbons.

### Step 4: Identify the Product
For n-propyl bromide (CH3-CH2-CH2-Br):
- The α-carbon is the one attached to the bromine (the first carbon).
- The β-carbon is the second carbon in the chain.
- When KOH abstracts a hydrogen from the β-carbon, the bromine leaves, leading to the formation of propene (CH3-CH=CH2).

### Step 5: Conclusion
Therefore, the product formed from the reaction of n-propyl bromide with alcoholic KOH is propene.

### Final Answer
The final product is **propene**.

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