n-propyl bromide reacts with aqueous KOH to form:-

Options

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

Answer

To solve the question of what n-propyl bromide forms when it reacts with aqueous KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is n-propyl bromide, which has the molecular formula C3H7Br. Its structure can be represented as CH3-CH2-CH2-Br.

2. **Understand the Reaction Conditions**:
- The reaction is taking place in the presence of aqueous KOH. Aqueous KOH is a strong base and can facilitate nucleophilic substitution reactions.

3. **Determine the Type of Reaction**:
- Since we are using aqueous KOH, the reaction will favor substitution rather than elimination. The bromine atom (Br) in n-propyl bromide will be replaced by a hydroxyl group (OH).

4. **Write the Reaction**:
- The reaction can be written as:
\[ \text{C}_3\text{H}_7\text{Br} + \text{KOH} \rightarrow \text{C}_3\text{H}_7\text{OH} + \text{KBr} \]
- Here, KBr is formed as a byproduct.

5. **Identify the Product**:
- The product formed is propanol (C3H8O), specifically n-propanol, since the bromine is replaced by a hydroxyl group.

6. **Final Answer**:
- The final product of the reaction between n-propyl bromide and aqueous KOH is **propanol**.