Treatment of sec-butyl bromide with alcoholic KOH forms__________.

Options

A. but-1-ene only
B. but-2-ene only
C. butan-2-ol
D. a mixture of but-1-ene and but-2-ene

Answer

To solve the question regarding the treatment of sec-butyl bromide with alcoholic KOH, we can follow these steps:

### Step 1: Identify the structure of sec-butyl bromide
Sec-butyl bromide is a secondary alkyl halide with the structure:
\[ \text{CH}_3\text{CHBrCH}_2\text{CH}_3 \]
This means that the bromine atom is attached to the second carbon in the butane chain.

### Step 2: Understand the reagent used
Alcoholic KOH (potassium hydroxide) is a strong base and is used in elimination reactions. It promotes dehydrohalogenation, which involves the removal of a hydrogen atom and a halogen atom (in this case, bromine) from adjacent carbon atoms.

### Step 3: Mechanism of the reaction
The reaction proceeds via an E2 elimination mechanism. In this mechanism:
- A hydrogen atom is removed from the carbon adjacent to the carbon bonded to bromine (the alpha position).
- The bromine atom is also removed simultaneously, leading to the formation of a double bond between the two carbon atoms.

### Step 4: Identify possible products
In sec-butyl bromide, there are two possible positions from which hydrogen can be removed:
1. Removing hydrogen from the carbon adjacent to the bromine leads to the formation of butyl-2-ene:
\[ \text{CH}_3\text{C}=\text{CH}\text{CH}_3 \]
2. Removing hydrogen from the other adjacent carbon leads to the formation of butyl-1-ene:
\[ \text{CH}_2=\text{C}\text{H}\text{CH}_2\text{CH}_3 \]

### Step 5: Conclusion
The treatment of sec-butyl bromide with alcoholic KOH results in the formation of two alkenes: butyl-1-ene and butyl-2-ene. Therefore, the answer to the question is:
**Butyl-1-ene and Butyl-2-ene.**