2-Bromopentane is treated with alcoholic KOH solution. Whatt will be the major product formed in this reaction and what is the type of elimination called?

Options

A. Pent-1-ene, `beta`-Elimination
B. Pent-2-ene, `beta`-elimination
C. Pent-1-ene, Nucleophilic substitution
D. Pent-2-ene, Nucleophilic substitution.

Explanation

To solve the problem of what happens when 2-bromopentane is treated with alcoholic KOH, we can follow these steps:

### Step 1: Identify the Reactants
The reactant is 2-bromopentane, which has the structure:
\[ \text{CH}_3\text{CHBrCH}_2\text{CH}_2\text{CH}_3 \]

### Step 2: Understand the Reaction Conditions
Alcoholic KOH is a strong base and is known to promote elimination reactions. In this case, it will induce a dehydrohalogenation reaction, where a hydrogen atom and a halogen atom (bromine) are removed.

### Step 3: Identify the Type of Elimination
The reaction that occurs is a beta elimination (E2 mechanism). In this type of elimination, a hydrogen atom from a beta carbon (the carbon adjacent to the carbon holding the halogen) is removed along with the halogen.

### Step 4: Determine the Beta Hydrogens
In 2-bromopentane, the carbon with the bromine (alpha carbon) is connected to two beta carbons:
- Beta carbon 1 (CH2) has 2 hydrogens.
- Beta carbon 2 (CH3) has 3 hydrogens.

### Step 5: Possible Products
1. **Elimination from Beta Carbon 1** (removing a hydrogen from CH2):
- The product will be:
\[ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3 \]
This is called 2-pentene.

2. **Elimination from Beta Carbon 2** (removing a hydrogen from CH3):
- The product will be:
\[ \text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}_3 \]
This is called 1-pentene.

### Step 6: Determine the Major Product
According to Zaitsev's rule, the more substituted alkene is generally more stable and thus favored. In this case, 2-pentene (formed by elimination from beta carbon 1) is more substituted than 1-pentene (formed by elimination from beta carbon 2). Therefore, 2-pentene is the major product.

### Step 7: Conclusion
The major product formed when 2-bromopentane is treated with alcoholic KOH is 2-pentene, and the type of elimination that occurs is beta elimination (E2 mechanism).

### Summary
- **Major Product**: 2-pentene
- **Type of Elimination**: Beta elimination (E2)