What will be the major product when 2-bromopentane is treated with alc. KOH:

Options

A. But-2-ene
B. Pent-2-ene
C. Pent-1-ene
D. 2-methylbut-ene

Explanation

To determine the major product when 2-bromopentane is treated with alcoholic KOH, we can follow these steps:

### Step 1: Identify the structure of 2-bromopentane
2-bromopentane has the following structure:
- It consists of a five-carbon chain (pentane) with a bromine atom attached to the second carbon.
- The structure can be represented as:

```
CH3-CH(Br)-CH2-CH2-CH3
```

### Step 2: Understand the reaction with alcoholic KOH
Alcoholic KOH is a strong base and will promote elimination reactions (E2 mechanism) rather than substitution reactions. This means that it will remove a hydrogen atom and the bromine atom to form a double bond.

### Step 3: Determine possible elimination sites
In 2-bromopentane, there are two possible sites from which the base can abstract a hydrogen atom:
1. From the carbon adjacent to the bromine (C1).
2. From the carbon on the other side of the bromine (C3).

### Step 4: Draw the potential products
1. If the hydrogen is removed from C1:
- The product will be pent-1-ene (double bond between C1 and C2):
```
CH2=CH-CH2-CH2-CH3 (pent-1-ene)
```

2. If the hydrogen is removed from C3:
- The product will be pent-2-ene (double bond between C2 and C3):
```
CH3-CH=CH-CH2-CH3 (pent-2-ene)
```

### Step 5: Analyze the stability of the products
According to Zaitsev's rule, the more substituted alkene is generally the major product. In this case:
- Pent-1-ene has a double bond between C1 and C2, making it less substituted.
- Pent-2-ene has a double bond between C2 and C3, making it more substituted (two alkyl groups attached to the double bond).

### Step 6: Identify the major product
Since pent-2-ene is more substituted and thus more stable than pent-1-ene, the major product of the reaction will be:
- **Pent-2-ene**.

### Conclusion
The major product when 2-bromopentane is treated with alcoholic KOH is **pent-2-ene**.

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