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 molecular formula is C5H11Br.

### Step 2: Understand the Reaction Conditions
Alcoholic KOH is a strong base that promotes elimination reactions (E2 mechanism) rather than substitution reactions. This means that the bromine atom will be eliminated along with a hydrogen atom from a neighboring carbon (beta carbon).

### Step 3: Identify Beta Carbons
In 2-bromopentane, the beta carbons are:
- Carbon 1 (C1) – adjacent to C2 (where bromine is attached).
- Carbon 3 (C3) – also adjacent to C2.

### Step 4: Elimination Reaction
When KOH is added, bromine (Br) will leave, and a hydrogen atom will be removed from one of the beta carbons:
- If hydrogen is removed from C1, the product will have a double bond between C1 and C2.
- If hydrogen is removed from C3, the product will have a double bond between C2 and C3.

### Step 5: Determine the Major Product
According to Zaitsev's rule, the more substituted alkene is favored as the major product.
- Removing hydrogen from C1 gives us a double bond between C1 and C2 (less substituted).
- Removing hydrogen from C3 gives us a double bond between C2 and C3 (more substituted).

Thus, the major product will be the alkene formed by removing the hydrogen from C3, resulting in the following structure:
- CH3-CH=CH-CH2-CH3 (pent-2-ene).

### Step 6: Naming the Product
The product is named based on the position of the double bond:
- Since the double bond starts at the second carbon, the name of the compound is 2-pentene.

### Final Answer
The major product when 2-bromopentane is treated with alcoholic KOH is **2-pentene**.

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