In dénydrohalogenation of tert-pentyl bromide using alc. KOH, major product obtained is

Options

A. 2-Methylbut-1-ene
B. 2-Methylbut-2-ene
C. Pen! 1-ene
D. Pent-2 ene

Answer

To solve the problem of determining the major product obtained from the dehydrohalogenation of tert-pentyl bromide using alcoholic KOH, we can follow these steps:

### Step 1: Identify the Structure of tert-Pentyl Bromide
- **Structure**: Tert-pentyl bromide (also known as neopentyl bromide) has the following structure:
```
CH3
|
CH3 - C - Br
|
CH2 - CH3
```
This indicates a tertiary carbon (the central carbon) attached to three other carbon atoms.

**Hint**: Remember that "tert" indicates a tertiary carbon, which is bonded to three other carbon atoms.

### Step 2: Understand the Reaction Mechanism
- When tert-pentyl bromide is treated with alcoholic KOH, a dehydrohalogenation reaction occurs. The bromine atom (Br) is eliminated, leading to the formation of a carbocation.

**Hint**: Dehydrohalogenation involves the elimination of a halogen (in this case, bromine) and a hydrogen atom.

### Step 3: Formation of Carbocation
- The elimination of bromine leads to the formation of a carbocation. Since the starting material is a tertiary halide, the carbocation formed is initially a primary carbocation, which is not stable.
- A methyl shift occurs, where a methyl group from an adjacent carbon migrates to stabilize the carbocation, resulting in a more stable tertiary carbocation.

**Hint**: Carbocation stability increases with the degree of substitution; tertiary carbocations are more stable than primary ones.

### Step 4: Determine the Possible Products
- The stable carbocation can now undergo elimination to form a double bond. There are two possible positions for the double bond to form:
1. Between the first and second carbon.
2. Between the second and third carbon.

**Hint**: Consider the location of the double bond and how it affects the stability of the product.

### Step 5: Apply Zaitsev's Rule
- According to Zaitsev's rule, the more substituted alkene is favored as the major product.
- In this case, if the double bond forms between the second and third carbons, it will be more substituted than if it forms between the first and second carbons.

**Hint**: Zaitsev's rule states that the more substituted alkene will be the major product.

### Step 6: Naming the Major Product
- The major product formed is 2-methylbut-2-ene. The structure can be represented as:
```
CH3
|
CH3 - C = C
|
CH2 - CH3
```
- The longest carbon chain containing the double bond has four carbons, and the double bond is located at the second carbon.

**Hint**: When naming alkenes, identify the longest carbon chain that includes the double bond and number the chain to give the double bond the lowest possible number.

### Final Answer
The major product obtained from the dehydrohalogenation of tert-pentyl bromide using alcoholic KOH is **2-methylbut-2-ene**.