An alkyl halide of formula `C_6 H_13 Br` on treatment with potassium t-butoxide gives two isomeric alkenes dimethy butane. Isomeric alkene are :

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to determine the isomeric alkenes that can be formed from the alkyl halide with the formula `C6H13Br` when treated with potassium t-butoxide.

### Step-by-Step Solution:

1. **Identify the Alkyl Halide Structure**:
The alkyl halide has the formula `C6H13Br`. This indicates that it is a six-carbon compound with a bromine atom. The possible structures for this alkyl halide can be:
- 1-bromo-hexane (linear structure)
- 2-bromo-pentane (branched structure)
- 3-bromo-pentane (branched structure)
- 2-bromo-2-methylbutane (more branched structure)

For the purpose of forming dimethyl alkenes, we will focus on the branched structures.

2. **Determine the Reaction with Potassium t-Butoxide**:
Potassium t-butoxide is a strong, bulky base that favors elimination reactions (E2 mechanism). The elimination will lead to the formation of alkenes.

3. **Possible Elimination Products**:
- From 2-bromo-pentane:
- Elimination can occur to form 2-methyl-2-pentene (major product due to more substituted double bond).
- Elimination can also occur to form 2-methyl-1-pentene (minor product).
- From 3-bromo-pentane:
- Elimination can lead to the formation of 2-methyl-2-pentene (major product).
- Elimination can also lead to 3-methyl-1-pentene (minor product).

4. **Identify the Isomeric Alkenes**:
The two isomeric alkenes that can be formed from the elimination reaction are:
- 2-methyl-2-pentene
- 2-methyl-1-pentene

5. **Conclusion**:
Therefore, the isomeric alkenes formed from the alkyl halide `C6H13Br` upon treatment with potassium t-butoxide are 2-methyl-2-pentene and 2-methyl-1-pentene.

### Final Answer:
The isomeric alkenes are **2-methyl-2-pentene and 2-methyl-1-pentene**.