Q. 2-Bromopentane is heated with potassium in ethanol. the major product obtained is

Options

A. 2-Ethoxypentane
B. Pentane-1-
C. cis-Pentene-2
D. trans-Pentene-2

Explanation

To solve the question regarding the reaction of 2-bromopentane with potassium in ethanol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is 2-bromopentane (C5H11Br), which has the structure:
```
CH3-CHBr-CH2-CH2-CH3
```

2. **Understand the Reaction Conditions**:
- The reaction involves heating 2-bromopentane with potassium in ethanol. Potassium acts as a reducing agent, and ethanol serves as a solvent.

3. **Determine the Reaction Type**:
- This reaction is a dehydrohalogenation reaction, where the bromine atom (Br) is eliminated along with a hydrogen atom from the adjacent carbon. This leads to the formation of an alkene.

4. **Identify the Elimination Mechanism**:
- Since 2-bromopentane is a secondary haloalkane, the reaction will likely proceed via an E2 elimination mechanism, which is favored under strong bases or metals like potassium.

5. **Draw the Product**:
- The elimination of HBr from 2-bromopentane will lead to the formation of pent-2-ene. The structure of pent-2-ene is:
```
CH3-CH=CH-CH2-CH3
```

6. **Consider Isomerism**:
- Pent-2-ene can exist in two geometric isomers: cis and trans. The trans isomer is more stable due to less steric hindrance between the substituents.

7. **Identify the Major Product**:
- The major product obtained from this reaction is **trans-2-pentene**.

### Final Answer:
The major product obtained when 2-bromopentane is heated with potassium in ethanol is **trans-2-pentene**.

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