How many beta-elimination products are possible for 2-bromobutane?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine how many beta-elimination products are possible for 2-bromobutane, we can follow these steps:

### Step 1: Identify the Structure of 2-Bromobutane
2-bromobutane has the following structure:
```
CH3-CH-CH2-CH3
|
Br
```
In this structure, the carbon atom to which the bromine (Br) is attached is the alpha carbon, and the adjacent carbon atoms are the beta carbons.

### Step 2: Identify the Beta Carbons
For 2-bromobutane, there are two beta carbons:
- The carbon on the left (C1)
- The carbon on the right (C3)

### Step 3: Perform Beta-Elimination
Beta-elimination involves the removal of the bromine atom and a hydrogen atom from the beta carbon.

1. **Elimination from the Left Beta Carbon (C1)**:
- Removing Br from C2 and H from C1 results in:
```
CH3-CH=CH2 + HBr
```
This product is butene.

2. **Elimination from the Right Beta Carbon (C3)**:
- Removing Br from C2 and H from C3 results in:
```
CH3-CH=CH-CH3 + HBr
```
This product is also butene, but it can exist in two geometric isomers: cis and trans.

### Step 4: Count the Products
- From the left beta carbon (C1), we get 1 product (butene).
- From the right beta carbon (C3), we get 2 products (cis and trans butene).

### Final Count of Products
Adding these together:
- 1 product from C1 + 2 products from C3 = 3 total products.

### Conclusion
Thus, the total number of beta-elimination products possible for 2-bromobutane is **3**.