When 2-methyl butyl bromide is treated with sodium ethoxide and ethanol, what will be the major product?

Options

A. 2-methyl but-2-ene
B. 3-methyl but-1-ene
C. 2-methyl but-1-ene
D. 2-methyl-sodium-butoxide

Answer

To determine the major product when 2-methyl butyl bromide is treated with sodium ethoxide and ethanol, we can follow these steps:

### Step 1: Identify the Reactants
We have 2-methyl butyl bromide (C5H11Br) and sodium ethoxide (C2H5ONa) in ethanol (C2H5OH). The structure of 2-methyl butyl bromide is as follows:

```
CH3
|
CH3-CH-CH2-CH2-Br
```

### Step 2: Understand the Reaction Type
The reaction between an alkyl halide and a strong base like sodium ethoxide typically leads to an elimination reaction (E2 mechanism) where a hydrogen atom and a halogen atom (bromine in this case) are removed, forming a double bond.

### Step 3: Determine Possible Products
When 2-methyl butyl bromide undergoes elimination, the following products can be formed:
1. **2-methyl butylene** (more substituted double bond)
2. **3-methyl butylene** (less substituted double bond)

### Step 4: Apply Zaitsev's Rule
According to Zaitsev's rule, the more substituted alkene is favored as the major product in elimination reactions. In this case:
- **2-methyl butylene** has a double bond between the second and third carbons, which is more substituted.
- **3-methyl butylene** has a double bond between the third and fourth carbons, which is less substituted.

### Step 5: Identify the Major Product
Since 2-methyl butylene is more substituted than 3-methyl butylene, it will be the major product of the reaction.

### Conclusion
The major product when 2-methyl butyl bromide is treated with sodium ethoxide and ethanol is **2-methyl butylene**.

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