The order of `E_(2)` elimination for alkyl halide is

Options

A. `1^(@) gt 2^(@) gt 3^(@)`
B. `3^(@) gt 2^(@) gt 1^(@)`
C. `2^(@) gt 3^(@) gt 1^(@)`
D. `3^(@) gt 1^(@) gt 2^(@)`

Answer

To determine the order of E2 elimination for alkyl halides, we need to consider the structure of the alkyl halides involved. The E2 elimination mechanism is influenced by the steric hindrance around the carbon atom that is bonded to the leaving group (the halide).

### Step-by-Step Solution:

1. **Understanding E2 Mechanism**:
- E2 (bimolecular elimination) is a concerted reaction where the base removes a proton (H) from the β-carbon while the leaving group (halide) departs from the α-carbon simultaneously.

2. **Influence of Alkyl Halide Structure**:
- The structure of the alkyl halide significantly affects the rate of the E2 reaction. The order of reactivity is influenced by the steric hindrance around the carbon atoms.

3. **Order of Reactivity**:
- **Tertiary Alkyl Halides**:
- Tertiary alkyl halides are the most reactive in E2 eliminations. The bulky nature of the tertiary carbon makes it easier for the base to abstract a proton, as the base is less hindered by surrounding groups.
- **Secondary Alkyl Halides**:
- Secondary alkyl halides are less reactive than tertiary but more reactive than primary. They have moderate steric hindrance, allowing for some E2 elimination to occur.
- **Primary Alkyl Halides**:
- Primary alkyl halides are the least reactive in E2 eliminations due to significant steric hindrance. The base finds it more difficult to approach the β-hydrogen to abstract it.

4. **Final Order**:
- Therefore, the order of E2 elimination for alkyl halides is:
- **Tertiary > Secondary > Primary**

### Summary of the Order:
- Tertiary Alkyl Halides > Secondary Alkyl Halides > Primary Alkyl Halides