Which of the following reactions will not give Hofmann alkene ?

Options

A.
B.
C.
D.

Explanation

To determine which of the following reactions will not give Hofmann alkene, we need to analyze the conditions that favor the formation of Hofmann alkenes. Hofmann alkenes are typically formed under specific conditions, which include:

1. **Use of a bulky base**: Bulky bases, such as tert-butoxide, favor the formation of less substituted alkenes (Hofmann alkenes) due to steric hindrance.

2. **Poor leaving groups**: Reactions involving poor leaving groups (e.g., NR3+, SR2+, F-) also lead to the formation of Hofmann alkenes.

3. **Steric hindrance at the gamma carbon**: If there is steric hindrance at the gamma carbon, this can also lead to the formation of Hofmann alkenes.

4. **Presence of double bonds at the beta carbon**: If the alkyl halide contains one or more double bonds at the beta carbon, it can also lead to the formation of Hofmann alkenes.

Now, let's analyze the given cases:

1. **First Case**: The leaving group is chlorine (Cl), which is a poor leaving group. Therefore, Hofmann alkene will form.

2. **Second Case**: This case involves both a poor leaving group and a bulky base. Hence, Hofmann alkene will also form.

3. **Third Case**: Here, we have a good leaving group (Cl) and a non-bulky base (OME). In this situation, the reaction will favor the formation of the more substituted alkene (Zaitsev product) instead of Hofmann alkene.

4. **Fourth Case**: This case involves alkyl halides with double bonds at the beta carbon, which will favor the formation of Hofmann alkenes.

Based on this analysis, the reaction that will **not** give Hofmann alkene is the **third case**.

### Final Answer:
The correct answer is **C** (the third case).