A strong solution of alcoholic alkali will preferentially promote alkyl halide into an alkene by

Options

A. Addition
B. Elimination
C. Polymerisation
D. Substitution

Answer

To solve the question, "A strong solution of alcoholic alkali will preferentially promote alkyl halide into an alkene by," we need to understand the reaction mechanism involved when an alkyl halide reacts with a strong alcoholic base.

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactant is an alkyl halide, denoted as RX, where R is an alkyl group and X is a halogen (like Cl, Br, or I). For example, let's consider ethyl chloride (C2H5Cl).

2. **Understand the Reaction Conditions**: The question specifies a strong solution of alcoholic alkali. This means we are using a strong base (like KOH or NaOH) dissolved in alcohol (ethanol, for example).

3. **Determine the Type of Reaction**: In the presence of a strong alcoholic base, the reaction will favor elimination rather than substitution. This is because the alcoholic medium promotes the removal of a hydrogen atom and the halogen atom from the alkyl halide.

4. **Mechanism of Elimination**:
- The base (OH-) abstracts a hydrogen atom from the alkyl halide (C2H5Cl).
- The electrons from the C-H bond shift to form a double bond between the two carbon atoms.
- Simultaneously, the bond between the carbon and the halogen (Cl) breaks, leading to the release of Cl-.
- This results in the formation of an alkene (C2H4) and the by-products KCl (or NaCl) and water.

5. **Write the Overall Reaction**:
- The reaction can be summarized as:
\[ C2H5Cl + KOH \rightarrow C2H4 + KCl + H2O \]
- Here, C2H4 is the alkene formed through an elimination reaction.

6. **Conclusion**: Therefore, the strong solution of alcoholic alkali promotes the conversion of alkyl halide into an alkene by an elimination mechanism (specifically, an E2 mechanism).

### Final Answer:
The correct answer is that a strong solution of alcoholic alkali will preferentially promote alkyl halide into an alkene by **elimination**.