Which one of the following methods cannot yield alkyl halide ?

Options

A. `C_2H_5OH + HCl to `
B. `CH_3COOAg + Br_2 overset(C Cl_4)(rarr)`
C. `CH_2 = CH_2 overset(SOCl_2)(rarr)`
D. `CH_3CH_2OH + HX overset(ZnCl_2)(rarr)`

Explanation

To determine which method cannot yield an alkyl halide, we will analyze each option provided in the question.

### Step-by-Step Solution:

1. **Understanding Alkyl Halides**:
- Alkyl halides are compounds that contain an alkyl group (R) bonded to a halogen (X). The general formula is R-X.

2. **Analyzing Each Reaction**:
- **Option A: Alcohol + HCl**
- When an alcohol reacts with hydrochloric acid (HCl), it undergoes a substitution reaction where the hydroxyl group (-OH) is replaced by a halogen (Cl in this case). Water is eliminated, and an alkyl halide is formed.
- **Product**: Alkyl halide (e.g., C2H5Cl).
- **Conclusion**: This method can yield an alkyl halide.

- **Option B: CH3COAg + Br2/CCL4**
- In this reaction, decarboxylation occurs, leading to the removal of carbon dioxide (CO2). The remaining components will react to form an alkyl halide.
- **Product**: Methyl bromide (CH3Br).
- **Conclusion**: This method can yield an alkyl halide.

- **Option C: CH2O + CH2SOCl2**
- Here, SOCl2 is used, which typically reacts with alcohols to form alkyl halides. However, when reacting with an alkene (as indicated by the presence of CH2O), it does not yield an alkyl halide.
- **Product**: No alkyl halide is formed.
- **Conclusion**: This method cannot yield an alkyl halide.

- **Option D: CH3CH2OH + HX in ZnCl2**
- This reaction involves the Lucas reagent (HX in ZnCl2), which facilitates the conversion of alcohols to alkyl halides through a substitution reaction.
- **Product**: Alkyl halide (e.g., CH3CH2X).
- **Conclusion**: This method can yield an alkyl halide.

3. **Final Conclusion**:
- After analyzing all options, we find that **Option C** (CH2O + CH2SOCl2) is the method that cannot yield an alkyl halide.

### Answer:
**Option C cannot yield an alkyl halide.**

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