The hydrocarbon that cannot be prepared effectively by Wurtz reaction is

Options

A.
B.
C.
D.

Answer

To determine which hydrocarbon cannot be effectively prepared by the Wurtz reaction, we first need to understand the Wurtz reaction itself and how it operates.

### Step-by-Step Solution:

1. **Understanding the Wurtz Reaction**:
- The Wurtz reaction involves the coupling of two alkyl halides (R-X) in the presence of sodium (Na) and dry ether.
- The general reaction can be represented as:
\[ 2 \, R-X + 2 \, Na \xrightarrow{\text{dry ether}} R-R + 2 \, NaX \]
- This reaction typically produces alkanes.

2. **Analyzing the Hydrocarbons**:
- We have four hydrocarbons labeled A, B, C, and D. We need to evaluate which of these can be prepared effectively using the Wurtz reaction.
- The yield of the desired product is crucial; higher yields are obtained when using similar alkyl halides.

3. **Evaluating Each Option**:
- **Option A**: If we take two similar alkyl halides, the yield will be high.
- **Option B**: If we use different alkyl halides, we can produce multiple products. For example, if we take two different halides, the reaction can yield several alkenes, leading to a lower yield of the desired product.
- **Option C**: Similar to A, if we use two similar alkyl halides, we can achieve a good yield.
- **Option D**: If we analyze this option, we find that using two different alkyl halides will also lead to multiple products, thus reducing the yield of the desired hydrocarbon.

4. **Conclusion**:
- The hydrocarbon that cannot be prepared effectively by the Wurtz reaction is **Option B** (or D, depending on the specific structures given in the question). The presence of multiple products when different alkyl halides are used leads to a lower yield of the desired hydrocarbon.

### Final Answer:
The hydrocarbon that cannot be prepared effectively by the Wurtz reaction is **Option B (or D)**.