Which compound below has four stereoisomers?

Options

A.
B.
C. `H-N=N-N=N-N`
D.

Explanation

To determine which compound has four stereoisomers, we need to analyze the given compounds based on their structure and the presence of stereocenters or double bonds that can lead to geometric isomerism.

### Step-by-Step Solution:

1. **Identify the Compounds**:
We have four compounds to analyze. Let's denote them as A, B, C, and D.

2. **Analyze Compound A**:
- Structure: CH2=CH-CH=CH-CH2
- This compound has two double bonds.
- The formula for the number of stereoisomers when terminal groups are the same is:
\[ \text{Number of stereoisomers} = 2^{n-1} + 2^{n-2}/2 \]
- Here, \( n = 2 \) (number of double bonds).
- Calculation:
\[ 2^{2-1} + 2^{2-2}/2 = 2 + 0.5 = 3 \]
- **Conclusion**: Compound A has 3 stereoisomers.

3. **Analyze Compound B**:
- Structure: CH3-CH=CH-CH=CH-NOH
- This compound has two double bonds and different terminal groups.
- Using the formula:
\[ \text{Number of stereoisomers} = 2^{n} = 2^{2} = 4 \]
- **Conclusion**: Compound B has 4 stereoisomers.

4. **Analyze Compound C**:
- Structure: H-N=NH-N=NH
- This compound also has two double bonds but does not have the same terminal groups.
- Using the formula:
\[ \text{Number of stereoisomers} = 2^{n} = 2^{2} = 4 \]
- **Conclusion**: Compound C has 4 stereoisomers.

5. **Analyze Compound D**:
- Structure: CH3-CH=CH-OH-CH=CH-CH3
- This compound has two double bonds and the terminal groups are the same.
- Using the formula:
\[ \text{Number of stereoisomers} = 2^{n-1} + 2^{n-2}/2 \]
- Calculation:
\[ 2^{2-1} + 2^{2-2}/2 = 2 + 0.5 = 3 \]
- **Conclusion**: Compound D has 3 stereoisomers.

### Final Answer:
The compounds that have four stereoisomers are **Compound B** and **Compound C**.