How many geometrical isomers of `CH_(3)CH=CH-CH=NOH` exist?

Options

A. Five
B. Two
C. Three
D. Four

Explanation

To determine the number of geometrical isomers for the compound `CH₃CH=CH-CH=NOH`, we need to analyze the double bonds present in the molecule and check for the possibility of geometrical isomerism (E/Z isomerism).

### Step-by-Step Solution:

1. **Identify the Structure**:
The compound can be represented as:
```
CH₃-CH=CH-CH=NOH
```
Here, we have two double bonds: one between the second and third carbon atoms (C=C) and another between the fourth carbon and nitrogen (C=N).

2. **Analyze the First Double Bond (C=C)**:
- The first double bond is between the second carbon (C2) and the third carbon (C3).
- On one side of the double bond (C2), we have `CH₃` (methyl group) and on the other side (C3), we have `H` (hydrogen) and `CH` (the carbon connected to the nitrogen).
- Since `CH₃` and `H` are different groups, we can have geometrical isomers (E/Z) for this double bond.

3. **Analyze the Second Double Bond (C=N)**:
- The second double bond is between the fourth carbon (C4) and nitrogen (N).
- On one side of the double bond (C4), we have `H` (hydrogen) and on the other side, we have `OH` (hydroxyl group) and the carbon chain.
- Since `H` and `OH` are different groups, we can also have geometrical isomers (E/Z) for this double bond.

4. **Count the Geometrical Isomers**:
- For each double bond, we have 2 possibilities (E or Z).
- Since there are 2 double bonds in the molecule, the total number of geometrical isomers can be calculated as:
\[ \text{Total Isomers} = 2 \text{ (from C=C)} \times 2 \text{ (from C=N)} = 4 \]

5. **Conclusion**:
Therefore, the total number of geometrical isomers for the compound `CH₃CH=CH-CH=NOH` is **4**.