In the given reaction [X] will be :
`C_(6)H_(5)-underset(O)underset(||)C-Hoverset(NH_(2)OH//H^(o+))rarr [X]`

Options

A. only syn oxime
B. only anti oxime
C. mixture of syn and anti oxime
D. secondary amide

Answer

To determine the product [X] in the given reaction, we will analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactant is a carbonyl compound, specifically an aldehyde, which is represented as:
\[ C_6H_5 - C(=O) - H \]
This is benzaldehyde (C6H5CHO).

### Step 2: Identify the Reagent
The reagent used in the reaction is hydroxylamine (NH2OH) in the presence of an acid (H+). Hydroxylamine is known to react with aldehydes and ketones to form oximes.

### Step 3: Reaction Mechanism
When benzaldehyde reacts with hydroxylamine, the carbonyl group (C=O) of the aldehyde will react with the nitrogen of hydroxylamine. The reaction proceeds as follows:

1. The nitrogen atom of hydroxylamine attacks the carbon atom of the carbonyl group.
2. A tetrahedral intermediate is formed.
3. A molecule of water (H2O) is eliminated in a condensation reaction.

### Step 4: Product Formation
After the elimination of water, the product formed is an oxime. The structure of the oxime can be represented as:
\[ C_6H_5 - C(=N - OH) - H \]
This indicates that the carbon atom is now double-bonded to nitrogen instead of oxygen.

### Step 5: Geometric Isomers
The oxime can exist in two geometric isomers due to the presence of the double bond between carbon and nitrogen:
1. **Z (cis) isomer**: where the phenyl group (C6H5) and the hydroxyl group (OH) are on the same side.
2. **E (trans) isomer**: where the phenyl group (C6H5) and the hydroxyl group (OH) are on opposite sides.

### Conclusion
The product [X] will be a mixture of both the Z and E isomers of the oxime.

Thus, the final answer is:
**[X] is a mixture of Z and E oximes.**