`Ph-CH=NO_2H underset("3 days")overset("isomerises")(rarr) underset((50-50%))((x))`Isomer (x) is :

Options

A. `Ph- NO-CH_2OH`
B. `Ph-CH_2-NO_2`
C. `Ph-NH-CO_2H`
D. None

Explanation

To solve the question regarding the isomerization of Ph-CH=NO2H over three days, we need to identify the isomer (x) that forms in a 50-50% mixture. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Structure
The given compound is Ph-CH=NO2H, which can be represented as:
- Ph = phenyl group (C6H5)
- CH = carbon with a hydrogen atom
- =NO2H = a nitrogen atom double-bonded to an oxygen atom and single-bonded to a hydrogen atom.

### Step 2: Identify the Isomerization Process
The compound undergoes isomerization (tautomerization) over three days. Tautomerization typically involves the migration of a hydrogen atom and a shift of a double bond. In this case, the compound can tautomerize to form an enol and a keto form.

### Step 3: Determine the Tautomeric Forms
1. **Enol Form**: The initial structure can be represented as an enol:
- Ph-CH=NO2H (enol form)
2. **Keto Form**: The tautomerization will lead to a keto form:
- Ph-CH2-NO2 (keto form)

### Step 4: Identify the Final Isomer (x)
After three days of isomerization, we have a 50-50% mixture of the enol and keto forms. The keto form is:
- Ph-CH2-NO2

### Step 5: Conclusion
The isomer (x) that forms after three days of isomerization is the keto form, which is represented as:
- Ph-CH2-NO2

### Final Answer
Isomer (x) is: **Ph-CH2-NO2**

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