The enolic form of `CH_3 -overset(O ) overset(।। )C- CH_2 COOC_2 H_5` is

Options

A. `CH_2 = overset(OH) overset(|) C - CH_2 COOC_2 H_5`
B. `CH_3 - overset(OH)overset(|) C = CHCOOC_2 H_5`
C. `CH_3 - overset(O ) overset(||) C- CH = overset(OH) overset(|)C-COOC_2 H_5 `
D. none of the above.

Explanation

To find the enolic form of the compound \( CH_3 - C(=O) - CH_2 - COO - C_2H_5 \), we will follow these steps:

### Step 1: Identify the Keto Form
The given compound is in its keto form, which is characterized by the presence of a carbonyl group (C=O). The structure can be represented as:
\[ CH_3 - C(=O) - CH_2 - COO - C_2H_5 \]

### Step 2: Recognize the Active Methylene Group
In the compound, the methylene group \( CH_2 \) is situated between two carbonyl groups (C=O). This methylene group is referred to as an "active methylene group" because it has acidic hydrogen atoms that can participate in tautomerism.

### Step 3: Tautomerism Process
Tautomerism involves the transfer of a hydrogen atom and the rearrangement of double bonds. In this case, the hydrogen atom from the active methylene group will shift to the oxygen atom of the carbonyl group. The double bond will then shift to form a new double bond between the carbon and the adjacent carbon atom.

### Step 4: Write the Enolic Form
After the tautomeric shift, the enolic form of the compound can be represented as:
\[ CH_3 - C(OH) = CH - C(=O) - O - C_2H_5 \]

This structure shows that we now have a hydroxyl group (OH) and a double bond (C=C), which is characteristic of enols.

### Final Enolic Form
Thus, the enolic form of the compound \( CH_3 - C(=O) - CH_2 - COO - C_2H_5 \) is:
\[ CH_3 - C(OH) = CH - C(=O) - O - C_2H_5 \]

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