The enolic form of given compound contians
`CH_(3)-CH_(2)-overset(O)overset(||)C-CH_(3)`
`x, sigma` bonds, y `pi` bonds, z lone pairs. The sum value of `(x+y+z)` is

Answer

To solve the problem, we need to analyze the enolic form of the given compound, which is represented as `CH3-CH=CHOH-CH3`. We will count the number of sigma bonds (x), pi bonds (y), and lone pairs (z) in this structure.

### Step-by-Step Solution:

1. **Identify the Structure**:
The enolic form of the compound is `CH3-CH=CHOH-CH3`. This structure consists of:
- Two methyl groups (CH3)
- One carbon-carbon double bond (C=C)
- One hydroxyl group (OH)

2. **Count Sigma Bonds (x)**:
- Each single bond (C-H, C-C, C-O) is a sigma bond.
- In the structure:
- Between the two CH3 groups and the adjacent carbons, there are 6 C-H bonds (3 from each CH3).
- There are 2 C-C bonds (one between the two CH3 and the next carbon, and one between the double bonded carbon and the last CH3).
- There is 1 C-O bond from the carbon to the hydroxyl group.
- Total sigma bonds = 6 (C-H) + 2 (C-C) + 1 (C-O) = 9 sigma bonds.

3. **Count Pi Bonds (y)**:
- A pi bond is formed in a double bond.
- In the structure, there is one double bond (C=C).
- Therefore, there is 1 pi bond.

4. **Count Lone Pairs (z)**:
- The hydroxyl group (OH) has one oxygen atom that typically has two lone pairs.
- In our structure, we have 2 lone pairs on the oxygen atom.

5. **Calculate the Sum (x + y + z)**:
- x (sigma bonds) = 9
- y (pi bonds) = 1
- z (lone pairs) = 2
- Therefore, x + y + z = 9 + 1 + 2 = 12.

### Final Answer:
The sum value of (x + y + z) is **12**.