the enol form of hept-5-en-2-one contains

Options

A. 17sigma-bonds and 2π-bonds
B. 19sigma-bonds and 2π-bonds
C. 15sigma-bonds and 1π-bond
D. 21sigma-bonds and 2π-bonds

Answer

To determine the number of sigma and pi bonds in the enol form of hept-5-en-2-one, we can follow these steps:

### Step 1: Understand the structure of hept-5-en-2-one
- The name "hept" indicates that there are 7 carbon atoms in the chain.
- The "5-en" indicates that there is a double bond between the 5th and 6th carbon atoms.
- The "2-one" indicates that there is a ketone functional group at the 2nd carbon atom.

### Step 2: Draw the structure of hept-5-en-2-one
The structure can be represented as follows:
```
CH3-CH-CO-CH=CH-CH2-CH3
|
H
```
In this structure, the double bond is between the 5th and 6th carbon atoms, and the ketone group (C=O) is attached to the 2nd carbon.

### Step 3: Convert to the enol form
In the enol form, the double bond shifts to the position of the carbonyl group, and the carbonyl oxygen becomes an alcohol (OH) group. The enol form of hept-5-en-2-one is:
```
CH3-CH=CH-CHOH-CH2-CH2-CH3
```

### Step 4: Count the sigma bonds
- Each single bond (C-C or C-H) is a sigma bond.
- Count the number of single bonds:
- C-C: 6 single bonds (between each pair of adjacent carbon atoms)
- C-H: Each carbon is bonded to enough hydrogens to satisfy the tetravalency (4 bonds total).
- The first carbon (CH3) has 3 H
- The second carbon (CH) has 1 H
- The third carbon (CH) has 1 H
- The fourth carbon (CHOH) has 1 H
- The fifth carbon (CH2) has 2 H
- The sixth carbon (CH2) has 2 H
- The seventh carbon (CH3) has 3 H
- Total C-H bonds: 3 + 1 + 1 + 1 + 2 + 2 + 3 = 13 H
- Total sigma bonds = 6 (C-C) + 13 (C-H) = 19 sigma bonds.

### Step 5: Count the pi bonds
- The enol form has one double bond (C=C), which contributes 1 pi bond.
- Therefore, total pi bonds = 1.

### Final Answer
The enol form of hept-5-en-2-one contains:
- **19 sigma bonds**
- **1 pi bond**