In the hydrocarbon `underset(6)underset()(CH_(3))-underset(5)underset()(CH)=underset(4)underset()(CH)-underset(3)underset()(CH_(2))-underset(2)underset()Cequivunderset(1)underset()(CH)`
The state of hybridization of carbons 1, 3 and 5 are in the following sequence

Options

A. `sp^(3), sp^(2), sp`
B. `sp^(2),sp, sp^(3)`
C. `sp , sp^(3), sp^(2)`
D. `sp, sp^(2), sp^(3)`

Answer

To determine the hybridization states of carbons 1, 3, and 5 in the hydrocarbon structure CH3-CH=CH-CH2-C≡CH, we can follow these steps:

### Step 1: Draw the Structure
First, we need to visualize the structure of the hydrocarbon. The structure can be drawn as follows:

```
CH3
|
CH
||
CH
|
CH2
|
C≡CH
```

### Step 2: Identify the Carbons
Label the carbons in the structure:
- Carbon 1 (C1) is the first carbon in CH3.
- Carbon 2 (C2) is the second carbon in CH.
- Carbon 3 (C3) is the carbon involved in the double bond (CH).
- Carbon 4 (C4) is the CH2 carbon.
- Carbon 5 (C5) is the carbon involved in the triple bond (C≡CH).
- Carbon 6 (C6) is the terminal carbon in the triple bond (CH).

### Step 3: Determine the Hybridization of Carbon 1
For Carbon 1 (C1):
- It forms 2 sigma bonds (one with CH3 and one with C2).
- There are no lone pairs.
- The steric number = Number of sigma bonds + Lone pairs = 2 + 0 = 2.
- Therefore, C1 is **sp hybridized**.

### Step 4: Determine the Hybridization of Carbon 3
For Carbon 3 (C3):
- It forms 4 sigma bonds (two with C2 and C4, and one with the double bond).
- There are no lone pairs.
- The steric number = 4.
- Therefore, C3 is **sp³ hybridized**.

### Step 5: Determine the Hybridization of Carbon 5
For Carbon 5 (C5):
- It forms 3 sigma bonds (one with C4 and two with the triple bond).
- There are no lone pairs.
- The steric number = 3.
- Therefore, C5 is **sp² hybridized**.

### Final Answer
The hybridization states of carbons 1, 3, and 5 are:
- Carbon 1: **sp**
- Carbon 3: **sp³**
- Carbon 5: **sp²**

Thus, the sequence of hybridization is **sp, sp³, sp²**.