`-CONH_(2) overset("Reduction")to -CH_(2)NH_(2)`
In above reaction hybridisation state of carbon changes from

Options

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

Answer

To solve the problem of determining the change in hybridization state of carbon during the reduction of -CONH2 to -CH2NH2, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Compound**:
The initial compound is -CONH2, which is an amide. The carbon in this structure is bonded to a carbonyl group (C=O) and an amine group (NH2).

2. **Determine the Hybridization of Carbon in -CONH2**:
- The carbon in -CONH2 is bonded to one oxygen atom via a double bond (C=O) and to one nitrogen atom via a single bond (C-N).
- This carbon is also bonded to another hydrogen atom.
- Therefore, it has a total of three regions of electron density (one double bond and two single bonds).
- According to VSEPR theory, three regions of electron density correspond to **sp² hybridization**.

3. **Identify the Final Compound**:
The final compound after reduction is -CH2NH2, which is an amine.

4. **Determine the Hybridization of Carbon in -CH2NH2**:
- In -CH2NH2, the carbon is bonded to two hydrogen atoms and one nitrogen atom.
- This carbon has four regions of electron density (four single bonds).
- Therefore, it corresponds to **sp³ hybridization**.

5. **Conclusion on Hybridization Change**:
- The hybridization of carbon changes from **sp²** in -CONH2 to **sp³** in -CH2NH2.
- Thus, the change in hybridization state of carbon during the reduction process is from sp² to sp³.

### Final Answer:
The hybridization state of carbon changes from **sp² to sp³**.

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