Acetaldehyde reacts with semicarbazide product will be

Options

A. `CH_3CH = NNH–CO–NH_2`
B. `CH_3CH = NCONHNH_2`
C. `CH_3CH = NHNH_2`
D. `CH_3-oversetunderset(||)(O)C-NH-CONH_2`

Answer

To determine the product formed when acetaldehyde reacts with semicarbazide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Acetaldehyde (CH3CHO) is an aldehyde with a methyl group.
- Semicarbazide (NH2NHCONH2) is a compound containing an amine and an amide group.

2. **Understand the Reaction Mechanism**:
- Acetaldehyde acts as an electrophile due to the carbonyl group (C=O), while semicarbazide acts as a nucleophile because of the lone pair on the nitrogen atom.

3. **Nucleophilic Attack**:
- The nitrogen atom in semicarbazide attacks the carbon atom of the carbonyl group in acetaldehyde. This forms a tetrahedral intermediate.

4. **Formation of the Intermediate**:
- The tetrahedral intermediate has a negatively charged oxygen atom and a positively charged nitrogen atom. The structure can be represented as:
\[ CH3C(OH)(NH2)NHCO \]

5. **Proton Transfer**:
- The negatively charged oxygen can abstract a proton (H+) from the solution, leading to the formation of a hydroxyl group (OH) and a positively charged nitrogen.

6. **Elimination of Water**:
- The intermediate can lose a water molecule (H2O) when the nitrogen atom forms a double bond with the carbon atom, resulting in the formation of a double bond between carbon and nitrogen.

7. **Final Product**:
- The final product is known as semicarbazone, specifically acetaldehyde semicarbazone. The structure can be represented as:
\[ CH3CH=N(CONH2) \]
- This can also be written as:
\[ CH3C(=N(CONH2))H \]

### Final Answer:
The product formed when acetaldehyde reacts with semicarbazide is acetaldehyde semicarbazone, which can be represented as:
\[ CH3CH=N(CONH2) \]