Hybridisation of carbon atoms present in the smallest ester are:

Options

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

Answer

To determine the hybridization of carbon atoms present in the smallest ester, we can follow these steps:

### Step 1: Identify the Smallest Ester
The smallest ester is formed from the simplest carboxylic acid and alcohol. The simplest carboxylic acid is formic acid (HCOOH) and the simplest alcohol is methanol (CH3OH). The reaction between these two gives us methyl formate (HCOOCH3), which is the smallest ester.

### Step 2: Write the Structure of the Smallest Ester
The structure of methyl formate can be drawn as follows:
```
O
||
H-C-O-CH3
```
In this structure, there are two carbon atoms: one in the carbonyl group (C=O) and one in the alkoxy group (O-CH3).

### Step 3: Determine the Hybridization of Each Carbon Atom
1. **Carbon in the Carbonyl Group (C=O)**:
- This carbon is bonded to one oxygen atom via a double bond and one other carbon atom via a single bond.
- The hybridization for this carbon can be determined by the number of sigma bonds and lone pairs:
- 1 sigma bond (to the other carbon) + 1 pi bond (to the oxygen) = 1 sigma and 1 pi bond.
- Therefore, the hybridization is **sp²** (since it has three regions of electron density: one double bond and one single bond).

2. **Carbon in the Alkoxy Group (O-CH3)**:
- This carbon is bonded to three hydrogen atoms and one oxygen atom.
- The hybridization for this carbon is:
- 4 sigma bonds (to three hydrogens and one oxygen) = 4 sigma bonds.
- Therefore, the hybridization is **sp³**.

### Step 4: Conclusion
The hybridization of the carbon atoms present in the smallest ester (methyl formate) are:
- Carbon in the carbonyl group: **sp²**
- Carbon in the alkoxy group: **sp³**

### Final Answer
The hybridization of carbon atoms present in the smallest ester are sp² and sp³.

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