In the esterification of propanoic acid with methanol in the presence
of a mineral acid which of the following are intermediate species?

Options

A. `C_(2)H_(5)-overset(o+)overset(OH)overset(||)C-OH`
B. `C_(2)H_(5)-underset(o+)overset(OH)overset(|)C-OH`
C. `C_(2)H_(5)-underset(OCH_(3))underset(|)overset(OH)overset(|)C-OH`
D. `C_(2)H_(5)-underset(o+)overset(OH)overset(|)C-OCH_(3)`

Answer

To solve the question regarding the intermediate species formed during the esterification of propanoic acid with methanol in the presence of a mineral acid, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants involved in the reaction are propanoic acid (C2H5COOH) and methanol (CH3OH).

2. **Protonation of the Carboxylic Acid**:
- In the presence of a mineral acid (H+), the carbonyl oxygen of the propanoic acid gets protonated, forming a protonated carboxylic acid. This step increases the electrophilicity of the carbonyl carbon.

3. **Formation of the Tetrahedral Intermediate**:
- The nucleophilic oxygen of methanol attacks the electrophilic carbon of the protonated carboxylic acid. This results in the formation of a tetrahedral intermediate, which has both the methoxy group (OCH3) and the hydroxyl group (OH) attached to the carbon.

4. **Loss of Water**:
- The tetrahedral intermediate can then lose a water molecule (H2O), leading to the formation of a carbocation intermediate.

5. **Proton Transfer**:
- A proton transfer occurs, stabilizing the carbocation and resulting in the formation of the final ester product (propyl methanoate or propanoate).

6. **Final Product Formation**:
- After the proton transfer and loss of a proton, the final product, which is the ester, is formed.

### Conclusion:
The intermediates formed during the reaction include:
- Protonated carboxylic acid
- Tetrahedral intermediate
- Carbocation intermediate

Thus, the answer to the question is that the intermediate species are all the mentioned forms.