In the following reaction, trace the position of isotopic `O^(18)`
`CH_(3)-overset(O)overset(||)C-O-H+CH_(3)CH_(2)-overset(18)O-HhArrCH_(3)-overset(O)overset(||)C-O-CH_(2)CH_(3)`
`+H_(2)O`

Answer

To trace the position of the isotopic \( O^{18} \) in the given reaction, we will analyze the reaction step-by-step, focusing on how the isotopic oxygen is incorporated into the product.

### Step-by-Step Solution:

1. **Identify the Reactants:**
The reactants are:
- Acetic acid: \( CH_3C(=O)OH \)
- Ethanol with \( O^{18} \): \( CH_3CH_2O^{18}H \)

2. **Write the Reaction Mechanism:**
The reaction involves the formation of an ester from the acid and alcohol. The mechanism can be outlined as follows:
- The acetic acid \( CH_3C(=O)OH \) donates a proton (H+) to form a more reactive species.
- The hydroxyl group (OH) of acetic acid becomes a better leaving group.

3. **Formation of the Reactive Intermediate:**
- The acetic acid forms a protonated form \( CH_3C(=O)OH_2^+ \).
- This intermediate is more susceptible to nucleophilic attack.

4. **Nucleophilic Attack by Ethanol:**
- The \( O^{18} \) from ethanol \( CH_3CH_2O^{18}H \) acts as a nucleophile and attacks the carbonyl carbon of the acetic acid.
- This results in the formation of a tetrahedral intermediate where the \( O^{18} \) is now attached to the carbon that was originally part of the carbonyl group.

5. **Formation of the Ester:**
- The tetrahedral intermediate collapses, leading to the elimination of water and the formation of the ester.
- The final product is \( CH_3C(=O)OCH_2CH_3 \) with \( O^{18} \) incorporated into the ester.

6. **Identify the Position of \( O^{18} \):**
- In the final ester product \( CH_3C(=O)OCH_2CH_3 \), the \( O^{18} \) is located in the ether part of the ester, specifically in the \( OCH_2CH_3 \) portion.

### Conclusion:
The isotopic \( O^{18} \) is present in the ester formed from the reaction, specifically in the ether linkage of the product.

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