`CH_(3)-overset(O)overset(||)(C)-overset(16)(O)-H+R-overset(18)(O)-Hoverset("dil." H_(2)SO_(4))underset(Delta)to`
Major product.
Product of the reaction is

Options

A. `CH_(3)-overset(O)overset(||)(C)-overset(18)(O)-R`
B. `CH_(3)-overset(O)overset(||)(C)-overset(16)(O)-R`
C. `R-overset(O)overset(||)(C)-overset(18)(O)-CH_(3)`
D. `R-underset(O)underset(||)(C)-underset(16)(O)-R`

Answer

To solve the given question, we need to understand the reaction taking place between the carboxylic acid and the alcohol, which is an esterification reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Reactants
We have a carboxylic acid and an alcohol. The carboxylic acid is represented as CH₃C(=O)OH (where the carbon with the double bond to oxygen is the carbonyl carbon), and the alcohol is represented as R-OH.

### Step 2: Understand the Esterification Reaction
In an esterification reaction, a carboxylic acid reacts with an alcohol to form an ester and water. The general reaction can be represented as:
\[ \text{Carboxylic Acid} + \text{Alcohol} \rightarrow \text{Ester} + \text{Water} \]

### Step 3: Remove Water
During the reaction, a water molecule is eliminated. The hydroxyl group (OH) from the carboxylic acid and a hydrogen atom (H) from the alcohol combine to form water (H₂O).

### Step 4: Form the Ester
The remaining parts of the reactants combine to form the ester. The carbonyl carbon from the acid will bond with the oxygen from the alcohol, leading to the formation of the ester.

### Step 5: Identify the Major Product
In this specific case, we have:
- The carboxylic acid part: CH₃C(=O)OH
- The alcohol part: R-OH

After the reaction, the ester formed will be:
\[ \text{Ester} = CH₃C(=O)O-R \]

### Step 6: Consider the Isotopes
The problem mentions specific isotopes (O-16 and O-18). The O-16 from the acid will be removed during the formation of water, while O-18 from the alcohol will be retained in the product. Therefore, the ester will contain O-18.

### Final Product
The final product of the reaction will be:
\[ \text{Ester} = CH₃C(=O)O-^{18}R \]

### Conclusion
Thus, the major product formed from the reaction is:
\[ \text{CH}_3C(=O)O-^{18}R \]