Concept:Through isotopic labeling experiments (using Oxygen-18), it is proven that during the Fischer esterification process, the carboxylic acid contributes the \(-\text{OH}\) group, while the alcohol contributes only the \(-\text{H}\) atom to form water.
Formula:$$ \text{RCO-OH} + \text{H-OR}' \longrightarrow \text{RCO-OR}' + \text{H}_2\text{O} $$
Solution:- The nucleophilic oxygen of the alcohol attacks the electrophilic carbonyl carbon of the carboxylic acid.
- To stabilize the resulting intermediate, the alcohol loses its proton (\( \text{H}^+ \)).
- Therefore, the bond that breaks in the alcohol molecule is exclusively the O-H bond (Oxygen and Hydrogen).
Why other options are incorrect:Breaking the C-O bond in alcohol would imply the alcohol provides the \(-\text{OH}\) group, which is factually incorrect for standard carboxylic acid esterification.
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