In the formation of an ester, which bond of a carboxylic acid is broken, \(\mathrm{C}-\mathrm{OH}\) or \(\mathrm{CO}-\mathrm{H}\) ?

Short Answer

Expert verified
In the formation of an ester, the bond that is broken in a carboxylic acid is the C-OH bond. This occurs during the esterification reaction between a carboxylic acid (R-CO-OH) and an alcohol (R'-OH), resulting in an ester (R-CO-O-R') and water.

Step by step solution

01

Write the general reaction for esterification

Esterification is the chemical reaction between a carboxylic acid and an alcohol, which results in the formation of an ester and water. The general reaction can be represented as follows: Carboxylic Acid + Alcohol ⟶ Ester + Water
02

Write the structural formulas for the reactants

In order to analyze which bond is broken during este formation, let's write the structural formulas of carboxylic acid (RCOOH) and alcohol (R'OH): Carboxylic Acid: R-CO-OH Alcohol: R'-OH Here, "R" and "R'" are alkyl groups which can vary in length and structure.
03

Analyze the esterification reaction and identify the broken bond

During the reaction, the hydrogen atom from the alcohol (R'-OH) and the hydroxyl group (OH) from the carboxylic acid (RCOOH) will combine to form water (H2O). This will result in the ester, formed from the remaining parts of carboxylic acid and alcohol: Ester: R-CO-O-R' The bond that is broken during ester formation is the C-OH bond of the carboxylic acid. The CO-H bond is not broken during this reaction. So, the answer to the question is: The C-OH bond of the carboxylic acid is broken during the formation of an ester.

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