In the esterification of alcohol:
A. H from alcohol and OH from acid is removed as water.
B. OH from alcohol and H from acid is removed as water.
C. H is replaced by chlorine.
D. H is replaced by sodium metal.

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Hint:. As we know that esterification is the process in which formation of ester from carboxylic acid and alcohol in the presence of concentrated sulphuric acid takes place.

Complete step by step answer:
- As we know that esterification is the process in which reaction between alcohol and carboxylic acid takes place to form ester. We can see the reaction that take place as:
\[R-COOH\text{ }+\text{ }H-O-{R}'\rightleftharpoons \text{ }R-COO{R}'+\text{ }{{H}_{2}}O\text{ }\]

- We can see here that in this reaction elimination of water molecules takes place. It is found that this reaction can be acid catalysed or base catalysed.
- And the reactivity of the alcohols is found to be dependent on the C-OH bond stability, that is lower the bond strength higher will be the ability or tendency of the esterification reaction.
- Hence, we can conclude that the correct option is (a), that is in the esterification of alcohol H from alcohol and OH from acid is removed as water. So, the correct answer is “Option A”.

Note: - It is found that in esterification carbonyl oxygen of carboxylic acid is protonated. Due to this it increases the positive character of carbon atoms. And nucleophile alcohol attacks the carbonyl carbon.
- Then further intramolecular hydrogen transfer takes place followed by loss of water molecules. And in the last step deprotonation takes place that gives ester.