Question

What is true regarding the following acid catalyzed esterification (Fischer) reaction?
$C{{H}_{3}}C{{H}_{2}}COOH+CH3CH2OH\xrightarrow{{{H}^{+}}}C{{H}_{3}}C{{H}_{2}}COO{{C}_{2}}{{H}_{5}}.$
(A) The reaction involves production of $s{{p}^{3}}$ oxygen of acid in the first step.
(B) The reaction involves protonation on $s{{p}^{2}}$ oxygen of acid followed by formation of acylium ion $\left( RO{{C}^{+}} \right)$ in the first step.
(C) The slowest step of reaction is the nucleophilic attack of alcohol on $s{{p}^{2}}$ carbon of acid.
(D) Alcohol undergoes protonation in the first step.

Answer 1

Hint : We know that when carboxylic acid and alcohol are refluxed in the presence of an acid catalyst is called Fischer esterification. The acid catalyst acts as a dehydrating agent and increases the rate of the reaction.

Complete Step By Step Answer:
In the name of an ester is obtained from carboxylic acid, which takes part in the esterification reaction. An example of nucleophilic acyl substitution reaction is Fischer esterification. It is based on the electrophilicity of the carbonyl carbon and the nucleophilicity of the alcohol. As electrophiles, the reactivity of carboxylic acid is less than the reactivity of esters. They have the tendency to deprotonate anions in dilute neutral solutions. Pure esters have a tendency to instinctively hydrolyze in water even though they are kinetically slow in the absence of catalysts. High yields of products aren’t favorable for Fischer esterification.
1. Transfer of proton from acid catalyst to the carbonyl oxygen raises the electrophilicity of carbonyl carbon.
2. Nucleophilic oxygen atom of the alcohol attacks the carbonyl carbon.
3. An activated complex is formed, when the proton is transferred from oxonium ion to a second molecule.
4. A new oxonium ion is formed by the protonation of one of the hydroxyl groups of the activated complex.
5. A molecule of water is lost from oxonium ion and deprotonation produces ester.
Esterification process also occurs between alcohol and acid chloride. A type of esterification, when carboxylic acid and an alcohol are refluxed in the presence of an acid catalyst is called Fischer esterification.
A. Not true, in the first step reaction occurs on $s{{p}^{2}}$ oxygen of acid.
B. Not true, after protonation of $s{{p}^{2}}$ oxygen, the next $RO+H$ is formed.
C. True, nucleophilic attack on $s{{p}^{2}}$ carbon is the slowest step.
D. Not true, the carboxyl group gets protonation in the first step.
Therefore, the correct answer is option A and C.

Note :
Remember that we must know that several of the carboxylic acids are appropriate for Fischer esterification reaction, but alcohol used must be either primary (or) secondary. Usage of tertiary alcohols is liable to elimination. Some of commonly used catalyst in Fischer esterification process