
Which ester given below can most easily be produced by acid catalysed esterification (Fischer’s esterification)?
A. \[C{{H}_{3}}C{{H}_{2}}COO{{C}_{6}}{{H}_{5}}\]
B. \[C{{H}_{3}}C{{H}_{2}}COOCH(C{{H}_{3}})C{{H}_{2}}C{{H}_{3}}\]
C. \[C{{H}_{3}}C{{H}_{2}}COOC{{(C{{H}_{3}})}_{2}}C{{H}_{2}}C{{H}_{3}}\]
D. \[C{{H}_{3}}C{{H}_{2}}COO{{C}_{2}}{{H}_{5}}\]
Answer
163.5k+ views
Hint: Fischer esterification, also known as Fischer-Speier esterification, is a specific type of esterification that is accomplished by refluxing an alcohol and a carboxylic acid in the presence of an acid catalyst. When an alcohol reacts with a carboxylic acid, it results in esterification.
Complete Step by Step Answer:
Fischer esterification mechanism:
In general, a slow reaction is carried out in reflux using strong acids like sulfuric or phosphoric acids. Both phases in the Fischer esterification process are reversible. A Lewis acid or a proton results in a more reactive electrophile (for example, p-TsOH or\[{{H}_{2}}S{{O}_{4}}\]). Two identical hydroxyl groups are present in a tetrahedral intermediate that is produced by the alcohol nucleophilic attack. One of these hydroxyl groups is eliminated following a proton shift (tautomerism) to produce water and the ester.
The acid catalyst initially protonates the carbonyl oxygen, and ethanol then activates it in the direction of a nucleophilic attack. The alcohol then launches a nucleophilic attack against the carbonyl. The alcohol's pi link with the other oxygen is broken by the bond with the carbonyl carbon, which creates a lone pair of electrons from the alcohol's oxygen atom. The oxygen's positive charge is neutralised by the pi bond electrons as they pass up to it. This produces an oxonium ion.
All those esters which contain primary alcohol fraction can mostly be easily synthesised by Fischer’s esterification.
Hence, the correct option is D. \[C{{H}_{3}}C{{H}_{2}}COO{{C}_{2}}{{H}_{5}}\]
Note: The thing to note in the question is the mechanism of the Fischer synthesis. The presence of more reactive electrophile fastens the reaction. Esters with primary alcohols are easily synthesised as compared to the secondary and tertiary alcohols.
Complete Step by Step Answer:
Fischer esterification mechanism:
In general, a slow reaction is carried out in reflux using strong acids like sulfuric or phosphoric acids. Both phases in the Fischer esterification process are reversible. A Lewis acid or a proton results in a more reactive electrophile (for example, p-TsOH or\[{{H}_{2}}S{{O}_{4}}\]). Two identical hydroxyl groups are present in a tetrahedral intermediate that is produced by the alcohol nucleophilic attack. One of these hydroxyl groups is eliminated following a proton shift (tautomerism) to produce water and the ester.
The acid catalyst initially protonates the carbonyl oxygen, and ethanol then activates it in the direction of a nucleophilic attack. The alcohol then launches a nucleophilic attack against the carbonyl. The alcohol's pi link with the other oxygen is broken by the bond with the carbonyl carbon, which creates a lone pair of electrons from the alcohol's oxygen atom. The oxygen's positive charge is neutralised by the pi bond electrons as they pass up to it. This produces an oxonium ion.
All those esters which contain primary alcohol fraction can mostly be easily synthesised by Fischer’s esterification.
Hence, the correct option is D. \[C{{H}_{3}}C{{H}_{2}}COO{{C}_{2}}{{H}_{5}}\]
Note: The thing to note in the question is the mechanism of the Fischer synthesis. The presence of more reactive electrophile fastens the reaction. Esters with primary alcohols are easily synthesised as compared to the secondary and tertiary alcohols.
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