Hydrolysis of Esters

Reaction type: Nucleophilic Acyl
Substitution
Summary
- Carboxylic esters hydrolyse to the parent carboxylic acid and
an alcohol.
- Reagents : aqueous acid (e.g.
H2SO4) / heat,or aqueous NaOH / heat (known
as "saponification").
- These mechanisms are among some of the most studied in organic
chemistry.
- Both are based on the formation of a tetrahedral intermediate
which then dissociates.
- In both cases it is the C-O bond between the
acyl group and the oxygen that is cleaved.
Reaction under BASIC conditions:
- The mechanism shown below leads to acyl-oxygen
cleavage (see step2).
- The mechanism is supported by experiments using 18O
labeled compounds and esters of chiral alcohols.
- This reaction is known as
"saponification" because it is the basis
of making soap from glycerol triesters in fats.
- The mechanism is an example of the reactive system type.
MECHANISM OF THE BASE HYDROLYSIS OF ESTERS
|
Step 1:
The hydroxide nucleophiles attacks at the electrophilic
Cofthe ester C=O, breaking the π
bond and creating thetetrahedral
intermediate. |
 |
Step 2:
The intermediate collapses, reforming the
C=O
results in the loss of the leaving group the alkoxide,
RO-, leading to the carboxylic
acid. |
Step 3:
An acid / base reaction. A very rapid equilibrium where the
alkoxide,RO- functions as a base
deprotonating the carboxylic acid,
RCO2H, (an acidic work up would allow
the carboxylic acid to be obtained from the reaction). |
|
Reaction under ACIDIC conditions:
- Note that the acid catalysed mechanism is the reverse of the
Fischer esterification.
- The mechanism shown below also leads to acyl-oxygen
cleavage (see step 5).
- The mechanism is an example of the less reactive system
type.
MECHANISM OF THE ACID CATALYSED HYDROLYSIS OF
ESTERS
|
Step 1:
An acid/base reaction. Since we only have a weak nucleophile and a
poor electrophile we need to activate the ester. Protonation of the
ester carbonyl makes it more electrophilic. |
 |
Step 2:
The water O functions as the nucleophile attacking
the electrophilic C in the C=O,
with the electrons moving towards the oxonium ion, creating the
tetrahedral intermediate. |
Step 3:
An acid/base reaction. Deprotonate the oxygen that came from the
water molecule to neutralise the charge. |
Step 4:
An acid/base reaction. Need to make the
-OCH3 leave, but need to convert it
into a good leaving group first by protonation. |
Step 5:
Use the electrons of an adjacent oxygen to help "push out" the
leaving group, a neutral methanol molecule. |
Step 6:
An acid/base reaction. Deprotonation of the oxonium ion reveals the
carbonyl C=O in the carboxylic acid product and
regenerates the acid catalyst. |
Hope you get the answer properly and you can get all the detail
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refferences -
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch20/ch20-3-3-1.html