Allahyar ( Shahzaib ) Shar QUEST Nawabshah Chemical Engineering

1. MECHANISM FOR REACTION FOR ACID
catalyzed ESTERIFICATION
Conversion of carboxylic acids to esters
using acid and alcohols (Fischer
Esterification)
Allahyar ( Shahzaib ) Shar
Chemical Engineering
Quest, Nawab Shah
QUEST N.Shah

2. Mechanism:
• Mechanism:
• For such a seemingly simple reaction (replacement of OH
by OR) there are actually a lot of steps.
• Protonation of the carbonyl oxygen by acid (Step 1, arrows
A and B) makes the carbonyl carbon a much better
electrophile.
• It undergoes 1,2-addition by the alcohol (Step 2, arrows C
and D) whereupon the proton from the alcohol is
transferred to one of the OH groups (Step 3, arrows E and
F). Subsequent 1,2-elimination of water (Step 4, arrows G
and H) leads to the protonated ester, and the ester is then
deprotonated (Step 5, arrows I and J).

3. Mechanism
• Description: When a carboxylic acid is treated
with an alcohol and an acid catalyst, an ester
is formed (along with water). This reaction is
called the Fischer esterification.

4. Mechanism
• The reaction is actually an equilibrium. The
alcohol is generally used as solvent so is
present in large excess. Many different acids
can be used; it’s common to see just “H+”,
although H2SO4 (sulfuric acid) and TsOH (tosic
acid) are also often used.

6. Step 1.An acid/base reaction. Protonation of
the carbonyl makes it more electrophilic.
Step2.The alcohol 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
alcoholic oxygen.
Step 4. An acid/base reaction. Need to make an -
OH leave, it doesn't matter which one, so
convert it into a good leaving group by
protonation.
Step 5.Use the electrons of an adjacent oxygen
to help "push out" the leaving group, a neutral
water molecule.
Step 6. An acid/base reaction. Deprotonation
of the oxonium ion reveals the carbonyl in the
ester product.