Chem 216 S11 Notes - Dr. Masato Koreeda
Topic: __Experiment 1____ page 1 of 2.
Date: May 3, 2011
Experiment 1: Synthesis of Acetamides from Aniline and Substituted Anilines
Many of the acetylated [CH3–C(=O)-] derivatives of aromatic amines (aka anilines) and phenols are
pharmacologically important compounds. Some of these exhibit distinct analgesic activity. Two of the
most representative examples are:
H
N
HO
CH3
O
O
HO
O
CH3
O
acetylsalicylic acid (Aspirin)
acetaminophen (Tylenol)
=======================================================================
The reaction to be carried out in this experiment is:
Acetylation of aniline
δH3C δ+ O
O
N H
O
acetic
anhydride
(electrophile)
H
N
CH3
H
CH3
HO
+
O
CH3
acetanilide
aniline (nucleophile)
O
acetic acid
Both aniline and acetic anhydride are somewhat viscous liquids. So, simply mixing them together does
not result in the efficient formation of acetanilide. Therefore, a solvent (in this case water) is used to
dissolve and evenly disperse two reactants in it.
O
Note:
R'
CH3
acetyl group
R
N
R"
R'
R
N+
R"
R'
O-
O
amide
R
N
CH3
O
acetamide
The amide N is usually not nucleophilic
because of a significant contribution of
this resonance form.
Reaction mechanism:
δH H3C δ+ O
H
N
O
H
H H3C
N
O
O
CH3
O
CH3
tetrahedral
(sp3)
H
N
H
O
intermediate
O
CH3
pKa ~ -5
B (including
H
N
O
+
CH3
O
acetanilide
CH3
O
O
CH3
O
)
Chem 216 S11 Notes - Dr. Masato Koreeda
Topic: __Experiment 1____ page 2 of 2.
Date: May 3, 2011
Additional comments on the reaction mechanism:
1. Aniline is a strong nucleophile (much stronger than water).
2. Acetic anhydride is a relatively unstable reagent, but does not react with water that easily.
3.
H
H
N
O
H3C
O
O
CH3
A direct substitution process at the C=O carbon does not take place.
No direct SN2 reaction at the C=O carbon is known. This is not feasible
on the basis of the orbital consideration.
Experimental procedure:
Aniline is not soluble in water; so 1 mol. equiv of conc. HCl (37% HCl by weight in water) is added in
order to dissolve the aniline in water.
aniline
add 1 mol equiv
of conc HCl
H2O
add 1 mol equiv
of acetic anhydride
PhNH3+Cl(all dissolved)
Erlenmeyer flask
Don't use a beaker!
H2 O
H2O
PhNH3+Cl-, acetic anhydride
(still all dissolved in H2O)
no reaction yet!
homogeneous solution
add 1.2 mol equiv of
NaCl, acetic acid and
CH3C(=O)O- dissolved
in H2O
oo
o o o oo o oo
o oo o o o o o
acetanilide as white
precipitates (collect by
suction filtration)
H3C
By the action of sodium acetate,
a small amount of free
aniline (PhNH2)
is regenrated.
Free aniline has virtually no water solubility.
But before aniline comes out of the H2O solution,
it quickly collides (i.e., reacts) with acetic
anhydride dissolved in H2O. As soon as the
acetamide product (acetanilide) is formed,
it will precipitate out of the H2O solution and
more PhNH3+ gets converted to free aniline....
O- Na+
O
(sodium acetate)
- weak base
H2O
Mostly PhNH3+
and NaCl, acetic anhydride, and
CH3C(=O)O(dissolved in H2O)
Questions:
(1) What would happen if the order of additions of acetic anhydride and sodium acetate is reversed?
(2) What would be the outcome if 1 mol equiv of NaOH is used instead of sodium acetate?
(3) What would you have to do in order to dissolve p-nitroaniline into water by adding conc. HCl? The
pKa of the conjugate acid of p-nitroaniline is 1.00 (see the note on pKa).