Chemistry 211
Fall 2013
Equilibrium Controlled Reactions:
Acid Base Reactions and Electron Energies – 7
Out of Class Applications
ANALYSIS
A.
Reading: CGW pp. 192-194 Get a Feel for pKas & 196 Electron-donating groups decrease acidity
B.
Activities:
1.
The compounds below have pKas ranging from 3.7 to 19.2. Draw the structures of the conjugate bases of these acids and provide warrants
necessary to account for the differences in acidity. The warrants should be based on structural effects we have discussed in Acid-Base 1-7.
CH3
O
O
CH
3
H
C
O
H
pKa
O
H
C
O
CH3
4.8
3.7
15.5
O
H
H
CH3
O
formic acid
pKa 3.7
CH3
O
acetic acid
pKa 4.8
H
O
H
O
methanol
pKa 15.5
O
ethanol
pKa 16.0
H
CH
CH3
O
17.1
H
O
H
isopropanol
pKa 17.1
H
C
CH3
O
CH3
19.2
O
H
tert-butanol
pKa 19.2
O
O
O
H
H
CH2
16.0
O
H
CH3
O
CH3
O
O
O
O
As always, we invoke the Relative Effect Assumption to remind us that the pKas of acids are determined by the relative energies of
their conjugate bases. So we will consider the structural effect on the relative energies of the conjugate bases to help us understand the
differences in pKas of these acids.
All of the conjugate bases have their HEE on a negative oxygen atom. But this group of acids presents two structural effects: The red
divider separates the conjugate bases (on the left) that have significant delocalization effects on their HEE from those (on the right)
that have no delocalization of their HEE. The large jump in pKa from the left-hand group to the right-hand group is due to the
relatively large decrease in energy of the left-hand conjugate bases due to delocalization on their HEE over two oxygen atoms.
Delocalization of the HEE is illustrated by the resonance structures below:
Acid-Base Rxns & Electron Energy-7
2
H
H
O
O
O
O
O
CH3
O
CH3
O
formate ion
O
acetate ion
Within the two groups, the variable is the number of methyl groups that replace a hydrogen atom of the simplest structure (formic acid
or methanol). As the number of hydrogen -> methyl group replacements increases the pKas of the acids increase indicating that the
energy of the conjugate bases increase. This effect can be understood if we assume that although carbon and hydrogen atoms have
very similar electronegativities, a methyl group has more electrons than does a hydrogen atom and thus can release (or donate) more
electron density to its bonded atoms than can a hydrogen atom. So methyl groups exert an electron donating inductive effect on
attached atoms. As with the electron withdrawing inductive effects (with Cl & F) we saw in Acid-Base-7, the methyl vs. H change
polarizes bonds to attached atoms; but this polarization is in the opposite direction, toward HEE. (See Figure 4 in the Summary of
Class Discussion of Acid-Base-7) This polarization is transmitted through the -bonded system and will ultimately increase the
electron density near the HEE, causing increased e--e- repulsion raising the energy of the conjugate base and therefore the pKa of the
acid.
O
O
pKaH 15.5
O
O
pKaH 17.1
pKaH 16.0
pKaH 19.2
Note that this effect is very small compared with other effects we have investigated in Acid-base 3-6.
2.
Indicate the site of the highest energy electrons on each ion. Then use the theory that you developed in class to predict the relative energies
(lowest to highest) of the highest energy electrons of following anions.
CH3
O
CH3NH
F
O
O
O
Explain how your predictions were made.
In each structure, the HEE seem to be on the negatively charge oxygen atom; however they are actually delocalized into the ring.
O
O
O


O
O

Delocalization spreads the excess electron density over the ring; however more of it is located on the oxygen and the carbons in the
2, 4 & 6 positions of the ring. Although the negative environment is delocalized, increasing or decreasing the excess electron
density in the ring will change the extent of e-- e- repulsion and raise or lower the energy. So, electron withdrawing inductive
Acid-Base Rxns & Electron Energy-7
3
effects (successive polarization of sigma bonds) of F and N lower the energy of the HEE in the -system compared with the
unsubstituted ion by decreasing e-- e- repulsion in the ring (site of the highest energy e-'s), while the e- donating inductive effect of
CH3 raises the energy of the ion by increasing the e--e- repulsion in the ring. F has a stronger effect than N it is more
electronegative (higher effective nuclear charge) than N. So the order of energies should be:
CH3NH
CH3
>
O
>
O
O
>
F
Highest energy
Lowest energy
3.
For each of the following compounds, circle the most acidic proton or group of protons and explain your choice on the basis of an appropriate
theory.
CH3
N
O
CH3
CH3
CH3
N
CH3
CH
CH3
CH3
C
CH3
CH3
CH3
CH2
CH3
O
F
H
H
O
H
O
CH3
CH3
H
O
H
H
O
O
C
CH3
O
CH3
H
H
CH3
H
O
H
O
All protons are attached to neutral sp3 carbon atoms. Loss of one of the
circled protons would create a lone pair of e-'s on a carbon adjacent to
an oxygen atom. Since O is the highest effective nuclear charge atom in
the molecule, it causes the largest decrease in e- - e- repulsion effect of the
new lone pair e-'s by its inductive e- withdrawing effect. Consequently,
loss of one of the circled protons should produce the lowest energy
conjugate base possible. So, the circled protons should be the most acidic
group of protons in the molecule.
CH3
CH
CH3
F
O
N
O
H
H
O
H
O
O
CH3
H
C
CH3
F
CH3
CH
CH3
CH3
O
There are two protons on oxygen atoms in the molecule. Loss of these
protons should produce lower energy lone pair e-'s due to the higher
nuclear charge of O vs. C (higher e- - nuclear attraction.) However, of
the two protons attached to O, the circled proton is few -bonds away
from the fluorine atom than is the other. Since the only difference in the
environments of the protons on oxygen atoms is their bonded distance
from the F, the loss of the one that is closer to the F will yield the lower
energy conjugate base. The inductive effect of the F will decrease the edensity of the HEE on the conjugate base more since inductive effects
decrease with increased bonded distance between the HEE and the
electronegative atom. So the circled proton is the most acidic because its
Acid-Base Rxns & Electron Energy-7
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loss yields the lowest energy conjugate base.
Acid-Base Rxns & Electron Energy-7
4.
5
Predict which of the following reactions should produce more product at equilibrium. Explain your choice on the basis of an appropriate
theory.
O
O
CH2
C CH
3
-
C
+ C
Cl
Cl
-O
CH2
Cl
C
C CH C
3
Cl
O
H
-O
O
CH2
O
C CH
3
H
CH2
C
C
H
C
C CH
3
C
O
H
The difference between the two reactions is the replacement of two hydrogen atoms in the second reaction with two chlorine atoms in
the first reaction. The structural change is adjacent to the HEE in the reactants, the lone pair of electrons on the negative carbon atom
and three atoms removed from the HEE in the products, a lone pair of e-'s on the negative oxygen atom. So, the largest effect should
occur in the HEE of the reactants. The inductive effect of the extra chlorine atoms should decrease the electron density around the
carbon holding the HEE. The decrease in electron density decreases e--e- repulsion, lowering the HEE energy in the reactants of the
first reaction relative to that in the reactants of the second reaction. Thus, the reactants of the second reaction should have higher
energy than those in the first reaction. Since the energy differences in the products should be smaller because inductive effect should be
smaller because of the increased distance between the HEE and the electronegative atoms than in the reactants. So the higher energy
of the reactants of the second reaction should lead to a more negative change in free energy for that reaction than the first reaction. So
the second reaction should produce more products at equilibrium.
D. Nomenclature of Alcohols and Aromatic Compounds
3. Assignment
a. Name the following:
References:
Acid-Base Rxns & Electron Energy-7
b. Draw structural formulas for the following compounds:
6