CHEM 109A
CLAS
Acids and Bases: Identification and Reactions - KEY
1. Predict the products of the following reactions. Label the acids, bases (Lewis or
Bronsted-Lowry) and conjugates (if applicable). Determine the side that is
favored at equilibrium for the reactions between Bronsted-Lowry acids and
bases).
H
O
H
a.
N
H
H
H
H
O
O
H
N+
N
H
H
H
H
B-L A
B-L B
~15 wkr A
pKa
H
-
CB
CA
~9
Keq = 10^(9-15) = 10-6
or Keq = Ka(CH3OH) < 1 (pKa prot amine < pKa alcohol) so R
Ka (NH4+)
favored
H
F
b.
N
B
F
H
F
H
F
H
B
N
H
H
F
F
F
Lewis A
Lewis B
F
H
B-
N+
F
H
H
Lewis A-B complex
H
HO
N
c.
O
H
H
H
H
O
O
N
O
pKa
B-L A
~5
B-L B
N+
H
O
H
H
-
CB
CA
~9 wkr A
Keq = 10^(9-5) = 10+4
Or Keq = Ka(CH3COOH) > 1 (pKa prot amine > pKa carbox. acid)
Ka(N+H3CH3) so P favored
Cl
O
d.
H
Al
Cl
Cl
Cl
Cl
O
H
Cl
Lewis B
O+
Al
Cl
Lewis A
H
Al-
Cl
Cl
Lewis A-B complex
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CHEM 109A
CLAS
Acids and Bases: Identification and Reactions - KEY
H
H
e.
N+
O-
H
H
H
H
N+
O-
H
H
H
H
H
N
O
H
B-L B
B-L A
~10
pKa
CA
15.7~16 wkr A
CB
Keq = 10^(16-10) = 10+6
Or Keq = Ka(N+H3CH3) > 1 (pKa prot amine < pKa water)
Ka(H2O)
so P favored
2. Put the following in order of relative acidity and list the major effect(s)
a. H2S, HCl, H2O, NH3, HBr
H
S
H
O
Cl
H
H
N
H
H
H
H
H
Br
Start by grouping according to atom attached to the proton and look for
element effects
H
O
N
H
H
H
H
S
H
H
H
<
N
H
H
<
O
H
H
S
H
H
H
Cl
<
H
Cl
H
Br
O
element effect EN
element effect size
Cl
element effect EN
Br
element effect size
H
H
S
H
<
H
H
H
H
H
confirm w/ pKas
Br
>
-9
H
-7
Cl
S
>H
H
7
O
H
>
H
15.7
N
H
H
36
F
Cl
HO
>
HO
HO
HO
F
b.
H is attached to an O in each compound, so NO element effects
No resonance structures possible, but there is an EN atom attached to
some of the compounds, let’s look at the inductive effect…
F
Cl
HO
<
HO
<
HO
HO
F
Page 2 of 6
CHEM 109A
CLAS
Acids and Bases: Identification and Reactions - KEY
F is most EN element, so will stabilized the CB the most, making the
compounds with F the more acidic
F
HO
<
HO
F
Compound w/ F closest to the acidic proton will be the more acidic b/c can
better stabilize the O- of the CB.
F
Cl
HO
<
<
HO
<
HO
HO
F
c.
H is attached to a C in each compound, so NO element effects
No resonance structures possible and no EN atoms. Only effect left is
hybridization and these compounds have single, double and triple bonds!
H
If H attached to C-C single bond was removed
have lone pair in an sp3 orbital
H
H
, the anion would
H
If H attached to C-C double bond was removed
have lone pair in an sp2 orbital
, the anion would
H
H
If H attached to C-C triple bond was removed
, the anion would
have lone pair in an sp orbital
The anion with the lone pair in an sp orbital would be the most stable since the
e-s would be held closer to the nucleus due to the large s-character of the
hybrid orbital, making it the most stable CB and the acid with the triple bond
the strongest.
H
H
>
H
>
H
H
H
3. Put the following in order of relative stability
Br
O-
H
O-
O
O
O
O
HO
O-
Cl
O-
In each compound there is a COO- group, so while there is resonance
stabilizing these anions, it will not allow us to distinguish between them.
There is an EN atom/group attached to some of the compounds, let’s look
at the inductive effect… EN groups inductively stabilize the anion, so the
Page 3 of 6
CHEM 109A
CLAS
Acids and Bases: Identification and Reactions - KEY
more EN the group, the more stabilizing (and the closer to the negative
charge).
O
O
H
OO-
HO
EN of H 2.1
O
EN of O 3.5
EN of Cl 3.0
O-
Cl
O
EN of Br 2.8
all EN groups are the same
Br
O
distance from the COO-, so the only difference is in the EN.
-
>
>
HO
O-
Cl
O
O
O
O
O
-
>
Br
O-
O-
H
4. Determine the favored structure at equilibrium AND the relative ratio or
percentage for each of the following
a. CH3COOH at pH = 6.75
pKa = 4.75, pH > pKa so A- favored by 102 (b/c pH – pKa = 2)
A-:HA is 100:1 or 99%:1%
b. ClCH2COOH at pH 6.76
pKa = 2.85, pH > pKa so A- favored by 104 (b/c pH – pKa = 3.9 = 4)
A-:HA is 10000:1 or 99.99%:0.01%
c. CH3NH3+ at pH = 7
pKa = 10, pH < pKa so HA favored by 103 (b/c pH – pKa = 3)
HA:A- is 1000:1 or 99.9%:0.1%
d. HCN at pH = 9.1
pKa = 9.1, pH = pKa so [HA]=[A-], 1:1 or 50% ea.
e. H2S at pH = 9.1
pKa = 7.04, pH > pKa so A- favored by 102 (b/c pH – pKa = 2)
A-:HA is 100:1 or 99%:1%
Additional Information:
Definitions of A/Bs
Bronsted-Lowry: H+ donor (any species w/ a proton)/H+ acceptor (any species
w/ lone pair of e- (i.e., O, N, anion))
Lewis: e- pair acceptor (species w/ unfilled orbitals, usu a metal atom)/ e- pair
donor (species w/ lone pair of e-, usu a nonmetal/halogen) – use this
definition more later in CHEM 109B.
A/B strength
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CHEM 109A
CLAS
Acids and Bases: Identification and Reactions - KEY
From Gen Chem: 7 strong A, all others weak….
Acid strength
↑
pH
Ka
pKa
↓ ↑ ↓
vry strng A <1
med strng A 1-3
wk A 3-5
vry wk A 5-15
extrm wk A >15
Factors affecting A strength*
Element Effects – when the atom that the proton is bound to is changing
1. Size – when atoms are related vertically in the PT
As the size of the atom bound to the proton inc, the length of the atom-H
bond inc making it weaker and easier to break and therefore a stronger A.
2. Electronegativity – when atoms are related horizontally in the PT
As the EN of the atom bound to the proton inc, the e- being shared
between the atom and the proton spend more time around the atom and
make the atom-H bond weaker and easier to break and therefore a stronger
A.
Structural Effects – have to do with the structure of the compound
3. Inductive effect – occurs when there is an EN element somewhere on the
compound (NOT bound to the acidic proton)
The EN element withdraws electrons from adjacent atoms and creates a
Coulombic interaction that stabilizes the charge on the CB, making the A
stronger.
Must take into account the EN of the element (more EN substituent, more
stabilization of CB, stronger the A) AND its proximity to the acidic proton
(closer EN substituent is to the negative charge of CB, more stabilizing,
stronger A).
4. Resonance effect – occurs when you can draw resonance structures for the CB
If the CB of the A is stabilized by resonance, it makes the A stronger.
5. Hybridization – when atom bound to the proton has different hybridization
The CB of the A is more stable when a lone pair of electrons is in a hybrid
orbital that has more s-character (sp3 < sp2 < sp) b/c the e-s are closer to
the nucleus and therefore held tighter and therefore the stronger the A.
* Using the relationships: Stronger the A, weaker the CB (or stronger the B,
weaker the CA) and vice versa the relative B strength can also be determined
using the above factors.
Table of pKas in Appendix II of text or my website
Aue’s prescription for solving A/B problems
Calculate the equilibrium constant and Gibb’s free energy for the reaction of acetic acid
and ammonia.
Page 5 of 6
CHEM 109A
CLAS
Acids and Bases: Identification and Reactions - KEY
1. Write a balanced chemical equation (You may ask you to use Kekule-Lewis
structures)
O
O
H
O-
O
NH 3
acetic acid
2.
3.
4.
5.
ammonia
acetate ion
H
NH 3+
ammonium ion
Label
A
B
CB
CA
Label pKas ~5
~9
ID relative strength: stronger A (smaller pKa)
wkr A (larger pKa)
Determine direction favored at equilibrium – “Meek shall inherit the Earth”,
weaker acid wins
Keq = 10±∆pKa
[if Ps favored sign on ∆pKa term should be +
6. Find Keq:
= Keq very large, if Rs favored sign on ∆pKa term should be - = Keq very small]
Keq = 10^(9-5) = 10+4
∆Go
= -1.4log(Keq) kcal/mol
7. Find ∆Go298K:
**
298K
o
+4
∆G 298K = -1.4log(Keq) = -1.4log(10 ) = -1.4 (+4) = -5.6 = -6 kcal/mol
** Comes from ∆Go = -RTln(K) [from Gen Chem]
R = 0.00199 kcal/mol K, room temp = 298-300 K
∆Go = -RTln(K)
o
ln = 2.303 log
∆G 298K = -0.597ln(K)
o
∆G 298K = -1.4log(K)
Acid form (HA) or base form (A-) in solution?
Depends on pH of solution and pKa of the acid:
pKa = pH + log([HA]/[A-]) better known as…
pH = pKa + log([A-]/[HA]) Henderson-Hasselbach [from Gen chem]
[HA or A-] = 10pH-pKa
pH < pKa: HA form
pH > pKa: A- form
pH = pKa: [HA] = [A-], 50:50
Page 6 of 6