Acid/Base stuff
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Important preliminary questions before looking at specific acids and bases.
1. What is an acid? What is a base? What are the Bronsted and Lewis definitions of acids and bases?
Bronsted definitions (1924) - proton transfer is emphasized
acid: a proton donor (no reference to the solvent)
base: a proton acceptor (no reference to the solvent)
This is the way we
look at protic acids.
Lewis definitions (1924) - electron pair transfer is emphasized
acids: substances which accept a pair of electrons (= electrophiles in organic)
bases: substances which donate a pair of electrons (= nucleophiles in organic)
This is the way
we look at
organic reactions
2. Write an equation using water as the base with generic acid, H-A. Use curved arrows to show how the reaction
occurs between an acid and a base (water)? Always push electrons with your arrows! Use full-headed arrows for
two electron movement (in acid/base chemistry), and half-headed arrows for one electron movement (in free
radical reactions). Remember how you used arrows in resonance. Practice this skill at every opportunity.
This is the organic way of looking at reactions.
It can be qualitative (which side does the
equilibrium favor?) or quantitative (what is
the value of Keq?). Water is the reference base
in reactions with various acids. Arrow pushing
shows how the reactions work.
Ka
A
H
O
H
A
H
O
pKa
H
H
This is the freshman chemistry way of looking at acid/base
chemistry. Symbols are mainly used for quantitative numerical
calculations. We won't write H+ by itself. We will always
attach the proton to some pair of electrons.
[C2H3O2H]
Ka =
H
H
C2H3O2
[ C2H3O2 ] [ H+]
[C2H3O2H]
3. Write an equilibrium expression for the reaction of acid ionization in water.
Keq =
(A )(H3O )
(HA)(H2O)
4. How does Ka differ from Keq? What is the Ka and what does it tell us about an acid? What magnitude is Ka for a
strong acid? What magnitude is Ka for a weak acid?
We almost always use Ka instead of Keq in acidity tables. They only differ by the concentration of water,
which is essentially constant ( 55.6 M), so Ka is about 56 times bigger than Keq. An acid with a large Ka
value (Ka > 1) is called a strong acid (up to 1020). An acid with a small Ka value (Ka < 1) is called a weak
acid (as low as 10-50). Acid strengths span a remarkable range of 1070!
Ka = Keq (H2O) =
(A )(H3O )
= #
(HA)
bigger number (>1) = strong acid (up to 1020)
smaller number (<1) = weak acid (low as 10-50)
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5. What is the pKa and what does it mean? (strong acids = ?)(weak acids = ?) What’s an order of magnitude?
pKa is another way to look at Ka It's a little confusing because it is the negative log of the Ka, (the negative of the tens
exponent). Every power of 10 is an order of magnitude (103 is an order of magnitude larger than 102 and 10-4 is two
orders of magnitude smaller than 10-2). (pKa = negative number for strong acids and pKa = positive number for weak acids)
pKa = - log (Ka) = pH - log [A ]
[HA]
Ka = 10-pKa
A very useful way to think of pKa is as (G)x(1.4) kcal/mole. Very approximately: pKa = (G)x(1.4) kcal/mole
G = -2.3RT (log Ka) = 2.3RT (-log Ka) = (1.4)(pKa) kcal/mole  pKa
R = 2 cal/mole-K
R = 8.3 joule/mole-K
G  pKa
Another useful way to think of pKa is that it is the pH when an acid is 50% ionized, [A
] = [HA]
assume T  300oK
6. Can you think of a better base that could be used in water (but similar looking)? What are the limits of basicity in
water? Using a less acidic solvent can allow for more basic environments. Some solvents are essentially
nonacidic and can tolerate very strong bases. This is often necessary in organic chemistry.
A
H
O
H
A
H
O
H
-16
Ka = 10
pKa = 16
Ka = varies
pKa = varies
Hydroxide is a stronger (better) base than
water, although water is still most likely
the solvent. One cannot go higher in pH
than the pKa of water, where 50% of the
water would be ionized to hydroxide and
no longer liquid.
7. Write an equation with water as the acid with generic base, B:. We won’t consider Kb or pKb.
Kb
H
O
H
-16
Ka = 10
pKa = 16
H
B
pKb
O
H
B
Ka = varies
pKa = varies
Water is the reference acid in reactions
with various bases. Arrow pushing
shows the mechanisms of how the
reactions work. Proton transfers are
simple, one step reactions.
8. Can you think of a better acid that you could use in water (but similar looking)? What are the limits of acidity in
water? Using a less basic solvent can allow for more acidic environments. This is sometimes necessary in
organic chemistry.
Hydronium ion is a stronger (better) acid
than water, although water is still most
H
B
O
H
B
H
O
H
likely the solvent. One cannot go lower in
Ka = varies
pH than the pKa of H3O+, where 50% of
H
+2
H Ka = 10
the water would be ionized to H3O+ and
pKa = varies
no longer liquid.
pKa = -2
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9. How does one draw an energy diagram (PE vs. POR) for strong acid ionization equation?
+
B
H
A
TS
B
H
A
stronger acid & base
(less stable)
B
H
A
weaker acid & base
(more stable)
H
PE
A
stronger
G =
acid
The equilibrium shifts towards the weaker conjugate acid and base
because they are more stable (away from the stronger acid and base).
POR
10. How does one draw an energy diagram (PE vs. POR) for weak acid ionization?
B
H
Y
weaker acid & base
B
H
A
+
B
H
Y
TS
Y
stronger acid & base
PE
weaker
acid
H
Y
The equilibrium shifts towards the weaker conjugate acid and base
because they are more stable (away from the stronger acid and base).
Y
G =
POR
11. What makes an acid stronger? What makes an acid weaker? In our course, we present two reasons: a.
inductive effects (related to electronegativity) and b. resonance effects through 2p orbitals. These are mainly
used to explain stabilities of the ‘less stable’ conjugate bases. Do stronger acids have more stable or less stable
conjugate bases (more stable)? What about weak acids (they have less stable bases)? At our peril, we ignore
solvation effects in our presentation and focus mainly on the stability of the conjugate bases.
+
+
B
H
A
B
H
Y
As
the
conjugate
TS
TS
base, A: gets more
stable,
the
acid,
HA,
H
A
gets stronger.
Y As the conjugate
weaker
stronger
PE
base, Y: gets less
PE
acid
G =
acid
G =
stable, the acid, HY,
A
H
Y
gets weaker.
POR
POR
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12. Consider the base, instead of the acid. What makes a base stronger (less stable)? What makes a base weaker
(more stable)? Turn acidity (pKa or Ka) around to evaluate basicity. Electron donating ability is related to the
reasons provided for relative acidities in questions 13 and 14? We often use available pKa tables of acidities to
determine relative basicities of the conjugate bases from their inverse relationships with acidities. (In acidity
equations the stronger acid pairs is on the same side as the stronger base and the weaker acid is on the same side
as the weaker base.) We can calculate Keq for a proton transfer reaction using Ka1 and Ka2. We can quantitatively
calculate G using pKa (= pKa1 - pKa2).
equilibrium lies
completely to the right
H
H
O
stronger acid
stronger base
-16
Ka = 10
pKa = 16
Keq =
H
H
weaker acid
O
weaker base
Ka1
=
Ka2
10-16
10-37
Ka = 10-37
pKa = 37
= 10+21
G = (pKa1 - pKa2) x 1.4 = (pK) x 1.4 = (16 - 37) x 1.4 = (-21) x 1.4 = -29 kcal/mole
H3C
H
H
C
H2
weaker acid
equilibrium lies
completely to the left
H3C
Keq =
-50
Ka = 10
pKa = 50
H
H
stronger acid
CH2
weaker base
Ka1
=
Ka2
10-50
10-37
stronger base
Ka = 10-37
pKa = 37
= 10-13
G = (pKa1 - pKa2) x 1.4 = (pK) x 1.4 = (50 - 37) x 1.4 = (13) x 1.4 = +18 kcal/mole
13. Having a negative charge (or lone pair of electrons) on a more electronegative atom makes it more stable (F-- is
more stable than HO-- is more stable than H2N-- is more stable than H3C--) when the atoms are approximately the
same size. Electron withdrawing inductive effects (due to higher electronegativities) can also help stabilize the
negative charge on a conjugate base and make an acid stronger. In an opposite way, electron donating effects can
destabilize the negative charge on a conjugate base and make an acid weaker.
Electronegativity Effect
+
B
H
F
+
B
F
TS
F
PE
weaker
acid
H
F
pKa = +3
G = +4
H
TS
CH3
H3C
is more stable than H3C
because F has a higher Zeff PE
(+7 > +4) making it more
electronegative than C. Both
are similar size second row
elements.
pKa = +50
G = +70
much
weaker
acid
H3C
H
POR
POR
Inductive Effects
O
H3C
O
H
C
C
H2
OH
pKa = 4.8
R group (CH3) inductively donates
electron density and makes conjugate
base less stable, so acid is weaker.
O
C
C
H2
Cl
OH
pKa = 4.7
H is the reference
substitution.
C
C
H2
OH
pKa = 2.8
Cl substituent inductively withdraws electron
density and makes conjugate basemore stable,
so acid is stronger.
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14. More delocalized electrons are more stable than less delocalized electrons. This delocalization could be due to the
size of the atoms: I-- > Br-- > Cl-- > F-- or due to resonance on carbon, nitrogen and/or oxygen atoms.
I
is more stable than
I
F
minus charge is
more concentrated
minus charge is
more delocalized
...so...
H
F
pKa = -10
stronger acid
H
pKa = 3
weaker acid
Because iodide is a larger anion and its electrons are more delocalized than fluoride's electrons, while both have the same
Zeff of +7. There is only one iodide and one fluoride, so this is a very limited example of delocalization. The more general,
and important example for our course is resonance (next)
Delocalization of electrons in organic chemistry usually refers to "resonance".
delocalized
charge
(resonance)
O
O
O
O
is more stable than
O
localized
charge
O
H
H
...so...
O
O
pKa = 5
stronger acid
pKa = 16
weaker acid
15. Many examples follow, providing opportunities to use organic logic of points 13 and 14.
Generic acid/base equilibrium equation, the organic way, with curved arrows. Keq and G can be estimated
for a proton transfer reaction involving two generic acids, as shown below.
An estimate of the equilibrium constant, Keq, can A1
be calculated by dividing the Ka1 of the acid on the
left (reactant) by the Ka2 of the acid on the right
(product). An estimate of G for the reaction
can be calculated by subtracting the pKa2 of the
acid on the right from the pKa1 of the acid on the
left, and multiplying by 1.4 kcal/mole.
Lewis definitions
acid = electron pair acceptor
base = electron pair donor
Bronsted definitions
acid = proton donor
base = proton acceptor
H
A2
A1
H
A2
Ka1
Keq = (:A2 )(HA1 ) =
(HA2)(:A1 )
Ka2
G = (pKa) x 1.4 = (pKa1 - pKa2) x (1.4) kcal/mole
G = (1.4)(pKa) kcal/mole  pKa
Acid ionization reactions use full headed arrows to show two electron movement. Water is the reference base in usual Ka
and pKa tables.
Ka
H
O
H
A
H
O
H
A
pKa
H
H
Problems – You should be able to match a pKa value with its acid in each group below and explain the differences.
You should be able to draw an arrow-pushing mechanism with general base, B:- for any of the acids, H-A. Include
resonance structures whenever appropriate. If there was a reaction shown between any two conjugate acids and
bases, you should be able to qualitatively and quantitatively indicate which side of the equilibrium is favored, and
what an approximate G is for the reaction. This is a big deal! You are learning the logic of organic chemistry,
right here.
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Acid/Base stuff
Beauchamp
acidic proton varies (on C, N, O or F)
H
1
H
C
H
6
H
H
N
pKa = 50
H
O
H pKa = 35
H
H
H
F
pKa = 16
Strongest acid, negative
charge is on the most
electronegative atom
because they are all about
the same size.
pKa = 3
What is Keq and G for the following reactions?
H
H
H
equilibrium lies
completely to the left
C
N
H
H
weaker acid
H
H
H
weaker base
-50
Ka1 = 10
pKa1 = 50
Ka1
Keq =
10-50
=
N
H
H
H
stronger base
stronger acid
Ka2 = 10-16
pKa2 = 16
-13
= 10
10-37
Ka2
H
C
G = (pKa1 - pKa2) x 1.4 = (50 - 37) x 1.4 = (13) x 1.4 = +18 kcal/mole
H
H
N
O
stronger acid
equilibrium lies
completely to the right
H
Ka1 = 10
pKa1 = 16
N
H
O
H
stronger base
-16
H
H
H
weaker acid
weaker base
Ka1
Keq =
10-16
=
10-37
Ka2
Ka2 = 10-37
pKa2 = 37
= 10+21
G = (pKa1 - pKa2) x 1.4 = (16 - 37) x 1.4 = (-21) x 1.4 = -29 kcal/mole
2
H2
C
H2
C
H
H3C
C
H2
C
H
H3C
N
H3C
H
H
H2
C
H
pKa = 50
H3C
O
pKa = 16
pKa = 37
F
pKa = ?
H
3
I
O
Br
H
O
Cl
H
O
pKa = 8.7
H
F
O
H
not stable
4
H
Cl
H
O
Cl
H
pKa = 7.5
Cl
H
pKa's = 16
Cl
Cl
Cl
5
H
H
C
H
H
H
C
Cl
H
H
C
Cl
H
C
pKa  40*
6
pKa  32*
* = my estimate
CH3
CH3
H3C
H
O
H3 C
pKa's = 19
H 3C
H2
C
CH
H
O
H3C
C
H
O
pKa = 50
pKa = 25
H
Cl
H
pKa's = 12.8
pKa's = 12.2
Cl
Cl
Cl
pKa's = 14.3
O
O
O
pKa = 11
pKa = NA
H 3C
H
O
pKa's = 17
pKa's = 15.5
pKa's = 15.8
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7. What is the expected order of stability of these reactive organic intermediates? (most stable = 1)
a. free radicals
H
R
R
relative energies
H
C
electron poor
C
H
H
b. carbocations
H
c. carbanions
very electron rich
R
R
H
R
H
R
H
H
R
R
C
C
C
C
H
H
H
R
H
R
relative energies
(most stable = 0)
70 kcal/mole
35
15
0
C
C
C
H
H
R
R
H
C
very electron poor
R
R
H
R
H
R
(most stable = 0)
12 kcal/mole
5
2
0
C
C
relative energies
(most stable = 0)
? kcal/mole
R
R
8
pKa = -7
pKa = -9
F
H
Cl
H
Br
I
H
pKa = -10
pKa = 3
H
9
H
O
S
H
pKa = 4
H
H
Se
H
H
Te
H
pKa = 3
pKa = 16
pKa = 7
H
O
10
H
O
O
pKa = 16
H
H
O
pKa = 4
O
11
O
C
H
H3C
H3C
H
O
C
H2C
H
Ref. pKa = 4.7
I
H
O
O
C
H2C
O
C
H
H2C
O
pKa = 15
pKa = 5
O
H2C
O
O
C
H
Cl
F
H
O
C
H
H2C
O
Br
CH3
O
pKa = 0.7
other pKa's = 4.9, 3.2, 2.90, 2.85, 2.59
O
O
O
H Cl
C
H3C
pKa = 20
H
H3C
O
C
H
O
13
C
H
O
H2C
H
N
H
12
O
C
C
pKa = 10
O
C
C
H2
H
O
Cl
H
C
CH
Cl
O
H
C
C
O
pKa = 1.3
Cl
Cl
Cl
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pKa = 2.8
pKa = 5
Acid/Base stuff
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8
O
O
Cl
H
H
H
H
O
O
pKa = 4.8
O
O
O
Cl
pKa = 4.5
O
pKa = 2.8
pKa = 4.0
Cl
15
O
N
H
H
H
H
H
C
H
H
H
C
H
N
H
pKa's = 10, 18, 28, 37, 41, 50
O
H
H
16. Provide an explanation for the general trend in acidities indicated by the pKa values.
H
H3 C
N
F
O
O2N
O
C
C
O
H
O
O
C
O
Ref. pKa = 4.2
C
O
H
C
O
C
H
O
O
pKa = 3.9
pKa = 4.3
H
H
pKa = 3.5
pKa = 3.6
N
17
C
pKa = 24.8
pKa = 28
N
C
N
H
N
H
N
H
pKa = 23.6
H
19
H
O
O
O
20
O
O
pKa = 50
F3C
H H
H
H H
H H
O
O
H
H
N
H
O
O
H
O 2N
21
pKa = 5
H H
O
O
O
O
H
pKa = 9
pKa = 20
CF3
H
Cl
pKa's = 8.4, 7.1, 10.0, 10.2
O
O
H
H
Cl
Cl
pKa's = 85, 9.0, 9.4, 10.0
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O
22
O
C
H
H2 C
O
C
H
O
C
C
O
C
O
O
pKa = 4.7
O
pKa = 15*
H
H
H
pKa = 50
24
H
%s = 33%
H
C
H
H
%s = 50%
H
C
H
H
H
pKa = 5.70
pKa = 2.85
O
H
H 3C
O
23
H
H2 C
pKa = 42
* also aromatic
H
%s = 25%
pKa = 50
C
C
H
H
C
H
H
C
H
pKa = 44
H
pKa = 25
Electrons in 2s orbitals are held tighter than electrons in 2p orbitals. Orbitals which have a greater %s character are more
electronegative than lesser %s character. Therefore the electronegativity of hybrid orbitals is: sp (50% s) > sp2 (33% s) > sp3 (25% s).
Greater electronegativity is better able to stabilize the negative charge in the conjugate base, so sp C-H bonds are the most acidic of
these hydrocarbons, then sp2 C-H and lastly sp3 C-H (lowest acidity of any acid in our course).
25
H
H
H
C
pKa = 9
N
H
H
H
C
N
H
H
H
C
H
N
H
pKa = 5
pKa = -10
H
Bascity
Use full headed arrows to show two electron movement. Water is the reference acid in usual Kb tables. We won't
use pKb values. Instead, we will compare pKa values and judge bases to be stronger when their acids are weaker
and bases to be weaker when their acids are stronger. In the examples that follow pair up each base with the pKa
value of its conjugate acid.
B
H
judge the strength of the base
from the weakness of the acid
HO3S
O
one of our
weakest bases
(most stable)
H2O
H
H
a
a
O
H
O
O
H
O
b
H
H3N
H
pKa = +9
H
H3C
one of our
strongest bases
(least stable)
H
pKa = -2
O
B
H
use this acid's pKa value to judge
the electron donating power of B:
to the reference acid, H2O.
O
A
NH2
NH2
O
NH2
NH2
b
NH3
The most basic site of an amide represents an especially ambiguous problem. Using H3O+ and NH4+ as examples, it seems logical
to use the nitrogen of an amide as the basic site, putting the positive charge on the nitrogen instead of the oxygen. However, if we
protonate both and look at the conjugate bases more carefully, we see that when we protonate the nitrogen the charge is fully
localized there. If we protonate the oxygen, the charge is delocalized onto the carbon and nitrogen, as well, so the more basic
site is actually the oxygen atom. If a C=O (or C=N) is present in a functional group, use that site as the best base.
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Here are a few qualitative examples. Where is the most basic site in each molecule below? Can an order of
basicity be explained for some or for all of the bases (approximate pKa’s of some of the conjugate acids are
provided)?
O
O
H
C
H
H
C
C
C
O
CH3
H
H
H
N
H
C
N
N
H
H
R
O
N
C
H
pKa  5 (guess)
N
C
N
H
H
pKa  -0.5
H
N
H
H
pKa  13.6
pKa  7
H
C
H
C
H
C
H
H
pKa  0.2
pKa  -6
H
C
C
N
C
N
H
H
H
C
C
H
pKa  -7
H
O
H
C
H
pKa  -7
H
H
H
C
O
H
H
O
O
Problem – What is the order of basicity among the following molecules of each group (1 = most basic)? Explain
your reasoning. Match the given pKa values with the conjugate acids of the indicated bases. Write arrow-pushing
mechanisms with general acid, H-A to illustrate the reactions. Include resonance structures whenever appropriate.
Where is the most basic site in each molecule? Explain your reasoning using arguments of inductive effects
(electronegativity), resonance effects (electron delocalization) or both. For any reaction between two conjugate
acids and bases, you should be able to qualitatively and quantitatively indicate which side of the equilibrium is
favored, and what is an approximate G is for the reaction.
1
H
H
F
N
H
H
2
H2
C
H2
C
H3C
H3C
N
O
H
H
H2
C
C
H3C
H
R
H2
C
F
H3C
O
N
C
H
H3C
pKa's of the bases'
conjugate acids = 50, 37, 16
O
N
O
C
H3C
H
N
C
H3C
H
pKa's of the bases'
conjugate acids = -7, 0, 7
O
H
H
4 H
H
C
H
pKa's of the bases'
conjugate acids = 50, 37, 16, 3
H
H
3
C
H
N
H
C
N
H
C
H
N
H
pKa's of the bases'
conjugate acids = 9, 5, -10
H
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Acid/Base stuff
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11
R
5
N
N
C
H
H
N
N
H
H
N
pKa's of the bases'
conjugate acids = 5, 7, 13
N
H
O
6
O
O
7
O
O
O
O
O
F3C
8
H
O
O
H
C
O
H3C
N
H
C
H
C
H3C
H
9
pKa's of the bases'
conjugate acids = 50, 20, 9, 5
H
O
C
H3 C
pKa's of the bases'
conjugate acids = 5, 10, 16
CF3
H
H
O
CH2
pKa's of the bases'
conjugate acids = 20, 15, 5
H
C
H
H
C
C
C
H
H
pKa's of the bases'
conjugate acids = 50, 44, 25
C
H
10
H
H
H
pKa's of the bases'
conjugate acids = 50, 42, 15
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Acid/Base stuff
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12
Acid/Base arrow pushing worksheet
1. These proton transfer reactions are the first step in multistep mechanisms to be studied later in the course.
Supply the necessary curved arrows, lone pairs of electrons and/or formal charge to show how the first
step each reaction proceeds. Starting at e are simple proton transfer reactions generating a carbanions
(very important for organic chemistry). Generally, there is some stabilizing feature that allows a
carbanion to form via acid/base chemistry, such as inductive and/or resonance effects, but this is not
always the case. In working the problem below, show any important resonance structures or identify the
inductive effect that makes the reaction possible. Estimate a Keq for each reaction using the Ka’s.
a.
O
H
Na
sodium hydride
R
H
alcohols
Ka = 10-16
H-H
Ka = 10-37
b.
S
H
R
H
Na
O
H
sodium hydroxide
thiols
Ka = 10-9
O
H
Ka = 10-16
c.
O
C
H
H
Na
O
H
sodium hydroxide
R
O
carboxylic acids
Ka = 10-5
O
H
-16
Ka = 10
d.
Li
N
H
Formation of lithium diisopropyl
amide (LDA) using butyl lithium.
CH2CH2CH2CH3
H
n-butyl lithium is
commercially
available
Ka = 10-37
CH2CH2CH2CH3
Ka = 10-50
e.
O
R
H
N
LDA
C
H2
Li
R
-78 oC
C
resonance
R
ketone Ka = 10-20
R2NH
Ka = 10-37
f.
O
R
H
N
LDA
R
Li
C
R
-78 oC
resonance
C
O
H2 ester
Ka = 10-25
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R2NH
Ka = 10-37
Acid/Base stuff
Beauchamp
13
g.
O
R
-78 oC
N
H
Li
R
C
R
C
N
H
R
resonance
H
LDA
R2NH
tertiary amide
Ka = 10-30
Ka = 10-37
h.
H
R
N
H
Li
R
C
C
N
-78 oC
resonance
R2NH
nitrile
Ka = 10-30
H
LDA
Ka = 10-37
i.
H
O
O
C
C
C
H3 C
Na
H
H
sodium hydride
-78 oC
CH3
1,3-dicarbonyl
resonance
resonance
H-H
Ka = 10-37
Ka = 10-9
j.
R
N
H
Na
R
sodium amide
C
C
-33 oC
R
R2NH
terminal alkyne
Ka = 10-25
Ka = 10-35
ammonia
terminal acetylide
k.
S
H
Li
H2C
H
S
dithiane
Ka  10
H2
C
-35
CH3
C
H2
-78 oC
possible resonance
with sulfur 3d orbitals,
but violates the octet rule
H
CH2CH2CH2CH3
Ka = 10-50
n-butyl lithium
l.
H
Ph
S
Br
C
Li
H
Ph
H
sulfur ylid
-35
Ph = phenyl Ka  10
H2
C
H2 C
CH3
C
H2
-78 oC
possible resonance
with sulfur 3d orbitals,
but violates the octet rule
H
CH2CH2CH2CH3
Ka = 10-50
n-butyl lithium
m.
Ph
Ph
Br H
P
C
Li
H
Ph
H
phosphorous ylid
(Ph = phenyl) Ka  10-35
H2C
H2
C
CH3
C
H2
-78 oC
possible resonance
with phosphorous 3d
orbitals,but violates
the octet rule
H
CH2CH2CH2CH3
n-butyl lithium
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Ka = 10-50
Acid/Base stuff
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14
2. Lone pair donors to a very strong acid
All of the following functional groups react with protic acid as the first step of a reaction studied in organic
chemistry. Often subsequent chemistry occurs after that initial step and you will study most of those reactions later
in the course. Show how they react in the first step by including all lone pairs, curved arrows to show electron
movement and formal charge.
Acid/Base arrow pushing worksheet
lone pair donors
lone pair acceptor
O
Keq = ?
H
O
H
H
O
S
OH
O
H
N
H
H
H
ammonia
R
N
N
H
H
H
O
H
H
O
equilibrium
Keq = ?
equilibrium
H
H
Keq = ?
pKa's = -2, 9
H
H
1o amine
R
O
equilibrium
pKa's = -5, -2
pKa's = -2, 10
H
Keq = ?
equilibrium
R
2o amine
R
N
pKa's = -2, 10
H
R
H
O
H
Keq = ?
equilibrium
R
3o amine
R
O
pKa's = -2, 10
H
H
H
alcohol
O
H
Keq = ?
pKa's = -2, -3
H
Keq = ?
R
O
R
H
ether
O
epoxide (ether)
H
O
pKa's = -2, -3
H
H
O
H
H
Keq = ?
pKa's = -2, -3
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additional
chemistry
possible
additional
chemistry
possible
additional
chemistry
possible
Acid/Base stuff
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15
O
H
C
O
R
H
aldehyde
H
Keq = ?
resonance
(2)
pKa's = -2, -7
H
additional
chemistry
possible
O
H
H
O
Keq = ?
C
R
H
R'
pKa's = -2, -7
ketone
R
C
N
H
nitrile
O
resonance
(2)
SO3H
Keq = ?
resonance
(2)
additional
chemistry
possible
resonance
(3)
additional
chemistry
possible
resonance
(3)
additional
chemistry
possible
pKa's = -10, -10
sulfuric acid
additional
chemistry
possible
O
H
O
H
Keq = ?
H
C
R
O
carboxylic acid
H
pKa's = -2, -6
O
H
C
O
H
Keq = ?
R'
R
O
H
pKa's = -2, -6
ester
O
H
O
H
Keq = ?
C
R
H
NH2
resonance
(3)
pKa's = -2, 0
amide
additional
chemistry
possible
H
N
H
C
R
NH2
O
H
H
Keq = ?
resonance
(3)
pKa's = -2, 12
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additional
chemistry
possible
Acid/Base stuff
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16
Carbon-carbon pi bonds as weak electron pair donors to very strong acid
H
H
C
R
alkene
CH2
O
Keq = ?
SO3H
additional
chemistry
possible
pKa's = -5, -10
sulfuric acid
NR2
H
O
H
Keq = ?
C
R
enamine
H
CH2
resonance
(2)
additional
chemistry
possible
resonance
(2)
additional
chemistry
possible
pKa's = -2, 5
R
O
H
O
H
C
R
R
H
CH2
enol ether
C
C
H
H
pKa's = -2, -7
Keq = ?
SO3H
O
sulfuric acid
alkyne
Keq = ?
additional
chemistry
possible
pKa's = -5, -10
H
C
H
C
H
Keq = ?
C
E
C
H
C
C
H
aromatic
H
resonance
(3)
pKa's = -10, -10
additional
chemistry
possible
E+ = electrophile
(Lewis acid = electron
pair acceptor)
At this point we are mainly interested in understanding acid/base proton transfers, curved arrow
pushing, formal charge, recognizing resonance structures and using the logic arguments of inductive
effects and resonance effects to explain relative stabilities of acids and bases. If you can do these things,
you are well on your way to understanding organic chemistry and biochemistry.
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Acid/Base stuff
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17
Ka/pKa table for a variety of acid types
The sign and magnitude of an acid’s pKa represents the approximate energy change to form the conjugate base from
the acid with water as the general base. Remember a difference of 1 pKa unit is the same as the difference between
a 6’ person and a 60’ person, a pretty obvious difference.
A pKa table provides us with immediate access to an acid’s proton donating ability and indirectly to its
conjugate base electron pair donating ability. You can decide from the values in the pKa table whether an acid is
strong or weak and its relative acidity (or basicity) compared to other acids (or bases) in the table. If it is weak
(most of them are, pKa > 1 to very large), you can evaluate approximately how large an energy input is necessary to
form the conjugate base. Remember, water is the reference base for all of the listed pKa’s of the acids even though
as a solvent for many acids, it is meaningless. The pH limits for water are about -2 to 16 because of the pKa values
of its conjugate acid, H3O+, (pKa = -2) and itself (pKa = 16).
pKa Table for a Variety of Acids – Approximately equal to Gacid ionization (in kcal/mole = (1.4)x(pKa))
Carbon Acids – There is a fair amount of uncertainty in the higher pKa values.
H = acidic hydrogen atom
O
O
O2N
NO2
CHO
OHC
CH
CH
H
H
pKa = 4
RO2C
CH
C
CH
O
H
CH2
CH
pKa = 20
Cl
C
C
R
H
pKa = 25
Ph
N
R
C
OR
pKa = 24
C
S
Ph
H
C
H
Cl
H
Ph
pKa = 25
pKa = 30
pKa = 31
H
H Ph
C
S
H
H
R
H
H
C
H
C
H2
R
H
R
H
pKa = 44
pKa = 43
C
H
H2 C
pKa = 42
CH2
pKa = 32
H
pKa = 40
pKa = 35
R
H
H
S
H
pKa = 19
Cl
H
R
pKa = 23
pKa = 34
H
H
pKa = 16
pKa = 15
O
H
C
H
O
O
R
Ph
CH
H
H
pKa = 13
S
pKa = 23
H
R
CH
H
pKa = 13
H
O
R
CH
H
H
H
CO2R
CH
pKa = 11
Ph
pKa = 11
O
SO2R RO2C
OR
H
H
pKa = 10
O
RO2S
C
H
pKa = 9
pKa = 9
N
C
CH
H
H
pKa = 6
N
NO2
R
N
H
pKa = 5
O
NO2
CH
pKa = 46
pKa = 50 - 60
Oxygen Acids
H = acidic hydrogen atom
O
O
H
Cl
O
pKa = -10
O
H
O
pKa = -1
H
C
R
O
pKa = +5
HO
S
H
H
O
H
O
O
R
S
H
O
pKa = -3
pKa = -3
H
H
O
O
pKa = -2
pKa = -1
O
O
H
O
C
R
O
H
O
O
O
N
H R
aldehydes, ketones
O
O
esters, acids
R
pKa = -8 to -6
R
R amides pK = 0
pK
a
a = -3
O
O
R
H
O
pKa = +8
O
H
pKa = +10
O
O
H
O
H H3C
pKa = +11.6
H H
H
R
O
H
H
pKa = +15.5 pKa = +15.7 pKa = +16-19 pKa = +25
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Acid/Base stuff
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18
Compare the following groups.
R
N
R
O
N
O
O
H
H
C
H
C
O
R
O
O
H
C
H3C
O
H Cl
C
O
H2C
O
CH
O
O
H
C
H
C
C
Cl pK = +1.3
a
O
H2C
O
O
H
C
O
F
Cl pK = +0.7
a
pKa = +2.8
pKa = +4.0
O
Cl
Cl pK = +2.9
a
pKa = +4.7
H Cl
H
C
Cl
pKa = +4.5
O
C
H
O
Cl
pKa = +4.8
pKa = +5
O
C
O
O
pKa = +1
O
Cl
H
H2C
O
Cl
pKa = +2.6
O
H
C
H2C
O
H
C
H2C
pKa = +2.9 Br pKa = +3.0
O
I pKa = +3.1
Nitrogen Acids
Ph
H
N
H
H
H
H
Ph
C
Ph
N
R
H Ph
N
Ph
pKa = -10
H Ph
N
H
pKa = -5
H
N
N
N
H
H
H
N
H
O
H
pKa = +1
pKa = +5
pKa = +6
pKa = +5
pKa = +7
O
H
H
H
N
H
H
N
H
H
N
H
R
N
H
O
O
R
H2N
N
H
H
H
C
H2N
O
R
pKa = +8 pKa = +9.2 pKa = +9-11 pKa = +9
H
C
H2 N
N
H
C
H3 C
N
N
H
H
H
H
pKa = +14
pKa = +13
N
pKa = +15
pKa = +17
R
H
N
N
H
R
H
pKa = +35
H
pKa = +37
Other Miscellaneous Acids
Compare the following groups.
R
H
H
H
H
H
N
N
H
S
H
H
S
H Ph
S
H R
S
H
H
P
H
H
pKa = +7
pKa = +8
pKa = +10
Compare the following groups.
F5SbF
H pKa = -20
I
H pKa = -10 HTe
FSO3
H pKa = -15
Br
H pK = -9 HSe
a
F4B
H
O3ClO
H
H pKa = 4
H pKa = -7
HS
H pKa = 7
pKa = -10
H pKa = +3
HO
H pKa = 16
F
ClO
H
pKa = +7.5
HO3SO
BrO
H
pKa = +8.7
O3SO
IO
H
pKa = +11
H
H
pKa = -3
pKa = +2
HO2SO
O2SO
H
O
pKa = +0
H pKa = 3 H2PO4
pKa = -15 Cl
H
HPO4
-2
PO4
pKa = +9
H
H2 N
H
H pKa = +2.1
H pKa = +7.2
O2NO
H pKa = -1
ONO
H pKa = +3
HO2CO
H pKa = +12.4
pKa = +2 HOO
pKa = +7
HO
pKa = +8.1
pKa = +13.7
O2CO
H
H
H
H
pKa = -5
N
O3ClO
H pKa = -10
O3ClO
H pKa = -1
H pKa = +6.4 O ClO
3
H pKa = +2
H pKa = +10.3
H
pKa = +12
H
pKa = +16
ClO
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H pKa = +7.5
Acid/Base stuff
Beauchamp
19
Table of acidities of some phenols
S
H
O
-Cl
8.48
-Br
8.42
-OH
9.98
-CH3 10.48
7.23
-NO2
-CHO 6.79
O
S
o = ortho
S
S = H (reference), pKa = 10.0
m = meta
p = para
9.02
9.11
9.44
10.08
8.35
8.00
9.38
9.34
9.96
10.19
7.14
7.66
Table of acidities of some very strong acids (= 100% ionization in water)
HF / SbF5
FSO3H
pKa  -20
pKa  -15
HF / BF3
HF / BF3
CF3SO3H
HI
HClO4
pKa  -15
pKa  -15
pKa  -14
pKa  -10
pKa  -10
R C N H pKa  -10
HBr
HCl
H
R S H
R
R S H
H
R O H
R
R O H
pKa  -9
pKa  -7
pKa  -7
pKa  -5
pKa  -3
H2SO4
O
S OH pKa  -2
O
H
H O H
pKa  -2
H2MnO4
pKa  -1
H2CrO4
pKa  -1
HNO3
pKa  -1
HClO3
pKa  -1
O
R
O
R
C
O
R
C
O
R
HH
C
C
O
pKa  -3
H
pKa  -8
R
C
CN
H
pKa  -8
H
NC
H
O
H
O
C
H
NC
H
NO2
O
H
N H
N
R
H
O
NO2
pKa  -10
H
pKa  -6
O
pKa  -11
R N
R
pKa  0
H
N
pKa  -10
CN
pKa  -7
pKa  -7
N
O H
R
R
pKa  -3
The above pKa tables dramatically demonstrate how much Bronsted acids can vary in strength. The magnitude
of the numbers is really beyond our comprehension. The strongest acid in the table has a Ka  10+20 (pKa = -20),
while the weakest acid has a Ka  10-50 (pKa = 50). That’s 70 orders of magnitude! What does 1070 mean? Even
so, we will only use two simple arguments to rationalize the differences in acidity (…and basicity): 1. inductive
effects (based on relative electronegativity) and 2. charge delocalization effects (usually based on resonance
through 2p orbitals). We will not emphasize steric effects, hydrogen bonding or solvation effects, which can also
modify relative acidities, sometimes greatly.
The Tautomer Game – an arrow-pushing training reaction in acid and in base, forward and reverse
Tautomers are isomers that differ by the location of a proton and a pi bond. Official tautomers usually have a
heteroatom or atoms (different than carbon, often oxygen or nitrogen or both) as part of the system. In the simplest
case, there are at two isomers in equilibrium with one another (there may be many, many more tautomers possible
in more complex systems). The tautomers are interchangeable by 1. proton transfer, 2. resonance intermediates and
3. proton transfer. The “keto” isomer, has a heteroatom in a pi bond and in the “enol” tautomer has two carbons
forming a pi bond. This simple pattern can occur in an infinite number of systems, from very simple to very
complex. A possible approach to figuring out what to do in keto/enol tautomer problems is provided below. More
complex tautomer relationships are shown in later examples. They are more complicated mainly because there are
more than two tautomers and interconversions may require one or more simple tautomer interconversions.
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Acid/Base stuff
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20
Tautomers in acid (simplest examples)
C
H
H
Use H3O /H2O or HO /H2O to accomplish the given transformations. Every
transformation will always follow a 3 step sequence. This is as simple as it gets.
O
H
C
H
H
C
H
acid
or
base
C
O
H
C
C
H
H
H
H
H
keto tautomer
enol tautomer
best acid
1. proton transfer (in acid = proton on) (in base = proton off)
2. resonance delocalized intermediates
3. proton transfer (in acid = proton off) (in base = proton on)
best base
in acid
H3O
H2O
in base
H2O
HO
a. “keto”  “enol”
H
O
H
H
H
C
C
H
H
O
H
C
H
H
H
keto tautomer
H
O
H
O
H
H
H
C
H
O
C
C
H
H
resonance
H
H
H
O
H
H
H
O
C
C
C
H
H
H
H
H
H
in acid = proton on
in acid = proton off
same resonance as b
H
C
C
C
H
H
enol tautomer
b. “enol”  “keto”
H
H
H
O
H
H
H
C
C
O
C
H
H
H
enol tautomer
O
H
H
in acid = proton on
resonance
C
C
C
H
H
H
O
H
H
H
O
O
H
H
H
H
H
O
H
H
same resonance as a
C
C
C
H
H
H
H
H
C
H
C
H
C
H
H
keto tautomer
in acid = proton off
Tautomers in base (simplest examples)
c. “keto”  “enol”
H
O
H
C
O
C
C
H
H
H
keto tautomer
O
H
H
in base = proton off
resonance
H
H
H
O
H
H
O
O
H
H
H
O
H
C
C
C
H
H
H
enolate
C
H
H
C
C
H
H
in base = proton on
same resonance as d
H
C
H
C
C
H
H
enol tautomer
d. “enol”  “keto”
H
H
O
H
H
O
H
H
C
C
O
H
C
H
H
enol tautomer
H
O
H
in base = proton off
H
C
C
H
H
H
O
H
H
H
C
O
O
resonance
H
enolate
C
C
H
H
H
H
C
H
in base = proton on
same resonance as c
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C
C
C
H
H
keto tautomer
H
Acid/Base stuff
Beauchamp
21
More Examples
H
H
H
N
N
H
C
acid
or
base
H
C
C
H
H
H
C
C
C
H
H
H
H
H
H
keto tautomer
enol tautomer
H
H
acid
or
base
H
C
C
N
H
O
C
H
C
H
acid
or
base
N
C
H
C
H
N
H
H
H
keto tautomer
H
H
keto tautomer
H
enol tautomer
N
N
acid
or
base
C
H
H
O
O
C
H
H
C
C
H
H
H
keto tautomer
enol tautomer
A slightly more complicated keto/enol tautomer problem – keto/enol with an additional pi bond
also possible
H
O
H
H
H
O
H
H
H
O
base
H
H
H
H
H
base
H
1
H
H
H
H
H
H
H
1
H
H
H
4
H
2
H
base
base
O
H
H
O
H
H
H
base
3
H
H
H
H
H
H
H H
5
3
H H
The tautomer interconversions shown above are possible in one step in base because of shared resonance intermediates. The
total number of tautomer changes required to change any tautomer into any other tautomer is shown below for base (on the
left) and acid (on the right). The number of tautomer changes in parentheses was worked out in my head, not on paper, so there
may be some wrong estimates.
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Acid/Base stuff
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22
Number of tautomer changes to transform
one tautomer into another in base.
1
2
2
3
4
5
3
Number of tautomer changes to transform
one tautomer into another in acid.
4
(1x)
(1x)
(1x)
(2x)
1
3
4
5
(1x)
(1x)
(2x)
(2x)
1
2
4
5
(1x)
(1x)
(2x)
(1x)
1
2
3
5
(1x)
(2x)
(2x)
(3x)
1
2
3
4
(2x)
(2x)
(1x)
1
5
2
3
4
5
(1x)
(2x)
(1x)
(1x)
1
3
4
5
(1x)
(1x)
(1x)
(1x)
1
2
4
5
(2x)
(1x)
(2x)
(1x)
1
2
3
5
(1x)
(1x)
(2x)
(2x)
1
2
3
4
(2x)
(1x)
(1x)
(2x)
2
3
4
5
(3x)
1. Circling the protons that change is always a good idea, because you know those protons are going to have to
move. However, these may not show every tautomer change because sometimes a necessary change is reversed in
a later step. The circled protons have to be moved, either taken off (with the best base available) or put on (with
the best acid available) and there is always resonance delocalization in the intermediate.
Best acid in H3O+/H2O is H3O+, best base in H3O+/H2O is H2O
Best base in H2O/HO-- is HO-- and the best acid in H2O/HO-- is H2O
2. Always work from a "keto" (CH-C=O or CH-C=N-) part or "enol" (C=C-OH or C=C-NH-) part of the
molecule. Do not use isolated pi bonds (C=C) to initiate change in the structure. With an allowed tautomer
change, an isolated pi bond may become conjugated with a “keto” or “enol” part of another tautomer. Any keto
or enol part will be the better base or the better acid, as is indicated because it will form a resonance stabilized
intermediate with the oxygen (or nitrogen) assisting in the intermediate resonance structures.
Not possible in in base in one keto/enol cycle.
OH
H
OH
H
H
H
OH
H
H
H
H
H
2
H
O
H
H
H
H
Changing tautomer 2 into tautomer 4 is possible in base by first converting tautomer 2
into tautomer 1 and then changing tautomer 1 into tautomer 4. On the other hand
tautomer 2 could be converted to tautomer 4 in a single tautomer change in acid. Do
not use isolated C=C pi bonds to interconvert tautomers because they represent a higher
energy path, which would be slower reacting. On paper, it may seem easier to write,
but such transformation are energetically out of reach.
H
base
H
H
H
4
H
This is an isolated
C=C bond. Don't
begin here in acid
or base.
H
H
OH
H
H
H
H
H
H
H
H
2
H
4
H
H
H
1
acid
H
base
H
This is an isolated
C=C bond. Don't
begin here in acid
or base.
4
H
This is an isolated
C=C bond. Don't
begin here in acid
or base.
3. If in acid, use the strongest acid (H3O+ in our examples) to put on a "gained" proton first and take off a "lost"
proton second, with a weak base (usually the solvent = H2O in our examples).
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Acid/Base stuff
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23
4. If in base, use the strongest base (HO- in our examples) to take off a "lost" proton first and put on a "gained"
proton second with a weak acid (usually the solvent = H2O in our examples).
5. In all tautomer mechanisms there will be resonance structures in the intermediate formed. The intermediate
structure will show the way to all other reasonable tautomers from that intermediate. You may have to repeat
the tautomer process once, twice, etc. until you accomplish an overall indicated transformation. Counting the
number of protons lost or the number of protons gained will give you an indication of a minimum number of
times you may have to repeat the tautomerization process. This may not always match however because
sometimes a tautomer sequence is reversed and hidden from the overall change indicated (See rule 2 for
isolated pi bonds.).
The following transformations can be done in base or acid. Intermediate resonance structures lead to stable structures
O
OH
H
H
acid
or
base
H
H
H
H
H
H
H
H
H
H
Use generic acid, H-A, or generic base, B:
to accomplish the given transformations.
For every transformation there will be
resonance delocalized intermediates that
lead toward the path desired.
H
1
H
H
O
acid
or
base
H
H
acid
or
base
H
H
This is the most thermodynamically
favored keto/enol structure because
it retains the C=O and has conjugated
pi bonds.
H
H
H
H
3
H
H
A
1
2
2
2
2
3
2
3
2
A
H
1
3
1
3
H
A
1
3
3
3
3
B
A
A
2
B
3 changes in base
OH
1
H
B
B
1
A
1
B
B
1
H
2
2
OH
H
Additional
tautomeric
structures.
H
H
H
H
3 changes in base
2 changes in acid
H
H
H
Additional tautomeric
interconversions
(40 different problems).
H
H
H
H
4
2 changes in acid
H
H
5
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Acid/Base stuff
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24
Tautomerism Arrow-Pushing Practice – Fill in missing formal charge, lone pairs and curved arrows.
H
O
H
H
H
H
O
H
H
H
H
O
H
O
H
H
H
H
H
H
B
H
H
H
H
H
1
H
H
H
H
H
H
H
O
H
H
H
B
H
O
H
H
H
H
H
H
H
O
A
H
2
H
H
H
A
O
H
H
H
H
Steps in acid for each tautomeric change:
1. proton transfer (proton on, best acid = H3O+)
2. resonance intermediates
3. proton transfer (proton off, best base = H2O)
O
H
O
H
O
H
H
H
H
H
H
H
H
Remember: each tautomer has the same overall formal
charge and the same total number of pi bonds.
H
H
H
H
H
4
H
H
C
H
D
D
C
H
H
O
H
H
H
H
H
H
H
H
H
H
H
H
H
H
3
H
H
H
H
H
5
H
O
H
H
H
H
O
H
O
H
H
H
H
H
O
O
O
H
H
H
H
H
H
H
H
A
H
C
O
H
O
B
O
H
H
H
H
O
H
H
H
H
H
H
H
H
H
H
H
G
A
O
C
H
H
H
D
H
D
H
O
H
H
H
H
H
H
O
H
H
H
2
H
H
H
G
O
H
H
H
O
H
F
H
H
H
O
H
H
O
E
H
O
O
H H
H
H
H
H
H H
H
H
H
H
3
H
O
Remember: each tautomer has the same overall formal
charge and the same total number of pi bonds.
H
O
O
H
Steps in acid for each tautomeric change:
1. proton transfer (proton off, best base = HO )
2. resonance intermediates
3. proton transfer (proton on, best acid = H2O)
H
E
H
H
H
H
H
H
H
H
H
O
O
H
H
H
H
H
H
O
H
H
H
H
H
H
1
O
H
H
H
B
H
O
H
H
O
4
H
H
H
H
F
H
H
H
H
5
H
H
H
H
H
H
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H
H
Acid/Base stuff
Beauchamp
25
Possible Key for arrow-pushing in “tautomer” problems
H
O
H
H
H
H
H
O
H
H
H
H
O
H
O
H
H
H
H
H
H
B
H
H
H
1
H
H
H
H
H
H
H
O
H
H
H
B
H
O
H
H
H
H
H
O
H
H
H
A
H
2
H
H
H
A
O
H
H
H
O
Steps in acid for each tautomeric change:
1. proton transfer (proton on, best acid = H3O+)
2. resonance intermediates
3. proton transfer (proton off, best base = H2O)
H
H
O
H
H
O
C
H
H
H
C
H
O
H
H
5
H
H
H
H
H
H
H
H
H
H
3
H
O
H
H
H
H
H
O
H
H
H
H
H
H
O
H
H
H
H
H
H
O
H
H
4
H
D
O
H
H
H
H
H
H
H
H
H
H
H
Remember: each tautomer has the same overall formal
charge and the same total number of pi bonds.
D
O
H
H
H
H
H
H
H
A
H
C
O
H
O
B
H
H
H
H
O
O
H
H
H
H
H
H
H
H
H
H
H
1
G
H
H
O
H
H
H
H
H
H
O
H
H
F
H
H
O
H
O
H
H
O
3
E
H
O
O
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
2
O
O
H
H
H
H
G
O
H
H
Remember: each tautomer has the same overall formal
charge and the same total number of pi bonds.
O
D
H
H
H
H
Steps in acid for each tautomeric change:
1. proton transfer (proton off, best base = HO )
2. resonance intermediates
3. proton transfer (proton on, best acid = H2O)
H
E
O
H
H
D
H
H
H
H
H
H
H
H
C
O
O
H
H
H
H
A
H
H
H
H
H
H
O
O
H
H
H
B
H
O
H
H
O
4
H
H
H
H
H
F
H
H
H
H
5
H
H
H
H
H
H
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H
H
Acid/Base stuff
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26
Problem 1 – What happens if we change one carbon in the ring above to a nitrogen atom? Take a look below. There
might be more, but these are all of the tautomers I could think of.
Every tautomer can be converted into every other tautomer in acid or in base (= 13 x 12 x 2 = 312 possibilities). Use
the tautomer maps below to see how many steps it takes and a path for interconversion between any two tautomers.
H
O
H
O
H
H
H
H
O
H
O
H
H
H
H
H
H
H
H H
H
H
N
N
H
H
N
H
H
H
H
1
N
H
H
H
H
2
3
O
H
H
H
N
H
H
5
H
H
N
N
H
H
N
H
H
10
H
H
N
H
H
11
2
5
2
9
5
4
H
9
1
1
H
8
H2O
3
3
N
H3O
8
HO
H
Acid - tautomer
interchange map
13
10
H2O
12
H
H
13
10
13
H
H
Base - tautomer
interchange map
H
H
H
O
H
H
H
H
H
H
H
O
H
H
H
9
H
H
H
O
H
H
H
6
H
O
H
8
H
H
H
H
N
N
H
H
4
H
H
H
H
O
H
H
H
N
H
H
H
O
7
O
H
H
H
H
O
4
7
7
11
11
12
12
6
6
In the examples on the following pages, the starting tautomer can be converted into the following tautomers
in two tautomer changes under the conditions specified at the top of the page, as also shown in the tautomer
maps above. A complete mechanism for 1  13 in acid and in base is provided on the last page of this topic.
Two step changes in either direction in base
O
H
O
H
H
O
H
H
H
H H
H
N
H
4
H
H
H
H
N
H
N
H
H
H
H
H
H
10
13
H
H
H
H
N
H
H
3
H
H
H
H
9
H
H
H
H
H
O
N
N
H
N
8
H
H
H
H
H
H
O
H
H
O
H
O
H
H
N
H
H
H
H
1
2
H
H
O
H
N
H
H
H
O
H
H
7
H
H
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H
Acid/Base stuff
Beauchamp
H
H
H
O
O
O
O
27
H
H
H
H
H
H
H
H
H
H
3
H
N
H
H
H
H
5
H
H
H
H
O
H
H
N
H
H
H
4
H
H
1
N
H
H
13
12
H
H
H
O
H
H
H
H
H
H
H
H
H
H
H
H
H
O
H
H
H
H
H
H
N
H
H
O
H
H
H
H
11
H
O
O
H
N
H
7
3
H
H
H
N
H
H
H
H
H
N
H
O
H
H
N
H
N
H
H
N
H
H
H
H
H
H
H
H
O
O
O
H
H
H
H H
H
H
10
H
H
O
N
H
9
H
4
H
H
H
H
N
H
6
O
H
6
H
H
H
H
H
H
H
O
5
H
H
N
H
O
H
H
N
H
H
H
O
H
H
H
H
H
O
O
H
H
H
11
2
H
H
N
N
N
H
H
H
H
H
H
H
O
O
H
H
H
H
N
H
H
7
2
H
H
H
O
N
H H
H
H
H
N
N
H
H
H
H
H
H
H
H
H
H
11
12
H
H
H
H
N
N
N
N
H
O
O
H
8
H
H
H
H
H
8
5
H
H
N
H
H
H
H
H
1
H
H
H
7
H
H
H
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Acid/Base stuff
Beauchamp
H
H
O
O
O
H
H
H
H
H
H
H
H
N
N
H
H
H
H
H
6
H
H
H
H
N
H
H
H
H
H
H
1
H
H
4
H
H
N
H
H
H
H
H
N
N
N
H
H
12
H
H
H
H
H
H
H
H
N
N
N
H
13
H
H
H
1
O
H
H
H
H
H
H
H
H
H
H
11
O
O
O
H
H
H
H
H
H
N
H
7
H
H
H
H
4
H
H
H
N
H
H
13
12
H
H
H
H
H
N
O
O
O
H
H
H
H
H
H
H
H
N
H
7
5
H
H
H
H
H
O
H
N
N
H
H
H
H
H H
3
H
H
H
H
O
O
H H
N
H
H
O
H
H
H
H
H
H
H H
H
11
H
6
O
O
H
N
H
H
O
H
H
H
H
H
H
H
N
N
H
O
H
H
H
H
H
H
O
O
O
10
H
H
1
H
H
N
H
H
9
28
H
H
4
H
H
11
H
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H
Acid/Base stuff
Beauchamp
Two step changes in either direction in acid
O
H
H
N
H
1
H H
H
H
H
H
N
H
H
H
H
N
N
H
H
3
H
7
H
H
H
H
O
O
O
H
H
H
H
H
H
H
H
H
H
H
H
H
N
H
H
H
H
H
5
H
H
H
H
H
H
H
H
H
H
H
H
1
H
6
H
H
H
H
H
H
H
H
H
7
H
H
N
H
H
H
H
H
H
H
H
H
O
H
H
H
N
H
12
H
H
N
2
H
10
O
H
N
H
H
O
H
H
H
H
H
N
H
H
O
H
H
H
9
H
4
H
N
H H
N
H
H
H
H
H
H
N
O
O
H H
H
H
H
O
H
H
H
H
H
H
O
H
H
11
H
H
H
H
O
H
N
H
H
3
H
H
H
N
N
H
H
H
H
H
N
H
13
12
O
H
H
N
H
H
O
O
H
10
H
H
N
H
H
H
O
H
H
N
H
H
H
H
H
N
H
H
H
O
H
H H
H
N
N
O
H
H
H
H
O
O
H
H
H
O
O
H
H
H
11
2
H
H
H
N
N
N
7
H
H
H
H
H
H
H
H
O
6
13
H
N
H
5
H
H
10
H
H
H
H
4
N
O
O
H
H
H
H
3
N
H
H
H
H
H
9
8
H
H
H
H
H
H
4
O
H
O
H
H
N
H
H
2
H
O
H
H
N
H
H
O
O
H
H
H
H
H
O
H
H
29
H
H
H
N
H
H
H
5
H
11
H
H
y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\1a acid-base list, 315, 27p.DOC
Acid/Base stuff
Beauchamp
Two step changes in either direction in acid
H
O
H
O
O
H
H
H
H
H
H
H
H
N
N
H
8
N
H
H
H
H
H
9
H
H
1
H
H
30
H
H
H
H
O
O
O
O
H
H
H
H
H
H
H
H
H
H
H
N
H
H
H
4
H
N
H
7
H
H
H
H
H
H
H
H
H
H
N
N
H
N
H
H
H
6
H
4
H
H
H
H
H
H
H
H
H
H
H
H
N
H
H
1
H
O
H
H
H
N
N
H
H
O
O
O
H
H
H
12
H
H
O
H
13
H
H
O
H
H
H
H
5
H
H
H
H
N
3
H
H
H
H
N
H
11
H
H
H
H
O
H
H
H
13
H
H
H
N
H
6
H
H
H
O
H
H
N
H
H
H
H
H
N
H
N
O
H
H
O
H
H
H
H
O
H
N
H
O
H
H
O
H
H
H
H
H
H
H
H
1
H
8
O
H
H
H
10
N
H
H
6
O
O
H
H
H
H
H
1
H
H
N
H
H
H
9
H
H
N
N
H
H
N
H
4
H
H
H
10
H
y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\1a acid-base list, 315, 27p.DOC
Acid/Base stuff
Beauchamp
31
Problem 2 – Another small change using three carbon atoms. These changes can occur in acid or in base.
OH
OH
H2N
O
O
H2N
H
OH
OH
H2N
OH
H
E/Z possibilities
OH
H2N
H
OH
HN
H
E/Z possibilities
OH
H
E/Z possibilities
The DNA and RNA bases have a lot of tautomeric possibilities.
H
H
H
N
N
N
H
H
O
N
O
O
H
Adenine
A
H
N
N
N
H
N
N
N
H
N
N
H
N
N
N
Guanine
H
T
G
2 hydrogen bonds
N
O
O
N
H
H
H
Cytosine
Thymine
DNA
Uracil
RNA
O
C
3 hydrogen bonds
Another example of keto/enol tautomer problems. (It never ends.)
H
H
H
O
H
O
acid or base
H
H
Circle the protons lost
in the starting structure.
H
Circle the protons gained
in the product structure.
1. The circled protons have to be moved, either taken off (with the best base available) or put on (with the best
acid available).
Best acid in H3O+/H2O is H3O+, best base in H3O+/H2O is H2O
best base in H2O/HO -- is HO -- and the best acid in H2O/HO -- is H2O
2. Always work from a "keto" or "enol" part of the molecule. Do not use isolated pi bonds to change the
structure. The keto or enol part will be the better base or the better acid, as is indicated because it will form a
resonance stabilized intermediate with the oxygen (or nitrogen, or…) assisting in the resonance structures.
3. If in acid, use the strongest acid to put on a "gained" proton first and take off a "lost" proton second, usually by
a weak base (solvent).
4. If in base, use the strongest base to take off a "lost" proton first and put on a "gained" proton second usually by
a weak acid (solvent).
y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\1a acid-base list, 315, 27p.DOC
Acid/Base stuff
Beauchamp
32
5. In all mechanisms there will be resonance structures in the intermediate formed. The intermediate structure
will show the way to all reasonable tautomers of that intermediate. You may have to repeat the process once,
twice, etc. until you accomplish the overall indicated transformation. Counting the number of protons lost or
the number of protons gained will give you an indication of how many times you may have repeat the
tautomerization process. This may not always match however. Finally, there is always the same number of pi
bonds in each tautomer and same overall formal charge.
Example1 (in strong acid): In acid a proton goes on first, then another proton comes off. There is no enol, only a
keto, so begin there and make into an enol. In this example it is difficult to see this because the proton added to the
oxygen will be lost at the end and will not show up in the overall transformation. Rule 2 requires that we begin at a
keto or an enol, so this our only option. In the second step we remove the proton that allows conjugation with the
middle, isolated pi bond with the newly formed enol. Once conjugated, the middle, formerly isolated double bond
can be protonated forming a new cationic intermediate having resonance structures, including the oxygen with full
octets. This then allows you to take off one of the lost protons to form an extended enol. The enol can gain a
proton at the other "gained" position and then the "invisible" proton transfer occurs when a proton is taken off the
oxygen. Try to think these steps through on your own after studying the problem.
H
H
H
H
O
overall transformation
O
H
H
H
H
H
O
H
Note that all proton transfers in
this problem occur between weak
bases and strong acids.
H
H
H
H
H
O
H
H
H
H
H
H
O
H
H
H
O
H
H
O
H
H
O
H
H
resonance
H
H
OH2
resonance
H
H2O
H
H
O
H
H
H
O
H
H
H
O
O
H
H
H
H
resonance
H
H
H2O
H
H
H
H
resonance
H
H
O
H
resonance
H
H
H
H
O
H
resonance H
H
OH2
O
H
H
H
H
O
H
H
H
H
H
y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\1a acid-base list, 315, 27p.DOC
Redo this problem
using a base solution,
or work it backwards
in acid or base.
Acid/Base stuff
Beauchamp
33
Example 2 (strong base): Begin at the enol portion and remove one of the lost protons (on oxygen) and generate
an enolate using the best base, HO-. Draw the other resonance form and reprotonate at the carbanion site. Pull off a
proton from the other "lost" position, which is now tied into the keto structure (negative charged pushed to oxygen).
This will form a resonance delocalized extended enolate. There are several "negative" sites that could be
protonated. You need to protonate the site that the problem tells you picked up a new proton. That generates the
final structure.
O
O
H
overall transformation
H
H
O
H
H
H
O
H
Note that all proton transfers
in this problem occur between
strong bases and a weak acid.
H
H
H
H
O
O
O
resonance
O
H
H
H
O
H
H
H
H
H
resonance
resonance
H
H
H
H
H
H
O
O
O
Redo this problem
using an acid solution,
or work it backwards
H
in base or acid.
O
H
H
H
Additional problems: Try to approach each tautomer problem in some systematic manner similar to the above (in
acid and/or base).
H
O
H
O
O
H
O
H
O
many steps
H
H
H
O
H
O
O
many steps
H
H
H
H
H H
Same molecule redrawn with H's
lost and added in the next sequence.
Note: There are always the same number of pi bonds and the same overall charge in each tautomer.
y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\1a acid-base list, 315, 27p.DOC
Acid/Base stuff
Beauchamp
34
Example tautomer reactions in acid and in base.
H
O
H
H
O
N
H
H
H
O
H
H
H
O
O
H
H
H
H
H
H
H
H
N
H
H
resonance
N
H
N
H
H
13
H
H
O
H
H
H
1
H
O
H
H
H
tautomer reactions in base
resonance
step 1
H
H
H
H
H
H
resonance
N
H
H
H
H
H
H
H
H
H
O
H
H
O
H
H
1
H
13
H
H
H
resonance
N
H
H
H
H
H
O
H
H
H
H
H
N
H
H
H
O
H
H
resonance
N
H
H
O
H
H
resonance
H
O
H
N
H
H
H
H
H
H
resonance
N
H
tautomer reactions in acid
O
H
H2O
H
H
H
H
H
H
H
H
step 1
H
H
O
H
H
H
H
H
H
N
H
H
O
H
N
H
H
H
H
H
H
N
O
O
H
H
H
H
H
H
O
H
O
N
H
H
H
H
N
H
H
H
H
H
O
H
H
H
resonance
N
H
H
O
O
O
O
H
H
O
O
H
H
H
H
H
N
H
step 2
H
H
y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\1a acid-base list, 315, 27p.DOC
H
N
H
H
H