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96 CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
3.4 BRØNSTED–LOWRY ACIDS AND BASES
A. Definition of Brønsted Acids and Bases
Although less general than the Lewis concept, the Brønsted – Lowry acid – base concept pro-
vides another way of thinking about acids and bases that is extremely important and useful in
organic chemistry. The Brønsted–Lowry definition of acids and bases was published in 1923,
the same year that Lewis formulated his ideas of acidity and basicity. A species that donates a
proton in a chemical reaction is called a Brønsted acid; a species that accepts a proton in a
chemical reaction is a Brønsted base.
The reaction of ammonium ion with hydroxide ion (see Eq. 3.5) is an example of a Brøn-
sted acid–base reaction.
On the left side of this equation, the ammonium ion is acting as a Brønsted acid and the hy-
droxide ion is acting as a Brønsted base; looking at the equation from right to left, water is act-
ing as a Brønsted acid, and ammonia as a Brønsted base.
The “classical” definition of a Brønsted acid–base reaction given above focuses on the
movement of a proton. But in organic chemistry, we are always going to focus on the move-
ment of electrons. As Eq. 3.9a illustrates, any Brønsted acid–base reaction can be described
with the curved-arrow notation for electron-pair displacement reactions. A Brønsted
acid–base reaction is nothing more than an electron-pair displacement reaction on a hydro-
gen. It’s the action of the electrons that causes the net transfer of a proton from the Brønsted
acid to the Brønsted base. In terms of electrons, a Brønsted base, then, is merely a special case
of a Lewis base: a Brønsted base is a Lewis base that donates its electron pair to a proton. A
Brønsted acid is the species that provides a proton to the base.
(3.9a)
H L L H
H
H
H L 21
H
H
_
N + OH 21 N + HL OH
|
ammonium ion
(a Brønsted acid)
hydroxide ion
(a Brønsted base)
ammonia
(a Brønsted base)
water
(a Brønsted acid)
the transferred
proton
and that the unshared electron pair (and negative charge) is shared equally by the two ter-
minal carbons.
(c) Using the curved-arrow notation, derive a resonance structure for benzene (shown here)
which shows that all carbon–carbon bonds are identical and have a bond order of 1.5.
benzene
i?
allyl anion
?
L H 2
C CH
_
CH 2
A
1
3.4 BRØNSTED–LOWRY ACIDS AND BASES 97
When a Brønsted acid loses a proton, its conjugate base is formed; when a Brønsted base
gains a proton, its conjugate acid is formed. When a Brønsted acid loses a proton, it becomes
a Brønsted base; this acid and the resulting base constitute a conjugate acid – base pair. In any
Brønsted acid–base reaction there are two conjugate acid–base pairs. Hence, in Eq. 3.9,
|
NH
4
and NH
3
are one conjugate acid–base pair, and H
2
O and
_
OH are the other.
Notice that the conjugate acid–base relationship is across the equilibrium arrows. For
example,
|
NH
4
and NH
3
are a conjugate acid–base pair, but
|
NH
4
and
_
OH are not a conju-
gate acid–base pair.
The identification of a compound as an acid or a base depends on how it behaves in a spe-
cific chemical reaction. Water, for example, can act as either an acid or a base. Compounds
that can act as either acids or bases are called amphoteric compounds. Water is the arche-
typal example of an amphoteric compound. For example, in Eq. 3.10, water is the conjugate
acid in the acid–base pair H
2
OÜ
_
OH; in the following reaction, water is the conjugate base in
the acid–base pair H
3
O
|
ÜH
2
O:
PROBLEMS
3.5 In the following reactions, label the conjugate acid–base pairs. Then draw the curved-arrow
notation for these reactions in the left-to-right direction.
(a)
(b)
3.6 Write a Brønsted acid–base reaction in which and act as conju-
gate acid–base pairs.
2 3
2
CH
3
OH CH
3
O
_
2
2
2 3!
2
H
2
O 2
2
OH
_
!
NH 2
3
NH 2
3
2
_
NH
4
|
2
NH
3
3 2
2
OH
_
_
2
NH
2
H 23
2
O
(3.11)
L L
acid
H
H H
|
H L L H O
H
acid
H
|
N
base
L H
H
3 O
L
base
H 3
H
H
+ N +
L L (3.10)
H 1
H
H
H L L H H OH
H
H
|
N OH
_
+ 1 N +
conjugate acid–base pair
conjugate base–acid pair
(3.9b)
H
L L
H
H
H
3 H 3 21
L
H
H
_
H
L
N + OH 21 N OH
|
Brønsted acid:
a species that provides
the proton
Brønsted base:
a Lewis base that donates
an electron pair
to a proton
the transferred
proton
Brønsted acid–base reaction:
an electron-pair displacement
reaction on hydrogen
STUDY GUIDE LINK 3.
Identification of
Acids and Bases
3.4 BRØNSTED–LOWRY ACIDS AND BASES 99
The group that receives electrons from the breaking bond, in this case the —Br, is called a
leaving group. A group becomes a leaving group when one of its atoms accepts an electron
pair from a breaking bond. One of the author’s students has put it less formally, but perhaps
more descriptively: a leaving group is a group that takes a pair of bonding electrons and runs.
A leaving group in one direction of a reaction becomes a nucleophile in the other direction,
as the following example demonstrates.
The term leaving group can also be used in Lewis acid–base dissociation reactions. Note
the absence of a nucleophile in a Lewis acid–base dissociation, and the absence of a leaving
group in a Lewis acid–base association.
The terms you’ve learned here are important because they are used throughout the world of
organic chemistry. Your instructor will use them, and we’ll use them throughout this book. To
summarize:
- A Lewis base can act as either a Brønsted base or a nucleophile. A Brønsted base is a
Lewis base that donates a pair of electrons to a hydrogen and removes the hydrogen as
a proton. A nucleophile is a Lewis base that donates a pair of electrons to an atom other
than hydrogen.
(3.17)
.. ..
..
..
B F
..
..
..
F
.. ..
..
F
.. ..
..
F
.. ..
..
F
.. ..
.. ..
..
..
..
..
..
B F
F
F
a nucleophile in the
association reaction;
there is no leaving group
a Lewis acid–base
association
a Lewis acid–base
dissociation
a leaving group in the
dissociation reaction;
there is no nucleophile
H (3.16)
2
O H
3
C Br CH
3
..
..
..
Br
.. ..
..
..
..
..
H
2
O
..
a nucleophile in the
forward direction
a leaving group in the
forward direction
a leaving group in the
reverse direction
a nucleophile in the
reverse direction
HO H (3.15)
3
C Br
..
..
..
..
..
..
a leaving group
HO H Br (3.14)
..
..
..
..
..
..
HO H
3
C Br
..
..
..
..
..
..
a Lewis acid and
an electrophile
a Lewis acid and
an electrophile
a Brønsted acid
100 CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
- An electrophil e is the atom that receives an electron pair from a Lewis base.
- A leaving group is a group containing an atom that accepts an electron pair from one of
its bonds, which is broken as a result.
This section is about analyzing the roles of the various species in reactions. The definitions
developed here describe these roles. You will find that most of the reactions you will study can
be analyzed in terms of these roles, and hence an understanding of this section will prove to
be crucial in helping you to understand and even predict reactions. The first step in this under-
standing is to apply these definitions in the analysis of reactions. Study Problem 3.5 points you
in this direction.
Study Problem 3.
Following is a series of acid–base reactions that represent the individual steps in a known organic
transformation, the replacement of —Br by —OH at a carbon bearing three alkyl groups. Consid-
ering only the forward direction, classify each reaction as a Brønsted acid–base reaction or a
Lewis acid–base association/dissociation. Classify each labeled species (or a group within each
species) with one or more of the following terms: Brønsted base, Brønsted acid, Lewis base,
Lewis acid, nucleophile, electrophile, and/or leaving group.
Solution Classify each reaction first, and then analyze the role of each species. Reaction 3.18a
is a Lewis acid–base dissociation. (Notice that a single curved arrow describes the dissociation.)
In compound A, Br is the leaving group.
Reaction 3.18b is a Lewis acid–base association reaction. Cation B (specifically, its electron-
deficient carbon) is a Lewis acid and an electrophile. Water molecule D is a Lewis base and a nu-
cleophile.
Reaction 3.18c is a Brønsted acid–base reaction. Ion E and compound G constitute a conjugate
Brønsted acid–base pair, and compound F and compound H are a conjugate Brønsted base–acid
pair. The water molecule F is both a Lewis base and a Brønsted base. The proton of E that re-
ceives an electron pair from water is a Lewis acid and is also an electrophile. The part of E that
becomes G is a leaving group.
(3.18c)
E
F H
G
H OH
.. 2
OH
.. 2
..
O
H
..
CH
3
CH
3
H
3
C C
..
O H
H
..
CH
3
CH
3
H
3
C C +
O H (3.18b)
H
..
..
CH 3
CH 3
H 3
C C O H
H
..
CH 3
CH 3
H 3
C C
B
D
E
(3.18a)
..
Br
..
..
..
CH 3
CH 3
H
3
C C
CH 3
CH 3
H
3
C C Br
..
..
..
A B
C
102 CHAPTER 3 • ACIDS AND BASES. THE CURVED-ARROW NOTATION
Each acid has its own unique dissociation constant. The larger the dissociation constant of an
acid, the more H
3
O
|
ions are formed when the acid is dissolved in water at a given concentra-
tion. Thus, the strength of a Brønsted acid is measured by the magnitude of its dissociation
constant.
Because the dissociation constants of different Brønsted acids cover a range of many pow-
ers of 10, it is useful to express acid strength in a logarithmic manner. Using p as an abbrevi-
ation for negative logarithm, we can write the following definitions:
p K
a
= -log K
a
(3.22a)
pH = -log [H
3
O
|
] (3.22b)
Some p K
a
values of several Brønsted acids are given in Table 3.1 in order of decreasing p K
a
Because stronger acids have larger K
a
values, it follows from Eq. 3.22a that stronger acids
have smaller p K
a
values. Thus, HCN (p K
a
= 9.4) is a stronger acid than water (p K
a
In other words, the strengths of acids in the first column of Table 3.1 increase from the top to
the bottom of the table.
It is important to understand the difference between pH and p K
a
. The pH is a measure of proton con-
centration and is an experimentally alterable property of a solution. The p K
a
is a fixed property of a
Brønsted acid. If this difference isn’t secure in your mind, be sure to consult Study Guide Link 3.
for further help.
Three points about the p K
a
values in Table 3.1 are worth special emphasis. The first has to
do with the p K
a
values for very strong and very weak acids. The direct p K
a
determination of
an acid in aqueous solution is limited to acids that are less acidic than H
3
O
|
and more acidic
than H
2
O. The reason is that H
3
O
|
is the strongest acid that can exist in water. If we dissolve
a stronger acid in water, is immediately ionizes to H
3
O
|
. Similarly,
_
OH is the strongest base
that can exist in water, and stronger bases react instantly with water to form
_
OH. However,
p K
a
values for very strong and very weak acids can be measured in other solvents, and
through various methods these p K
a
values can in many cases be used to estimate aqueous p K
a
values. This is the basis for the estimates of the acidities of strong acids such as HCl and very
weak acids such as NH
3
in Table 3.1. These approximate p K
a
values will suffice for many of
our applications.
The second point is that much important organic chemistry is carried out in nonaqueous
solvents. In nonaqueous solvents, p K
a
values typically differ substantially from p K
a
values of
the same acids determined in water. However, in some of these solvents, the relative p K
a
val-
ues are roughly the same as they are in water. In other nonaqueous solvents, though, even the
relative order of p K
a
values is different. (We’ll learn about solvent effects in Chapter 8.) De-
spite these differences, aqueous p K
a
values such as those in Table 3.1 are the most readily
available and comprehensive data on which to base a discussion of acidity and basicity.
The last point has to do with the K
a
of water, which is 10
_15.
, from which we obtain p K
a
15.7. Don’t confuse this with the ion-product constant of water, which is defined by the
3.11 What is the p K
a
of an acid that has a dissociation constant of
(a) 10
_ 3
(b) 5.8 X 10
_ 6
(c) 50
3.12 What is the dissociation constant of an acid that has a p K
a
of
(a) 4 (b) 7.8 (c) - 2
3.13 (a) Which acid is the strongest in Problem 3.11?
(b) Which acid is the strongest in Problem 3.12?
PROBLEMS
STUDY GUIDE LINK 3.
The Difference
between p K
a
and pH
3.4 BRØNSTED–LOWRY ACIDS AND BASES 103
expression K
w
= [H
3
O
|
][
_
OH] = 10
_ 14
M
2
, or - log K
w
= 14. The ionization constant of water
is defined by the expression
K
a
_15.
M
This expression has the concentration of water itself in the denominator, and thus differs from
the ion-product constant of water by a factor of 1/55.5. The logarithm of this factor, - 1.7, ac-
counts for the difference between p K
a
and p K
w
p K
a
of H
2
O = - log K
w
D. Strengths of Brønsted Bases
The strength of a Brønsted base is directly related to the p K
a
of its conjugate acid. Thus, the
base strength of fluoride ion is indicated by the p K
a
of its conjugate acid, HF; the base strength
of ammonia is indicated by the p K
a
of its conjugate acid, the ammonium ion,
|
NH
4
. That is,
when we say that a base is weak, we are also saying that its conjugate acid is strong; or, if a
_ 14
M
2
55.5 M
K
w
[H
2
O]
[H
3
O
|
][
_
OH]
[H
2
O]
Relative Strengths of Some Acids and Bases
Conjugate acid p K
a
Conjugate base
(ammonia) ; 35
†
(amide)
(alcohol) 15–19* (alkoxide)
(water) 15.7 (hydroxide)
(thiol) 10–12* (thiolate)
(trialkylammonium ion) 9–11* (trialkylamine)
(ammonium ion) 9.25 (ammonia)
HCN (hydrocyanic acid) 9.40 (cyanide)
(hydrosulfuric acid) 7.0 (hydrosulfide)
(carboxylic acid) 4–5* (carboxylate)
(hydrofluoric acid) 3.2 (fluoride)
(hydronium ion) - 1.7 (water)
H
2
SO
4
(sulfuric acid) - 3
†
HSO
4
_
(bisulfate)
(hydrochloric acid) - 6 to - 7
†
(chloride)
(hydrobromic acid) - 8 to - 9.
†
(bromide)
(hydroiodic acid) - 9.5 to - 10
†
(Iodide)
HClO
4
(perchloric acid) - 10 (?)
†
ClO
4
_
(perchlorate)
*Precise value varies with the structure of R.
†
Estimates; exact measurement is not possible.
I 2
2
3 3
_
HI 2
2
3
Br 2
2
3 3
_
HBr 2
2
3
Cl 2
2
3 3
_
HCl 2
2
3
H
2
O 2
2
H 3
O 2
|
( p -toluene-
sulfonate, or
“tosylate”)
H 3
C SO 3
LcL
_
( p -toluene-
sulfonic acid)
H 3
C SO 3
H
LcL
F 2
2
3 3
_
HF 2
2
3
_
L L 3
S
2
2
C
O
L L R O
S
2
2
C
O
R OH
HS 2
2
3
_
H
2
S 2
2
3 CN
_
H 3
N 3 NH
4
|
R 3
N 3 R
3
NH
|
RS 2
2
3
_
RSH 2
2
HO 2
2
3
_
HOH 2
2
RO 2
2
3
_
ROH 2
2
NH 2
_
NH 32
3
2
TABLE 3.
GREATER ACIDITY
GREATER BASICITY
3.4 BRØNSTED–LOWRY ACIDS AND BASES 105
the p K
a
of the acid on the left of Eq. 3.23 (HCN) from the one on the right (H
2
O) gives the loga-
rithm of the desired equilibrium constant K
eq
. (The relevant p K
a
values come from Table 3.1.)
log K
eq
= 15.7 - 9.4 = 6.
The equilibrium constant for this reaction is the antilog of this number:
K
eq
= 10
= 2 X 10
6
This large number means that the equilibrium of Eq. 3.23 lies far to the right. That is, if we dis-
solve HCN in an equimolar solution of NaOH, a reaction occurs to give a solution in which there
is much more
_
CN than either
_
OH or HCN. Exactly how much of each species is present could
be determined by a detailed calculation using the equilibrium-constant expression, but in a case
like this, such a calculation is unnecessary. The equilibrium constant is so large that, even with
water in large excess as the solvent, the reaction lies far to the right. This also means that if we
dissolve NaCN in water, only a minuscule amount of
_
CN reacts with the H
2
O to give
_
OH and
HCN.
Sometimes students confuse acid strength and base strength when they encounter an am-
photeric compound (see p. 97). Water presents this sort of problem. According to the defini-
tions just developed, the base strength of water is indicated by the p K
a
of its conjugate acid,
H
3
O
|
, whereas the acid strength of water (or the base strength of its conjugate base hydrox-
ide) is indicated by the p K
a
of H
2
O itself. These two quantities refer to very different reactions
of water:
Water acting as a base:
(3.26a)
Water acting as an acid:
(3.26b)
3.15 Write an equation for each of the following equilibria, and use Table 3.1 to identify the p K
a
value associated with the acidic species in each equilibrium.
(a) ammonia acting as a base toward the acid water
(b) ammonia acting as an acid toward the base water
Which of these reactions has the larger K
eq
and therefore is more important in an aqueous so-
lution of ammonia?
PROBLEM
B 3 BH
_ _
H OH
2
O
p K a
= 15.
AH
p K
a
= - 1.
A 3
_
H
2
O H
3
O
|
3.14 Using the p K
a
values in Table 3.1, calculate the equilibrium constant for each of the following
reactions.
(a) NH
3
acting as a base toward the acid HCN
(b) F
_
acting as a base toward the acid HCN
PROBLEM
