Chapter 3
Acids and Bases
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 3. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• A Brønsted-Lowry acid is a proton ________, while a Brønsted-Lowry base is
a proton ________.
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 3. The answers appear in the section entitled
SkillBuilder Review.
SkillBuilder 3.1 Drawing the Mechanism of a Proton Transfer
HOH
DRAW THE CURVED ARROWS FOR THE FOLLOWING ACID-BASE REACTION
+ CH
3
O HO + CH
3
OH
CHAPTER 3
41
SkillBuilder 3.4 Using pK
a
Values to Predict the Position of Equilibrium
OO
OH
O O
H
O
H
pKa = 9.0 pKa = 15.7
CIRCLE THE SIDE OF THE EQUILIBRIUM THAT IS FAVORED:
SkillBuilder 3.5 Assessing Relative Stability. Factor #1: Atom
H
N O H
COMPARE THE TWO PROTONS SHOWN
IN THE FOLLOWING COMPOUND, AND
CIRCLE THE ONE THAT IS MORE ACIDIC.
USE THE EXTRA SPACE TO DRAW THE
TWO POSSIBLE CONJUGATE BASES.
SkillBuilder 3.8 Assessing Relative Stability. Factor #4: Orbital
COMPARE THE TWO PROTONS SHOWN IN
THE FOLLOWING COMPOUND, AND
CIRCLE THE ONE THAT IS MORE ACIDIC.
USE THE EXTRA SPACE TO DRAW THE
TWO POSSIBLE CONJUGATE BASES.
H
H
SkillBuilder 3.9 Assessing Relative Stability. Using all Four Factors
N
HH
COMPARE THE TWO PROTONS SHOWN IN
THE FOLLOWING COMPOUND, AND CIRCLE
THE ONE THAT IS MORE ACIDIC. USE THE
EXTRA SPACE TO DRAW THE TWO
POSSIBLE CONJUGATE BASES.
42
CHAPTER 3
SkillBuilder 3.12 Identifying Lewis Acids and Lewis Bases
O
HBH
H
IDENTIFY THE LEWIS ACID
AND THE LEWIS BASE
IN THE FOLLOWING REACTION:
+
Solutions
3.1.
a)
OH
+OH +
O
HOH
Acid Base Conjugate
A
c
i
d
Conjugate
B
a
s
e
3.2.
a) There is only one arrow, and it is going in the wrong direction. The tail has been
placed on the hydrogen atom, and this is incorrect. Curved arrows do not show the
motion of atoms, but the motion of electrons. The tail of this curved arrow should be on
the lone pair of the nitrogen atom, and the head of the curved arrow should be on the
proton. In addition, a second curved arrow is also required. It should look like this:
Cl
+
N N H
H Cl
CHAPTER 3
43
3.3.
O
O
H
O
O
OO
H
3.5.
O
O
N
H
H
pK
a
~ 16 pK
a
~ 38
(more acidic)
44
CHAPTER 3
3.7.
a)
H
N
H
b)
O
c)
H
O
H
d)
O
e)
C C
H
H
H
H
H
f)
O
H
3.8.
H
strongest base
3.9.
a) A lone pair on a nitrogen atom will be more basic than a lone pair on an oxygen atom.
b) The lone pair on the nitrogen atom is thirteen orders of magnitude more basic than the
lone pair on the oxygen atom.
3.10.
a) left side
3.11. The equilibrium does not favor deprotonation of acetylene by hydroxide, because
water is more acidic than acetylene. The equilibrium will favor the weaker acid
(acetylene). A suitable base would be one whose conjugate acid is less acidic than
acetylene. For example, H
2
N‾ would be a suitable base, because ammonia (NH
3
) is less
acidic than acetylene.
3.13.
a)
O
H
b)
H C N
H
H
H
H
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45
3.14. A proton connected to a sulfur atom will be more acidic than a proton connected to
3.15.
a)
H
b)
N
HN
H
c)
H
3.16.
O
OHO
O
HO
HO
H
The proton highlighted above is the most acidic proton in the structure, because
deprotonation at that location generates a resonance-stabilized anion, in which the
negative charge is spread over two oxygen atoms and one carbon atom:
O
OHO
O
HO
HO
O
OHO
O
HO
HO
O
OHO
O
HO
HO
3.17.
OH
O
O
H
more acidic
3.18.
a) The highlighted proton is more acidic. When this location is deprotonated, the
conjugate base that is formed is stabilized by the electron-withdrawing effects of the
46
CHAPTER 3
b) The highlighted proton is more acidic. When this location is deprotonated, the
conjugate base that is formed is stabilized by the electron-withdrawing effects of the
electronegative chlorine atoms, which are closer to this proton than the other proton:
HO OH
O
O
Cl Cl
3.19.
a) The compound with two chlorine atoms is more acidic, because of the electron-
withdrawing effects of the additional chlorine atom, which help stabilize the conjugate
base that is formed when the proton is removed:
3.20.
a) In the compound below, one of the chlorine atoms has been moved closer to the acidic
proton, which further stabilizes the conjugate base that is formed when the proton is
removed:
OH
O
C
l
C
l
CHAPTER 3
47
3.21. Both protons are the same distance from the fluorine atom, and both protons are
the same distance from the chlorine atom. Accordingly, these protons are expected to be
of equivalent acidity.
3.22. The compound below (acetylene) is more acidic. The conjugate base of this
C
C
H
H
3.23.
H
H
H
3.24. Most imines will have a pK
a
below 35, because imines are expected to be more
3.25.
a)
OH
b)
OO
H
c)
HO NH
2
O
N
3.26.
a)
H
B
r
b)
H
2
S
c)
NH
3
d)
H
H
e)
Cl
3
C CCl
3
OH
48
CHAPTER 3
3.27.
a) When the proton is removed, the resulting conjugate base is highly resonance
stabilized because the negative charge is spread over four nitrogen atoms and seven
oxygen atoms. In addition, the inductive effects of the trifluoromethyl groups (-CF
3
)
Alternatively, the conjugate base could be further stabilized by spreading the charge over
a larger number of nitrogen and oxygen atoms, for example:
SN N
O
CF
3
S
N
OF
3
C
S
O
N
F
3
C
S
N
O CF
3
S
O
N
CF
3
H
SS
CF
3
O
O
F
3
C
O
O
The additional structural units (highlighted above) would enable the conjugate base to
spread its negative charge over six nitrogen atoms and nine oxygen atoms, which should
be even more stable than being spread over four nitrogen atoms and seven oxygen atoms.
3.29.
a) the right side
b) the left side
c) the right side
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49
3.30.
O
O
H
O
O
OH
O
The equilibrium favors the right side because the negative charge is resonance stabilized.
3.31.
a) Yes, because a negative charge on an oxygen atom will be more stable than a
negative charge on a nitrogen atom.
3.32.
3.33. Water is more acidic than ethanol. Indeed, the pK
a
of water (15.7) is lower than the
pK
a
of ethanol (16).
3.34.
a)
+Cl Al Cl
Cl
O O Al
Cl
Cl
Cl
L
e
w
i
s
B
a
s
e
L
e
w
i
s
A
c
i
d
50
CHAPTER 3
c)
+Br Al Br
Br
Br Br
Br
Al Br
Br
Br Br
L
e
w
i
s
B
a
s
e
L
e
w
i
s
A
c
i
d
3.35.
O N
3.36.
O
O
N
O
3.37.
O
N
H
O
H
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51
3.38. Compound A is 1000 times more acidic than compound B.
3.39. In each reaction below, identify the Lewis acid and the Lewis base:
a)
OH +O
H
L
e
w
i
s
B
a
s
e
L
e
w
i
s
A
c
i
d
H
3.40.
OH
+
+
H
OH
HNHHNH
H
3.41. No, because the leveling effect would cause the deprotonation of ethanol to form
ethoxide ions, and the desired anion would not be formed under these conditions.
3.42. No, water would not be a suitable proton source in this case. This anion is the
3.43.
a)
H Br
HOHHOH
H
Br
+ +
52
CHAPTER 3
c)
HOHHOH
H
++
O
HO
3.44.
a)
HOHOH
++
3.45.
a) The second anion is more stable because it is resonance stabilized.
b) The second anion is more stable because the negative charge is on a nitrogen atom,
rather than an sp
3
hybridized carbon atom.
c) The second anion is more stable because the negative charge is on an sp hybridized
carbon atom, rather than an sp
3
hybridized carbon atom.
3.46.
a)
N
H
b)
O
C
l
Cl Cl
3.47.
SH
OH
OH
Cl
Cl
Cl
Cl
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53
3.48.
H A B
Na
+
H B A
Na
+
3.49.
a)
HOH
O
++
OH
HO
d)
3.50.
HO OO
O
O
Increasing base strength
3.51.
54
CHAPTER 3
d)
O
H
e)
HO
O
OH
NH
2 f)
H
O
H
3.52.
a)
O
O
HOH O
O
HOH
a
c
i
d
b
a
s
e
3.53.
Increasing acidity
S
H
S
CHAPTER 3
55
3.55.
a) There is only one sp
3
hybridized carbon atom in cyclopentadiene.
b) The most acidic proton in cyclopentadiene is highlighted below:
H
3.56. When salicylic acid is deprotonated, the resulting conjugate base is further
stabilized by intramolecular hydrogen bonding:
O
O
O
H
3.57.
OH
O
or OH
O
3.58. The four constitutional isomers are shown below.
NO
2
NO
2
NO
2
NO
2
56
CHAPTER 3
The last compound is expected to have the highest pK
a
because its conjugate base is not
resonance stabilized. The other three compounds have resonance-stabilized conjugate
bases, for example:
N
O
ON
O
O
3.59. Compare the conjugate bases. Both are resonance stabilized. But the conjugate
base of the first compound has a negative charge spread over two nitrogen atoms and two
3.60.
a) The two most acidic protons are labeled H
a
and H
b
:
N
N
N
H
a
N
H
b
3.61.
a) When R is a cyano group, the conjugate base is resonance stabilized:
C N
H
N C N
H
N