Chapter 6
Chemical Reactivity and Mechanisms
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 6. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• ________________ reactions involve a transfer of energy from the system to the
surroundings, while ______________ reactions involve a transfer of energy from
the surroundings to the system.
• Each type of bond has a unique ________________________ energy, which is
the amount of energy necessary to accomplish homolytic bond cleavage.
• Entropy is loosely defined as the ___________ of a system.
• In order for a process to be spontaneous, the change in ____________________
must be negative.
CHAPTER 6
93
Review of Skills
Fill in the empty boxes below. To verify that your answers are correct, look in your
textbook at the end of Chapter 6. The answers appear in the section entitled SkillBuilder
Review.
SkillBuilder 6.2 Identifying Nucleophilic and Electrophilic Centers
HC
H
H
Li
δ
δδ
δ–
H
O
H
HC
H
H
Cl
CLASSIFY EACH OF THE HIGHLIGHTED REGIONS BELOW AS EITHER A NUCLEOPHILIC CENTER OR AN ELECTROPHILIC CENTER
δ
δδ
δ+
SkillBuilder 6.3 Identifying an Arrow Pushing Pattern
94
CHAPTER 6
SkillBuilder 6.4 Identifying a Sequence of Arrow Pushing Patterns
O
H
H Br O
H H
Br
Br
– H
2
O
IDENTIFY EACH OF THE FOLLOWING ARROW PUSHING PATTERNS
SkillBuilder 6.5 Drawing Curved Arrows
CHAPTER 6
95
Solutions
6.1.
a)
Bonds Broken
kJ/mol
Bonds Formed
kJ/mol
H—CH(CH
3
)
2
+ 397 (CH
3
)
2
CH—Br – 285
Br—Br + 192 H—Br – 368
c)
Bonds Broken
kJ/mol
Bonds Formed
kJ/mol
(CH
3
)
3
C—Br + 272 (CH
3
)
3
C—OH – 381
H—OH + 498 H—Br – 368
Sum = + 21 kJ/mol.
∆Hº for this reaction is positive, which means that the system is gaining energy.
It is receiving energy from the environment, so the reaction is endothermic.
96
CHAPTER 6
6.2.
The C-C bond of CH
3
—CH
3
has a bond dissociation energy of = +368 kJ/mol. If
a C=C bond has a total bond dissociation energy of +632 kJ/mol, then the pi
component of the double bond can be estimated to be (632 kJ/mol) – (368 kJ/mol)
6.3.
a) ∆S
sys
is expected to be negative (a decrease in entropy) because two molecules
are converted into one molecule.
b) ∆S
sys
is expected to be negative (a decrease in entropy) because an acylic
6.4.
a) There is a competition between the two terms contributing to ∆G. In this case,
the reaction is endothermic, which contributes to a positive value for ∆G, but the
second term contributes to a negative value for ∆G:
G = H T S
–
+
The sign of ∆G will therefore depend on the competition between these two
terms, which is affected by temperature. A high temperature will cause the
second term to dominate, giving rise to a positive value of ∆G. A low
CHAPTER 6
97
temperature will render the second term insignificant, and the first term will
dominate, giving rise to a negative value of ∆G.
b) In this case, both terms contribute to a negative value for ∆G, so ∆G will
definitely be negative (the process will be spontaneous).
6.5. A system can only achieve a lower energy state by transferring energy to its
surroundings (conservation of energy). This increases the entropy of the
surroundings, which more than offsets the decrease in entropy of the system. As a
result, ∆S
tot
increases.
6.6.
a) A positive value of ∆G favors reactants.
6.7.
a) Process D will occur more rapidly because it has a lower energy of activation
than process A.
b) Process A will more greatly favor products at equilibrium than process B,
because the former is exergonic (the products are lower in energy than the
98
CHAPTER 6
d) In process A, the transition state resembles the reactants more than products
because the transition state is closer in energy to the reactant than the products
(the Hammond postulate).
6.8.
a)
L
i
b)
OH
NH
2
c)
N
d)
6.9.
a)
OH
O
b)
O
6.10.
L
i
B
Electrophilic
center
Nucleophilic
center
6.11.
a)
H
O
b)
O
c)
O
CHAPTER 6
99
6.13. The pi bond functions as a nucleophile and attacks the electrophilic carbocation.
This step is therefore a nucleophilic attack.
6.14.
a) proton transfer; nucleophilic attack; proton transfer
6.15. Both reactions have the same sequence: 1) nucleophilic attack, followed by 2)
loss of a leaving group. In both cases, a hydroxide ion functions as a nucleophile
and attacks a compound that can accept the negative charge and store it temporarily.
The charge is then expelled as a chloride ion in both cases.
6.16.
a)
HOH
O
H
H
+
6.17.
a)
Br
Cl
Cl + Br
Cl +
100
CHAPTER 6
c)
Cl
H
RO
+ROH +Cl
6.18.
a)
H
b) This carbocation is tertiary and will not rearrange
c)
tertiary tertiary allylic
d) This carbocation is secondary, but it cannot rearrange to form a tertiary
carbocation.
6.19.
CHAPTER 6
101
6.20.
a) a carbon-carbon triple bond is comprised of one sigma bond and two pi bonds,
and is therefore stronger than a carbon-carbon double bond (one sigma and one pi
6.21.
a)
Bonds Broken
kJ/mol
Bonds Formed
kJ/mol
RCH
2
—Br + 285 RCH
2
—OR – 381
RCH
2
O—H + 435 H—Br – 368
6.22.
a) A reaction for which K
eq
> 1 will favor products.
6.23. K
eq
= 1 when ∆G = 0 kJ/mol (See Table 6.2).
6.24. K
eq
< 1 when ∆G has a positive value. The answer is therefore “a” (+1 kJ/mol)
6.25.
a) ∆S
sys
is expected to be negative (a decrease in entropy) because two moles of
reactant are converted into one mole of product.
102
CHAPTER 6
6.26.
a)
Reaction
C
o
o
r
d
i
n
a
t
e
Free
Energy
b)
Reaction
C
o
o
r
d
i
n
a
t
e
Free
Energy
c)
Reaction
C
o
o
r
d
i
n
a
t
e
Free
Energy
6.27.
a) B and D
b) A and C
6.28. All local minima (valleys) represent intermediates, while all local maxima (peaks)
represent transition states:
Transition States
CHAPTER 6
103
6.29.
a) Rate = k[nucleophile][substrate].
b) The rate will be tripled, because the rate is linearly dependent on the
6.30.
a) loss of a leaving group
6.31.
Increasing Stability
6.32.
Cl Al Cl
Cl
Cl Al
Cl
Cl
Cl Cl Al
Cl
Cl
Cl
O
Cl
O
NUCLEOPHILIC
ATTACK
LOSS OF A
LEAVING GROUP
O
104
CHAPTER 6
6.33.
H
O
S
O
O
O
O
6.34.
S
O O
OH
O S O
O
O
HH HS
O
O
OH
HS
O
O
OH
HS
O
O
O
H
HOH
SO
3
PROTON
TRANSFER PROTON
TRANSFER
6.35.
NO
2
OH
NO
O
Br
CHAPTER 6
105
6.36.
OH
OEt
O
Me
EtOH
OH
O
O
H
Et
Me
EtOH
OMe
OH
H O
Et
H
OMe
O
NUCLEOPHILIC
ATTACK
PROTON
TRANSFER
PROTON
TRANSFER
PROTON
TRANSFER
6.37.
R Cl
OH O H O
O
Cl
R
H
H
Cl
R O
O
H
H
H O H
R OH
O
–
NUCLEOPHILIC
ATTACK
LOSS OF A
LEAVING GROUP
PROTON
TRANSFER
6.38.
O
O
O
O
NUCLEOPHILIC
ATTACK
LOSS OF A
LEAVING GROUP
NUCLEOPHILIC
ATTACK
106
CHAPTER 6
6.39.
OH
OR
OH A
OH
R
OH
OH
O
H
R
A
NUCLEOPHILIC
ATTACK
PROTON
TRANSFER
PROTON
TRANSFER
PROTON
TRANSFER
6.40.
OH
NR
R
R
NRH
H A A
OH
OOH
N
H
RR
NUCLEOPHILIC
ATTACK PROTON
TRANSFER
PROTON
TRANSFER
CHAPTER 6
107
6.41.
NN
H
H N N
H
NN
H
HNN
H
PROTON
TRANSFER
PROTON
TRANSFER
OH HOH
6.42.
N
O
S
O
O
H
3
C
NO
2
HO
O
O
OH
N
O
S
O
O
H
3
C
O
O
OH
– CH
3
SO
3
108
CHAPTER 6
6.43.
N
NH
2
HO
O
H
H
N
NH
2
H
O
OSO
3
H
OSO
3
HH
N
NH
2
O
OSO
3
HH
O O
– H
2
O
CHAPTER 6
109
6.44.
N
N H
N H
OH
N H
O
H
H
H
O
H
H
H
O
HH
O
H
H
6.45.
H
O
H
O
H
O
H
O
H
H
HO
H
O
O
H
H
H
O
H
H
H
O
H
O
110
CHAPTER 6
6.46.
N
N
H
OH
N
H
O
H
H
N
H
H
O
H
H
O
H
H
H
O
H
H
6.47.
OO OHH
H
OH
HO OH HO O
HHO
HO O
HO
HO OH
6.48.
a)
b)
H
CHAPTER 6
111
H
6.49.
a)
H
3
C
C
H
HBr
CH
3
C
H
H
OBrOH ++
H
b) Nucleophilic attack and loss of a leaving group.
c) A CH
3
CH
2
—Br is broken, and a CH
3
CH
2
—OH is formed. Using the data in
Table 6.1, ∆H for this reaction is expected to be approximately (285 kJ/mol) –
(381 kJ/mol). The sign of ∆H is therefore predicted to be negative, which means
that the reaction should be exothermic.
d) Two chemical entities are converted into two chemical entities. Both the
reactants and products are acyclic. Therefore, ∆S for this process is expected to
f) The position of equilibrium is dependent on the sign and value of ∆G. As
mentioned in part e, ∆G is comprised of two terms. The effect of temperature
appears in the second term (-T∆S), which is insignificant because ∆S is
approximately zero. Therefore, an increase (or decrease) in temperature is not
expected to have a significant impact on the position of equilibrium.
g) This transition state corresponds with the peak of the curve, and has the
following structure:
112
CHAPTER 6
6.50.
a) K
eq
does not affect the rate of the reaction. It only affects the equilibrium
concentrations.
b) ∆G does not affect the rate of the reaction. It only affects the equilibrium
concentrations.
6.51. In order to determine if reactants or products are favored at high temperature, we
must consider the effect of temperature on the sign of ∆G. Recall that ∆G has
two components: (∆H) and (-T∆S). The reaction is exothermic, so the first term
(∆H) has a negative value, which contributes to a negative value of ∆G. This
6.52. Recall that ∆G has two components: (∆H) and (-T∆S). We must analyze each
term separately. The first term is expected to have a negative value, because three
pi bonds are being converted into one pi bond and two sigma bonds. A sigma
bond is stronger (lower in energy) than the pi component of a double bond (see
CHAPTER 6
113
second term becomes more significant. At high enough temperature, the second
term (-T∆S) should dominate over the first term (∆H), generating a positive value
for ∆G. Therefore, the reaction will favor reactants at high temperature.
6.53. The nitrogen atom of an ammonium ion is positively charged, but that does not
render it electrophilic. In order to be electrophilic, it must have an empty orbital
that can be attacked by a nucleophile. The nitrogen atom in this case does not
have an empty orbital, because nitrogen is a second row element and therefore
N
R
H
R
R
Nuc
6.54.
NNH
Cl
NN
Cl