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Chapter 5 193
Copyright © 2017 Pearson Education, Inc.
18.
a.
Because the equilibrium constants for all the monosubstituted cyclohexanes in Table 3.9 on page 128
of the text are greater than 1, all of the equilibria have negative
∆
G
°
values.
(Recall that
∆
G
°
=
-
RT
ln
K
eq
.)
b.
tert
-butylcyclohexane
c.
tert
-butylcyclohexane, because it is the largest substituent
d.
∆
G
°
=
-
RT
ln
K
eq
K
eq
=
18
∆
G
°
=
-
1.986
*
10
-
3
kcal
>
mol K
*
298 K
*
ln 18
1
recall that
T
=
°
C
+
273
2
∆
G
°
=
-
0.59
*
ln 18 kcal
>
mol
∆
G
°
=
-
0.59
*
2.89 kcal
>
mol
∆
G
°
=
-
1.7 kcal
>
mol
19.
Solved in the text.
20.
a.
∆
G
°
=
-
RT
ln
K
eq
-
2.1
=
-
1.986
*
10
-
3
*
298
*
ln
K
eq
ln
K
eq
=
3.56
K
eq
=
35
K
eq
=
3
isopropylcyclohexane
4
equatorial
3
isopropylcyclohexane
4
axial
=
35
1
,
of equatorial
isopropylcyclohexane
=
3
isopropylcyclohexane
4
equatorial
3
isopropylcyclohexane
4
equatorial
+
3
isopropylcyclohexane
4
axial
*
100
=
35
35
+
1
*
100
=
35
36
*
100
=
97
,
b.
Isopropylcyclohexane has a greater percentage of the conformer with the substituent in the equatorial
position because the isopropyl substituent is larger than the fluoro substituent. The larger the sub-
stituent, the less stable is the conformer in which the substituent is in the axial position because of the
1,3-diaxial interactions.
21.
∆
S
°
is more significant in reactions in which the number of reactant molecules and the number of product
molecules are not the same.
a.
1.
A + B
C
2.
A
B + C
b.
Reaction
2.
has a positive
∆
S
°
.
In order to have a positive
∆
S
°
, the products must have greater freedom of motion than the reactants.
(In other words, there must be more molecules of products than molecules of reactant.)