Instructor Supplemental Solutions to Problems
© 2010 Roberts and Company Publishers
Chapter 5
Addition Reactions of Alkenes
Solutions to In-Text Problems
5.1
(a)
The iodine is the electrophile, because it adds to the carbon of the double bond with fewer alkyl substituents.
This result does seem to fit the electronegativity pattern for electrophilic addition, as nitrogen is considerably
more electronegative than iodine.
5.2
(b)
The reaction of Br2 with 2-methyl-1-hexene follows the pattern in Eqs. 5.12 and 5.13 of the text.
5.4
The reaction of bromide ion as a nucleophile at the bromine of the bromonium ion would give a species that violates
the octet rule.
As we’ll learn, atoms in periods beyond the third can have more than an octet of electrons, and a structure such as
this is not impossible. However, the octet rule can still be used as a guide to reactivity. A reaction of the bromide ion
at the carbon of the bromonium ion gives a much more stable structure.
5.5
(b)
(d)
In this part and in part (d), the double bond is symmetrically placed in the molecule; so, regioselectivity is not
an issue.
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 5
5.6
When 3-methyl-1-butene undergoes acid-catalyzed hydration, rearrangement occurs because carbocation
intermediates are involved. (See the solution to Problem 4.35.) Oxymercuration–reduction gives addition without
rearrangement because the intermediate is a cyclic mercurinium ion, which does not rearrange.
5.7
(b)
5.8
(b)
2
In this part and in part (d), the double bond is symmetrically placed in the molecule; so, regioselectivity is not
an issue.
(d)
5.9
The products of hydroboration–oxidation are the same as the ones in parts (b) and (d) of Problem 5.5. In both, the
alkene is symmetrical; consequently, regioselectivity has no meaning. In part (c), the carbons of the double bond
have the same number of alkyl substituents; consequently, there is no reason to expect significant regioselectivity.
However, the relative amounts of the products in each case could be different.
5.10
(b)
5.12
(a)
Because cis-2-butene is a symmetrical alkene, it reacts to give the same product in either oxymercurationreduction or hydroboration-oxidation.
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 5
(b)
5.13
Because 1-methylcyclohexene has different numbers of alkyl branches at the carbons of its double bond, it
gives different products in the two reactions.
(b)
(d)
5.14
3
2-Methylpentane is an alkane; because it has no double bond, it does not undergo ozonolysis.
(b)
(d)
2-Methylpentane is an alkane; because it has no double bond, it does not undergo ozonolysis.
5.15
(b)
5.16
As the solution to Problem 5.15(b) illustrates, E and Z isomers of an alkene give the same ozonolysis products.
Therefore, ozonolysis cannot be used to determine the stereochemistry at the alkene double bond.
5.17
(a)
(d)
5.20
(a)
The first initiation step is shown in Eq. 5.49, text p. 203. In the second initiation step, the radical produced in
the first initiation step abstracts a hydrogen atom from HBr:
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 5
4
The bromine atom then adds to the p bond of cyclohexene in the first propagation step (1); the resulting
radical then reacts with HBr in the second propagation step to give the addition product.
5.21
(a)
5.22
(b)
The free-radical intermediates in the propagation steps of the peroxide-promoted HBr addition to 1methylcyclopentene:
5.23
(b)
Two Cl—CH2 bonds are formed (355 kJ mol–1 each); a Cl—Cl bond (239 kJ mol–1) and a carbon–carbon p
bond ( 243 kJ mol–1) are broken.
H° = 239 + 243 – 2(355) = –228 kJ mol–1
5.25
The second propagation step in which Br is abstracted rather than H involves formation of a secondary C—Br bond
rather than a secondary C—H bond; the bond broken (H—Br) is the same. Hence, we use the calculation in the
solution of Problem 5.24(a), except that the bond dissociation energy of a C—Br bond is substituted for that of a
C—H bond. The closest value in Table 5.3 is the bond dissociation energy of (CH3)2CH—Br, 309 kJ mol–1. The
required H° is therefore
H° = 368 – 309 = 59 kJ mol–1
Hence, abstraction of Br is a very unfavorable process energetically (H° >> 0), whereas the abstraction of H is a
very favorable process (H° < 0). As is usually the case in free-radical reactions, the relative enthalpies of the
processes governs their relative rates. Hence, abstraction of H is much more favorable, and therefore is much faster,
than abstraction of Br.
5.26
(b)
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 5
Solutions to Additional Problems
5.28
(a)
(b)
(f)
(g)
(k)
5.30
(a)
(d)
5.31
(b)
5.32
(b)
(d)
(c)
(d)
(h)
(l)
(i)
(m)
(b)
(e)
(j)
(n)
(o)
(c)
(e)
(f)
(d)
(g)
5
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 5
6
(f)
(h)
(j)
5.33
(b)
An oxymercuration–reduction on the alkene used in part (a) gives the desired compound.
5.36
(a)
Only the first two products are formed in the absence of peroxides, and only the third is formed in the
presence of peroxides. Different products are formed because different mechanisms and reactive
intermediates are involved under the different conditions.
The mechanism for the formation of the first two products is identical to that shown for reaction of the same
alkene with HCl in Eqs. 4.26 and 4.27a–b, text p. 155, except that HBr is used instead of HCl. The first
product results from a carbocation rearrangement, and the second from normal regioselective
(“Markovnikov”) addition. The third product is the consequence of a free-radical addition mechanism, the
propagation steps of which are as follows:
(b)
(The initiation steps are shown in Eqs. 5.48 and 5.50 on p. 203 of the text.)
(c)
Peroxide-promoted addition is in competition with normal addition and rearrangement. The normal processes
occur at the same rate at which they occur in the absence of peroxides. The fact that only the product of
peroxide-promoted addition is observed, then, means that this process is much faster than the other,
competing, processes.
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 5
5.37
(b)
5.42
(a)
(b)
5.45
7
(d)
The H—CN bond is stronger than the O—H bond, the abstraction of a hydrogen atom from HCN by the tertbutoxy radical is endothermic by 528 – 438 = 90 kJ mol–1. (These numbers are from Table 5.3 with the bond
energy of CH3O—H as an approximation for that of (CH3)3CO—H.) The first initiation step, formation of the
tert-butyloxy radical from a peroxide, is also endothermic. Because both initiation steps are highly
endothermic, the reaction is not likely to generate a high enough concentration of radicals to initiate a chain
reaction.
This propagation step involves breaking an H—CN bond (528 kJ mol–1) and formation of a secondary
carbon–hydrogen bond (–412 kJ mol–1). The H° of this step is 528 – 412 = +116 kJ mol–1. This step is
highly endothermic and therefore not reasonable as a propagation step.
(a)
The structure of polystyrene:
(b)
Because both “ends” of 1,4-divinylbenzene can be involved in polymer formation, addition of 1,4divinylbenzene serves to connect, or crosslink, polymer chains. Such a crosslink is shown with bolded bonds
in the following structure:
Notice that because only a small amount of 1,4-divinylbenzene is used, divinylbenzene does not polymerize
with itself.
Crosslinks are introduced into polymers to increase their strength and rigidity.
5.47
Compound A has the connectivity of octane; because it has an unsaturation number U = 1 and reacts with bromine, it
is an alkene. The ozonolysis results show that compound A is 4-octene, CH3CH2CH2 CHACHCH2 CH2CH3. Because
the double bond is located symmetrically, only one ozonolysis product is formed. Ozonolysis cannot determine
whether the compound is cis or trans.
5.48
(d)
The oxygen of a hydroxy group introduced in the oxymercuration reaction of one double bond serves as the
nucleophile in the opening of the mercurinium ion formed at the second double bond within the same
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 5
8
molecule. Two products are formed because the oxygen can react with the mercurinium ion at either of two
carbons.
[The acetate ion is a by-product of the first step of oxymercuration; its reaction with the protonated ether is
shown only for (b).] Treatment with NaBH 4 replaces the mercury with a hydrogen.
(f)
The initiation step, reaction of the thiol with an alkoxy radical formed by homolysis of a peroxide, is shown
in Eq. 5.55a, text p. 206. The radical produced in that step adds to the p bond of the alkene so as to produce
the tertiary free radical, and this radical reacts with the thiol to propagate the chain.
Be sure that you did not form the product by a recombination of two radicals; see the discussion of Eq. 5.56
in Study Problem 5.4, text p. 206.
5.50
These are both examples of steric effects. Placing three highly branched groups around a central boron results in van
der Waals repulsions. These repulsions are severe enough that only two groups can be bound to boron in the case of
disiamylborane, and only one group in the case of thexylborane. (Notice the greater number of alkyl substitutuents
in the “thexyl” group.) Now, alkyl branches stabilize sp2-hybridized boron (see the solution to Problem 4.48 on p. 62
of the Study Guide and Solutions Manual) just as they stabilize sp2-hybridized carbon; but if the alkyl branches are
themselves branched, they form a thicket of methyl groups that interact repulsively with each other as more of these
branches are accumulated. These repulsive interactions reduce the relative stability of the trialkylboranes—
evidently, so much so that they cannot form.