O SlideShare utiliza cookies para otimizar a funcionalidade e o desempenho do site, assim como para apresentar publicidade mais relevante aos nossos usuários. Se você continuar a navegar o site, você aceita o uso de cookies. Leia nosso Contrato do Usuário e nossa Política de Privacidade.
O SlideShare utiliza cookies para otimizar a funcionalidade e o desempenho do site, assim como para apresentar publicidade mais relevante aos nossos usuários. Se você continuar a utilizar o site, você aceita o uso de cookies. Leia nossa Política de Privacidade e nosso Contrato do Usuário para obter mais detalhes.
Chapter 4 127
Solutions to Problems
3- > > CH3CH2CH2çH2
b. The halogen atoms decrease the stability of the carbocation because, since they are more
electronegative than a hydrogen, they are more effective than a hydrogen at withdrawing
electrons away from the positively charged carbon. This increases the concentration of
positive charge on the carbocation which makes it less stable.
Because fluorine is more electronegatíve than chloiine, the Huorine-substituted
carbocation is less stable than the chloiine-substituted carbocation.
CH3ClJHCH2çH2 > cmencHzçnz > crgtltncnzçn¡
CH3 c¡ F
2. a. The bond orbitals of the carbon that is adjacent to the positively charged carbon are the bond
orbitals that are available for overlap with the vacant p orbital. Because the methyl cation does
not have a carbon adjacent to the positively charged carbon, there are no bond orbitals
available for overlap with the vacant p orbital.
b. An ethyl cation is more stable because it has three carbon-hydrogen bond orbitals available for
overlap with the vacant p orbital, while a methyl cation does not have any carbon-hydrogen
bond orbitals available for overlap with the vacant p orbital.
3. a. products b. reactants c. reactants d. products
a. cH3CH2çHCH3 c- @Ám e. @Br
? Hs ? Hs
b. CH3CH3çCH3 d. CH3('3CH2CH2CH2CH3 f. CH3CH2('3HCH3
Br Br Br
Chapter 4 129
?Ha ? Hs ? Hs ? Hs
1. cH3<I: cH3 2. CH3CIZCH3 3. CH3(|3CH3 4. CH3(| '.'CH3
c¡ Br OH OCH3
b. The first step in all the reactions is addition of an electrophilic proton (H+) to the carbon of
the CH¡ group.
The tert-butyl carbocation is fomied as an intermediate in each of the reactions.
c. The nucleophile that adds to the tert-butyl carbocation is different in each reaction.
In reactions #3 and #4, there is a third step-a proton is lost from the group that was the
nucleophile in the second step of the reaction.
3. (12113 ? H3 4_ ? H3 ([3113
CHgçcHg --> CH3C| ZCH3 + n30* CH§CH3 --> CH3(ECH3 + 113o*
63H OH 53cm OCH3
Hzõ_ H cH3§H H
Q + CH3OH --)› OCH3
(5113 + C| JH3
b. CH2=CCH3 + CH3OH À» CH3O'CCH3
c. CH3CH= CHCH3 + CH3CH2OH _Hs_ CH3CH¡('2HCH3
CH2=CHCH2CH3 + CH3CH2OH A» cH3cH2çHcH3
CH2=CH2 + CH3CH2CHCH3 43+ cHaCHzçHCHa
130 Chapter 4
d. CH3CH= CHCH3 + Hzo -›- CH3CHCH2CH3
CH2=CHCH2CH3 + Hzo _Ha- CH3(|2HCH2CH3
e. c; + L» Ó
r. CH3CH2CH= CHCH2CH3 + Hzo -H-›- CH3CH2C| ZHCH2CHZCH3
+ CH3(|3HCH2CH2QH ? H3
_ _L *_
CH3C-CH2 -H-›- CH3CCH3 cmcncnzcnzpà CGH? ,
l 3 H CH¡
@H3 C143 CH
B: is any base that is $
present in the reaction mixture
HB* + cH3c¡: HcH2cH2o<| :cH3
13. Solved in the text.
Chapter 4 131
H** + shift + B; lr
a. CH CHCH= CH -› CH CHCHCH -› CH CCH CH -› CH CCHQCH
3' 2 3' 3 3' 2 3 3' 3
CH3 CH3 CH¡ CH3
CH2 CH: Br CH3
+ . -
c. CH3C|2HCH2CH= CH2 -H-› CH3(IJHCHZCHCH3 L CH3CHCH2CHCH3
CH3 CH3 CH3 BT
d. GHz CH; Br CH¡
H+ + Br'
CH CH 1,2-methyl CH CH Br
13 H+ + I 3 shift 13+ Br- I 3¡
e. CH = CHCCH -› CH CHCCH _› CH CHCCH _› CH CH-CCH
2 l 3 3 | 3 3 l 3 3 | 3
CH3 CH3 CH3 CH3
f. CH3 cHa
CH? , + Br_
H/ r QE
“dl CH3 CH3
Ó e gb
1-Bromo-3-methylcyclohexane and 1-bromo-4-methylcyclohexane will be obtained in
approximately equal amounts because in each case the initially formed carbocation is secondary.
C113 Addition of H* would form a
CH &CH CH carbocation that could rearrange.
3 2 3
àr Addition of Br* forms a cyclic
bromonium ion rather than a carbocation
so there is no rearrangement.
a. The first step in the reaction of ethene with Br; forms a cyclic bromonium ion, whereas the
first step in the reaction of ethene with I-IBr forms a carbocation.
b. If the bromide ion were to attack the positively charged bromine atom, a highly unstable
compound (with a negative charge on carbon and a positive charge on bromine) would be
WB¡ co_ - 'Í'
HC E + : Br: CHz-CHz-Br-Br
Notice that the electrostatic potential maps of the cyclic bromonium íons on p. 158 of the text
show that the ring carbons are the least electron dense (most blue) atoms in the intermediate.
The nucleophile that is present in greater concentration is more apt to collide with the
carbocation intermediate. Therefore, if the solvent is a nucleophile, the major product will come
from reaction of the solvent with the carbocation or cyclic bromonium (or chloronium) ion
intermediate, because the concentration of the solvent is much greater than the concentration of
the other nucleophile. (For example, in "a" the concentration of CH3OH is much greater than the
concentration of ClÍ)
a. ClCH¡Cl2CH3 and ClCH2(|2CH3 c. CH3CH2CHCH3 and CH3CH2C| HCH3
QCH; Cl OH Cl
b- CH3CHCH3 and CH3CHCH3 d- CH3CH2çHCH3 and CH3CH2CIIHCH3
I Br 0CH3 Br
As elements, sodium and potassium achieve an outer shell of eight electrons by losing the
single electron they have in the 3s (in the case of Na) or 4s (in the case of K) orbital, thereby
becoming Na+ and K+. In order to form a covalent bond, they would have to regain electrons in
these orbitals, thus losing the stability associated with having an outer shell of eight electrons
and no extra electrons.
Chapter 4 133
19. Because chlorine is more electronegative than iodine, iodine will be the electrophile. Therefore,
it will become attached to the sp? carbon that is bonded to the greater number of hydrogens.
CH2CH2CH= CH2 + I-Cl _› CH3CH2CH-CH2_› CH3CH2(| IHCHZI
k/ ,Ç-J; Cl
a. CH2CHCH2CH3 b. CH2CHCH2CH3 c. CH2CHcH2CH3 d. CH2CHCH2CH3
Br Br Br OH Br OCH2CH3 Br OCH3
21. Notice that the addition of water or the addition of an alcohol to an alkene can be carried
out using an acid Catalyst (Section 4.5) or by mercuration/ demercuration (Section 4.8).
H+ OCH-, CH3
a. -_-_› -
1. Hg(O¡CCF¡)¡, CH¡CH2OH
b. CH2 H, CH3 CH3 CH;
É* OH OI' H3., OH
1. Hg(OAc)2, H2O, THF
3- NaBHA 2. NaBH4
c. CH3CH= CHCH3 __i-_› CH3CHCH2CH3
1. Hg(O¡CCF3)2, CH¡CH2OH
134 Chapter 4
? Hs H+ ? Ha
d. CH¡ = CCHZCH3 -_›- cH3ccH2cH3
1. Hg(O¡CCF¡)2, CH¡OH
CH3 + CH¡ l,2-hydn'de CH3 CH3
I H I shift I Hzo |
a. CH2=CHCHCH3 +- CH3CHCHCH3 ? ›- CH3CH2CCH3 -›- CH3CH2C|3CH3
I 1. Hg(OAc)2, H2O, THF I
b. cH2=cHcHcH3 44a- CH3CHCHCH3
2. NaBH4 |
In a, a carbocation rearrangement is required to get the desired product from the given starting materia
In b, the desired product must be obtained from a reaction that will not form a carbocation íntermediat
23. Because one mole of BH3 reacts with three moles of an alkene, one third of a mole of BH3 is
needed to react with each mole of alkene. Therefore, two thirds of a mole of BH3 is needed to
react with two moles of an alkene (in this case, l-pentene).
a. CH3C= CHCH3 CH3CHC| jHCH3
2 2 2 OH
b_ 1. BH¡ OH
2. HOÍ H2O2, Hzo
Chapter 4 135
25. Addition of a proton to the carbon that is bonded to the greater number of hydrogens forms
a carbocation intermediate. The alcohol group in the same molecule is the nucleophile that
reacts with the carbocation.
CIHB HQ* ? H3 H3C
CH2=CCH2(l2HCH3 É» CH3-CCH¡C| ZHCH3 -›- H (QO/ CH:
CH CH 3 o
l 2 LJ I 2 CZ¡
OH = <.>. H sua
HB* is any acid present in the solution H3C
B: is any base present in the solution CH3
H3C O + HB+
+ Br- : ->
° BT Br
+ HBr _'-)' + Br-
27. Because alkene A has the smaller heat of hydrogenation, it is more stable.
This alkene is the most stable bcacause it has the greatest
number of alkyl substituents bonded to the Spz carbons.
° a This alkene is the least stable bcacause it has the fewest
number of alkyl substituents bonded to the sp2 carbons.
' This alkene has the smallest heat of hydrogenation
because it is the most stable of the three alkenes.
136 Chapter 4
H C / CH3 H3cx
H CHzCHz CH3CH2 CHZCH¡ 3 *CH cH/ Cns
/ / /
Cíc > Cíc > Cíc >
/ x / / x
CH3CH2 H H H H H
2 trans substituents 2 cis substituents 2 CiS SUbSÚÍUCnÍS that
cause greater steric strain
30. Solved in the text.
31. The reaction of 3-methylcyclohexene with HBr and peroxide would form approximately equal
amounts of l-bromo-2-methylcyclohexane (the desired product) and l-bromo-B-methyl-
cyclohexane because the bromine radical could add to either the l-position or the 2-position of
the alkene since in both cases a secondary radical would be formed. Thus, only half as much of
the desired product would be formed from 3-methylcyclohexene than from l-methyl-
CH: CH; CH;
Br. Br _
l-bromo-Z-methyl- 1 -bromo-3-methyl-
Chapter 4 l 37
If 3-methylcyclohexene were treated with HBr in the absence of peroxide, little l-bromo- 2-
methylcyclohexane (the desired product) would be formed because the secondary carbocation
with a positive charge at the 2-position would rear-range to a more stable tertiary carbocation.
CH: CH3 CH; CH3
$135 *Br* *BF
CH3 CH3 Br CHE]
l-bromo-3-methyl- l-bromo-2-methyl- 1'bT°m°'1'm°thY¡*
cyclohexane cyclohexane Cycbhexane
a. CH3CH= CH2 . _.H_›- CH3CHCH3
b. CH3CH2CH= CHCH3 + Br, --›- CH3CH2(|3HCHCH3
°' 1. BH¡
2. HO , 11202, HZO
GHz CHa CH3O CH3
GHz CH3 Br CH3
138 Chapter 4
f H+ OCH2CH2CH3
' Ô + CH3CH2CH20H -› Cr
Bl' ? H3 CH2CH3
. b. CH CCH CH C-
a dcHzcHa 3¡ 2 3 @Br
a. CH3CHCH3 + : õif -_› CH3(| ZHCH3
+ . .
w : ç_1:
nucleophile electrophile __
b. CH3CH= CH --> CH3ÇHCH2BI=
electrophile ? Hiu
(ljH3 nucleophile _--› CH3(!2-(I)CH3
c_ CH3C| + + CH3QH CH3H
? Hs ? Hs
a. CH3C= CHCH3 + HBr : -› CH3(l2-CH2CH3
? Hs . ?Hs
b. CH3C= CHCH3 + HBr Períwíí CH3CH-(|3HCH3
?Hs ? Hs
C. + HI a
m. CH3C= CHCH3 --_-›- CH3CH-(EHCH3
2 2 OH
? H3 . ?H3
d. cH3c= cHcH3 + HI ENAdS_ CH3C-CH2CH3
? Ha ? Hs
e. CH3C= CHCH3 + IC] : ›- CH3C-CHCH3
f- cH3c': =cHcH3 _ff- CH3CH-CH2CH3
g. CH3C= CHCH3 + Br¡ + excess NaCl -a- CH3(lj-(IjHCH3
? H3 1. Hg(OAc)2, HZO ? H3
h. CH3C= CHCH3 cH3ç-cH2cH3
i. cH3c= cHcH3 + HZO “Link CH3(I3-CH2CH3
? Hs ? Ha
j- CH3C= CHCH3 + Br, __›. CH3C-CHCH3
CHZCIZ Ér 1|3r
? H3 H o ? H3
k. cH3c= cHcH3 + Br¡ -2-›- cH3c-çHcH3
l. CH3Cl= CHCH3 + Br¡ 395- CH3C-CHCH3
CH3(| ) i3:
? H3 1. BH, ?H3
140 Chapter 4
i 3 l. Hg(O2CCF3)2, CH3OH l 3
n. CH3C= CHCH3 CH3Ç"CH2CH3
' d 4 OCH3
? H3 ? H3 ? H3
CH3CH= C(| :HCH2CH3 CH3C= CCH2CH2CH3 CH3CH= CHCH(| IHCH3
CH3 CH3 CH3
3,4-dimethyl-2-hexene Z3-ÓÍmCÍhYÍ-Ê-Í1CXCHC 4v5'dimeth)'l'z'hexene
2,3-Dimethyl-2-hexene is the most stable because it has the greatest number alkyl substituents
bonded to the sp¡ carbons. Because it is the most stable, it has the smallest heat of
4,5-Dimethyl-2-hexene has the fewest alkyl substituents bonded to the spz carbons. making it
the least stable of the three alkenes. It, therefore, has the greatest heat of hydrogenation.
A H 1,2-hydride
CH cHcH-CH + m SW* *
3 l - 2 H-Br e CH3C-CHCH3 --> CH3C| ZCHZCH3
CH; CH3 CH3
l: 1:3; lzlâi-Ê
3- ? H3 b. CH2=CHCH2CH2CH2CH3 CH2=CHCH2CHCH3
CH3C= CCH3 5B
CH ? H3 3
' 3 CH2=CHCHCH2CH3 cH3(| jcH= -cH2
This compound is most stable. CH3 CHS
It has 4 alkyl substituents
bonded to the spz carbons.
These compounds are the least stable.
Each has only one alkyl substituem
bonded to the sp? carbons.
: õ:' : Õ
i) II _
a. CH3-CÚCH3 à» CH3-C-CH3 + CH3O
b, + Êú-Hz É* CH3CÉC_ + : NH3
c. CH3CH2-Br + ong? -› CH3CHZ-ÇZCH3 + Br'
l' B“3 B M? ? 0/
2. Hzoz, Ho' r?
CHzfHCHg H+ CH2CH= CH2 CH2çHCH3
OH 'Hzo __›HBT Br
*ECÉOH Br CH2CH2CH2Br
2 pêroxide g
142 Chapter 4
b. CH3CH2CH2CH= CH2 _-›.
c. à, r. CH3CH2CH= CHCH2CH3 é.
C- no reaction
Cl uses a 3 sp3 orbital in bond formation, while Br uses
a 4 sp3 orbital. Cl, therefore, forms a shorter and stronger
bond with carbon.
A primary radical is less stable than a secondary
radical, so it is harder to break a bond that results in
the formation of a primary radical than it is to break a
bond that results in the formation of a secondary radical.
A methyl radical is less stable than a primary radical,
so it is harder to break a bond that results in the formation
of two methyl radicals than it is to break a bond that results
in the formation of a methyl radical and a primary radical.
I forms a weaker bond than Br because I uses a 5 sp3
orbital in bond formation, while Br uses a 4 sp3 orbital.
without an OH
acid Catalyst r c¡
e. CH3CH2CH= CHCH2CH3 _gâ-
CH3CH2CH= CHCH2CH3 _BL
+ HB, peroxide
, HB, &E; Ó/ B'
CH: H3C Cl
+ HCl f* Õ
CH3 H3C Cl
d. Peroxide has no effect on the addition of HCl, so both compounds will give the same
product (the product shown in c).
? H3 en,
a CH3C= CHCH2CH3
d. CH3CH2CH2CHCH3 Cl
144 Chapter 4
46. a. To determine their relative rates of hydration, the rate constant of each alkene is divided by
the smallest rate constant of the series (3.51 x 10's).
propene = 4.95 x 10'3/3.5l x l0'3 = 1.41
cis-Z-butene = 8.32 x lO'8/3.5l x l0'3 = 2.37
trans-Z-butene = 3.51 x l0'3/3.51 x l0'3 = l
2-methyl-2-butene = 2.15 x 104/351 x 10-8 = 6.12 x 103
2,3-dimethyl-2-butene = 3.42 x l0'4/3.51 x 10's = 9.74 x 103
b. Both compounds form the same carbocation, but since (Z)-2-butene is less stable than
(E)-2-butene, (Z)-2-butene has a smaller free energy of activation.
c. 2-Methyl-2-butene reacts faster because it forms a tertiary carbocation in the rate-limiting
step, while cis-Z-butene forms a less stable secondary carbocation.
d. Both compounds form tertiary carbocation intermediates. However, 2,3-dimethyl-2-butene
has two sp¡ carbons that can react with a proton to form the tertiary carbocation whereas
2-methyl-2-butene has only one sp2 carbon that can react with a proton to form the tertiary
carbocation. Therefore there will be more collisions with the proper Orientation that lead to
a productive reaction in the case of 2,3-dimethyl-2-butene.
a H H b. Cl H c. Cl H
/ ¡ l
IC= C / C=C IC:
H c¡ H CH3 H CH¡
48. Only a symmetrical alkene will form the same alkyl halide when it reacts with HBr in the
presence of peroxides that it forms when it reacts with HBr in the absence of peroxides.
b. There are many more symmetrical alkenes in the case of an alkene with an even
number of carbon atoms.
CH3CH2 / CH2CH3 H3C / CH3 CH3
/ c=c / C=c
H HaC CH3 me CH
CH3 CH; CH3 3
CH3CH2 / H
Chapter 4 145
No, he should not follow the student's advice. Markovnikov's rule would indicate that the
secondary carbocation is more stable than the primary carbocation. However because of the
electron-withdrawing fluoro substituents in this compound, the primary carbocation is more
stable than the secondary carbocation, since in the latter the positive charge is closer to the fluoro
substituents. So the major product will be 1,1,l-trifluoro-3-iodopropane, not l, l,1-trif| uoro-2-
iodopropane, the compound that would be predicted to be the major product by Markovnikov's
a. CH3CH2CH= CH2 __› cmcnzcncns f» CH3CH2CHCH3
. . ,
H C? CH3
b. the first step e. the sec-butylcation
c. H* f. methanol
a. Both l-butene and 2-butene react with HCl to form Z-chlorobutanc.
b. Both alkenes form the same carbocation, but because 2-butene is more stable than l-butene,
2-butene has the greater free energy of activation.
c. Because l-butene has the smaller free energy of activation, it will react more rapidly with
d. Both compounds form the same carbocation, but since (Z)-2-butene is less stable, it will
react more rapidly with HCl.
a CH¡ CH3 CH3 CH3
b. l-Methylcyclopentene is the most stable.
c. Because 1-methylcyclopentene is the most stable, it would have the smallest heat of
146 Chapter 4
Í-_s CH3 _ Br CH3
a. C-CHz D + 3 + CHS _âr_ CH
lj “A -+ -›-' 3
CH3 CH; ___ CH¡ CH;
+ . à ÚBr
b. The initially formed carbocation is tertiary.
c. The rearranged carbocation is secondary, which then undergoes another rearrangement to a
more stable tertiary carbocation.
d. The initially formed carbocation rearranges in order to release the strain in the four-
membered ring. (A tertiary carbocation with a strained four-membered ring is less stable
than a secondary carbocation with an unstrained five-membered ring. )
54. It tells us that the first step of the mechanism is the slow step. If the first step is slow, the
carbocation will react with water in a subsequent fast step, which means that the carbocation will
not have time to lose a proton to reform the alkene.
CH-CD --›-H+ É: -›-H2O
_ 2 slow H CHDZ fast ? H-CHD?
148 Chapter 4
c. The weak Oxygen-Oxygen bond breaks homolytically, forming a radical that abstracts a
hydrogen atom from HCCl3. The radical that is formed adds to the double bond. A n
electron of the other double bond pairs with the unpaired electron, forming a bond that
divides the ring. The new radical abstracts a hydrogen atom from HCCl3, and the
propagation steps are repeated.
RCÇ-“OR -? -> 2RO°
RO-g-CCI¡ _a ROH + ›cc13
+ -CC13 à» -: ›
r CÓ +
c¡ A Cl Cl
! Q . . _ l _ I _
a Cl-(E H + H9: 2+ Cl-C: :-› Cl-Ct + Cl
c¡ o c¡ c¡
I n . . - l I _
b. ci-c-'lc-fg: A» Cl-Cz' : › Cl-C: + c¡
c¡ Na* c1