Manual soluções - cap.04 Paula Bruice

Resolução de exercícios de química orgânica do capítulo 4 de Paula Bruice vol.01.

1. Chapters 127
Solutions to Problems

1
cn,
a CHICH,CCH, > CH,CH.CHCH, > CHICHACH,CH,

b. The halogen atoms decrease the stability ofthe 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 onthe carbocation which makes i os stale,

Because fluorine is more electronegative than chlorine, the fuorine-substtuted
carbocation is less stable than the chlorine substituted carbocation,

CHJCHCH,CH, > CHCHCH,CH, > CH,CHCH,CH,
CA a F

‘8. The bond orbitals ofthe carbon that is adjacent tothe positively charged carbon are the bond
‘orbitals hat are available for overlap with the vacant p orbital, Because the methy cation docs
‘ot have a carbon adjacent to the positively charged carbon, there are no bond orbitals
available for overlap withthe vacant p or

Bb. An ethyl cation is more stable because it has three carbon-hydrogen bond orbitals availabe for
‘overlap with the vacant p orbital while a methyl cation does not have any earbon hydrogen
bond orbitals available for overlap with the vacant orbital

3 apro bares crus podas
4
cH, cn
senengiem DÉ . O
be
CH q
b.cmyciiec, de CALCIO, 2 CIC,

Br Br Br

2. 128

9.

Chapter 4

eo

CH
a. CH,CH,C=CH,

In both a and b the compound that is more highly regioselective is the one
do CH tire the choice is between forming terry carbocation ora primary
‘carbocation,
{nthe les regioselective compound, the choice is between forming a tertiary
{Curbocation ora secondary carbocation, so the difference in he stability of
the two possible carbocations is not as great.

“As long as the pH is greater than 2.5 and less than about 15, more than 50% of 2-propanol
‘Would be in its neutral, nonprotonated form.

Bosause when the pH = pas hal the compound is in its acid form and haf is in its basic form, at
à pH less than —2.5, more than half of the compound wil be in its positively charge protonated
fg. rt st 5, mora al of te ‘Compound wil exist as the negatively
charged anton,

3 transition states, €: the neutral alcohol
2 intermediates the second and third steps in the forward direcion

se

2 CHACHCH,CHCH, — €. CHCHHCIRCHLCHCH, and CH,CH,CH;CHCH,CHs
on ón on

"oO" "og
o

3. Chapter 129

10 a
gts CH, q en
1. CSC 2. CIEN, 3 cHgcHy 4 CHCCH,
a Br on oc

Bb. The fist step in al he reactions is addition of an electrophilic proton (H') tothe carbon of
the CH, group.
‘The fert-butyl carbocation is formed as an intermediate in each ofthe reactions.

‘© The nucleophile that adds to he er-buty carbocation is different in each reaction,

In reactions #3 and #4, there is a third step—a proton is ost from he group that was the
nucleophie in the second step of the reaction,

3 ch gts 4 Oi gts
geh m CHI, + 110° CHAN) —> cHECH + 110°
oH on och, och;

H CH À
ur

1.

® + cron A Docs

qu A Gis
D ci=êcn, + cron He mobi
CH

« CHscH=cHOH + cHcHon HZ cH,crcHcty
or denen,
Ei=enenen, + croton HY CHLCHLÇHCH,
ae OCR CH

ccm + emengnen He eineignch,
A Seen,

4. à + mo He amenemen,
bn
+ mo He cmpuenon,
M
on
«O + no
E cuomcu=cucu en + MO CICHGNCILCILCH,

on

2

Y agencia ch

eng=ch, He con a —- SS

En, ds ds Pay
Bs

Bay base hats |

prenden tre

cu,

HE" + CICLOS

y Gy

13, Solved inthe text

5. 14.

Chapter4 131

Latte pr
A cd cud
a. emgnen=cn, ES csonóncn FH engoncn, Bie cm fome,
En, En, En, En,
secondary tertiary
ge
& cngremcu=en, Ee cmeneménem BE cneneiscuch,
En, En, bay de
gs
ts Hy Lm Cr, Gt Be
e emzenten, Abe cmenden, BE cinduden BS omeu-ten,
En En, En, En,
secondary cer
t CH cH,
Hy Bs
EU)

1-Bromo-3-meihyleyelohexane and 1-bromo-4-methyleyelahexane will be obtained in
approximately equal amounts because in each cas the intlly formed carbocation is secondary.

6. 12

15.

16.

m.

Chapter

Addon of H would form à
GH coccion ta cou reamange
emeanch,

Addition of Br* forms acyclic

Br ‘bromonium ion rather han a carbocation
SO there sno rearrangement.
a. The first step inthe reaction of ethene with Bro forms a eycliebromonium ion, whereas the

fist step in the reaction of ethene with HBr forms a carbocation.

b. Ifthe bromide ion were to attack the positively charged bromine atom, a highly unstable
compound with gate charge o cañon and postive car on bromine) woud be
formed.

Notice that the electrostatic potential maps ofthe eyeli bromonium ions on p. 158 ofthe text
show that he ring carbons are he least electron dense (most blue) atoms inthe intermediate,

“The nucleophile that is present in greater concentration is more apt to collide withthe
carbocation intermediate. Therefore, ifthe solvent is a nucleophile, the major product will come
{rom reaction ofthe solvent with the carbocation or eyelic bromonium (or chloronium) ion
intermediate, because the concentration ofthe solvent is much greater than the concentration of
the other nucleophile. (For example, in "3" the concentration of CH,OH is much greater than the
concentration of CY.)

oth gts

a CIC md CICHÉCHS € CHLCHAEHCH, and. CHJCH,GHCH,
Sch, & on &
major major

D. CHCHCH, and CICHCH, a. CHCHÇHCIR and. CHCH,CHCH,
i 5 den, br
major major

As elements, sodium and potassium achieve an outer shel of eight electrons by losing the
Single electron they have in Ihe 35 in the case of Na) or 4s (inthe ease 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 hell of eight electrons
and no extra électrons.

7. Chapters 133

19. Because chlorine is more electronegative than iodine, iodine wil be the electrophile. Therefore,
it will become attached tothe sp? carbon that is bonded tothe greater numberof hydrogens

A
Ca —> CH,CH,CH—CH, —> (CH CH,CHCH,L
à

CHCHCHÉCH, + N

a. CHÇHCHLCH b. CHLGHCH,CH, © CHLÇHCHLCH, 4. CHCHCHLCI,
Br Br Br OH Br OCHCH, Br OCH,

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 mercuratio/demercuraion (Section 48)

. y oct,
mr ©

LH(O:CC I», CH,CHLOH
NABH
cn, cH
Sa. O Oo
HO
à 1: Hg(OAc),, H,O, THF
2 NDR,
© CHcH=cHeHn —H LCHCH;
TS

= OcH,CH,

1. Hg(O,CCF,),, CH,CHLOH
2. NaBH,

8. 14 Chapter 4

acu=tenon Hoe andenen,
RE
1 nyo.cer9, CON
Fran,
=
(eth Mae a qu
a aecnduen, > eienbuen, Es cuen, 0 cuendo,
be
2H OO a
semoule, EURO NOTE cacon,

3 DNaBH,
on

Ina, a carbocation rearrangements required to get the desired product from the given starting materit
In b the desired product must be obtained from a reaction that will not form carbocation intermedia:

23. Because one mole of BH reacts with three moles of an alkene, one third of a mole of BH is
‘needed to react with each mole of alkene. Therefore, two thirds of a mole of BH; is needed to
ect with two moles of an alkene (inthis case, -pentene)

CH,
emencuch,
oH

10,40

Hy CH

b. 1. BH, au

2. HO’ HO, HO

9. Chapter 135

25. Addition of proton tothe carbon that is bonded to the grater number of hydrogens forms
‘carbocation Intermediate. The alcohol group in the sar molecule isthe nuceophile that
Feacts with the carbocation,

1.0 Hy HAC,
CH =
Buy

HB? is any acid present inthe solution

Bs is any base present inthe solution CH,
meh, + HB"

Bu E
Chim + EY es

27. Because alkene A has the smaller heat of hydrogenation, it is more sable

28.

HCH,
This alkene is the most stable beacause it has the greatest
‘number of alkyl substituents bonded to the sp? carbons.
CH,CH;

é ACH,
d “This alkene isthe least table beacause it has the fewest
number of alkyl substtvens bonded tothe sp? carbons

‘CH,CH

a .CH¿CHy
‘This alkene has th smallest heat of hydrogenation
because it isthe most stable of the thee alkenes.

CH,

10. 136 Chapter

>.
ne. EC,

mem crc, jenen ed cn
=. \ _/ ek

F A > z A > e: € $
amenf Nu dl a ia,

ne raté Acce

‘cause greater steric stain

cHCHLCH CH A

vt

one substituent

30. Solvedin the text

‘31. The reaction of 3-methyleyclohexene with Hr and peroxide would form approximately equal
amounts of 1-bromo-2-methyleyclohexane (the desired product) and I-bromo-3-methy
‘yclohexane because the bromine radical could add to ether the I-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-methyleyclohexene than from I-methyl-

Br Br .
or. oO
ju |

Hy

HBr

BG,

‘bromo-2-methyl- 1-bromo--meihyl-
eyelohexane cyelohenane

11. 2

Chapter 137

If 3.methyleyclohexene were treated with HBr in the absence of peroxide, litle 1-bromo- 2-
methylcyclohexane (the desired product) would be formed because the secondary carbocation
‘with a positive charge at the 2-position would rearrange toa more stable tertiary carbocation.

Sec, ? OS
pe ler lar
à © ©

CH;

1-bromo-3-meihyl- 1-bromo-2-methyl- 1-bromo-1-methyl-
® ‘OCH,
.
bo cneescien, + ay —- xeric
f
CH; CH,OH
$ L BH,
qe CH CHO, CH;
‘O-0 &
CH;OH
CR Hy Br, CHs

12. 138 Chapter 4

O + mann 2
pr qu CH;CH; qu

clectophile nucleophile
a. CHCHCH, +

— amench,

QS ds

rucleophile eleirphile

noms | — cuca
elecoptile Gh
Gi tee > cant cr
ñ q
© cH* + CHOH CH;H
EN
=
qa ge
a cusene + mar — cnf-emen,
de
qu, cm

b. CHyÉ=CHCI + br LE cu n—encn,
Br

13. q
© cugecutey + m

CH

Chapters

Hy
i

—- ame-cich,

I
CH;

i £
d. cmescnen, + m Eté CH (=CH¿CH,

CH

I
CH

D

at

cH gis
f CH,C=CHCH, > CH,CH-CH,CH,
cm cu
E CHE=CHCH, + Be + NC —= CH CH,
ES
CH CH, a
¿ 1. Hg(OAc) HO i
ho A ds
cH=cucn, OR CHÖ-CH;CH,
ón
cm q
1 cHc=cucn, + mo HCL CH-CHCH,

HH, + E na
h

qa

k. cHC=CHCH, + Br, HO

ony
cud=cuce, + Br, HOH

BH
2. HO, HO”

Sn
ge
anne,
dede
en,
cube,
1b de
qa
engecncn,
cu de

ge
emenecnen,
Sn

19

14. 140. Chapter 4

1. HK(OCCES), CHOW
> ab,

gt CH
CHC=CCH,CH,CH, CHCHSCHEHEHCH,
CA CH,
34-dimethyl2hexene 23-dimethyl2hesene 4S-dimehyl-2herene

123-Dimethyl-2-hexene isthe most stable because it has the greatest number alkyl substituents
‘bonded tothe sp? carbons. Because it isthe most stable, it as the smallest heat of
hydrogenation.

45:Dimety.2-exen has the fest aly substituents bonded totes? cans. making it
the least stable ofthe tree alkenes. I, therefore, has the greatest heat of hydrogenation,

EN
D 12-hydride
a shift, ë

M + Br —> cie=cnon, He CHÉCHCH

ony Gh
pis |
Br Br

i ?
CHGHCHCHS CHÉCHCHS
CH Hy

gs b. CIRECHCH;CHCH,CH, CHECHCHLCHCH,
CHyÓ=CcH, a ORs
.

En, i
CH=CHCHCHCH, CHÉCH=CR,
“This compound is most stable. ¿
It has 4 alkyl substituent y ‘Hy
Teas be pi cab ‘These compound ar the lest tbe.
Esch has only one aig suben
bonded 0 he 32 carbons.

15. ».

Chapter

5 i
A
PS

A — come +,

— manch + BF

cest
1 han AO!

Icii,cl,
= CY“ engnen,

ACC . cic
ou
non Fe En -CH,CH,CH,Br

ora

1

16. 12 — Chapter 4

a

cl Cluses 23 sp orbital in bond formation, while Br uses
4 sp orbital Cl, therefore, forms a shorter and stronger
‘bond with carbon

1,CH, A primary radical is less sable than a secondary
MACS al sot arder to break a bond that ess

a the formation ofa primary rdical ihn tis to break a
bond that results inthe formation of à secondary radica,

e che A met radical ses stable tan primary acl
CHT CHy is hard to break a bond that els nthe formation
of to meth rail than io brea a bond that ets

inthe formation of amy radical and primary ade,

de Br—Br 1 forms a weaker bond than Br because uses a 5 sp?
orbital in bond formation, while Br uses a 4 sp orbital

:
=
a a ®
i
cu,c,cn,cu=cu, ECL Br,
CH úl >

u, 1. CHCH,CH=CHCH,CH, >
Nate

Br,
HCH;CH, Bie
Nec

CHICH,CH:

17. Chapters 143

CH, CH;
ws
O
CH; CH
+ HBr _peroxide
x CH;
5
+ Me =.
sa
a mea
.— Ó
CH ic cl
Km e

7

. Peroxide has no effect on the addition of HC, so both compounds il give the same
product (the product shown in €). dá id

E

. CÉNCI¿CHy LN

CHCHCH,
E. CHICHA,
cHcHcHËRCH, à

18. 144

46,

a

48.

Chapter 4

a. To determine their relative rates of hydration the rate constant ofeach alkene i divided by
the smallest rate constant of the series 3.51 x 10°),

propene 4.95 x 1093.51 x 108
832x 108/351 x 108
butene = 3SIX 10°83.51x108 = 1

Dmethyl-2-butene = 2.15x 10-7351 x 108 = 612 x 103
23-dimethyl-2-butene = 342x 1043.51 x 10% = 9.74 x 103

141
237

b. Both compounds form the same carbocation, but since (2)2-butene is less stable than
(B)-2-butene, (Z)-2-butene has a smaller fee energy of activation.

€. 2-Methyl-2-butene reacts faster because it forms a tertiary carbocation in the rate-limiting
‘Step, while cs-2-butene forms a less stable secondary carbocation.

d. Both compounds form tertiary carbocation intermediates. However, 23-dimethy]-2-butene
has two sp? earbons that can react witha proton o form te tertiary carbocation whereas
2-methyl-2-butene has only one sp? carbon that can reat with a proton to form the tertiary
‘carbocation, Therefore there will be more collisions withthe proper orientation that lead to

the case of 2,3-dimethyl-2-butene

“un van +
ES, PS,
a wo,

Only asymmetrical alkene will form the same alky halide when it reis with HBr in the
presence of peroxides that lt forms when it reacts with HBr in the absence of peroxides

b. There are many more symmetrical alkenes inthe case of an alkene with an even
number of carbon atoms.

CHCHX „CHEN,
aa
Y

19. 4.

si.

se

Chapter 145

No, he should not follow the students advice. Markownikov' rule would indicate thatthe
secondary carbocation is more stable than the primary carbocation. However because ofthe
clectron-withdrawing fluor substituents in this compound, the primary carbocation is more
‘able than he secondary carbocation, since in the latter the postive charge i loser to the fluoro
substituents, So the major product will be I,1,1-tifluoro-3-iodopropane, not 1, 1-tifluoro-2-
odopropane, the compound that would be predicted tobe the major product by Markovnikovs
rule,

a anemenden, “HS cun, = coque,
o Si
it
aneignen,

Son

be he tsp te ty en

ea es

Sos

a. Both I-butene and 2-butene react with HCI to form 2<hlorobutane,

b. Both alkenes form the same carbocation, but because 2-butene is more stable than I-butene,
2-butene has the greater free energy of activation,

+. Because I-butene has the smaller fre energy of activation, it will reset more rapidly with
Bec nergy pi

4d. Both compounds form the same carbocation, but since (2-2-butene is Less stable, twill
act more rapidly with HCI

- mo

4 alkenes Ô

b. 1-Meihyleyelopentene is the most stable.

+. Because I-methyleyclopenten i the most stable, it would have the smallest heat of
hydrogenation.

20. b. The initially formed carbocation is tertiary,

+. The rearranged carbocation i secondary, which then undergoes another rearrangement to à
more stable tear carbocation

4. The initially formed carbocation rearranges in order o release the strain in the four.
‘membered ring. (A tertiary carbocation with a strained four-membered ring is es stable
than a secondary carbocation with an unstrained five-membered ring.)

tells us thatthe first step of the mechanism is the slow step. Ifthe first step is slow. the
Carhocation wil react sith water in a subsequent fast sep, which means thatthe carbocation will
ot have time to ose a proton to reform the alkene,

(\-cu=co, > (Nenn; Zr CN -qu-cun,

"OH;

|

21. Chapter d 147

If the first step were not he slow step, an equilibrium would be set up between the alkene and

the carbocation, and because the carbocation could ose either H* or D* when it reformed the
alkene, all the deuterium (D) would not be retained inthe alkene

Ouen, Le ne, — (Den

.
| + DO

2. A proton adds to he alkene, forming a secondary carbocation,
‘expansion rearangement to form a more stable ferry carboc

HC. CHECH, HC. CHCH, EN ied
u PAD A

CH

hich undergoes a ring

HO,
AS

b. At first glance this appears to be a difficult mechanism, but examination of the reaction shows
‘that the only electophile avaiable to the alkene adds tothe sp? carbon bonded tothe most
hydrogen (the one that results in the formation of the most stable carbocation). This 5
followed by the addition ofthe nucleophilic nitrogen to the other sp? carbon.

cucHeden, + ci;
CH;

22. 148 Chapter

+. The weak onygen-oxygen bond breaks homoltically forming a radical that abstracts a
hydrogen atom from HCCI. The radical that is formed adds tothe double bond. A x
lect ofthe other double bond pairs with the unpaired electron, forming a bond that
‘vides the ring. The new radial abstracts a hydrogen atom from HCCI3 and the
propagation steps are repeated

Ro Htc, — ROH + -CCl

Ha

och,

os

— att a