36 pág.

Vollhardt Capítulo 7 (Haloalcanos 2)

Disciplina:Química Orgânica II963 materiais31.941 seguidores
Pré-visualização12 páginas
is a good solvent for SN1
reactions (Section 7-4).

2. We learned in Section 7-4 that different nucleophilic power has no effect on the rates of unimo-
lecular processes. The rates should be (and, experimentally, are) identical.

3. This part requires a bit more thought. According to Table 7-4 and Section 7-8, elimination by
the E1 pathway always accompanies SN1 displacement. However, increasing the base strength
of the nucleophile may “turn on” the E2 mechanism, increasing the proportion of elimination
product. Referring to Tables 6-4 and 6-7, we see that chloride is more basic (and less nucleo-
philic) than iodide. More elimination is indeed observed with chloride than with iodide. The
mechanism is as shown in Figure 7-7, with chloride acting as the base to remove a proton from
the carbocation.

b. The table in the margin presents data for the reactions that take place when the chloro compound
shown here is dissolved in acetone containing varying quantities of water and sodium azide, NaN3:

A
H3C CH

Cl

A
OH

A
N3

CH3

CH3 H3C�

H2O, NaN3,
acetone

CH CH3H3C CH

In the table, H2O% is the percentage of water by volume in the solvent, [N3
2] is the initial concentra-

tion of sodium azide, RN3% is the percentage of organic azide in the product mixture (the remainder
is the alcohol), and krel is the relative rate constant for the reaction, derived from the rate at which
the starting material is consumed. The initial concentration of substrate is 0.04 M in all experiments.
Answer the following questions.

1. Describe and explain the effects of changing the percentage of H2O on the rate of the reaction
and on the product distribution.

2. Do the same for the effects of changing [N3
2]. Additional information: The reaction rates shown

are the same when other ions — for example, Br2 or I2 — are used instead of azide.

H2O [N3
2] RN3 krel

 % %

 10 0 M 0 1

 10 0.05 M 60 1.5

 15 0.05 M 60 7

 20 0.05 M 60 22

 50 0.05 M 60 *

 50 0.10 M 75 *

 50 0.20 M 85 *

 50 0.50 M 95 *

*Too fast to measure.

 C h a p t e r 7 277

SOLUTION
1. We begin by examining the data in the table, specifi cally rows 2 – 5, which compare reactions in

the presence of different amounts of water at constant azide concentration. The rate of substitution
increases rapidly as the proportion of water goes up, but the ratio of the two products stays the
same: 60% azide and 40% alcohol. These two results suggest that the only effect of increasing
the amount of water is to make the solvent environment more polar, thereby speeding up the
initial ionization of the substrate. Even when the proportion of water is only 10%, it is present
in great excess and is trapping carbocations as fast as it can, relative to the rate at which azide
ion reacts with the same intermediates (Section 7-2).

2. We note from rows 1 and 2 in the table that the reaction rate rises by about 50% when NaN3 is
added. Without further information, we might assume that this effect is a consequence of the
occurrence of the SN2 mechanism. Were that to be the case, however, other anions should affect
the rate differently. However, we were told that bromide and iodide, far more powerful nucleo-
philes, affect the measured rate in exactly the same way as does azide. We can explain this obser-
vation only by assuming that displacement is entirely by the SN1 mechanism, and added ions affect
the rate only by increasing the polarity of the solution and speeding up ionization (Section 7-4).
 In rows 5 – 8 of the table, we note that increasing the amount of azide ion increases the amount
of azide-containing product that is formed. At the higher concentrations, azide, a better nucleo-
phile than water, is better able to compete for reaction with the carbocation intermediate.

New Reactions
1. Bimolecular Substitution — SN2 (Sections 6-2 through 6-9, 7-5)

Primary and secondary substrates only

}0
GC CI

Direct backside displacement with 100% inversion of configuration

H3C CH3

CH2CH3 CH2CH3

I��
H H

O NuO }0
GNu�ð

2. Unimolecular Substitution — SN1 (Sections 7-1 through 7-5)

Secondary and tertiary substrates only

A
OCH3

CH3

A
CH3

Through carbocation: Chiral systems are racemized

CBr
A

OCH3

CH3

A
CH3

CNuOCH3
�Br�

Nu�ð
CH3

CH3

C�
G

D

3. Unimolecular Elimination — E1 (Section 7-6)

Secondary and tertiary substrates only

A
OCH3

CH3

A
CH3

Through carbocation

CCl OCH3
�Cl�

B�ð
CH2

CH3H3C
C
B

GD
BH�

CH3

CH3

C�
G

D

4. Bimolecular Elimination — E2 (Section 7-7)

� �BH I�PCH3CH2CH2I CH3CH CH2
Simultaneous elimination of leaving group and neighboring proton

B�ð

Important Concepts
1. Secondary haloalkanes undergo slow and tertiary haloalkanes undergo fast unimolecular substitution

in polar media. When the solvent serves as the nucleophile, the process is called solvolysis.

2. The slowest, or rate-determining, step in unimolecular substitution is dissociation of the C – X
bond to form a carbocation intermediate. Added strong nucleophile changes the product but not
the reaction rate.

I m p o r t a n t C o n c e p t s

278 C h a p t e r 7 F u r t h e r R e a c t i o n s o f H a l o a l k a n e s

3. Carbocations are stabilized by hyperconjugation: Tertiary are the most stable, followed by sec-
ondary. Primary and methyl cations are too unstable to form in solution.

4. Racemization often results when unimolecular substitution takes place at a chiral carbon.

5. Unimolecular elimination to form an alkene accompanies substitution in secondary and tertiary
systems.

6. High concentrations of strong base may bring about bimolecular elimination. Expulsion of the
 leaving group accompanies removal of a hydrogen from the neighboring carbon by the base.
The stereo chemistry indicates an anti conformational arrangement of the hydrogen and the
leaving group.

7. Substitution is favored by unhindered substrates and small, less basic nucleophiles.

8. Elimination is favored by hindered substrates and bulky, more basic nucleophiles.

Problems
25. What is the major substitution product of each of the following solvolysis reactions?

H3C

H3C

Br

H

CH3OH

/∑

/∑

(a)
A

A

CH3

CH3

CH3CBr
CH3CH2OH

 (b)
A
Br

(CH3)2CCH2CH3
CF3CH2OH

 (c)

CH2CH3
CH3OH

Cl

(d)
A

A
O

B

C

Br

CH3

CH3

HCOH

O

 (e)
A

A

CH3
CH3CCl

CH3

D2O
 (f)

A

A

CH3
CH3CCl

CH3

H

OD

26. For each reaction presented in Problem 25, write out the complete, step-by-step mechanism
using curved-arrow notation. Be sure to show each individual step of each mechanism
 separately, and show the complete structures of the products of that step before going on to
the next.

27. Write the two major substitution products of the reaction shown in the margin. (a) Write a
mechanism to explain the formation of each of them. (b) Monitoring the reaction mixture reveals
that an isomer of the starting material is generated as an intermediate. Draw its structure and
explain how it is formed.

28. Give the two major substitution products of the following reaction.

CH3CH2OH
H3C

OSO2CH3
C6H5

C6H5H3C
H

29. How would each reaction in Problem 25 be affected by the addition of each of the following
substances to the solvolysis mixture?

(a) H2O (b) KI
(c) NaN3 (d) CH3CH2OCH2CH3 (Hint: Low polarity.)

30. Rank the following carbocations in decreasing order of stability.

�

�

CH3CH3H CH2�H

 C h a p t e r 7 279

31. Rank the compounds in each of the following groups in order of decreasing rate of solvolysis in
aqueous acetone.

(a)
A
CH3

CH3CHCH2CH2Cl
A

A

CH3

Cl

CH3CHCHCH3
A

A

CH3

Cl

CH3CCH2CH3

(b)

B
OCCH3

O

OHCl

(c)

H3C ClClBr

32. Give the products of the following substitution reactions. Indicate whether they arise through the
SN1 or the SN2 process. Formulate the detailed mechanisms of their generation.

 CH3CH2OH
(a) (CH3)2CHOSO2CF3 uuuy (b)

CH3 Excess CH3SH, CH3OH

Br
 (C6H5)3P, DMSO NaI, acetone
(c) CH3CH2CH2CH2Br uuuuuy (d) CH3CH2CHClCH2CH3 uuvuy

33. Give the product of each of the following substitution reactions.