Vollhardt  Capítulo 6 (Haloalcanos)
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Vollhardt Capítulo 6 (Haloalcanos)

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charge on the nucleophile has spread
partly onto the leaving group. As the reaction comes to completion, the leaving group evolves
to a fully charged anion. In the electrostatic potential map of the transition state, this process
is refl ected in the attenuated red color around the two halogen nuclei, compared with the full
red visible in the starting and ending halide ions. Note that in the reaction schemes preceding
the electrostatic potential renditions, we use the mechanistic color scheme of green for the
leaving group, and not red. In other respects, the colors match, as you would expect.

} }
@

C Br BrI − I

H

S R

H3C
CH3CH2

C +I

H

CH3
CH2CH3

C

H

CH3 CH2CH3

‡

δ − δ −

Stereochemistry of the Backside Displacement Mechanism for SN2 Reactions

Backside displacement(Chiral and
optically active)

(Chiral and optically
active; configuration

inverted)

Br −

MECHANISM

MODEL BUILDING

Exercise 6-11

Draw representations of the hypothetical frontside and backside displacement mechanisms for the
SN2 reaction of sodium iodide with 2-bromobutane (Table 6-3). Use arrows like those shown in
Figures 6-2 and 6-3 to represent electron-pair movement.

ANIM
ATION ANIMATED MECHANISM:

Nucleophilic substitution (SN2)

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Há inversão da configuração do carbono quiral, que passa de (R) para (S), uma vez que única substituição que ocorre de verdade é que acontece por trás.

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 C h a p t e r 6 227

‡

‡

I I IBr Br Br

The stereochemistry of displacement at a primary carbon is more diffi cult to observe
directly, because a primary carbon atom is attached to two hydrogens in addition to the
leaving group: It is not a stereocenter. This obstacle may be overcome by replacing one of
the two hydrogen atoms by deuterium, the hydrogen isotope with mass 5 2. The result is
a stereocenter at the primary carbon and a chiral molecule. This strategy has been employed
to confi rm that SN2 displacement at a primary carbon atom does indeed occur with inversion
of confi guration, as the example below illustrates.

S-1-Chloro-1-deuteriobutane
(Chiral and optically active)

Stereochemistry of SN2 Displacement at a Primary Carbon Atom

R-1-Azido-1-deuteriobutane
(Chiral and optically active;

configuration inverted)

i
∑

[

CH
D

ClO
CH3CH2CH2

70%

i

∑

[
N3 H

D
CO

CH2CH2CH3
A “primary stereocenter”

NaN3, CH3OH, H2O
SN2 displacement

with 100% inversion

The nucleophile, azide ion (N3
2), gives rise to stereospecifi c backside displacement of chloride,

giving the azidoalkane product with the inverted confi guration at the chiral carbon.

6 - 5 F r o n t s i d e o r B a c k s i d e A t t a c k ?

Exercise 6-12

Write the products of the following SN2 reactions: (a) (R)-3-chloroheptane 1 Na
12SH; (b) (S)-2-

bromooctane 1 N(CH3)3; (c) (3R,4R)-4-iodo-3-methyloctane 1 K
12SeCH3.

Exercise 6-13

Write the structures of the products of the SN2 reactions of cyanide ion with (a) meso-2,4-
dibromopentane (double SN2 reaction); (b) trans-1-iodo-4-methylcyclohexane.

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228 C h a p t e r 6 P r o p e r t i e s a n d R e a c t i o n s o f H a l o a l k a n e s

6-6 Consequences of Inversion in SN2 Reactions
What are the consequences of the inversion of stereochemistry in the SN2 reaction? Because
the reaction is stereospecifi c, we can design ways to use displacement reactions to synthesize
a desired stereoisomer.

}

}

@

C

C

Nu:− +

Nuδ
− δ −

X

X

CNu

+ X−

‡

Reaction coordinate

E

Figure 6-5 Potential energy
 diagram for an SN2 reaction. The
process takes place in a single
step, with a single transition state.

ANIM
ATION ANIMATED MECHANISM:

Nucleophilic substitution (SN2)

‡

R′′
R R′

C
C

Nu

XXX

Nu
sp2 hybridization at carbon

Nu
δ −

δ −

R′′
R′

R

−

−

C
R′′

R
R′

Figure 6-4 Orbital description of
backside attack in the SN2 reac-
tion. The process is reminiscent of
the inversion of an umbrella ex-
posed to gusty winds.

The transition state of the SN2 reaction can be described in an
orbital picture
The transition state for the SN2 reaction can be described in orbital terms, as shown in
Figure 6-4. As the nucleophile approaches the back lobe of the sp3 hybrid orbital used by
carbon to bind the halogen atom, the rest of the molecule becomes planar at the transition
state by changing the hybridization at carbon to sp2. As the reaction proceeds to products,
the inversion motion is completed and the carbon returns to the tetrahedral sp3 confi guration.
A depiction of the course of the reaction using a potential energy – reaction coordinate
diagram is shown in Figure 6-5.

A former president experiences
 inversion of confi guration.

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O Estado de transição é, obviamente, o estado de mais alta energia.

 C h a p t e r 6 229

We can synthesize a specifi c enantiomer by using SN2 reactions
Consider the conversion of 2-bromooctane into 2-octanethiol in its reaction with hydrogen
sulfi de ion, HS2. If we were to start with optically pure R bromide, we would obtain only
S thiol and none of its R enantiomer.

} ~
O O

0 0

D
HS � C

CH3(CH2)4CH2 CH3 CH3

H

CH2(CH2)4CH3

H
Br HS

Inversion of Configuration of an Optically Pure
Compound by SN2 Reaction

(R)-2-Bromooctane
([ ] � �34.6)

(S)-2-Octanethiol
([ ] � �36.4)

G C ��

� �

Br �

But what if we wanted to convert (R)-2-bromooctane into the R thiol? One tech-
nique uses a sequence of two SN2 reactions, each resulting in inversion of confi guration
at the stereocenter. For example, an SN2 reaction with iodide would fi rst generate (S)-2-
iodooctane. We would then use this haloalkane with an inverted confi guration as the
substrate in a second displacement, now with HS2 ion, to furnish the R thiol. This double
inversion sequence of two SN2 processes gives us the result we desire, a net retention
of confi guration.

} ~
O O

0 0

DC
CH3(CH2)4CH2 CH3 CH3

H

CH2(CH2)4CH3

H
II

�

�Br�

First
inversion of

configuration

Second
inversion of

configuration

HS�

�I�

Using Double Inversion to Give Net Retention of Configuration

(R)-2-Bromooctane
([ ] � �34.6)

(S)-2-Iodooctane
([ ] � �46.3)

(R)-2-Octanethiol
([ ] � �36.4)

G
} O0

C
CH3(CH2)4CH2 CH3

H
SHGC

� ��

Br

6 - 6 C o n s e q u e n c e s o f I n v e r s i o n i n S N 2 R e a c t i o n s

Color code for priorities
(see Section 5-3)
Highest: red
Second highest: blue
Third highest: green
Lowest: black

Exercise 6-14

As we saw for carvone (Chapter 5, Problem 43), enantiomers can sometimes be distinguished by
odor and fl avor. 3-Octanol and some of its derivatives are examples: The dextrorotatory compounds
are found in natural peppermint oil, whereas their (2) counterparts contribute to the essence of
lavender. Show how you would synthesize optically pure samples of each enantiomer of 3-octyl
acetate, starting with (S)-3-iodooctane. (The conversion of acetates into alcohols will be shown in
Section 8-5.)

O

3-Octyl acetate

OCCH3

CH3CH2CHCH2CH2CH2CH2CH3
A

B

Exercise 6-15

MODEL BUILDING

Working with the Concepts: Stereochemical Consequences of SN2 Displacement

Treatment of (S)-2-iodooctane with NaI in solution causes the optical activity of the starting
organic compound to disappear. Explain.
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