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Vollhardt Capítulo 7 (Haloalcanos 2)

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o a l k a n e s

Figure 7-8 Orbital description of the E2 reaction of 2-chloro-2-methylpropane with hydroxide ion.

−

−

Hydroxide
as base

Proton transfer

Chloride
leaving

H

H
H

H

CH3
CH3

Cl Cl
Cl

‡

H
H

H CH3

CH3H

H

H

H

CH3
CH3

O

O

HO

C C C C C C

δ −

δ −

ANIM
ATION ANIMATED MECHANISM:

Elimination (E2) reaction of
2-chloro-2-methylpropane Notice that the E1 (Figure 7-7) and E2 mechanisms are very similar, differing only in

the sequence of events. In the bimolecular reaction, proton abstraction and leaving-group
departure are simultaneous, as depicted in the transition state above (for a Newman projec-
tion, see margin). In the E1 process, the halide leaves fi rst, followed by an attack by the
base. A good way of thinking about the difference is to imagine that the strong base par-
ticipating in the E2 reaction is more aggressive. It does not wait for the tertiary or second-
ary halide to dissociate but attacks the substrate directly.

Experiments elucidate the detailed structure of the E2
transition state
What is the experimental evidence in support of a one-step process with a transition state
like that depicted in Figure 7-8? There are three pieces of relevant information. First, the
second-order rate law requires that both the haloalkane and the base take part in the rate-
determining step. Second, better leaving groups result in faster eliminations. This observa-
tion implies that the bond to the leaving group is partially broken in the transition state.

Newman Projection of
E2 Transition State

H

H

CH3H3C

H
Cl>) )

OH� >
>

)

‡

Relative Reactivity in
the E2 Reaction

RI . RBr . RCl . RF

Increasing

Exercise 7-13

Explain the result in the reaction shown below.

Cl ClI CH3O
�

Cl

The third observation not only strongly suggests that both the C – H and the C – X bonds
are broken in the transition state, it also describes their relative orientation in space when
this event takes place. Figure 7-8 illustrates a characteristic feature of the E2 reaction: its
stereochemistry. The substrate is pictured as reacting in a conformation that places the break-
ing C – H and C – X bonds in an anti relation. How can we establish the structure of the
transition state with such precision? For this purpose, we can use the principles of conforma-
tion and stereochemistry. Treatment of cis-1-bromo-4-(1,1-dimethylethyl)cyclohexane with
strong base leads to rapid bimolecular elimination to the corresponding alkene. In contrast,

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 C h a p t e r 7 269

under the same conditions, the trans isomer reacts only very slowly. Why? When we exam-
ine the most stable chair conformation of the cis compound, we fi nd that two hydrogens are
located anti to the axial bromine substituent. This geometry is very similar to that required
by the E2 transition state, and consequently elimination is easy. Conversely, the trans system
has no C – H bonds aligned anti to the equatorial leaving group (make a model). E2 elimina-
tion in this case would require either ring-fl ip to a diaxial conformer (see Section 4-4) or
removal of a hydrogen gauche to the bromine, both of which are energetically costly. The
latter would be an example of an elimination proceeding through an unfavorable syn transi-
tion state (syn, Greek, together). We shall return to E2 elimination in Chapter 11.

(CH3)3C (CH3)3C (CH3)3C

H

HH

H
Br

H

HH

H
BrNa��OCH3,

CH3OH
Fast

Na��OCH3,
CH3OH

Very slow

Anti Elimination Occurs Readily for cis- but Not for
trans-1-Bromo-4-(1,1-dimethylethyl)cyclohexane

cis-1-Bromo-4-
(1,1-dimethylethyl)cyclohexane

(Two anti hydrogens) (No anti hydrogens; only
anti ring carbons)

trans-1-Bromo-4-
(1,1-dimethylethyl)cyclohexane

Br

Br

7 - 7 B i m o l e c u l a r E l i m i n a t i o n : E 2

MODEL BUILDING

Exercise 7-14

Working with the Concepts: Elimination Rates and Mechanisms

The rate of elimination of cis-1-bromo-4-(1,1-dimethylethyl)cyclohexane is proportional to the
concentration of both substrate and base, but that of the trans isomer is proportional only to the
concentration of the substrate. Explain.
Strategy
In this problem we are given rate information that we need to explain. Throughout Chapters 6 and
7 we have seen how reaction kinetics can help defi ne mechanism. Apply the lessons: Consider the
kinetic order of each reaction and the consequences of the corresponding mechanism.
Solution
• Because the base-promoted elimination of cis-1-bromo-4-(1,1-dimethylethyl)cyclohexane occurs
with a rate proportional to both substrate and base, it must be following the E2 pathway.
• The E2 reaction strongly favors an anti orientation of the leaving group with respect to the
proton being removed. The graphic (above, left) shows that this orientation is already present in
the most stable chair conformation of the substrate molecule. Thus the E2 mechanism readily
occurs, with the base removing an H (blue) and initiating simultaneous expulsion of Br (green).
• In contrast, in the trans isomer (graphic, above right) the best conformation places the Br equatorial;
the C – Br bond is anti to two C – C bonds of the cyclohexane ring (blue). No hydrogen on either
adjacent carbon is anti with respect to the Br. E2 cannot readily occur from this conformation.
• For the trans isomer to undergo easy E2 reaction, it would fi rst have to undergo ring-fl ip in order
for the Br to become axial. The energy of this ring-fl ip is prohibitively unfavorable, however,
because the result would be a very high-energy conformation with the bulky tertiary butyl group
also in an axial position (see Table 4-3).

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270 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

In Summary Strong bases react with haloalkanes not only by substitution, but also by
elimination. The kinetics of these reactions are second order, an observation that points to
a bimolecular mechanism. An anti transition state is preferred, in which the base abstracts
a proton at the same time as the leaving group departs.

7-8 Competition Between Substitution and Elimination:
Structure Determines Function

The multiple reaction pathways — SN2, SN1, E2, and E1 — that haloalkanes may follow in
the presence of nucleophiles may seem confusing. Given the many parameters that affect the
relative importance of these transformations, are there some simple guidelines that might allow
us to predict, at least roughly, what the outcome of any particular reaction will be? The answer
is a cautious yes. This section will explain how consideration of base strength and steric bulk
of the reacting species can help us decide whether substitution or elimination will predominate.
We shall see that variation of these parameters may even permit control of the reaction
pathway.

Weakly basic nucleophiles give substitution
Good nucleophiles that are weaker bases than hydroxide give good yields of SN2 products
with primary and secondary halides and of SN1 products with tertiary substrates. Examples
include I2, Br2, RS2, N3

2, RCOO2, and PR3. Thus, 2-bromopropane reacts with both iodide
and acetate ions cleanly through the SN2 pathway, with virtually no competing elimination.

• Indeed, the rate of elimination from the trans isomer is proportional only to the substrate con-
centration and not that of the base, indicating that the mechanism is E1, not E2.
• Considering the possible options (below), the kinetic experiments tell us that the preferred path-
way is initial dissociation of the leaving group, leading to unimolecular elimination. Flipping the
chair to give the proper conformation for E2 does not compete.

Conformation required
for anti elimination: too

high in energy

H

�

(CH3)3C

(CH3)3C
Br

(CH3)3C

CH3O
�

Product forms via
E1 pathway instead

(CH3)3C HH

Br

(Caution: In the E2 mechanism