Vollhardt  Capítulo 7 (Haloalcanos 2)
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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.
\u2212
\u2212
Hydroxide
as base
Proton transfer
Chloride
leaving
H
H
H
H
CH3
CH3
Cl Cl
Cl
\u2021
H
H
H CH3
CH3H
H
H
H
CH3
CH3
O
O
HO
C C C C C C
\u3b4 \u2212
\u3b4 \u2212
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\ufffd >
>
)
\u2021
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
\ufffd
Cl
The third observation not only strongly suggests that both the C \u2013 H and the C \u2013 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 \u2013 H and C \u2013 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 \u2013 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\ufffd\ufffdOCH3,
CH3OH
Fast
Na\ufffd\ufffdOCH3,
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
\u2022 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.
\u2022 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).
\u2022 In contrast, in the trans isomer (graphic, above right) the best conformation places the Br equatorial; 
the C \u2013 Br bond is anti to two C \u2013 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.
\u2022 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 \u2014 SN2, SN1, E2, and E1 \u2014 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.
\u2022 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.
\u2022 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
\ufffd
(CH3)3C
(CH3)3C
Br
(CH3)3C
CH3O 
\ufffd
Product forms via
E1 pathway instead
(CH3)3C HH
Br
(Caution: In the E2 mechanism