The E2 Reaction
E2 Rate and Transition State
The rate law of the above
E2 reaction follows:
In the
E2 reaction, a base removes a
β-hydrogen, forms a double bond and kicks out the
leaving group. The reaction occurs through a concerted mechanism and
requires a
β-hydrogen. This mechanism is also called
β-elimination.
The
E2 rate law is first order for both reactants. Here's the simplest possible transition
state:
E2 Stereochemistry
As in the SN2 reaction, the mechanism of transition state formation can be deduced
using a stereocenter. There are two possible modes of reactivity that yield two different stereoisomers.
In
syn elimination, the base attacks the
β-hydrogen on the
same side as the leaving group.
In
anti elimination, the base attacks the
β-hydrogen on the
opposite side of the
leaving group.
It has been experimentally determined that E2 elimination occurs through an anti
mechanism.
Steric and Molecular Orbital Explanation of Anti Elimination
Just as with the SN2 backside attack mechanism, there are steric and molecular
orbital explanations for E2 anti elimination.
The steric argument notes that the syn transition state is an eclipsed
conformation. The anti state is in the more energetically stable
staggered conformation. Since the rate of a reaction is inversely
proportional to the stability of the transition
state, the anti route should be much more favorable.
The
E2 molecular orbital argument hinges on the
σ* C-LG antibond, just like
the
SN2 MO argument. In this case, the
σ* C-LG antibond is only
accessible to the electrons of the
σ C-H bond if th e two bonds are antiperiplanar. In other
words the C-H and C-LG bonds must point opposite directions in parallel planes. If this condition is
met, the electrons of the
σ C-H bond donate into the
σ* C-LG antibond.
The C-H
σ electrons attack from the "back," of the C-LG bond in a manner similar to
SN2's backside attack.
In the
syn elimination, the bonds are
not antiperiplanar, and the electrons of the
σ
C-H bond cannot attack the
σ* C-LG antibond. Thus
E2 cannot occur through
syn elimination.
Saytzeff's Rule
Recall the first
E2 reaction presented in this section:
There's no reason why the methoxide ion can't attack a
β hydrogen on the "right" methyl group.
The product of both
β-eliminations is the same for this reaction, but there are
many
E2 reactions that can yield enantiomers
or diastereomers depending on
which
β hydrogen is eliminated.
The three products that result from the β-elimination of the three underlined β-hydrogens.
Product 3 is the major product.
But wait! The observed products seem to contradict the antiperiplanar argument. There are three
products and three β-hydrogens, but only two of those hydrogens may be antiperiplanar to the C-OTs
bond at one time. How can the two hydrogens off carbon B give two products, even though only one can be
antiperiplanar?
While only one hydrogen off carbon B can be antiperiplanar at one time, both hydrogens spend some time in
the antiperiplanar position. Thus the elimination of the hydrogens off carbon B gives the cis and
trans products. This is possible because alkanes that are not inordinately hindered rotate about
their C-C bonds. For an example of a structure that cannot rotate freely about its C-C bond, see problem
#4. A deeper discussion of C-C bond rotation can be found under
conformational analysis.
Product 3 is the major product because an E2 elimination favors the formation of the most stable
alkene. The stability of an alkene is proportional to the degree of substitution at its double-bond
carbons. The double-bonds of products 1 and 2 are less branched than the double-bonds of product 1.
Thus there will be more of product 3 than products 1 and 2 after the reaction has run to completion.
E2's preference for the more stable (and more highly branched) alkene is called Saytzeff's Rule.
