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Elimination Reactions General FeaturesWhen
a nucleophile attacks an alkyl halide, two substitution pathways are
possible. There is however another reaction which can occur and in
some cases compete with or override the substitution pathway. This
is elimination, or removal of HX from an alkyl halide. The
general equation below shows a comparison of substitution and elimination:
It can be seen that elimination could well take place under the same reaction conditions as substitution, if there is a hydrogen atom b to the halogen. Furthermore, there is the question of regiochemistry in elimination of unsymmetrical alkyl halides. If there is a b-hydrogen on each side of the halogen, two alkene products are possible:
In the elimination of 2-bromo-2-methylbutane with a strong, unhindered base such as sodium ethoxide in ethanol, the two alkene products are formed in the ratio 7:3 respectively. We see that the more substituted alkene product is favoured. This is not surprising, since the thermodynamic stability of alkenes increases with substitution; although what really matters is that the more substituted alkene has a transition state with a lower energy. The more substituted alkene is termed the Saytzeff product (German spelling, Russian spelling = Zaitsev). With a sterically hindered, strong base such as potassium t-butoxide in t-butanol, the ratio is reversed and the less substituted alkene is favoured, on steric grounds. The less substituted alkene is termed the Hofmann product. Hofmann elimination is also favoured by substrates with bulky leaving groups such as Et3N+, Me2S+. As
in substitution reactions, eliminations can occur by a variety of
mechanisms, the most common being E2, E1 and E1CB.
The E2 PathwayKineticsThe
rate law for an E2 reaction is found to be analogous to that for SN2:
Rate = k [RX] [Base] MechanismIn
common with the SN2 mechanism, E2 elimination is a bimolecular
process, i.e. two species are involved in the rate-determining
step (r.d.s.). There are no intermediates, just one transition state.
The diagram below shows the general mechanism:
There is stereochemical evidence for the mechanism as shown above. Elimination must always occur from a periplanar geometry. This means that all four atoms H-C-C-X must all lie in the same plane. Further, antiperiplanar (staggered) geometry is favoured over synperiplanar (eclipsed):
Careful
choice of substrate can lead to formation of products which can only
form by antiperiplanar elimination. Molecular orbital interactions
also favour the antiperiplanar geometry. The E1 PathwayKineticsThe
E1 pathway shows first-order kinetics and the rate is dependent only
on the concentration of the alkyl halide:
Rate = k [RX] MechanismAs
the designation implies, the E1 pathway involves just one species
in the rate-determining step, the alkyl halide. The first (rate-determining)
step is exactly the same as for the SN1 reaction, heterolytic
fission of the C-X bond to give a carbocation and a halide ion. For
the second step, instead of the nucleophile/base adding to the carbocation,
a proton is removed, forming an alkene:
The best candidates for E1 elimination will therefore be those which can form relatively stable carbocations, i.e. tertiary halides. Also, if the base is weak, the E1 pathway may be followed. The
more substituted alkene is usually obtained, obeying Saytzeff's rule.
The E1CB PathwayThe
E1CB mechanism is another unimolecular elimination pathway.
However, a hydrogen is lost first, followed by the halogen.
The designation CB stems from the fact that it is the conjugate base
of the substrate that is losing the halide ion. The E1CB
pathway is likely to be followed by substrates with acidic hydrogens
and poor leaving groups.
The problems in this section concentrate on the E2, E1 and E1CB pathways described above. If you encounter an unfamiliar reagent or substrate, don't panic! Use the chemistry you know to propose a viable mechanism, and remember, you are looking for an overall elimination. |
