Organic Reaction Mechanisms.
2. Haloalkane reaction with electron rich species.
d. The E2 mechanism.

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Experimental observations.

The overall reaction involves the loss of an electron-rich atom or group (the leaving group) and a proton in the presence of a base with the following observations:

1. Kinetics:
Rate = [RX][base] (first order in haloalkane and in base, second order overall.)
2. The base must be:
a poor nucleophile other wise the competing substitution may dominate.
strong, and/or bulky, this latter making it difficult for it to act as a nucleophile in a substitution reaction.
3. The leaving electron-rich atom or group - the "leaving group":
is a good leaving group - that is it is a reasonably stable entity.
4. The beta-carbon site:
is the more substituted of the possible ones,leading to the more substituted alkene as the major product, though more than one product may be produced. When E- and Z-alkenes can be formed, the E-alkene (lower energy) predominates though both are formed.
5. The beta-hydrogen:
must be capable of alignment periplanar with the leaving group (anti is mostly found though syn periplanar eliminations are known. In cases where these alignments cannot occur, elimination by this mechanism does not take place.

Accounting for the experimental observations.

1. The kinetics.

The simplest explanation of the second order kinetics is that the reaction is a one step process involving both the substrate and the base. That is the base removes the beta-hydrogen, the double bond forms, and the leaving group leaves all simultaneously.

2. The base:

Must be designed to avoid the competing sustitution reaction.

3. The leaving group.

The less stable the leaving group, the more energy is required for it to leave and the reaction slows, or stops.

4. The beta-carbon site.

Because the reaction produces as the major product, the most substituted alkene, the transition state must have considerable double bond character: it is known that double bonds have lower energy the more substituted they are. Also that E-alkenes have a lower enrgy than the corresponding Z-alkene.

5. The stereochemistry.

If, as postulated in 4 above, there is considerable double bond character in the transition state, then the two orbitals which must overlap to form the pi bond must be aligned so that they are able to so overlap. Since the two orbitals forming the pi bond must be parallel to one another, then so must be the orbitals which are going to form the pi bond during the reaction. So the beta C-H bond and the alpha C-X bonds must lie in the same plane. Of the two possibilities, the base and the leaving group on the same side of the molecule, and the base and the leaving group on opposite sides of the molecule, this latter, for steric and electrical reasons is the lower in energy and the one which is found practically, the reaction occurs antiperiplanar whenever it can.

Whenever the above geometry constraints are not met, the reaction doesn't occur - the orbitals needed to form the pi bond cannot overlap to form it.

The animated mechanism. | Reaction summary.

Date created: 2005 06 26.