E1 and E2 Reactions
Elimination reactions in organic chemistry
From smore on SN1 and SN2:
What conditions favor elimination?
- Primary alkyl haldies + weak nucleophile/strong base
- secondary alkyl halides + strong Nuc/base AND/OR weak nucleophile/strong base
- most tertiary alkyl halides (die to steric hindrance)
As you can see, there are two different mechanisms (E1 and E2). Just as in SN1 and SN2, the "1" and "2" represent the number of species present in the rate determining step. A slightly more confusing way to think about it is SN1 and E1 are two steps with an intermediate whereas SN2 and E2 are one-step mechanisms with a transition state.
Weak nucleophiles favor E1
The reason why is that the nucleophile is not involved in the slow step. Once the carbocation is formed, the nucleophile is really acting only as a Lewis base at this point. As can be seen from the flowchart above, in the cases of secondary or tertiary halides favoring the E1 mechanism (see below for mechanism), it is likely to be the solvent that actually causes the alkene to be formed.
Strong Nucleophiles Favor E2
Primary alkyl halides combined with a weaker nucleophile that is a strong base that is too sterically hindered for substitution will undergo E2.
Secondary alkyl halides will do the same as primary given the same type of nucleophile. Secondary alkyl halides give the widest range of products due the possibility of almost every type of mechanism! See below for more complicated examples.
Tertiary alkyl halides favor elimination as can be seen in the flow chart with stronger nucleophiles favoring the E2 mechanism.
E2 Mechanism with transition state
Look at whether the alkyl halide is symmetrical or asymmetrical
In the case of propane, there is only one possible product. Because of the simplicity of this molecule, there is no chance for geometric isomers as well. Which mechanism is favored (E1 or E2) is dictated by the nucleophile.
Now things get more complicated because there are choices as to which beta-hydrogen is removed. See examples below. There is also a chance for geometric isomers as well. Again, choice of mechanism will be dictated by the type of nucleophile AND there is competition with substitution.
As with the secondary example, there are choices for which beta hydrogen is removed. The beta hydrogen must be on the opposite side of the molecule to the leaving group (Br in this case). As with the secondary example, the choice of mechanism is dictated by the nucleophile AND there is competition with substitution but to a lesser extent as compared to secondary (due to steric hindrance in tertiary alkyl halides).
Two unsymmetrical examples: secondary and tertiary
An assortment of elimination reactions
Comments on reactions above
(h) 3-methyl-2-hexene and 3-methyl-3-hexene have geometric isomers as well.