Mechanistic Comparison: SN vs. E

It was noted previously (here) that there were great similarities in SN and E in terms of the starting materials and reagents – both involve a molecule containing a leaving group, and a reagent that is either a nucleophile or a base (it was also noted in the Nucleophilic Substitution chapter that these two share similar properties). So why do some reactions go as substitutions and some as eliminations?

Unfortunately, not even the question can be that simple, let alone the answer! Reactions will not necessarily go 100% substitution, or elimination – they can follow both paths in different quantities. Obviously, some reactions will clearly be one or the other – for example in the substitution by OH of I in MeI where there are no β-hydrogens so 1,2- elimination cannot occur.

In the case of SN1 vs. E1, the rate determining step is the same in both cases (formation of a carbocationic intermediate) – from there the reaction may go either way. For SN2 vs. E2, the mechanisms follow different pathways from the start, but still the initial conditions are the same. A compound likely to undergo SN1 will also be likely to undergo E1, rather than E2, and vice versa – so in comparing the two reaction types, we must consider;

i. Which of E1/SN1 will be (or is) preferred.
ii. Which of E2/SN2 will be (or is) preferred.
iii. Whether anything will cause a shift in mechanism from unimolecular to bimolecular, or vice versa – shifting the mechanism does not necessarily leave the ratio of SN:E the same.

The overall products of such reactions can be analysed in terms of the percentage of alkene formed. Of course, similar factors to those which favoured SN1 over SN2 will favour E1/SN1 over E2/SN2 (or vice-versa) because of the mechanistic similarities (need of stabilised cation in the unimolecular process, benefit of not solvating the nucleophilic/basic species in the bimolecular process).

The proportion of elimination will rise as the carbon centres become more substituted – the reasons for this are summarized below;
1. This reflects the increased relative stabilisation of the TS as the number of alkyl groups on the developing double bond increases. In E2 this is clear in the TS – however the TS of E1 has not yet been mentioned, so this is less clear. The TS in question lies between the carbocationic intermediate and the product – so it contains some double bond character and hence favours increased substitution.
2. For E2 the increased steric hindrance towards nucleophilic attack will increase the rate of substitution; in E1 there are two steric benefits –  steric hindrance to attack of the positive centre of the intermediate will facilitate deprotonation (elimination) rather than nucleophilic attack, and the centre remains sp2hybridised allowing greater substituent separation rather than returning to spas in SN1. Also, therefore, elimination will be favoured by use of a bulky base (e.g. NEtvs. MeO).
3. A nearby group (such as the phenyl substituent used in earlier elimination examples) able to conjugate to the developing double bond will enhance the elimination rate in both cases.

There is a thermodynamic effect involved too – as the temperature is raised, the proportion of elimination will rise because of the entropy increase inherent in releasing more molecules (substitution starts with two molecules and ends with two; elimination starts with two and ends with three).

This can be seen using the simple equation ΔG = ΔH – TΔS .As DS gets larger, the TΔS term causes ΔG to be more negative.