Comparing the E1 vs SN1 Reactions
Last updated: November 30th, 2020 |
The Important Role of The Counter-Ion In Determining E1 vs SN1
- Alcohols Don’t Undergo Elimination Reactions Until OH Is Converted To A Better Leaving Group
- Adding Acid To Alcohols Results In A Better Leaving Group’
- When Elimination And Substitution Compete: E1 vs SN1
- In Order To Get E1 To Predominate vs SN1 In The Reaction Of Alcohols Use H2SO4, TsOH, or H3PO4 (And Heat)
- (Advanced) References and Further Reading
Imagine you’re starting with the alcohol on the left and you’d like to get to the alkene on the right.
What bonds are formed and broken here? We’re forming C-C (π), we’re breaking C-H, and we’re breaking C-OH. It’s an elimination reaction.
Notice a problem here? We need to have HO(-) as a leaving group. If you’ll recall, strong bases [like HO(-) ] are terrible leaving groups – which makes the E1 pathway unlikely. So what if we tried to use a strong base, maybe trying to promote an E2 reaction? Well, that would be even worse – we’d likely deprotonate OH before the C-H, and you can imagine that we’d have to have O(2-) as a leaving group here. Not good!
That means that the reaction, as written, is very unlikely to happen.
Yet, there is something very simple that we can do to make this reaction work. We’d need to have a better leaving group (a weaker base). How can we do this?
If we add a strong acid, we turn OH into H2O+, the conjugate acid is a better leaving group. Now, water can leave, forming a carbocation; and then a base can break the C-H bond, forming the alkene.
Notice that this is now a classic E1 reaction. The rate is going to be dependent on the stability of the carbocation. This one is tertiary, so it should proceed at a reasonably high rate.
A question arises here. What’s going to act as the base? As it stands, a C-H bond adjacent to a carbocation has an extremely high acidity (at least below -2, if you follow pKa). That means that just about any weak base (water, or the conjugate base of the acid) is sufficient to deprotonate the carbon. It’s possible that more than one species can act as a base here. I’ve shown water removing the proton, but it’s not unreasonable to show the conjugate base of the acid removing the proton in most circumstances.
Now comes one of the things about organic chemistry that often causes trouble for students. For one of the first times in our discussions here, we’re dealing with a situation where we can have competing reactions.
Let’s back up. The E1 reaction goes through a carbocation, correct? Well, if you’ll recall, so does the SN1 reaction.
We’ve already seen examples where a carbocation was formed from an alcohol by adding a strong acid like HCl, HBr, or HI, and we ended up with the alkyl halide.
Why? The halide ions (Cl- , Br-, I- ) are decent nucleophiles under the reaction conditions.
So how can we stack the deck in favor of the E1 process?
Use a strong acid with a conjugate base that is a poor nucleophile.
The usual choice is H2SO4. The HSO4(-) ion is a relatively poor nucleophile due to the negative charge of the oxygen being distributed throughout the molecule (resonance). Two other acids you might see for this purpose are p-toluenesulfonic acid (p-TsOH), which is essentially a cousin of H2SO4, and phosphoric acid (H3PO4).
Also, don’t forget that elimination reactions are favored by heat.
4. In Order To Get E1 To Predominate vs SN1 In The Reaction Of Alcohols Use H2SO4, TsOH, or H3PO4 (And Heat)
In summary, if you’d like E1 to predominate over SN1: choose an acid with a weakly nucleophilic counterion [H2SO4, TsOH, or H3PO4], and heat.
If you’d like SN1 to predominate over E1, choose an acid like HCl, HBr, or HI.
We’re almost done talking about elimination reactions. Next post – we’ll talk about rearrangements.
Since the E1 and SN1 reactions both proceed through a common carbocation intermediate, these pathways can and do compete with each other under a lot of reaction conditions.
- Ion Pairs in Elimination
M. Cocivera and S. Winstein
Journal of the American Chemical Society 1963, 85 (11), 1702-1703
Prof. Saul Winstein (UCLA) was a highly influential physical organic chemist in the early 20th century, and introduced several concepts that are now fundamental to organic mechanisms, such as anchimeric assistance, ion-pairing, internal return, and many others. In this paper, Prof. Winstein shows that the E1 reaction has a strong solvent dependence – as the dielectric constant (or polarity) of the solvent decreases, the amount of olefin obtained also decreases, since carbocations cannot be formed easily. In anhydrous ethanol, t-butyl chloride and bromide give 44% and 36% yield of alkene respectively, whereas in glacial acetic acid, 73% and 69% of olefin, respectively, is obtained.
- Mechanism of elimination reactions. Part XI. Kinetics of olefin elimination from tert.-butyl and tert.-amyl bromides in acidic and alkaline alcoholic media
M. L. Dhar, E. D. Hughes, and C. K. Ingold
J. Chem. Soc. 1948, 2065-2072
The E1 reaction is not very useful synthetically for olefin synthesis, because the ratio of elimination to substitution products is substantially lower than in the E2 reaction. For example, solvolysis of t-butyl bromide in dry ethanol only yields 19% isobutylene, whereas 93% yield of the alkene is obtained with 2M ethoxide.
- Chemical Effects of Steric Strains. II. The Effect of Structure on Olefin Formation in the Hydrolysis of Tertiary Aliphatic Chlorides
Herbert C. Brown and Roslyn Silber Fletcher
Journal of the American Chemical Society 1950, 72 (3), 1223-1226
This paper by Nobel Laureate Prof. H. C. Brown provides a perfect study of how the structure of the substrate can influence the E1-SN1 competition. With a series of t-alkyl chlorides, the proportion of olefin obtained on solvolysis in 80% ethanol rises as the alkyl groups become more highly branched.
- Mechanism of elimination reactions. Part VIII. Temperature effects on rates and product-proportions in uni- and bi-molecular substitution and elimination reactions of alkyl halides and sulphonium salts in hydroxylic solvents
K. A. Cooper, E. D. Hughes, C. K. Ingold, and B. J. MacNulty
J. Chem. Soc. 1948, 2049-2054
Prof. Ingold states: “[..] for any given pair of simultaneous bimolecular processes, the elimination has, in each of the investigated cases, an Arrhenius energy of activation which lies higher than that of the accompanying substitution by 1-2 kcal/g.-mol. The elimination thus has always the larger temperature coefficient, so that a rise of temperature increases the proportions in which olefin is formed”. An increase in temperature produces a modest increase in the proportion of olefin in E1-SN1 reactions.
- Mechanism of elimination reactions. Part VII. Solvent effects on rates and product-proportions in uni- and bi-molecular substitution and elimination reactions of alkyl halides and sulphonium salts in hydroxylic solvents
K. A. Cooper, M. L. Dhar, E. D. Hughes, C. K. Ingold, B. J. MacNulty and L. I. Woolf
J. Chem. Soc. 1948, 2043-2049
Another modest increase in the yield of olefin in E1-SN1 reactions is found on going from aqueous ethanol to anhydrous ethanol.
- The mechanism of elimination reactions. Part II. Unimolecular olefin formation from sec.-octyl halides in aqueous alcohol. A new criterion of mechanism
Edward D. Hughes, Christopher K. Ingold, and Uriel G. Shapiro
J. Chem. Soc. 1937, 1277-1280
This paper demonstrates how the E1 and SN1 reactions compete when secondary substrates are used. 2-bromo and 2-chlorooctane each give around 14% of olefin when heated to 100 °C in 60% aqueous ethanol.