The E1 Reaction

The E1 Reaction – Three Key Pieces of Evidence, and a Mechanism

Last time in this walkthrough on elimination reactions, we talked about two types of elimination reactions. In this post, we’re going to dig a little bit deeper on one type of elimination reaction, and based on what experiments tell us, come up with a hypothesis for how it works.

Table of Contents

1. How Do We Explain What Happens In This Elimination Reaction (Which We Will Call, “E1”)
2. First Clue About The Mechanism of the “E1 Elimination”: The Rate Only Depends On Concentration of Substrate (Not Base)
3. The Second Clue About The Mechanism Of The E1 Elimination Reaction – Rate Is Fastest For Tertiary Substrates
4. The Third Clue About The Mechanism Of The E1 Elimination Reaction: It Competes with the SN1 Reaction
5. Putting It Together: The E1 Mechanism Proceeds Through Loss Of A Leaving Group, Then Deprotonation
6. (Advanced) References and Further Reading

1. How Do We Explain What Happens In This Elimination Reaction (Which We Will Call, “E1”)

Here’s the reaction. First, look at the bonds that are being formed and broken. This is a classic elimination reaction – we’re forming a new C–C(π) bond, and breaking a C–H and C–leaving group (Br here) bond.

But now we want to know more than just “what happens”. We want to understand how it happens. What’s the sequence of bond-forming and bond breaking? To understand HOW it happens, we need to look at what the data tells us. That’s because chemistry is an empirical science; we look at the evidence, and then work backwards.

2. First Clue About The Mechanism of the “E1 Elimination”: The Rate Only Depends On Concentration of Substrate (Not Base)

Let’s look at the first important clue for this reaction. We can measure reaction rates quite readily. When we vary the concentration of the substrate, the reaction rate increases accordingly. In other words, there is a “first-order” dependence of rate on the concentration of substrate.

However, if we vary the concentration of the base (here, H₂O) the rate of the reaction doesn’t change at all.

What information can we deduce from this?  The rate determining step for this reaction (whatever it is) therefore does not involve the base. Whatever mechanism we draw will have to account for this fact.

3. The Second Clue About The Mechanism Of The E1 Elimination Reaction – Rate Is Fastest For Tertiary Substrates

Another interesting line of evidence we can obtain from this reaction is through varying the type of substrate, and measure the rate constant that results. So if we take the simple alkyl halide on the far left (below) where the carbon attached to Br is also attached to 3 carbons (this is called a tertiary alkyl halide), the rate is faster than for the middle alkyl halide (a secondary alkyl halide) which is itself faster than a primary alkyl halide (attached to only one carbon in addition to Br).

So the rate proceeds in the order tertiary (fastest) > secondary >> primary (slowest)

Any mechanism we draw for this reaction would likewise have to account for this fact. What could be going on such that tertiary substrates are faster than primary?

4. The Third Clue About The Mechanism Of The E1 Elimination Reaction: It Competes with the SN1 Reaction

A final interesting clue about the mechanism of this reaction concerns the by-products that are often obtained. For instance, when the alkyl halide below is subjected to these reaction conditions, we do obtain the expected elimination product. However, we also get substitution reactions in the product mix as well. Remember – substitution reactions involve breakage of C-(leaving group) and formation of C-(nucleophile). What makes this particular starting alkyl halide particularly interesting is that the carbon attached to Br is a stereocenter. And if we start with a single enantiomer of starting material here, we note that the substitution product formed is a mixture of stereoisomers. Note that both inversion and retention of stereochemistry at the stereocenter has occurred.

We’ve seen this pattern before – it’s an S_N1 reaction!

This last part is a very important clue. If an SN1 reaction is occurring in the reaction mixture, looking back at the mechanism of the SN1 could help us think about what type of mechanism might be going on in this case to give us the elimination product.

5. Putting It Together: The E1 Mechanism Proceeds Through Loss Of A Leaving Group, Then Deprotonation

Taking all of these clues into account, what’s the best way to explain what happens? This:

The reaction is proposed to occur in two steps: first, the leaving group leaves, forming a carbocation. Second, base removes a proton, forming the alkene. This nicely fits in with the three clues mentioned above. [Also note that the more substituted alkene is formed here, following Zaitsev’s rule].

Similar to the S_N1 mechanism, this is referred to as the E1 mechanism (elimination, unimolecular).

So what’s going on in the other type of elimination reaction? That’s the topic for the next post!

Next Post: The E2 Mechanism

(Advanced) References and Further Reading

1. Mechanisms of Elimination Reactions, by Saunders and Cockerill, 1973, p. 210.
   This book states, “There are three main factors that favor the E1 mechanism: a substrate that gives a relatively stable carbocation, an ionizing solvent, and the absence of strong bases or nucleophiles“.
2. Hydrolysis of Secondary and Tertiary Alkyl Halides
   Edward D. Hughes
   Journal of the American Chemical Society 1935, 57 (4), 708-709
   DOI: 10.1021/ja01307a033
   An early paper, and likely the first paper to feature the term “E1”. This paper also notes that the E1 reaction competes with the S_N1 reaction for several substrates.
3. Unimolecular Elimination and the Significance of the Electrical Conduction, Racemization and Halogen Replacement of Organic Halides in Solution.
   HUGHES, E., INGOLD, C. & SCOTT, A.
   Nature 1936, 138, 120–121
   DOI: 1038/138120b0
   This paper describes electrical conductivity experiments of tertiary and secondary substrates in SO₂ with HCl in support of the E1 mechanism.
4. Ion Pairs in Elimination
   Cocivera and S. Winstein
   Journal of the American Chemical Society 1963, 85 (11), 1702-1703
   DOI: 10.1021/ja00894a046
   In highly polar and dissociating solvents (like aqueous alcohols) the carbocation is relatively “free” and the amount of alkene product is independent of the nature of the leaving group. In this paper, Prof. Winstein shows that in less polar solvents, the carbocation is less “free” and tight ion pairs are formed. Consequently the elimination/substitution ratio can be strongly dependent on the basicity of the leaving group.  The key finding here is that more basic leaving groups (Cl-, Br-) lead to more E1 products than less basic leaving groups (I-, S(CH3)2 ). Furthermore, the proportion of E1 products in less polar solvents are higher than those in more dissociating solvents.
5. Mechanism of elimination reactions. Part X. Kinetics of olefin elimination from isopropyl, sec.-butyl, 2-n-amyl, and 3-n-amyl bromides in acidic and alkaline alcoholic media
   M. L. Dhar, E. D. Hughes, and C. K. Ingold
   J. Chem. Soc. 1948, 2058-2065
   DOI: 10.1039/JR9480002058
   Table III in this paper shows that secondary substrates are generally poor substrates for E1 reactions – yields ranged from 4.7% to 17.4%.
6. 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
   DOI: 10.1039/JR9480002049
   On how temperature effects elimination/substitution ratio, Ingold states here: “[..] 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.”