Introduction
Our first article on elimination reactions! This article will explain the key concepts in elimination reactions: the E1, E2 and E1cB mechanisms, the regiochemistry (major and minor products of elimination reactions), as well as the Zaitsev’s, Hofmann’s, and Bredt’s rules (yup, that’s 6).
So, before we start delving into the more detailed explanations, let us first take note of what exactly elimination reactions are, and how they are comparable to addition and substitution reactions (this has already been discussed in this post, but we felt it apt to raise it here again).
Simply speaking, an elimination reaction is the opposite of an addition reaction. In the elimination reaction, two substituents (including hydrogens) are removed and eliminated from the same molecule. This may or may not result in the formation of a pi bond.
Why, you ask, can a carbon bonded to a hydrogen substituent be able to participate in an elimination reaction? Remember that elimination reactions are not the same as substitution reactions, and thus we cannot consider substituents as leaving groups.
There are three main types of net elimination reactions, and these are alpha-elimination, beta-elimination and gamma-elimination. What all these reactions have in common is that two substituents will be removed from the molecule (although a pi bond is not necessarily formed).
Perhaps the most commonly seen is beta-elimination (Fig. 1). Beta-elimination occurs to two atoms adjacent to each other, usually carbon atoms, that each possess one viable substituent. The two substituents depart, resulting in the formation of a pi bond. Thus the product of a beta-elimination has to be unsaturated.
Fig. 1: Diagram of beta-elimination.
Before we proceed with a detailed treatment of beta-elimination (which will be the focus of this article), let us also touch on the other two more obscure elimination reaction types, alpha and gamma-elimination.
Alpha-elimination may also be termed carbene formation. A carbene is a neutral molecule containing a carbon atom with two unshared valence electrons. It also has to be bonded to two other substituents only. Thus, carbene formation may be performed by the elimination of two substituents from a carbon atom.
How does this work? We can understand this by regarding the alpha-elimination of chloroform (CHCl3), which is a simple method to generate carbenes. The reaction itself is a two step, first involving the extraction (or elimination) of a proton from CHCl3 by a base.
Fig. 2: Step 1 of the alpha-elimination of chloroform.
This results in the formation of the :CCl3 anion. The orbital structure of this anion is shown here. However, this is NOT a carbene, because it is an anion. Carbenes have to be neutral, thus this intermediate is still not a carbene. The second and final step is the elimination of the chlorine from :CCl3 (Fig. 3). In this step, the original two unshared electrons are still retained in the sp2 orbital, forming :CCl2, which is dichlorocarbene.
Fig. 3: Step 2 of the alpha-elimination of chloroform.
As for gamma-elimination, it is a ring-closing reaction, forming three-membered rings. A requirement is that the 1 and 3-carbons each possess viable substituents. The elimination of one substituent each from the 1 and 3-carbons results in the formation of a new bond between these two carbons, closing the ring.
What’s with the names of the different elimination reactions? Each elimination reaction involves the elimination of two substituents. In each of these reactions, one of the substituents will be located on the alpha carbon. The name of the carbon which the second substituent is located on will determine the name of the elimination reaction.
For example, carbene formation requires that elimination happen twice on the same carbon, i.e. the alpha carbon. Thus, the second substituent is still located on the alpha carbon and it is known as alpha-elimination. For beta-elimination, the first substituent is located on the alpha carbon while the second is located on the adjacent, or beta, carbon, and so it is thus named.
E1 Mechanism
In this article, we will be focussing mostly on beta-elimination, as it is the most common elimination reaction. Beta-elimination can occur via three different mechanisms. These are the E1, E2 and E1cB mechanisms. The difference between these three mechanisms may not be clear-cut, and we will explain this as well.
Firstly, we should note that the E1 and E2 mechanisms have parallels to their nucleophilic substitution counterparts, which are the SN1 and SN2 reactions. We begin discussing the E1 mechanism first. E1 is short for elimination unimolecular. It consists of two steps, with the first being rate-determining, much like the SN1 reaction.
In the first step of the reaction (Fig. 4), a substituent stable as a free atom will leave (think halogen), also known as the nucleofuge, and it usually takes both the electrons in its bond with carbon. This results in the formation of a carbocation, just like in the SN1 mechanism.
Fig. 4: Step 1 of the E1 mechanism.
For the second step (Fig. 5), the other substituent on the adjacent carbon will leave (think proton). This substituent is usually the less stable (or equally stable) as a free atom as compared to the other substituent (which is why it leaves later). That is also why the departure of this substituent is typically facilitated by a solvent. Even if the substituent is a hydrogen, adding base is optional, because the departure is driven by formation of a double bond between the two carbons.
Fig. 5: Step 2 of the E1 mechanism.
Let us consider the stereoselectivity of such a mechanism. Since there are two steps, and the carbocation is formed in the first one, the carbocation is able to choose to assume the most stable conformational isomer, which in turn affects the stereochemistry of the product. Thus in the case of the E2 mechanism, the product is not stereoselective.
Finally, let us consider evidence for the E1 mechanism. There are two notable ones. Firstly, the reaction is expected to be first order, since the rate-determining step involves only the substrate, and this has indeed been observed.
Secondly, carbocations are prone to rearrangements. When conditions suitable for the suspected E1 reaction are employed, evidence of rearrangement has been found (i.e. side products). Since only the E1 reaction results in carbocations, there is strong evidence for its existence.
E2 Mechanism
The E2 mechanism refers to an elimination, bimolecular mechanism. The mechanism is made up of just a single step, also the rate-determining step.
In this one step, there is a simultaneous elimination of the two groups. As we have noted earlier, it is typically that one of the substituents on the molecule is more stable as a free atom while the other is equally or less stable. A good example would be the elimination of an alkyl halide.
In the alkyl halide, there would be two adjacent carbons, with one of them possessing a halide substituent (relatively stable as a free atom, due to its electronegativity). The other possesses a hydrogen (less stable).
The E2 mechanism for this reaction proceeds with the departure of the halide substituent (without any facilitation), and the extraction of a proton on the adjacent carbon. The extraction of this proton is usually facilitated by a base. The elimination of the two substituents then results in the formation of the pi bond.
Fig. 6: Diagram of the E2 mechanism.
Why is this so? Remember that in the E1 mechanism, the departure of the hydrogen does not need to be facilitated by a base as it is driven by the formation of the much more stable double bond. However, there is no driving factor for the formation of the halide substituent, which means that a base will be necessary for the proton to be eliminated.
As for the kinetics of this reaction, it is second order, because this time a base is involved as well as the substrate. The E2 mechanism bears a large number of similarities to the SN2 mechanism; both involve one step, both have two reactants, for instance. Evidence for this mechanism is given by taking advantage of known second order kinetics.
Unlike the E1 mechanism, the E2 mechanism is stereoselective, because there is no opportunity for the most stable conformation to be assumed. We start by noting that there are two possible isomers of the alkyl halide; specifically, cis-trans isomers. Either the hydrogen and the halogen substituents face the same direction, or they face opposite directions.
In the case where the substituents face opposite directions (trans), the E2 mechanism is favored (anti-elimination), unless other factors (such as steric hindrance) are involved. The reasoning behind this is again due to constitutional isomers, with the trans isomer being staggered and thus requiring a lower energy to transition to the elimination product.
E1cB Mechanism
We can arrange the E1 and E2 mechanisms on a spectrum: the E1 mechanism involves the more stable substituent departing first followed by the less stable one, while the E2 mechanism involves both departing at the same time. Given this logic, there should be a mechanism where the less stable one departs first followed by the more stable one.
This is indeed so. The E1cB mechanism is also a unimolecular elimination, with the ‘cB’ referring to ‘conjugate base’. The mechanism consists of two steps and is quite similar to the E1 mechanism; however, instead of the formation of a carbocation, a carbanion is formed instead.
We mentioned earlier that it is more likely for the stable substituent to leave first, and this is usually a halide. However, cases also exist where the other substituent (usually the hydrogen) is able to leave first before the other substituent. These are special cases where the hydrogen is acidic or when the halide or other substituent is a poor leaving group.
In the first step of the E1cB mechanism (Fig. 7), a base extracts a proton from the molecule in an acid-base reaction, resulting in a carbanion (since both bonding electrons go to the carbon). Note that this is a reversible reaction, as with acid-base reactions. It is also rate-determining.
Fig. 7: Step 1 of the E1cB mechanism.
The second step of the E1cB mechanism (Fig. 8) is very similar to the second step of the E1 mechanism, but this time the halide or similar substituent leaves instead, and this results in the formation of the pi bond, forming the product similar to the two other mechanisms.
Fig. 8: Step 2 of the E1cB mechanism.
Evidence for the E1cB mechanism is similar to the E1 mechanism. It hinges on the fact that the first, and rate-determining, step is reversible.
Zaitsev’s Rule
There is still a regioselectivity issue we have not yet discussed. Consider a structure where the more stable substituent exists on a carbon that is adjacent to two carbons, both with available hydrogens that can be eliminated. Which carbon will have its hydrogen eliminated and form the major product?
In cases where both carbons are equivalent, of course, only one product can be formed, as the molecule would have to be symmetrical in that case. Important to look at are situations where both carbons are not equivalent, such as in 2-chlorobutane. After an elimination reaction takes place, would 1-butene or 2-butene be favored?
Zaitsev’s rule is a useful way for us to predict which product is the major product. It states that the more substituted alkene will be the major product in an elimination reaction. This means that the pi bond will be formed between two carbons which together have the least hydrogens.
Fig. 9: Diagram of Zaitsev’s rule.
Referring to our earlier example, this means that 2-butene would be favored, because between the two carbons forming the pi bond, there is the least number of hydrogens. The reason behind this is because the more substituted the alkene (the lower the number of hydrogens and the larger the number of alkyls), the more thermodynamically stable the alkene and therefore the more energetically favorable the elimination reaction.
Hofmann’s Rule
In certain cases, the elimination reaction does not obey the Zaitsev’s rule, and the double bond is formed with the less substituted carbon instead. In these cases, the elimination reaction is said to be following Hofmann’s rule, and the less substituted alkene product subsequently formed is known as a Hofmann product.
The Hofmann product is thermodynamically less stable than the Zaitsev product. So, what defines the boundary between which elimination reactions follow Zaitsev’s rule and which elimination reactions follow Hofmann’s rule? In general, there are two scenarios where Hofmann’s rule is followed and thus the Hofmann product is favored over the Zaitsev product.
The first scenario is where a sterically-hindered, or bulky base is used. We know that the more substituted the carbon, the more difficult for a base to approach and extract a proton, due to the larger size of alkyl groups bonded to the carbon atom as compared to just simple hydrogens. Thus, when a bulky base is employed, the less substituted carbon is more likely to be deprotonated, forming greater percentages of the Hofmann product.
The second scenario is where the nucleofuge (the more stable substituent as a free atom) is positively charged. This gives the nucleofuge much greater electron-withdrawing effects (it acts through the field effect), allowing it to stabilize the conjugate base of both beta carbons. This makes both beta carbons more acidic.
However, alkyl groups are electron-donating! This means that these groups will ‘counter’ the electron-withdrawing effects of the nucleofuge. Since more substituted carbons have more alkyl groups, it means that these carbons will be less acidic, and thus less likely to react with a base. Thus, the less substituted carbons will be more acidic and react with bases at higher rates, allowing for a larger proportion of the Hofmann's product.
Bredt’s Rule
This rule concerns itself with a relatively small class of carbons, which is bridgehead carbons. Bredt’s Rule suggests that a double bond cannot be formed between a bridgehead carbon and another carbon (in an elimination reaction), except when the ring size is large enough.
The reason behind this is that if the ring is too small, the resulting double bond would be too strained and thus too unstable. For a double bond to be formed to the bridgehead carbon, it needs to be sp2 hybridized. However, a molecule must be planar to be sp2 hybridized. Since small-ring bridgehead carbons cannot assume planarity, double bonds cannot be formed to them.