Now that weve gone through the mechanisms of theE1andE2reactions, lets take a moment to look at them side by side and compare them.Heres how each of them work:

Heres what each of these two reactions has in common: in both cases, we form a new C-C bond, and break a C-H bond and a C(leaving group) bond in both reactions, a species acts as a base to remove a proton, forming the new bond both reactionsfollowZaitsevs rule(where possible) both reactions arefavored by heat.Now, lets also look at how these two mechanisms are different. Lets look at this handy dandy chart:

The rate of the E1 reaction dependsonly on the substrate, since the rate limiting step is theformation of a carbocation. Hence, the morestablethat carbocation is, the faster the reaction will be. Forming the carbocation is the slow step; a strong base isnotrequired to form the alkene, since there is no leaving group that will need to be displaced (more on that in a second). Finally there is no requirement for the stereochemistry of the starting material; the hydrogen can be at any orientation to the leaving group in the starting material [although well see in a sec that wedorequire that the C-H bond be able to rotate so that its in the same plane as the empty p orbital on the carbocation when the new bond is formed].The rate of the E2 reaction depends onboth substrate and base, since the rate-determining step isbimolecular(concerted). Astrong baseis generally required, one that will allow for displacement of a polar leaving group. The stereochemistry of the hydrogen to be removedmust beantito that of the leaving group; the pair of electrons from the breaking C-H bond donate into the antibonding orbital of the C-(leaving group) bond, leading to its loss as a leaving group.Now were in a position to answer a puzzle that came up when we first looked at elimination reactions. Remember this reaction where one elimination gave the Zaitsev product, whereas the other one did not. Can you see why now?

So whats going on here? The first case is an E2 reaction. The leaving groupmustbeantito the hydrogen that is removed.

The second case is an E1 reaction.

In our cyclohexane ring here, the hydrogen has to be axial. Thats the only way we can form a bond between these two carbons; we need the p orbital of the carbocation to line up with the pair of electrons from the C-H bond that were breaking in the deprotonation step. We can always do a ring flip to make this H axial, so we can form the Zaitsev product. Heres that deprotonation step:

As you can see, cyclohexane rings can cause some interesting complications with elimination reactions! In the next post well take a detour and talk specifically about E2 reactions in cyclohexane rings.

Having gone through theE1 mechanism for elimination reactions, weve accounted for one way in which elimination reactions can occur. However, theres still another set of data that describes some elimination reactions that we havent adequately explained yet.Heres an example of the reaction Im talking about:

Whats interesting about this reaction is that it doesnt follow the same rules that we saw for the E1 reaction. Well talk about two key differences here.Clue #1 The Rate LawRemember that the E1 reaction has a unimolecular rate determining step (that is, the rateonlydepends on the concentration of the substrate?)Well, when we look at the rate law for this reaction, we find that it depends on two factors. Its dependent on the concentration ofbothsubstrateandthe base.That means that whatever mechanism we propose for this reaction has to explain this data.By the way, see howusefulchemical kinetics can be? Theyre such simple experiments measure reaction rate versus concentration and you get these nice graphs out of it. I cant even begin to stress how important this data can be in understanding reaction mechanisms. So simple, so elegant, and so useful.Another note you might notice that the base here (CH3O-) is a stronger base than we see for the E1 reaction (more on that later).Second Clue: StereochemistryHeres the second key piece of information and we didnt talk about this for the E1.The reaction below is very dependent on the stereochemistry of the starting material.When we treat this alkyl halide with the strong base, CH3ONa, look at this interesting result. Whats weird about this? Well, this seems to fly in the face ofZaitsevs rule, right? Why dont we get the tetrasubstituted alkene here?

The mystery gets a little deeper. If, instead of starting with the alkyl halide above, we label it withdeuterium that is, we replace one of the hydrogens with its heavy-isotope cousin that has essentially identical chemical properties we see this interesting pattern:

Note how the group that is on the opposite face of the cyclohexane ring to the leaving group (Br) is always broken.In fact, if we use the molecule above and make just one modification, now we actuallydoget the Zaitsev product!See whats going on? The hydrogen that is broken is always opposite, or anti to the leaving group.So how do we explain these two factors?Heres a hypothesis for how this elimination reaction works. It accounts for all the bonds that form and break, as well as the rate law, and crucially the stereochemistry.

In this mechanism, the base removes the proton from the alkyl halide that is orientedantito the leaving group, and the leaving group leaves all in one concerted step.Since its an elimination reaction, and the rate law is bimolecular, we call this mechanism theE2.In the next post, well directly compare the E1 and E2 reactions

Last time in this walkthrough on elimination reactions, we talked abouttwo types of elimination reactions.In this post, were 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.Heres the reaction. First, look at the bonds that are being formed and broken. This is a classic elimination reaction were forming a new CC() bond, and breaking a CH and Cleaving group (Br here) bond.

But now we want to know more than just what happens. We want to understandhowit happens. Whats the sequence of bond-forming and bond breaking?To understand HOW it happens, we need to look at what the data tells us. Thats because chemistry is an empirical science; we look at the evidence, and then work backwards.First Clue The Rate LawsLets 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 doesnt change at all.

What information can we deduce from this? The rate determining step for this reaction (whatever it is) thereforedoesnotinvolve the base. Whatever mechanism we draw will have to account for this fact.Second Clue Dependence of Rate on SubstrateAnother interesting line of evidence we can obtain from this reaction is through varying thetypeof 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 atertiaryalkyl halide), the rate is faster than for the middle alkyl halide (asecondaryalkyl halide) which is itself faster than aprimaryalkyl halide (attached to only one carbon in addition to Br).So the rate proceeds in the ordertertiary (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?Third Clue This Elimination Reaction Competes with the SN1 ReactionA 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 astereocenter.And if we start with a single enantiomer of starting material here, we note that the substitution product formed is amixture of stereoisomers.Note thatboth inversionandretention of stereochemistry at the stereocenter has occurred.Weve seen this pattern before its an SN1 reaction!

This last part is a very important clue. If an SN1 reaction is occurring in the reaction mixture, looking back at themechanism of the SN1could help us think about what type of mechanism might be going on in this case to give us the elimination product.Taking all of these clues into account, whats 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 Zaitsevs rule].Similar to the SN1 mechanism, this is referred to as theE1 mechanism(elimination, unimolecular).So whats going on in theother type of elimination reaction? Thats the topic for the next post

3. The E1 ReactionJust as there were two mechanisms for nucleophilic substitution, there are two elimination mechanisms. The E1 mechanism is nearly identical to the SN1 mechanism, differing only in the course of reaction taken by the carbocation intermediate. As shown by the following equations, a carbocation bearing beta-hydrogens may function either as a Lewis acid (electrophile), as it does in the SN1 reaction, or a Brnsted acid, as in the E1 reaction.

Thus, hydrolysis of tert-butyl chloride in a mixed solvent of water and acetonitrile gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration. The alcohol is the product of an SN1 reaction and the alkene is the product of the E1 reaction. The characteristics of these two reaction mechanisms are similar, as expected. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific.(CH3)3CCl + H2O> [ (CH3)3C(+)] + Cl() + H2O> (CH3)3COH + (CH3)2C=CH2 + HCl + H2OTo summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:1.The cation may bond to a nucleophile to give a substitution product.2.The cation may transfer a beta-proton to a base, giving an alkene product.3.The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.Since the SN1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place.Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered,the most important being the structure of the alkyl group and the nature of the nucleophilic reactant. The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).

1. The E2 ReactionWe have not yet considered the factors that influence elimination reactions, such as example3in the group presented at thebeginningof this section.(3) (CH3)3C-Br + CN()> (CH3)2C=CH2 + Br() + HCNWe know that t-butyl bromide is not expected to react by an SN2 mechanism. Furthermore, the ethanol solvent is not sufficiently polar to facilitate an SN1 reaction. The other reactant, cyanide anion, is a good nucleophile; and it is also a decent base, being about ten times weaker than bicarbonate. Consequently, a base-induced elimination seems to be the only plausible reaction remaining for this combination of reactants. To get a clearer picture of the interplay of these factors consider the reaction of a 2-alkyl halide, isopropyl bromide, with two different nucleophiles.

In the methanol solvent used here, methanethiolate has greater nucleophilicity than methoxide by a factor of 100. Methoxide, on the other hand is roughly 106times more basic than methanethiolate.As a result, we see a clear-cut difference in the reaction products, which reflectsnucleophilicity(bonding to an electrophilic carbon) versusbasicity(bonding to a proton). Kinetic studies of these reactions show that they are both second order (first order in RBr and first order in Nu:()), suggesting a bimolecular mechanism for each. The substitution reaction is clearlySN2. The corresponding designation for the elimination reaction isE2. An energy diagram for the single-step bimolecular E2 mechanism is shown on the right. We should be aware that the E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of CH and CX bond-breaking and C=C bond-making varies. For example, if the Rgroups on the beta-carbon enhance the acidity of that hydrogen, then substantial breaking of CH may occur before the other bonds begin to be affected. Similarly, groups that favor ionization of the halogen may generate a transition state with substantial positive charge on the alpha-carbon and only a small degree of CH breaking. For most simple alkyl halides, however, it is proper to envision a balanced transition state, in which there has been an equal and synchronous change in all the bonds. Such a model helps to explain an important regioselectivity displayed by these elimination reactions.If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.

By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1-hydrogens (red) and two 2-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1-beta-hydrogens compared with one 3-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called theZaitsev Rule.The main factor contributing to Zaitsev Rule behavior is the stability of the alkene. We noted earlier that carbon-carbon double bonds are stabilized (thermodynamically) by alkyl substituents, and that this stabilization could be evaluated by appropriateheat of hydrogenationmeasurements. Since the E2 transition state has significant carbon-carbon double bond character, alkene stability differences will be reflected in the transition states of elimination reactions, and therefore in the activation energy of the rate-determining steps. From this consideration we anticipate that if two or more alkenes may be generated by an E2 elimination, the more stable alkene will be formed more rapidly and will therefore be the predominant product. This is illustrated for 2-bromobutane by the energy diagram on the right. The propensity of E2 eliminations to give the more stable alkene product also influences the distribution of product stereoisomers. In the elimination of 2-bromobutane, for example, we find that trans-2-butene is produced in a 6:1 ratio with its cis-isomer.The Zaitsev Rule is a good predictor for simple elimination reactions of alkyl chlorides, bromides and iodides as long as relatively small strong bases are used. Thus hydroxide, methoxide and ethoxide bases give comparable results. Bulky bases such as tert-butoxide tend to give higher yields of the less substituted double bond isomers, a characteristic that has been attributed to steric hindrance. In the case of 2-bromo-2,3-dimethylbutane, described above, tert-butoxide gave a 4:1 ratio of 2,3-dimethyl-1-butene to 2,3-dimethyl-2-butene ( essentially the opposite result to that obtained with hydroxide or methoxide). This point will be discussed further once we know more about the the structure of the E2 transition state.Bredt's RuleThe importance of maintaining a planar configuration of the trigonal double-bond carbon components must never be overlooked. For optimum pi-bonding to occur, the p-orbitals on these carbons must be parallel, and the resulting doubly-bonded planar configuration is more stable than a twisted alternative by over 60 kcal/mole. This structural constraint is responsible for the existence ofalkene stereoisomerswhen substitution patterns permit. It also prohibits certain elimination reactions of bicyclic alkyl halides, that might be favorable in simpler cases. For example, the bicyclooctyl 3-chloride shown below appears to be similar to tert-butyl chloride, but it does not undergo elimination, even when treated with a strong base (e.g. KOH or KOC4H9). There are six equivalent beta-hydrogens that might be attacked by base (two of these are colored blue as a reference), so an E2 reaction seems plausible. The problem with this elimination is that the resulting double bond would be constrained in a severely twisted (non-planar) configuration by the bridged structure of the carbon skeleton. The carbon atoms of this twisted double-bond are colored red and blue respectively, and a Newman projection looking down the twisted bond is drawn on the right. Because a pi-bond cannot be formed, the hypothetical alkene does not exist. Structural prohibitions such as this are often encountered in small bridged ring systems, and are referred to asBredt's Rule.

Bredt's Rule should not be applied blindly to all bridged ring systems. If large rings are present their conformational flexibility may permit good overlap of the p-orbitals of a double bond at a bridgehead. This is similar to recognizing that trans-cycloalkenes cannot be prepared if the ring is small (3 to 7-membered), but can be isolated for larger ring systems. The anti-tumor agent taxol has such a bridgehead double bond (colored red), as shown in the following illustration. The bicyclo[3.3.1]octane ring system is the smallest in which bridgehead double bonds have been observed. The drawing to the right of taxol shows this system. The bridgehead double bond (red) has a cis-orientation in the six-membered ring (colored blue), but a trans-orientation in the larger eight-membered ring.

2. Stereochemistry of the E2 ReactionE2 elimination reactions of certain isomeric cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram.Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state.

The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers ananti orientationof the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference forequatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates.A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.An orbital drawing of the anti-transition state is shown on the right. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the SN2 mechanism. In this drawing the and carbon atoms are undergoing a rehybridization from sp3to sp2and the developing -bond is drawn as dashed light blue lines. The symbolRrepresents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints syn-elimination has been observed.It is also worth noting that anti-transition states were preferred in severaladdition reactions to alkenes, so there is an intriguing symmetry to these inverse structural transformations.