Elimination Reactions

By James Ashenhurst

The E2 Mechanism

Last updated: November 13th, 2020 |

E2 Mechanism – How The E2 (Elimination, Biomolecular) Reaction Works

Having gone through the E1 mechanism for elimination reactions, we’ve accounted for one way in which elimination reactions can occur. However, there’s still another set of data that describes some elimination reactions that we haven’t adequately explained yet.

Table of Contents

  1. Example Of An “E2” Reaction: How Do We Explain What Happens In This Reaction?
  2. Clue #1 About The Mechanism Of The E2 Reaction: The Rate Depends  on  Concentration of Both  Substrate and Base
  3. Clue #2 About The Mechanism Of The E2 Reaction: Stereochemistry Of The C–H Bond And The Leaving Group Is Always “Anti”
  4. Putting It Together: The Mechanism Of The E2 Reaction
  5. (Advanced) References and Further Reading

1. Example Of An “E2” Reaction: How Do We Explain What Happens In This Reaction?

Here’s an example of the reaction I’m talking about:

treatment of alkyl halide with naoch3 giving alkene elimination reaction how to explain reaction

What’s interesting about this reaction is that it doesn’t follow the same rules that we saw for the E1 reaction. We’ll talk about two key differences here.

2. Clue #1 About The Mechanism Of The E2 Reaction: The Rate Depends  on  Concentration of Both  Substrate and Base

Remember that the E1 reaction has a “unimolecular” rate determining step (that is, the rate only depends 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. It’s dependent on the concentration of both substrate and the base.

That means that whatever mechanism we propose for this reaction has to explain this data.

for e2 reaction rate depends both on concentration of alkyl halide and also of base second order overall

By the way, see how useful chemical kinetics can be? They’re such simple experiments – measure reaction rate versus concentration – and you get these nice graphs out of it.  I can’t 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).

3. Clue #2 About The Mechanism Of The E2 Reaction: Stereochemistry Of The C–H Bond And The Leaving Group Is “Anti”

Here’s the second key piece of information – and we didn’t 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. What’s weird about this? Well, this seems to fly in the face of Zaitsev’s rule, right? Why don’t we get the tetrasubstituted alkene here?

clue about mechanism of e2 reaction is anti relationship between hydrogen and halide

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

deuterium labelling in e2 elimination reaction shows only deuterium anti to leaving group is removed

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 actually do get the Zaitsev product!

e2 reaction on cyclohexyl bromide can give zaitsev product with right modification

See what’s going on? The hydrogen that is broken is always opposite, or “anti” to the leaving group.

So how do we explain these two factors? 

4. Putting It Together: The Mechanism Of The E2 Reaction

Here’s 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.

mechanism of the e2 reaction is one step deprotonation by base giving new alkene followed by loss of leaving group explains stereochemistry e2 mechanism

In this mechanism, the base removes the proton from the alkyl halide that is oriented anti to the leaving group, and the leaving group leaves – all in one concerted step.

Since it’s an elimination reaction, and the rate law is “bimolecular”, we call this mechanism the E2.

In the next post, we’ll directly compare the E1 and E2 reactions.

Next Post: Comparing the E1 and E2 Reactions


(Advanced) References and Further Reading

  1. —The nature of the alternating effect in carbon chains. Part XVIII. Mechanism of exhaustive methylation and its relation to anomalous hydrolysis
    Walther Hanhart and Christopher Kelk Ingold
    J. Chem. Soc. 1927, 997-1020
    DOI:
    10.1039/JR9270000997
    One of the first proposals for the mechanism of the E2 reaction. Prof. Ingold mentions in this paper, “It follows from the basic hypothesis that the ease of removal of the b-proton (reaction A) depends (a) on its vulnerability, (b) on the proton-avidity of the attacking anion
  2. Influence of poles and polar linkings on the course pursued by elimination reactions. Part XV. Dynamics of the elimination of olefins from quaternary ammonium compounds
    E. D. Hughes and C. K. Ingold
    J. Chem. Soc. 1933, 523-526
    DOI:
    10.1039/JR9330000523
    Depending on the structure of the substrate, either E1 (unimolecular) or E2 (bimolecular) eliminations are possible. This paper contains a kinetic experiment demonstrating that the bimolecular elimination is second order, first order in both base and R-X (where X = -NH3+ in this case).
  3. Electrophilic Substitution at Saturated Carbon. XIII. Solvent Control of Rate of Acid-Base Reactions that Involve the Carbon-Hydrogen Bond
    Donald J. Cram, Bruce Rickborn, Charles A. Kingsbury, and Paul Haberfield
    Journal of the American Chemical Society 1961, 83 (17), 3678-3687
    DOI:
    1021/ja01478a029
    E2 reactions require the use of a reasonably strong base, so solvents which can support the base in a dissociated form are best. Aprotic solvents are actually not ideal, since they can hydrogen bond with the base and ‘buffer’ it, reducing its activity. Fig. 4 and Table VI illustrate the dramatic dependence of base activity on the percentage of DMSO in the solvent system.
  4. Description of steric relationships across single bonds
    Klyne & V. Prelog
    Experientia 1960, 16, 521–523
    DOI:
    10.1007/BF02158433
    This is where the term ‘anti-periplanar’ is defined for the first time.
  5. A theoretical account for stereoselective E2 reactions
    Kenichi Fukui, Hiroshi Fujimoto
    Tetrahedron Lett. 1965, 6 (48), 4303-4307
    DOI:
    10.1016/S0040-4039(00)71091-4
    Kenichi Fukui received the Nobel Prize in Chemistry in 1981 for the development of Frontier Molecular Orbital theory. This paper uses FMO theory to explain the stereoselectivity of the E2 reactions in terms of orbital overlap between the anti-periplanar C-H bond and the C-X bond. Fukui calculates frontier electron densities of hydrogen atoms, and hydrogen atoms anti to chlorine atoms have the highest values.
  6. Studies in Stereochemistry. VII. Molecular Rearrangements During Lithium Aluminum Hydride Reductions in the 3-Phenyl-2-butanol Series
    Donald J. Cram
    Journal of the American Chemical Society 1952, 74 (9), 2149-2151
    DOI:
    1021/ja01129a003
    Classic paper by Nobel Laureate Prof. D. J. Cram (UCLA) demonstrating the anti stereochemistry of the E2 reaction, with erythro– vs threo 3-phenyl-2-butyl tosylate with NaOEt in EtOH.

Comments

Comment section

14 thoughts on “The E2 Mechanism

  1. I’m not sure I see what the “one modification” was to make the molecule follow Zaitsev’s rule? The substrate still looks the same as before?

    1. In the first molecule, the wedge was referring to a methyl group, while in the second molecule (the modification), the wedge showed a hydrogen, which was available to be broken off by OCH3

  2. Oh I see now. But, how can you just pus that H downward? Is that allowed? If I had a problem like this on the test, should I just redraw the cyclohexane with the H pointing downward and making it 180 from Br?

    1. Those are actually two different molecules. In the second case, the hydrogen on the more substituted beta carbon is anti. Thus, this molecule can give a Zaitsev product.a

  3. I was just wondering: for E2 and E1 we don’t have to worry about inversion at any chiral center or a racemic mixture of products the way we would have for SN2 and SN1 reactions, right?

  4. I have a question regarding the bonds – how come the wedges and dashes become flat bonds? Is it possible that some of them could stay wedges and dashes and others could become flat – in other words, what’s the criteria for when a bond stays a wedge/dash or when it becomes planar?

    1. I assume you mean the bottom mechanism? I forgot to mention one thing – that we rotated our “view” from a “side-view” to a “top view” . When we look at the alkene from the top, it’s flat, so it shouldn’t have dashes or wedges. When we look at the alkene from the side, some groups will point towards us and some will point away. I should update this figure.

    1. @Varun this is because the transition state has an alkene-like structure. I am sure of the fact that 3>2>1, but I cannot illustrate the diagram here. You should look it up in a reference book if you still doubt it

  5. If the Morse Potential allows bond breaking, by spectroscopic stretching and rotation; could this allow the disruption of inter and intra molecular attachments?
    In turn, could selection of an appropriate bond, application of correct downfield absorption, and suitable reduced mass point, allow complex bond breaking?
    Also, would the molecular resonance be tolerable and penetrable in organisms?
    If you follow my drift?

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