I've learnt that

The best reaction conditions for synthesis of alkene by dehydrohalogenation are those that promote an E2 mechanism.

(From "Organic Chemistry" by TW Graham Solomons)

But, in the same book it is given that when 2-bromo-2-methylbutane follows Hofmann elimination with tert-BuOK as follows:

Hofmann elimination

Clearly, it isn't following E2 mechanism.

When given such reactions, should I go with E2 mechanism or Hofmann rule, and why?


The difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium.

Let's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination.

Why not E1?

First let's consider the reaction as:

$$\ce{2-bromo-2-methylbutane + EtO^- (in EtOH) -> }$$

Clearly, this reaction is conducted in a highly polar solvent where the carbocation - formed when $\ce{Br^-}$ leaves - gets stabilised. The base then abstracts the hydrogen of $\ce{C3}$ to form 2-methylbut-2-ene by an E1 mechanism.

If you use the same substrate but this time in a highly non-polar solvent, the possibility of E1 decreases as carbocation formation won't get extra stability. But as the base is not sterically hindered, it can still abstract hydrogen to form the Saytzeff product, but there is a case where it can also form the Hoffmann product.

Why not E2?

The E2 mechanism follows an "anti-periplanar orientation of hydrogen and leaving group". Suppose the reactant's stereochemistry is such that there is no anti-$\ce{H}$ on the more substituted adjacent carbon for forming the Saytzeff product, but there is an anti-$\ce{H}$ on a less substituted adjacent carbon, which will lead to the Hoffmann product. In that case, the major product can be the less substituted alkene due to stereospecificity of E2 reaction.

Why E1cb?

Now consider your above given case. Here, the base $\ce{t-BuO^-K+}$ is sterically hindered and is a very strong base. So, prior to any carbocation formation, this base will abstract hydrogen from the less hindered position, and form a carbanion. Also, a less substituted carbanion is more stable than a more substituted one (due to a smaller +I effect). Both these effects lead to formation of a less substituted alkene via an E1cb mechanism.

So, depending upon which product is needed as a major one and the proper stereochemistry of reactant, you should choose whether to go with an E1, E2, or E1cb mechanism by choosing the proper base (bulky or less hindered) and proper solvent (polar or non-polar).


Hint: When the reactant is bulky , or it involves fluorine , then hoffmann rule is preffered over the normal saytzeff rule .

  • $\begingroup$ So, is there no relation between Hofmann rule and E2 mechanism? $\endgroup$ – CaptainIRS Feb 17 '18 at 18:49
  • $\begingroup$ @RinSam This is another type of reaction also better known as $E_1cb$ $\endgroup$ – Tanuj Feb 17 '18 at 18:55
  • $\begingroup$ Saytzeff and Hofmann eliminations are competitive, irreversible E2 processes with your bromide and in general. The Saytzeff/Hofmann ratio of elimination products decreases with increasing bulk of the base. The potassium alkoxide of 3-ethyl-pentan-3-ol gives even more Hofmann product than potassium tertiary butoxide. $\endgroup$ – user55119 Feb 18 '18 at 2:10
  • $\begingroup$ @user55119 It is understood that a bulky base triggers Hofmann elimination. But, if it was an E2 process, then won't it affect coplanarity? $\endgroup$ – CaptainIRS Feb 18 '18 at 7:10

The anti-elimination is not controlled by the bulk of the base. What is controlled by the bulk of the base is the accessibility of the bases to either of the two methylene hydrogens or the six methyl hydrogens and consequently the distribution of the Hofmann and Saytzeff alkenes. From the data below you can see that an increase in the bulk of the base favors the Saytzeff alkene.

The selectivity values correct for the number of hydrogens available for each mode of elimination. For potassium ethoxide, assume that the ratio is 70/30 to make the math simple. The percent 70 is divided by 2 (70/2 = 35) and 30 is divided by 6 (30/6 =5). The ratio 35/5 simplifies to 7/1 selectivity. This means that the rate constant for Hofmann elimination is seven times greater than for Saytzeff elimination, both of which are irreversible. In the presence of potassium t-butoxide the rates of the two eliminations are about equal. The base derived from 3-ethylpentan-3-ol favors Saytzeff over Hofmann elimination by a factor of 3.

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