In the following reaction, Hofmann elimination is preferred.

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This reaction takes place by E1 mechanism. I thought the Zaitsev product would be major due to 1 extra hyperconjugating structure. In the book, it is given that it is due to sterically compressed alkene, so Hoffmann product is major. I am unable to understand why the sterical hinderance would lead to the instability in alkene as the three methyl groups is one bond away from alkene.

  • $\begingroup$ The -ch3 bonds are all connected via single bonds. This means that they are free to rotate. Even though they are one carbon atom away, they can swing and rotate causing steric hinderence thus leading to the hoffmann product. $\endgroup$ Commented May 11, 2020 at 13:25
  • $\begingroup$ @boomavijay but why is steric hinderance causing instablitity in the alkene? $\endgroup$ Commented May 11, 2020 at 13:34
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    $\begingroup$ No , steric hinderence causes the base oh- to not be able to approach and abstract that hydrogen. So that alkene is not formed, but is infact the more stable alkene. $\endgroup$ Commented May 11, 2020 at 13:38
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    $\begingroup$ I agree, this is all about the tBu group hindering the approach of the hydroxide ion $\endgroup$
    – Waylander
    Commented May 11, 2020 at 13:41
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    $\begingroup$ tert-Bu group is enormous. Even $\ce{OH}$ grop cannot penetrate through because your carbocation is also tertiary. Thus, Hofmann product is preferred. $\endgroup$ Commented May 11, 2020 at 20:31

2 Answers 2


The Hoffmann product is formed as a major product generally in four cases. They are:

  1. When the ${\gamma}$ carbon is 4°
  2. When the leaving group is poor leaving group
  3. When the leaving group is bulky leaving group
  4. When the elimination is internal elimination (${E_i}$ mechanism)

Case 1:

It happens in the example given in the question the hindrance of base in abstraction of proton favors Hoffman product

Case 2:

It happens with poor leaving groups such as flourine (based on the carbanion character on the carbon from which proton is abstracted in the trasition state)

Case 3:

It happens in the Hoffmann's exhaustive methylation reaction where the bulky leaving group is a 3° amine group

Case 4:
It happens in the treatment of a 3° amine with Caro's acid followed by heating and in heat treatment of xanthate esters

Lastly I want to add that these are general results the yields may vary based on reactant.


Look at the carbon which has more α-hydrogens and thus will be stabilized more by resonance when it is attacked by the $\ce{OH-}$ ions. So the carbon atom on the right hand side that has two methyl groups has more α-hydrogens with it and is stabilized by resonance and $\ce{OH-}$ is attacked there. Hence the major product has carbon with the double bond.


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