I remember my teacher saying that during E2 (elimination bimolecular) reactions of alkanes, if the strong base that is used as a reagent is a bulky one, it has higher probability of giving a Hofmann (less stable) product as the major product even if there are possibilities for a Saytzeff (or Zaitsev) product to form.

How should I find if a base is bulky or not?

Some bases like t-butoxide ion are pretty obvious. I usually just guess it is based on how much steric-clouded it seems when I draw a representation on a paper. Is there any better way to figure this out?

  • 3
    $\begingroup$ Related: Is the -OCH3 group bulky? $\endgroup$ – andselisk Dec 23 '20 at 22:00
  • $\begingroup$ chemistry.stackexchange.com/questions/104019/… $\endgroup$ – Mithoron Dec 24 '20 at 1:05
  • 3
    $\begingroup$ user55119's answer has the details, but the underlying point is that it's empirical. There isn't some theoretical cutoff where you perform some calculations and then it gives you a bulk coefficient $\xi$ where if $\xi > 4$ it's bulky and if $\xi \leq 4$ it's not. Well, maybe you can calculate the size, but then one should ask, exactly what size is considered "bulky"? which brings us back to square one. $\endgroup$ – orthocresol Dec 24 '20 at 2:07

This question about the effect of the bulkiness of bases on the mode of E2 elimination has been studied and was published in 1956 by Brown, et al.[1] The elimination of 2-bromo-2-methylbutane was examined using solutions of potassium ethoxide, potassium t-butoxide, potassium t-amyloxide and potassium t-heptoxide in their respective alcohols. The clear result was that increased branching of the alkoxide increased the yield of the Hofmann product over the Zaitsev product. The bulkier the base the less selective for attacking methylene hydrogens over methyl hydrogens. The numbers in parentheses are the selectivities corrected for the number of hydrogens available in each elimination.

  1. H. C. Brown, I. Moritani and Y. Okamoto, J. Am. Chem. Soc., 1956, 78, 2193.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.