In a book they have given a statement,

In the process of pyrolysis of esters, esters having beta hydrogen are heated at high temperature and then alkenes are formed along with carboxylic acid. If there are two beta hydrogens then beta hydrogen having less hindrance will participate in the reaction.

In the last line, what do they mean by less hindered hydrogen?


Consider these following two structures from the same compound:

The two Newman projections are along the horizontal $\ce{C-C}$ bond which is, of course, freely rotatable. The two structures represent two possible syn-elimination transition states. Note that in the left-hand structure there is a steric clash between the tert-butyl group and the methyl group. The right-hand structure has no such clash. It is therefore the preferred transition state for this reaction.

‘Less hindered hydrogen’ basically means that hydrogen that leads to the more stable transition state.

  • $\begingroup$ I didn't get it . We will remove hydrogen from methyl or from 3rd carbon? $\endgroup$ – Dimenein Dec 21 '15 at 17:17
  • $\begingroup$ Hm, we probably could remove the methyl one, too, since that is technically also beta. Let me think about this … $\endgroup$ – Jan Dec 21 '15 at 18:25
  • 1
    $\begingroup$ @VinayakAgarwal I just realised a major blunder I had in my images which has now been fixed. You could also draw a transition state towards the terminal methyl group rather than along the chain. That would work equally. However, the double bond you create would be terminal. Therefore, under thermodynamic control (evidently, it’s a pyrolysis after all) the higher-substituted double bond should be created as shown here. I mean, I could turn that methyl into another tBu group but I’m too lazy now. $\endgroup$ – Jan Dec 21 '15 at 18:41
  • $\begingroup$ By less hindered hydrogen do you mean hydrogen to be removed, leads to the more stable transition state. $\endgroup$ – Dimenein Dec 22 '15 at 14:29
  • 1
    $\begingroup$ Although I like this analysis, the wording of 'less hindered hydrogen' makes it seem that the hydrogen from the beta-methyl would be participating, at least according to the book in question. Do you have any evidence that the more substituted alkene is formed? $\endgroup$ – jerepierre Dec 22 '15 at 16:41

The beta-hydrogen with less steric hindrance refers to the beta-hydrogen on the least substituted carbon. Your book is suggesting that the major product is the less-substituted alkene, which is not the most stable product (as pointed out by Jan).

Jerry March's book gives the example of the pyrolysis of sec-butyl acetate. The 'major' product is the less substituted alkene, although it appears to be statistically driven (there are three beta-hydrogens and only two-beta'-hydrogens), indicating that there is little energy difference in the transition states leading to the two products.

enter image description here

All else being equal, the less substituted alkene product (Hofmann's rule) should be predicted to be major. However, it appears that mixtures are typical and other factors must be considered for more complex substrates. The most important of these is that the reaction is a syn-elimination through a cyclic transition state, so the compound must be able to adopt the appropriate conformation. Other steric interactions may also influence the regioselectivity of the product away from the less substituted.

The Wikipedia page is a good summary of these intramolecular elimination reactions.

  • $\begingroup$ Does that mean that methyl group will have less hindered hydrogen than the carbon connected to tertiary butyl in the example given by Jan $\endgroup$ – Dimenein Dec 23 '15 at 18:15
  • 1
    $\begingroup$ Yes, in general the less substituted carbon will be less hindered. However, as I say on the answer, the effect is more likely statistical than steric in nature. $\endgroup$ – jerepierre Dec 24 '15 at 2:54
  • $\begingroup$ So if we want to say what do we mean by less hindered hydrogen in one line,we can say it as hydrogen on the least substituted carbon and it should lead to the more stable transition state. $\endgroup$ – Dimenein Dec 24 '15 at 13:05

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.