How can one choose which group has more shifting tendency in 1,2 carbocation rearrangement? The obvious order is via the stability of the carbocation of the group. But, phenylic groups have high shifting tendency--and a phenylic carbocation is unstable.

Is there a way of predicting these? I was thinking that it will be related to the delocalization in the triangular intermediate formed.

Update: I'm talking about comparing "migratory aptitude" of the $R$ groups in a system similar to $>C(+)-C(R_1R_2R_3)$

  • $\begingroup$ @CHM: The first case.. Eg if we have $C(+)-C(R_1,R_2,R_3)$, which $R$ will shift? $\endgroup$ – ManishEarth Apr 26 '12 at 4:22
  • $\begingroup$ "migratory aptitude" is the more established term, IIRC. $\endgroup$ – user95 Apr 26 '12 at 6:36

The obvious order is via the stability of the carbocation of the group.

I think you meant the migrating groups stability?
This is not what textbooks say.
Reactions are ruled by the $ΔG$ of the activated complex. This may be close to educts, or to products, or some halfway state or a real intermediate minimum is on the reaction path. So, in some cases the model of a separate carbenium ion (in a ionic pair maybe) is useful, sometimes you need to think of a three-center bond state.

| improve this answer | |
  • $\begingroup$ So there is no general rule? I thought maybe there might be a way to predict effectiveness of delocalization of the carbocation... $\endgroup$ – ManishEarth Apr 27 '12 at 10:29
  • $\begingroup$ Of course there is a rule, You mentioned it, but You took the stability of the hypotheticallly migrating group's carbcation as a criterion. It is the real cation that makes up the activated complex, that is all. $\endgroup$ – Georg Apr 27 '12 at 10:34
  • 1
    $\begingroup$ oh, I never took that--that was the first thing that popped into my mind, thats all. Many times stability of a system translates to stability of a single ion. In this case, its slightly more complex. Though I'm beginning to realise what it could be... Probably a mixture of +I and +R.. Not sure about hyperconjugation, does that come into play? $\endgroup$ – ManishEarth Apr 27 '12 at 10:39

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.