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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 $>C(+)-C(R_1R_2R_3)$

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  • $\begingroup$ @CHM: The first case.. Eg if we have $C(+)-C(R_1,R_2,R_3)$, which $R$ will shift? $\endgroup$ Commented Apr 26, 2012 at 4:22
  • $\begingroup$ "migratory aptitude" is the more established term, IIRC. $\endgroup$
    – user95
    Commented Apr 26, 2012 at 6:36

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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.

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  • $\begingroup$ So there is no general rule? I thought maybe there might be a way to predict effectiveness of delocalization of the carbocation... $\endgroup$ Commented Apr 27, 2012 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
    Commented Apr 27, 2012 at 10:34
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    $\begingroup$ 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? $\endgroup$ Commented Apr 27, 2012 at 10:39

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