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The author claims that the transition state leading to the formation of the 1,2 product from the allylic carbocation is lower in energy since the carbon (which is attached to Br in the 1,2 product) is more electrophilic.

This makes sense to me, however, doesn't the transition state represent the energies of both, the reactants, and the products? In which case the thermodynamically favored product (1,4 product) should also have a lower energy transition state.

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    $\begingroup$ Please add the citation to which the context is referring, i.e. who is the author with that claim? For addition reactions there are a multitude of different paths, it can be concerted or step wise, aided by the solvent or not, etc. In any case, the specific atomic arrangements in the transition states will certainly be of different energies, none of which can be predicted from the reactants by any easy measure. It is not uncommon, that the more stable thermodynamic product has a higher activation barrier than a different path. $\endgroup$ – Martin - マーチン Aug 8 '17 at 5:10
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    $\begingroup$ I'm not sure if it's an outright duplicate, but I covered this topic in a previous question: Addition of hydrogen bromide to 1,3-butadiene: thermodynamic and kinetic control $\endgroup$ – orthocresol Aug 8 '17 at 6:18
  • $\begingroup$ Without any other information than in your question it would seem that the 1,2- has the lower barrier as this is preferentially formed at very low temperatures. Is there not then a second reaction (the rearrangement you show after the low temperature one) and so there is a also a barrier for rearrangement between species 1,2- to 1,4- that is observable at the higher temperature only since the ratio of products changes? Direct reaction at 40 C would follow this scheme also, low barrier to 1,2- (80% product) then rearrangement. $\endgroup$ – porphyrin Aug 8 '17 at 6:41
  • $\begingroup$ @orthocrestol that answer was helpful, but I want to know why the accepted answer before Nordlander et al makes sense. Although I understand why a more electrophilic carbon will stabilize the transition state, doesn't the transition state represent the product as well as the reactant? In which case the thermodynamically more stable product should lower the energy of the transition state leading to its formation. $\endgroup$ – xasthor Aug 8 '17 at 7:31
  • $\begingroup$ @martin LG Wade. But that's exactly what I don't understand. The transition state represents the products and the reactants. So why wouldn't a more stable product imply a more stable transition state? $\endgroup$ – xasthor Aug 8 '17 at 7:33

A transition state does not ‘represent the energies of […] reactants and […] products’. A transition state is a certain arrangement of atoms somewhere between the most stable atom arrangement the reactants can have and the most stable atom arrangement the products can have. The key characteristic of a transition state is that it is directly en route from reactant to product and that it is the arrangement with the highest internal energy.

For any arrangement of atoms you can calculate a single point energy: reactants, products and the transition state. The energy difference between reactants and the transition state is colloquially known as activiation energy although that may not always be correct. It is, however, important to realise that each arrangement has its own energy that cannot a priori be deduced from any other arrangement. Thus, a transition state’s energy is not directly related to the energies of reactants or products (with the exception that by definition it must be higher in energy than both).

While it is a general rule that the transition state leading to the thermodynamically more stable products has a lower energy than any other transition state, this is not true for all cases. Indeed, for any reaction in which one can discuss kinetic (lower energy transition state) versus thermodynamic (lower energy products) control, the transition state energies must be ‘reversed’ with respect to the product energies.


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