[iScience]

http://i.imgur.com/9q2jcuO.jpg


1.) can the Br^- substitute directly to the cationic center that forms right when the H2O falls off?

2.) why does the H-atom from the  tertiary center move over to the secondary cationic center again? ie why/what makes this process energetically more favorable than for the H to just stay on the tertiary atom?

[danteOne]

1.) Yes. The bromide ion can react with the secondary carbonation before it undergoes a carbonation rearrangement. In reactions involving a carbonation rearrangement a mixture of different products are formed.

2.) Look at the stabilizes of each carbonation, the one before the rearrangement and the one after the rearrangement. Which one is more stable?

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[iScience]

Thanks;

for (2), so the after state is more stable because the tertiary cation has 2 methyl groups from which the H's can donate electron density to, whereas the initial secondary cation only has one methyl group from which the H's can donate electron density to right?

(assuming i'm correct), okay i can see how the after state is more stable than the first state, but, what makes the process happen? ie, i can set a ball on a table and if the ball falls down the table, that is an energetically more stable state than the ball being on the table, but something has to get the ball off the table or get the ball in a situation such that the ball CAN fall farther downward to achieve the more stable state. similarly, i can see how the final (tertiary carbon) state is more stable than the first (secondary carbon), but what is the driving force so to speak that makes this happen?

is one of the hydrogens from one of the methyl groups on the quatrinary carbon donating electron density to the secondary cation where then the hydrogen pulls the entire carbon (the methyl group that it's on), and is that how the transfer takes place?

[danteOne]

I don't know the exactly what happens during the rearrangement. What I do know is that there is an intermolecular reaction that is thermodynamically favorable and which has a low enough energy of activation to occur at room temperature.
The products of this intermolecular reaction are favored because (as you noticed) the tertiary carbocation is more energetically stable. The reason that tertiary carbons are more stable than secondary carbocations is due to the electron donating CH3 groups and also due to hyperconjugation. Delocalization of electrons is stabilizing (that is why resonance is so stabilizing).

[iScience]

hmm.. delocalization of electrons is stabilizing? if we view the electron cloud as a spring system (analogous to einstein solids), then we could say that the resting point is the electron cloud in its normal position (no effect). But when we add something to this system such that the spring gets pulled toward that something and gets stretched (ie delocalized), doesn't this mean a potential energy now exists for the electron cloud getting 'stretched' so to speak? There is a competing energetic result taking place however, and that would be that since the interaction between the electron cloud and the "something" is attractive, the potential is lower.

so the act of stretching the spring increases the potential, but the fact that the force upon the spring is attractive, the potential decreases. which one wins? i think i'll regret asking this question because i need QM to model the situation... but basically my question is: is the result of these two competing effects such that the magnitude of the attractive potential is always larger such that the net potential is decreased?


off the side question: if two species are brought close together by attractive interactions, then the potential always decreases. but what if there was some chemical group on a carbon atom. and another species came and attached itself to the same carbon by kinetics or something.. and the original group and the new group repel each other; even if the bond angle between them was allowed to get distorted, the net potential would still increase right?

[danteOne]

I have always thought that the reason delocalization is stabling in terms of the number of nodes. Delocalization = less molecular nodes = lower energy = stabilizing. That is the best answer I can give you because I don't know the exact QM reason for the stabilization.