# Epoxide Ring Opening - Paradoxical Statement

I am told that the nucleophilic chloride anion attacks the tri-substituted location because it has the most partial positive character due to its ability to stabilize positive charge.

How does that statement make any sense?

I'm not following his train of "logic," which is as follows:

1. The nucleophile will want to attack the location with the greatest partial positive charge. Okay, Coulomb's law. I'm good with this statement.

2. The carbon with the greatest partial positive charge is the one that can best stabilize positive charge. (?)

3. Alkyl groups are slightly electron-donating; the tertiary carbon will be able to better stabilize partial positive charge.

4. Thus, the chloride anion attacks the tertiary rather than the secondary carbon. I don't see how this follows from the previous assertions. Doesn't stability correspond negatively with reactivity? If so, what is the real reason the chloride anion attacks the tertiary carbon rather than the secondary carbon?

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The relationship between stability and reactivity is complicated. A compound might be more reactive under some conditions because it forms a "more stable" (i.e. lower energy) transition state. That being said, all of this language is meant to help the novice navigate something better represented as, "the nucleophile is more attracted to the carbon atom with the greatest partial positive charge, which is the one with the greater LUMO density." –  Ben Norris Aug 4 '14 at 0:56
Can you elaborate? –  Dissenter Aug 4 '14 at 1:30
I will make a try an attempt at an answer in the near future. –  Ben Norris Aug 4 '14 at 1:34
Essentially the same question: chemistry.stackexchange.com/questions/4305/… –  Martin - マーチン Aug 4 '14 at 4:36
It is important to state, that in asymmetric epoxides, the $\ce{C-O}$ bond lengths are not equal. Upon protonation, this effect is enhanced, giving rise to a weak and a stronger bond that can be broken. This means the carbon, that can better stabilise the charge is more weakly bonded to the oxygen. Thus this will be the carbon suffering the nucleophilic attack. –  Martin - マーチン Aug 4 '14 at 4:46

To the question mark in (2): this implication is simple: if the charge were on a neighboring carbon you'd have a state of higher energy. "Can better stabilize" is a synonym for "gives lower overall energy".

To the objection in (4): the stability ranking refers to various hypothetical electronic distributions in the intermediate. It settles the matter of where the positive charge will be.

The reactivity ranking refers to the fitness of each atom in the intermediate as a site of attack by the nucleophile. The most likely target is the one where the positive charge is concentrated.

So there is no contradiction. The carbon most able to stabilize a positive charge is also the one most likely to be the target of a nucleophilic attack (other things being equal; of course there could be factors, like sterics, changing the ranking).

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So you're saying that the most substituted carbon will be the one that can best stabilize the product, which will definitely have a partial positive charge on the carbon (after all, it is a nucleophile that will be attaching to the carbon). –  Dissenter Aug 4 '14 at 5:01
If that is the case I don't like the way the video phrased it. The video strongly implied that the carbon that was the most substituted existed with a strong partial positive charge. –  Dissenter Aug 4 '14 at 5:01
Well yes, it does, during the very brief existence of the carbocation! Otherwise, no. –  Silvio Levy Aug 4 '14 at 5:06
The "dilution" happens before the attack. We're getting into what Ben said: a full explanation would have to involve MOs. But basically, when O pulls an H+, the whole molecule gets a positive charge; in the diagram we put it on the O, but in fact it spreads around, unevenly. Of the two Cs on the ring, the one with 2 methyls can better accept some of that positive charge. When Cl$^-$ comes by, it prefers to go to that carbon. The only more desirable place for it to attach is the H on the O, but that would just reverse the initial step. We ignore that possibility as uninteresting. –  Silvio Levy Aug 4 '14 at 5:31
I think your original thought is valid in the following sense: of two ways to arrange the same atoms, the more reactive one is usually the one higher in energy (O$_3$ is more reactive than O$_2$, and atomic O is more reactive than both). Even then the correlation is not absolute since the lowest energy configuration may have a higher activation energy (higher-energy transition state) -- your mention of the activation barrier is very much to the point. –  Silvio Levy Aug 4 '14 at 5:45