2
$\begingroup$

Reaction of (S)-2,2-dimethyl-3-propyloxirane with hydrochloric acid

In the above reaction, I am told that the nucleophilic chloride anion attacks the trisubstituted carbon 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 the 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?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Essentially the same question: Regioselectivity of acid-catalyzed ring-opening of epoxides $\endgroup$ – Martin - マーチン♦ Aug 4 '14 at 4:36
  • 3
    $\begingroup$ 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. $\endgroup$ – Martin - マーチン♦ Aug 4 '14 at 4:46
4
$\begingroup$

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 electronic distributions in the possible transition states. In the transition state of an SN2 reaction, and when there is a positively charged leaving group, the carbon undergoing nucleophilic substitution will bear a partial positive charge in the transition state. So, the degree to which this positive charge is stabilised settles the matter of which carbon will be attacked preferentially.

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 which leads to the lower-energy transition state, i.e. has a lower activation energy barrier.

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).

Improve this answer
$\endgroup$
0

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.