4
$\begingroup$

Which of the 2-fluorobutan-3-yl and butan-2-ol-3-yl carbocations will be more stable? I know that fluorine has a greater electron-withdrawing inductive effect, but what about the resonance effect of those?

$\hspace{35 mm}$cations

$\endgroup$
  • $\begingroup$ Firstly, they don't have a +I effect? Also, resonance doesn't play a part in these carbocations. Try drawing resonance structures to stabilise the carbocations. You can't. But, to answer your original question, all you need to do is to look at the comparative reactivities of fluorobenzene and phenol. Which one is more reactive? So which one shows a greater +M effect? $\endgroup$ – orthocresol Oct 28 '15 at 16:30
  • 2
    $\begingroup$ They are stabilized by Neighbouring group participation (anchimeric assistance) not mesomeric effect. $\endgroup$ – Mithoron Oct 28 '15 at 16:59
  • $\begingroup$ I heavily edited your question, you can change it if you wish. $\endgroup$ – Mithoron Oct 28 '15 at 17:08
  • $\begingroup$ @Mith If the heteroatom is attached to the positive carbon, then that happens. But not if it is one carbon away. i.imgur.com/TSqejdS.png $\endgroup$ – orthocresol Oct 28 '15 at 17:34
  • $\begingroup$ @orthocresol Sorry, your link is wrong, what they call Ngp is simple mesomeric stabilisation. $\endgroup$ – Mithoron Oct 28 '15 at 17:55
2
$\begingroup$

The answer is actually quite straightforward. Fluorine is more electronegative than oxygen, thus it is less willing to share its lone pair with the carbon bearing the positive formal charge. Thus, the $\ce {-OH}$ group provides greater stabilisation. In fact, the donation of the lone pair by the $\ce {O}$ generates a familiar moiety, an epoxide. The donation by $\ce {F}$ is almost negligible since the positive formal charge on $\ce {F}$ would be highly unfavoured.

| improve this answer | |
$\endgroup$

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.