enter image description here

Why does the $\ce{O}$ take the $\ce{H+}$ instead of the $\ce{F}$?

What I took into consideration:

  • $\ce{O}$ pulls more $e^-$ from atoms near it: $\ce{O}$ is more basic
  • $\ce{F}$ has one more $e^-$ pair than $\ce{O}$: $\ce{F}$ is more basic
  • $\ce{F}$ is more electronegative than $\ce{O}$ : $\ce{O}$ is more basic(?)
  • $\ce{O}$ has more substituent group making it sterically hindered: doesn't matter(?)

Is it okay to think this way? What's the most important factor?


Because oxygen has a lower electronegativity than fluorine, oxygen is less stable with a negative charge (and more stable with a positive charge). This is why $\ce{OH^-}$ is more basic than $\ce{F^-}$ and why $\ce{HF}$ in water is an acid (hydrofluoric acid), while $\ce{H2O}$ in water is neutral.


The reaction mechanism is actually:



The $\ce{Si-F}$ bond is one of the most stable bonds in all of chemistry ($565\ \mathrm{kJ\ mol^{-1}}$). Therefore, $\ce{F-}$ easily displaces the oxygen, leaving it as an alkoxide ion.

Once in the tert-butyldimethylsilyl fluoride, the fluorine is a very poor base (due to its high electronegativity) and is very resistant to sharing any of its unbonded electrons with a hydrogen ion. The alkoxide, on the other hand, will readily accept the hydrogen ion to form the alcohol.

  • $\begingroup$ Thank you again^^ let me ask you a little bit more T_T. It's difficult for me to understand why higher bonding energy(kJ/mol) is associated with stability. because I always thought these high bonding E is usually from partial ionic bond b.w two atoms, thus making this bond susceptible to reaction. and as far as I know, OH bond is strong(467 kJ/mol) but seems to be broken pretty well. Si-F is higher but 'about 100kJ/mol' makes so much difference? $\endgroup$ – NK Yu Mar 10 '16 at 7:28
  • $\begingroup$ Are you sure that first reaction is an SN2 reaction? Silicon can (and often does) readily take a fifth bond. $\endgroup$ – SendersReagent Mar 10 '16 at 7:42
  • $\begingroup$ Yes, there would be an intermediate that places 5 bonds on the silicon, but from what I've seen it does break off into the silyl fluoride and an alkoxide. $\endgroup$ – ringo Mar 10 '16 at 8:12
  • $\begingroup$ @NKYu When bonds form, the more energy that is released, the higher the bond energy of the bond is, and by default, the more stable it is. Being more stable by $100\ \mathrm{kJ\ mol^{-1}}$ is a lot. This would be an extremely exothermic reaction. $\endgroup$ – ringo Mar 10 '16 at 8:23
  • $\begingroup$ isn't it right that the reason Si-F bond is so strong is because of difference of electronegativity b.w SI and F? if so, wouldn't the electronegativity that made this bond strong also makes the bond vulnerable to reaction at the same time because F can hog e- pair shared b.w Si and F? $\endgroup$ – NK Yu Mar 11 '16 at 2:35

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.