Relative basicity of fluorine and oxygen in silyl ether deprotection

[![enter image description here](https://i.stack.imgur.com/JlmqB.png)](https://i.stack.imgur.com/JlmqB.png)


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.


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The reaction mechanism is actually:


[![RXN1](https://i.stack.imgur.com/PSSEJ.png)](https://i.stack.imgur.com/PSSEJ.png)


[![RXN2](https://i.stack.imgur.com/deJjT.png)](https://i.stack.imgur.com/deJjT.png)


The $\ce{Si-F}$ bond is one of the most [stable bonds](http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html) 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.


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