3
$\begingroup$

Why is $\ce{OH-}$ more basic than $\ce{F-}$?

From what I understand, $\ce{O}$ has larger radius so it should be more stable (meaning less basic) than $\ce{F}$. When they form anions, since $\ce{O^{2-}}$ is larger, it can distribute its electrons over larger volume than $\ce{F-}$, meaning $\ce{OH-}$ should be less basic. Is this because of partial negtaive charge on oxygen?

In a polar aprotic solution, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

I learned that basicity (not nucleophilicity) is about stability. So can this be explained by electron density? Is explaining it with polarizability right?

In a lecture, my teacher said $\ce{RO-}$ is relatively stable because of the high electronegativity of $\ce{O}$.

I think it should be unstable because of high electron density caused by high electronegativity. What's right?

When people say something is stable, are they referring to its acid/base characteristics, or can stability also be referring to nucleophilicity/electrophilicity?

$\endgroup$
2
$\begingroup$

Why is $\ce{OH-}$ more basic than $\ce{F-}$?

To answer this question, let us consider the $\mathrm{p}K_\mathrm{a}$'s of the conjugate acids of these anions: $\ce{H2O}$ and $\ce{HF}$. $$\mathrm{p}K_\mathrm{a}(\ce{H2O})=15.7\ \\ \mathrm{p}K_\mathrm{a}(\ce{HF})=3.17$$

You didn't need me to tell you that $\ce{HF}$ is a stronger acid than water though, but it is -- $10^{12.53}$ times stronger to be exact. This means that this is also how many times stronger $\ce{OH-}$ is than $\ce{F-}$

In a polar aprotic solution, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

Small, electron-dense ions are always great nucleophiles, except when the solvent crowds the nucleophile with hydrogen bonds, as in the case of $\ce{F-}$. It is so heavily surrounded by solvent molecules in polar protic solvents that it can't make an attack very well. $\ce{I-}$ is not as heavily solvated in these solvents, and it is very easily polarizable, making it a better nucleophile in these solvents.

In polar aprotic solvents, the small $\ce{F-}$ ion can easily slip between solvent molecules and make attacks. Now in comparison, $\ce{I-}$ is a much poorer nucleophile.

In a lecture, my teacher said $\ce{RO-}$ is relatively stable because of the high electronegativity of $\ce{O}$.

Again, we can answer this by looking at $\mathrm{p}K_\mathrm{a}$'s. The $\mathrm{p}K_\mathrm{a}$ of ethanol, for example, is $15.9$! Only $10^{0.2}=1.58$ times weaker an acid than water! That means that ethoxide will deprotonate everything $\ce{OH-}$ will and then a little more, so it's not really that much more unstable.

When people say something is stable, are they referring to its acid/base characteristics, or can stability also be referring to nucleophilicity/electrophilicity?

The two are pretty well related. You have exceptions like how $\ce{F-}$ is a poor nucleophile in water despite being a good base, but as a rule of thumb, I would say people mean acid/base reactivity when they talk about "stability."

$\endgroup$
  • $\begingroup$ Thank you. it makes sense when you explain it using pKa. but can't you explain another way using movement of e- like I wrote in my question? $\endgroup$ – NK Yu Mar 8 '16 at 0:52
  • $\begingroup$ Yes, but both $\ce{F-}$ and $\ce{OH- /OR-}$ have unpaired electrons. What is more important is to understand how reactive these electron pairs are. $\endgroup$ – ringo Mar 8 '16 at 0:54
  • $\begingroup$ and you said "Small, electron-dense ions are always great nucleophiles", but I don't understand. high polarizability means good Nu, thus small, elctron-dense ion should be poor Nu. where am I confused? $\endgroup$ – NK Yu Mar 8 '16 at 0:54
  • $\begingroup$ Charge density actually makes for a better nucleophile, and $\ce{F-}$ is much smaller than $\ce{I-}$. If you think about it, which will be better at fitting into a molecules structure to force a reaction, something small or something big? $\endgroup$ – ringo Mar 8 '16 at 1:02
  • $\begingroup$ I actually thought bigger atom would be better Nu. because bigger ones pulls their e- less. If you are saying it's wrong, then where does the polarizability come from? $\endgroup$ – NK Yu Mar 8 '16 at 1:11

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.