[cseil]

CH3O- or CH3OH ?
My book says CH3O- but I don't agree.

In a protic solvent like water there would be stronger dipole dipole interaction with CH3O- than with CH3OH. So, methanol is a better nucleophile.
Am I wrong? Why?

And if I have CH3OH as solvent, which is the best nucleophile: NH3 or H2O?

[orgopete]

If methoxide has a stronger hydrogen bond, what might this tell you about the electrons of methoxide compared to methanol? How might that apply to its reaction as a nucleophile?

Re, ammonia v water
Which electrons do you think are more available to react? What effect do you think the additional proton in the nucleus of oxygen has on the electrons of oxygen (both have 10p & 10e)?

[cseil]

Methoxide is more basic, the electrons are more available. It has stronger hydrogen bond, so there are stronger interactions between the solvent and CH3O-.

Talking about ammonia,
I believe that the electrons of NH3 are more available, because ammonia is more basic than water. Then again, strongest interaction with the solvent and H2O is the better nucleophile.

[orgopete]

Since we know methoxide and ammonia are the better nucleophiles, then we can speculate on which effect must be stronger, nucleophilicity v hydrogen bonding. I agree the solvent effects are most easily explained by a hydrogen bonding effect and this would be the rationale you are using. Those solvent effects are normally argued with different anions. Fluoride is more basic than iodide, hence more hydrogen bonding.

In this case, you are asked whether the electron availability that leads to hydrogen bonding is greater than its ability to react as a nucleophile? That is, will there be a preference for a proton over a carbon? You can use reaction data to answer that. (The electron availability that leads to hydrogen bonding also makes those electrons better nucleophiles.)

If this is any consolation, indeed it is a paradox that iodide should be the best leaving group and the best nucleophile. A hydrogen bonding effect is the best(?) I've seen. I'm not sure it is a complete explanation as iodide seems to be the least affected by hydrogen bonding, but still remains quite fast in non-polar a protic solvents.

[Borek]

My orgo is weaker than my French, so could be I am missing something very obvious, but I don't understand what the question is about. Does methoxide exist in water in concentrations high enough to react as a nucleophile?

[cseil]

Since we know methoxide and ammonia are the better nucleophiles, then we can speculate on which effect must be stronger, nucleophilicity v hydrogen bonding. I agree the solvent effects are most easily explained by a hydrogen bonding effect and this would be the rationale you are using. Those solvent effects are normally argued with different anions. Fluoride is more basic than iodide, hence more hydrogen bonding.

In this case, you are asked whether the electron availability that leads to hydrogen bonding is greater than its ability to react as a nucleophile? That is, will there be a preference for a proton over a carbon? You can use reaction data to answer that. (The electron availability that leads to hydrogen bonding also makes those electrons better nucleophiles.)

If this is any consolation, indeed it is a paradox that iodide should be the best leaving group and the best nucleophile. A hydrogen bonding effect is the best(?) I've seen. I'm not sure it is a complete explanation as iodide seems to be the least affected by hydrogen bonding, but still remains quite fast in non-polar a protic solvents.

I don't understand what you mean with "will there be a preference for a proton over a carbon?", sorry.
What data should I use?

My orgo is weaker than my French, so could be I am missing something very obvious, but I don't understand what the question is about. Does methoxide exist in water in concentrations high enough to react as a nucleophile?

I don't know why I should answer to this question.
The question is which one is the better nucleophile and it is a theoretical question, then I have to assume that CH3O- can exist in the same concentration of CH3OH.
I am a little bit confused about the topic and I don't understand how your post could be of any help.

[Borek]

I am a little bit confused about the topic and I don't understand how your post could be of any help.

I am not trying to help - I am confused myself, so I am trying to clarify the problem.

[cseil]

I am a little bit confused about the topic and I don't understand how your post could be of any help.

I am not trying to help - I am confused myself, so I am trying to clarify the problem.

I can't express myself more clearly.
The book asks that.

[orgopete]

I don't understand what you mean with "will there be a preference for a proton over a carbon?", sorry.
What data should I use?

I am inferring that your question means you think

CH3O(-) + HOR  [CH3O•••HOR](-)

interferes with

CH3O(-) + CH3X  [CH3O•••CH3X](-)    CH3OCH3 + X(-)

As I look at this reaction, I see a parallel between a potential proton transfer [CH3OH + (-)OR] and the nucleophilic substitution reaction. I thought since the substitution reaction prevails, then you might conclude even though a hydrogen bond may well be present, it doesn’t prevent the reaction from taking place. They may slow a reaction compared to one in which they are not present though.

Perhaps I might have asked, which do you think can form more or stronger hydrogen bonds, methanol or methoxide? I expect you would recognize that methoxide would. I would argue the characteristics that allow methoxide to form stronger hydrogen bonds are the same characteristics that make it a better nucleophile.

[cseil]

I don't understand what you mean with "will there be a preference for a proton over a carbon?", sorry.
What data should I use?

I am inferring that your question means you think

CH3O(-) + HOR  [CH3O•••HOR](-)

interferes with

CH3O(-) + CH3X  [CH3O•••CH3X](-)    CH3OCH3 + X(-)

As I look at this reaction, I see a parallel between a potential proton transfer [CH3OH + (-)OR] and the nucleophilic substitution reaction. I thought since the substitution reaction prevails, then you might conclude even though a hydrogen bond may well be present, it doesn’t prevent the reaction from taking place. They may slow a reaction compared to one in which they are not present though.

Perhaps I might have asked, which do you think can form more or stronger hydrogen bonds, methanol or methoxide? I expect you would recognize that methoxide would. I would argue the characteristics that allow methoxide to form stronger hydrogen bonds are the same characteristics that make it a better nucleophile.

The reaction takes place but slowly, that's okay.
Do you agree with me that methoxyde forms stronger hydrogen bonds, right?

I do not understand why it is still a better nucleophile than CH3OH in water.
If there was an aprotic solvent I'd understand it, but the fact that there are strong hydrogen bonds doesn't make it a worse nucleophile than CH3OH (that form weaker hydrogen bonds)?

[orgopete]

In the diagram, I was trying to show the difference between a hydrogen bond and a covalent bond. If methoxide forms a hydrogen bond, it is weaker than a covalent bond. If it were stronger, it would be methanol.

I have gone through this before, but the arguments are different than in your textbook. Every pair of electrons is minus two. Hydroxide has three pairs of electrons. It can abstract a proton from an acid with pKa 14. Adding a proton does not change the charge of the non-bonded electrons. However, the remaining electrons cannot abstract a proton until the acid has a pKa -2.3.

Even though the net charge has changed, no change has occurred on the remaining non-bonded electrons or the nuclear charge of the oxygen. Remember, the non-bonded electrons of ammonia are more easily protonated than the electrons of a fluoride ion. It isn't net charge, it is the local environment of the electron pair. You can look at the bond lengths and determine the electrons extend further from the nucleus of a nitrogen than from fluorine. Also, even though ammonia has one more proton, the three protons are much further away compared to those in the nucleus of the fluoride ion. This is an inverse square relationship, 2x distance is 1/4 force.

If you are following my reasoning, then protonation must also shift the distance of the non-bonded electrons to account for the change in their basicity. Otherwise, all of the electrons of hydroxide could become protonated at a narrow(er) pH range. That doesn't happen. The bond lengths do change, but only by a small amount. But because we are comparing very small differences and the inverse square law, they have a big impact.

The net effect is that if we compare the non-bonded electrons of methoxide vs methanol, the non-bonded electrons extend further and are more basic. The process of converting a hydrogen bond to a covalent bond shifts all of the electrons. Methoxide has stronger hydrogen bonds because all of its electrons extend further from the nucleus. A very strong hydrogen bond is called a covalent bond.

The electrons of methoxide should be more reactive. If they react with water or another methanol, it converts methoxide to the less basic form, methanol. We can think of a hydrogen bond as an intermediate between a free methoxide and methanol. A reaction will occur most likely on electrons that are furthest from the oxygen nucleus (methoxide). Hydrogen bonding will compete with formation of the substitution transition state. They are equivalent in that sense. Hydrogen bonding will reduce the concentration of 'free' methoxide, hence a faster rate can be achieved in an aprotic solvent. Hydrogen bonding will be less strong for methanol because it's electrons are now closer to the nucleus and less reactive. Hydrogen bonding should only have an effect upon the rate by reducing the amount of free methoxide, but it will not alter the reactivity difference of methoxide v methanol.