Determining Acidities of Alcohol Structures. Need help!

[deleted244469]

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Can someone please explain this to me. I have no idea of how to approach it. The order is 2, 1, 3, 4. (The groups on the far right molecule are both CH3)

Thanks!

 

[aldol16]

Which conjugate base is the most stable? Strong acids have stable conjugate bases. Consider 1) resonance and then 2) hyperconjugation.

 

[deleted244469]

So, I think I figured it out.

The molecule with the carbonyl would be the MOST acidic because it would take the electrons from the Negatively charged Oxygen (After it is deprotonated and becomes a Conjugate Base) (because it is an Electron Withdrawing Group), thus removing the net negative charge from it and stabilizing it (Giving it a positive charge). The more electron density removed from a molecule, the more acidic it becomes? It would also be the most acidic because none of the other molecules have a Electron Withdrawing Group (Duh, lol).

Next, because the amount of total methyl groups on compounds #4 and #4 are the same, the more Methyl groups "directly" bonded to the Alcohol functional group on molecule #4 means that a higher electron density is being donated to the negatively charged oxygen conjugate base (Since CH3 is an Electron Donating Group), This means that the oxygen will become increasingly negative and as a result will be less acidic because of the increase in negative charge.

Am I understanding this correctly? If I am, then I feel stupid, because it was a pretty simple concept.

 

[Wolfpack93]

So, I think I figured it out.

The molecule with the carbonyl would be the MOST acidic because it would take the electrons from the Negatively charged Oxygen (After it is deprotonated and becomes a Conjugate Base) (because it is an Electron Withdrawing Group), thus removing the net negative charge from it and stabilizing it (Giving it a positive charge). The more electron density removed from a molecule, the more acidic it becomes? It would also be the most acidic because none of the other molecules have a Electron Withdrawing Group (Duh, lol).

Next, because the amount of total methyl groups on compounds #4 and #4 are the same, the more Methyl groups "directly" bonded to the Alcohol functional group on molecule #4 means that a higher electron density is being donated to the negatively charged oxygen conjugate base (Since CH3 is an Electron Donating Group), This means that the oxygen will become increasingly negative and as a result will be less acidic because of the increase in negative charge.

Am I understanding this correctly? If I am, then I feel stupid, because it was a pretty simple concept.

Yep you've got it. The key point here is electron donating versus withdrawing. When you're dealing with acids, electron withdrawing groups pull electron density towards them making them more acidic. Methyl groups will do the opposite, donating electron density and decreasing acidity. But something that I always mixed up after doing these problems is what happens when you're dealing with a cation instead of an anion. Basically its just the opposite effect as what happens with acids.

 

[Gulliver Dark]

Ochem as a second language has some good practice in this, and so does Khan academy, specifically with EDGs and EWGs.

I feel like at this point owes me advertising money.

 

[aldol16]

The molecule with the carbonyl would be the MOST acidic because it would take the electrons from the Negatively charged Oxygen (After it is deprotonated and becomes a Conjugate Base) (because it is an Electron Withdrawing Group), thus removing the net negative charge from it and stabilizing it (Giving it a positive charge). The more electron density removed from a molecule, the more acidic it becomes? It would also be the most acidic because none of the other molecules have a Electron Withdrawing Group (Duh, lol).

Yes, you got it. The point here, though, is that there's no positive charge. Remember that charge must be conserved - this is simply another statement of conservation of energy. So the carbonyl contributes to a resonance form that puts the negative density on the carbonyl oxygen as well, which stabilizes the conjugate base via delocalization.

 

[deleted244469]

Ochem as a second language has some good practice in this, and so does Khan academy, specifically with EDGs and EWGs.

I feel like at this point owes me advertising money.

LOL. I already have the EK 1001 Orgo book coming my way.

EDIT: I figured out which Ochem as a second language book I need I appreciate the suggestion!

 

[Gulliver Dark]

Did it edit out me saying Klein, or did I forget to put it? Either way, weird. I suggest the first semester book over the second semester book ... at least for me, it was better roi.

 

[aalamruad]

Did it edit out me saying Klein, or did I forget to put it? Either way, weird. I suggest the first semester book over the second semester book ... at least for me, it was better roi.

It probably edited out "Klein" for you, that word is almost always censored on these forums.

 

[Gulliver Dark]

I thought it might have been considered advertising or something. Nice sarcasm, though.