11:45 Why not though? On the carboxy group, there is only one additional place where the negative charge could go: on the other oxygen. It cannot enter the aromatic ring. For the xydroxy group, on the other hoof, it can not only enter the aromatic ring, making the carbanion there (as you showed), but this carbanion can travel further, either around the ring (giving 2 more alternative resonance structures), or displacing the π bond on the carboxy group's oxygen, landing on that oxygen, which is pretty much a similar configuration to that from the earlier resonance for the carboxy group's hydrogen. This is extra stability, if anything! And it's enough to copmare those two last configurations (with the negative charge on the oxygen, which is the common pattern for both) to see that the hydroxy group's oxygen has this resonant structure as well, plus a couple more extra around the aromatic ring. Extra stability means more acidic, doesn't it?
This is a great point. Yes, deprotonation of the simple OH group opens up MANY different resonance forms and potentially huge delocalization of the charge. The problem is that those additional resonance forms require disruption of the aromatic ring. So, while there are many additional resonance forms, they are all poor contributors and do not provide appreciable stabilization of that particular conjugate base.
@@ChemHelpASAP What do you mean by "disruption"? I don't remember that being any of the "rules of the game" introduced so far for figuring out acidity.
@@bonbonpony That's a fair comment, but that ring is special. It is a benzene ring and very stable with that arrangement of alternating double and single bonds. Any resonance forms that break up that pattern mess up the stability of the ring and therefore are not favorable resonance forms.
@@ChemHelpASAP So why is it that in problem 8 the exact same argument worked in favor of the carbanion to the right (having the additional resonance into the benzene ring), but for the salicylic acid's hydroxy proton it's suddenly bad for the carbanion to get involved into the benzene ring's resonance? :q This theory seems to be self-contradictory, which might be an indication that there's actually something else responsible for this phenomenon in the case of the salicylic acid, irrelevant from the resonance. My bet is on the hydrogen bonding opportunity due to the hydroxy group being in ortho position to the carboxy. The -OH's hydrogen is close enough to the carboxy's oxygen to be able to interact with its lone pair and make a hydrogen bond, increasing the stability of the hydroxy proton. Even more if the carboxy group gets deprotonated, because the negative charge on its oxygen will attract that proton even more towards itself, increasing the strength of that hydrogen bond. And even if the hydroxy group gets deprotonated first, it will probably be able to steal the proton from the carboxy group and reform that hydrogen bond, because the most comfortable place for that proton to be is between those two oxygens. OK, but having those discrepancies, I don't think that theory is able to resolve this. One source will explain this one way, another will explain it the other. So how can we ask the Nature itself what's the real answer and check EXPERIMENTALLY which of those protons is REALLY detaching first, and what is the REAL reason for that?
@@bonbonpony In question 8, the first structure only has resonance into the carbonyl. The second structure has resonance into the carbonyl AND resonance into the aromatic ring. Yes, resonance into the ring is not ideal, but it's better than nothing (and nothing is what the first structure has beyond the carbonyl). Going back to salicylic acid, internal hydrogen bonding likely does play a role here. I don't think it is a deciding role. In the end, all discussions in science end with experimental results. If theory does not match experimental results, then you don't typically question the experimental results (can happen). You refine the theory. In salicylic acid, the more acidic proton is the carboxylic acid proton because it leads to the more stable conjugate base. That is the reality. We develop models (theory) not only to match reality retrospectively (already know the outcome) but also to help predict the behavior of molecules that we might not be able to make or have not yet made. We use known outcomes to train the models so that the models are predictive for new situations.
could you please explain why in the 3rd problem the 'b" carbon wouldnt resonate with the sulphur atom beacuse sulphur has a vacant orbital? so b carbaonion would have resonance right?
If we remove the hydrogen off the 'B' carbon, we would get a carbanion. The sulfur next door still has its lone pairs, an S-H bond, and an S-C bond. That sulfur has a full octet. We cannot donate the carbon lone pair to sulfur because the sulfur already has a full valence shell.
@@shouryagoyal8820 You're right. Sulfur can expand beyond an octet. That 3d orbital, however, is big - much bigger than a 2p orbital on carbon. So, you can draw the resonance form, but the overlap between a 3d orbital and 2p orbital is very low. It is a size mis-match. So, can you draw this resonance form. Yes. Is it a resonance form that provides meaningful stabilization? No. Formally, you might consider it to be a very minor contributor to the structure of the anion.
This is a great point, but there is another factor at work here. So, yes, the OH on the benzene ring can be deprotonated. In one resonance form, the OH oxygen carries the negative charge. You can then draw several resonance forms into the benzene ring with different C minus options. These aren't great (C-minus vs O-minus), but they are regardless more resonance forms. Finally, it's possible to create one last resonance form with the negative charge on the carbonyl oxygen. This looks like a clear win for the OH on the ring. The "other factor" is the benzene ring. Resonance forms that disrupt the very stable benzene ring are strongly disfavored, so all these additional resonance forms are not as favorable as they appear, and they do not contribute much stability to the conjugate base.
The most acidic for cyclopentadiene is one of the CH2 hydrogens. When you deprotonate that carbon, the resulting carbanion is aromatic and surprisingly stable for a carbanion. See this video on aromatic ions... th-cam.com/video/m3OjQPHcFqc/w-d-xo.html
Why does in problem 6, b is more acidic..As much as i know OCH3 is electron releasing or +I group and same should apply for Oxygen in a ring and according to this B should be less stable and less acidic than A.. Plz try to resolve this query.
An oxygen is electron-releasing (an electron-donating group) when it is directly attached to an atom that can accept the electron density, normally by resonance. That requires the adjacent atom to have a pi bond or p-orbital. In this example, the charge we are trying to affect is on nitrogen and separated from the oxygen by atoms that are sp3 hybridized - no pi bonds or p-orbitals. With resonance not possible, the inductive effect of oxygen becomes the driving factor.
Heyy thank you for this it really helped :) but I have a doubt about problem 3 that why we didn't say that if electrons are more closer to the nucleus of sulfur it would be better like we did in problem 5 for p orbital...
In general, the effect of hybridization on acidity is only discussed when you are comparing two atoms of the same type (carbon vs. carbon, or nitrogen vs. nitrogen). That is the case in problem 5. When you are comparing two different atoms (like in problem 3 - sulfur vs. oxygen), you focus on size and electronegativity differences. Size is the more important factor if the two atoms are in the same period of the periodic table. This is a bit confusing because, in the discussion on hybridization, we do talk about the distance of lone pairs from the nucleus. This discussion feels like a "size" argument. Note that size in the sulfur vs oxygen is about comparing two atoms in a different row of the periodic table. With hybridization, we are comparing two of the same atom. In practice, people only talk about hybridization and acidity with carbon, nitrogen, and (much less often) oxygen.
@@ChemHelpASAP yes I got it but the part that we took the fact that if the distance is less between nucleus and lone pair only in case of hybridization is confusing
Can't we argue that for problem 2 because B has resonance and an inductive effect on the carbon it is more stable than C and hence the acidic proton comes from there?
You can make that argument. B does indeed have a stabilizing inductive effect from the OH on the left. It's not enough, however. The better resonance forms for C give it the lower pKa. The hydrogen next to a carboxylic acid carbonyl has a pKa of about 25. With the OH next door, I'd guess that the pKa drops to around 20 with the inductive effect. The carboxylic acid OH is around 5, much lower. The second resonance form with the charge delocalized across two oxygens is very stabilizing for the conjugate base.
1) You're a great teacher. Thank you.
2) You kind of sound like Owen Wilson and it makes the lecture so much better.
Thank you, but Owen Wilson may not share in the flattery.
@@ChemHelpASAP Being a teacher is far more noble profession than being a actor.
really wish my chem teacher was as clear in her explanations, thank you for saving my grades!
So glad the videos help you. Thank you for watching.
Oh god.. I had been searching since so long that I breathed a sigh of relief when I got to know about this... Thanks a lot... ☺️🤗
Wow. Glad you found what you were looking for!!
This really helped me out since I was so confused on a lot of these
Hang in there! You're doing great.
11:45 Why not though? On the carboxy group, there is only one additional place where the negative charge could go: on the other oxygen. It cannot enter the aromatic ring. For the xydroxy group, on the other hoof, it can not only enter the aromatic ring, making the carbanion there (as you showed), but this carbanion can travel further, either around the ring (giving 2 more alternative resonance structures), or displacing the π bond on the carboxy group's oxygen, landing on that oxygen, which is pretty much a similar configuration to that from the earlier resonance for the carboxy group's hydrogen. This is extra stability, if anything! And it's enough to copmare those two last configurations (with the negative charge on the oxygen, which is the common pattern for both) to see that the hydroxy group's oxygen has this resonant structure as well, plus a couple more extra around the aromatic ring. Extra stability means more acidic, doesn't it?
This is a great point. Yes, deprotonation of the simple OH group opens up MANY different resonance forms and potentially huge delocalization of the charge. The problem is that those additional resonance forms require disruption of the aromatic ring. So, while there are many additional resonance forms, they are all poor contributors and do not provide appreciable stabilization of that particular conjugate base.
@@ChemHelpASAP What do you mean by "disruption"? I don't remember that being any of the "rules of the game" introduced so far for figuring out acidity.
@@bonbonpony That's a fair comment, but that ring is special. It is a benzene ring and very stable with that arrangement of alternating double and single bonds. Any resonance forms that break up that pattern mess up the stability of the ring and therefore are not favorable resonance forms.
@@ChemHelpASAP So why is it that in problem 8 the exact same argument worked in favor of the carbanion to the right (having the additional resonance into the benzene ring), but for the salicylic acid's hydroxy proton it's suddenly bad for the carbanion to get involved into the benzene ring's resonance? :q
This theory seems to be self-contradictory, which might be an indication that there's actually something else responsible for this phenomenon in the case of the salicylic acid, irrelevant from the resonance. My bet is on the hydrogen bonding opportunity due to the hydroxy group being in ortho position to the carboxy. The -OH's hydrogen is close enough to the carboxy's oxygen to be able to interact with its lone pair and make a hydrogen bond, increasing the stability of the hydroxy proton. Even more if the carboxy group gets deprotonated, because the negative charge on its oxygen will attract that proton even more towards itself, increasing the strength of that hydrogen bond. And even if the hydroxy group gets deprotonated first, it will probably be able to steal the proton from the carboxy group and reform that hydrogen bond, because the most comfortable place for that proton to be is between those two oxygens.
OK, but having those discrepancies, I don't think that theory is able to resolve this. One source will explain this one way, another will explain it the other. So how can we ask the Nature itself what's the real answer and check EXPERIMENTALLY which of those protons is REALLY detaching first, and what is the REAL reason for that?
@@bonbonpony In question 8, the first structure only has resonance into the carbonyl. The second structure has resonance into the carbonyl AND resonance into the aromatic ring. Yes, resonance into the ring is not ideal, but it's better than nothing (and nothing is what the first structure has beyond the carbonyl).
Going back to salicylic acid, internal hydrogen bonding likely does play a role here. I don't think it is a deciding role.
In the end, all discussions in science end with experimental results. If theory does not match experimental results, then you don't typically question the experimental results (can happen). You refine the theory. In salicylic acid, the more acidic proton is the carboxylic acid proton because it leads to the more stable conjugate base. That is the reality. We develop models (theory) not only to match reality retrospectively (already know the outcome) but also to help predict the behavior of molecules that we might not be able to make or have not yet made. We use known outcomes to train the models so that the models are predictive for new situations.
I just wanna say thank you for all of the videos you are really getting me through orgo! Keep doing what you do.
Such kind feedback. Thank you! Keep at it - you'll make it through just fine.
this helped immensely, thank you!
So glad to have helped.
could you please explain why in the 3rd problem the 'b" carbon wouldnt resonate with the sulphur atom beacuse sulphur has a vacant orbital? so b carbaonion would have resonance right?
If we remove the hydrogen off the 'B' carbon, we would get a carbanion. The sulfur next door still has its lone pairs, an S-H bond, and an S-C bond. That sulfur has a full octet. We cannot donate the carbon lone pair to sulfur because the sulfur already has a full valence shell.
@@ChemHelpASAP I am sorry for asking again but actually I was taught that sulphur has a vacant d orbital so it can have more than 8 valence electrons.
@@shouryagoyal8820 You're right. Sulfur can expand beyond an octet. That 3d orbital, however, is big - much bigger than a 2p orbital on carbon. So, you can draw the resonance form, but the overlap between a 3d orbital and 2p orbital is very low. It is a size mis-match. So, can you draw this resonance form. Yes. Is it a resonance form that provides meaningful stabilization? No. Formally, you might consider it to be a very minor contributor to the structure of the anion.
@@ChemHelpASAP thankss a lot that just cleared a lot of my doubts😅
For problem 4, atom B would have more resonance structures, allowing it to be the most acidic proton.
This is a great point, but there is another factor at work here. So, yes, the OH on the benzene ring can be deprotonated. In one resonance form, the OH oxygen carries the negative charge. You can then draw several resonance forms into the benzene ring with different C minus options. These aren't great (C-minus vs O-minus), but they are regardless more resonance forms. Finally, it's possible to create one last resonance form with the negative charge on the carbonyl oxygen. This looks like a clear win for the OH on the ring. The "other factor" is the benzene ring. Resonance forms that disrupt the very stable benzene ring are strongly disfavored, so all these additional resonance forms are not as favorable as they appear, and they do not contribute much stability to the conjugate base.
this was very helpful, i really needed this thank you!!!
Glad to help. Thank you for watching. Hang in there!
great explanation, thank you!
You are welcome!
Great video! Thank you!
Very good video thank you
You are welcome.
great video sir.....it help me a lot..
Good to hear! Keep working at it.
Nice explanation!
Thank you for watching!
could you please explain which will be the most acidic H in cyclopentadiene
The most acidic for cyclopentadiene is one of the CH2 hydrogens. When you deprotonate that carbon, the resulting carbanion is aromatic and surprisingly stable for a carbanion. See this video on aromatic ions... th-cam.com/video/m3OjQPHcFqc/w-d-xo.html
thank you for the video
Why does in problem 6, b is more acidic..As much as i know OCH3 is electron releasing or +I group and same should apply for Oxygen in a ring and according to this B should be less stable and less acidic than A..
Plz try to resolve this query.
An oxygen is electron-releasing (an electron-donating group) when it is directly attached to an atom that can accept the electron density, normally by resonance. That requires the adjacent atom to have a pi bond or p-orbital. In this example, the charge we are trying to affect is on nitrogen and separated from the oxygen by atoms that are sp3 hybridized - no pi bonds or p-orbitals. With resonance not possible, the inductive effect of oxygen becomes the driving factor.
thank you so much
You are welcome. Keep working at it!
Heyy thank you for this it really helped :) but I have a doubt about problem 3 that why we didn't say that if electrons are more closer to the nucleus of sulfur it would be better like we did in problem 5 for p orbital...
In general, the effect of hybridization on acidity is only discussed when you are comparing two atoms of the same type (carbon vs. carbon, or nitrogen vs. nitrogen). That is the case in problem 5. When you are comparing two different atoms (like in problem 3 - sulfur vs. oxygen), you focus on size and electronegativity differences. Size is the more important factor if the two atoms are in the same period of the periodic table. This is a bit confusing because, in the discussion on hybridization, we do talk about the distance of lone pairs from the nucleus. This discussion feels like a "size" argument. Note that size in the sulfur vs oxygen is about comparing two atoms in a different row of the periodic table. With hybridization, we are comparing two of the same atom. In practice, people only talk about hybridization and acidity with carbon, nitrogen, and (much less often) oxygen.
@@ChemHelpASAP yes I got it but the part that we took the fact that if the distance is less between nucleus and lone pair only in case of hybridization is confusing
Omg you're a god
I can't live up to that standard, but I am certainly glad you enjoyed the video!
Thanks!
My pleasure.
Thank you very much
You are welcome. Thank you for watching.
god bless you
and you as well
Can't we argue that for problem 2 because B has resonance and an inductive effect on the carbon it is more stable than C and hence the acidic proton comes from there?
You can make that argument. B does indeed have a stabilizing inductive effect from the OH on the left. It's not enough, however. The better resonance forms for C give it the lower pKa. The hydrogen next to a carboxylic acid carbonyl has a pKa of about 25. With the OH next door, I'd guess that the pKa drops to around 20 with the inductive effect. The carboxylic acid OH is around 5, much lower. The second resonance form with the charge delocalized across two oxygens is very stabilizing for the conjugate base.
@@ChemHelpASAP Thank you for the explanation. Your video really helped me understand the concept.
@@isaacmtonga375 So glad to have helped.
oh my god this was great
So glad it helped! I hope your end of semester exams go well.
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