Thank you, thank you, thank you. This was the best explanation of ARIO I have been given, and that is after 2 years of general Chemistry, and one year of Organic Chemistry. I'm in my second year of Organic now, and I am having such a difficult time recalling the mountains of information from O-Chem 1 that I need for the second semester. I appreciate your examples and clear explanations. Nice job!
Hi. Can we use atom in the ARIO to find which is more acidic between c2h2 vs. Nh3. The result is nh3 is more acidic than c2h2 if atom is used? Although the pka and orbitals suggests that c2h2 is more acidic than nh3. So atom and orbital in this situation contradicts. Is atom can be used???
Hi Raba, that is a great question. I tell my students that if they can find a pka value on the table, they should use that (which you can do in your example). There are a few instances where ARIO doesn't predict acidity differences well. The Acetylene vs ammonia example you gave is one of them, also acetic acid vs phenol, where the acetate conjugate base only has 2 resonance structures, and the phenoxide conjugate base has 4. However, it is the acetate that is more stable, likely because both of its resonance structures have a negative charge on an oxygen.
Can I just ask, in the example at 10:25 when you are talking about the compound with a pi bond, why would the hydrogen not be taken from the end with the sigma bond as there are 3 there to play with? wouldn't that be another possible structure? And if that were the case, wouldn't they be more or less the same? I'm probably missing some key concept here so sorry in advanced haha.
Hi Henry, I'm going to assume you are speaking of the molecule in the middle of the page, with just one pi bond. Its called propene. You are right that there are more hydrogens on the left carbon, the methyl. And if we remove one of these, we get a resonance stabilized allylic anion. However, this anion that is formed would not be on an sp2 hybridized carbon that is part of a pi bond. That is what I was trying to show in this part of the video, the effect of hybridization on pka. Thus, I used an alkane, alkene and alkyne to show that effect. If we needed to, I could discuss the pka of the hydrogens in the methyl group on the propene in a separate part of the video. Hope that helps, its hard to reply without being able to make drawings!
In the first example you used electronegativty to decide which conjugate base is more stable. Oxygen was because of electronegativty. But the second problem you use atom size and sulfur was more stable? Carbon and Sulfur only differ in electronegativties by 0.03. So why were two different paths used?
That is a good question. When we compare two atoms in the same period, like , going across the periodic table, electronegativity is key. The more electronegative atom is more stable with the negative charge. When we compare atoms in the same group, like O and S, then size is the deciding factor. The larger atom, in this case S because its in the 3rd period, is more polariziable, meaning it is more stable with a negative charge than the O. I like to think of it like a classroom. If I am teaching a class with only 6 students in it, it will be really noticeable when one student brings her sister to observe class. But if my class has over 100 students in it, I may not even notice when one student brings her sister to observe. The same thing happens when atoms have more electrons. One more isn't much of a big deal.
@@nadinethegamer2382 I think you are referring to the comparison of say, N-H and S-H? In that case, one can't use the "Atom" in the ARIO concept to predict acidity. A pka table would be your best bet. I'm not sure about your professor's opinion on this, but in my class, I wouldn't ask students to make that comparison without also providing them with an appropriate pka table. Let me know if that doesn't answer your question.
Hi Brad, that's true, thanks for the comment. I should have been more specific about talking about sp vs sp2 vs sp3 orbitals (and not simply s vs p), and the shape of the sp orbital is more "round" or "squished" vs the elongated shape of an sp3 orbital. This more compact shape allows the electron density to reside closer to the nucleus, which offers a bit more stability compared to an electron pair in an sp2 or sp3 orbital.
Hi ranya x, I'm not sure what you mean by used for aromatic compounds. If you mean to discuss the acidity of phenol, then, yes, you can use the ARIO concept for that. As for your other comment, an electron withdrawing group typically has electronegative atoms, likely a carbonyl. This type of group would pull electrons towards itself. A electron donating group typically has a lone pair that can be used to "donate" electrons. These groups are often -OH or -OCH3. If you have an organic textbook, you can probably find a good description of electron withdrawing vs electron donating in the "aromatic reactions" chapter. In our book, by David Klein, that is chapter 19. Also, this video may help to explain the concept. ua-cam.com/video/eXoi0Q4k1O8/v-deo.html
Thank you, thank you, thank you. This was the best explanation of ARIO I have been given, and that is after 2 years of general Chemistry, and one year of Organic Chemistry. I'm in my second year of Organic now, and I am having such a difficult time recalling the mountains of information from O-Chem 1 that I need for the second semester. I appreciate your examples and clear explanations. Nice job!
Thank you so much, this saved my butt for my Organic lecture!
Appreciative of how clarified this explanation was :)
Thank you for the overview! It'll definitely help for my test tomorrow.
THANKS . AHHH. so simple and clear
Excellent video, easy to follow.
Great job man, brought much needed clarity.
thanks for the help
Thank you so much! Such a great video, so clear and well thought out.
I just came here because my name is Ario
You are amazing.
this was such a good explanation
brilliant good help, thank u
Thank you for the help!
very clear explanation 👍👍👍
Great video, thank you
Thank you!!! So helpful!
great video
Hi. Can we use atom in the ARIO to find which is more acidic between c2h2 vs. Nh3. The result is nh3 is more acidic than c2h2 if atom is used? Although the pka and orbitals suggests that c2h2 is more acidic than nh3. So atom and orbital in this situation contradicts. Is atom can be used???
Hi Raba, that is a great question. I tell my students that if they can find a pka value on the table, they should use that (which you can do in your example). There are a few instances where ARIO doesn't predict acidity differences well. The Acetylene vs ammonia example you gave is one of them, also acetic acid vs phenol, where the acetate conjugate base only has 2 resonance structures, and the phenoxide conjugate base has 4. However, it is the acetate that is more stable, likely because both of its resonance structures have a negative charge on an oxygen.
Ok thank you
Can I just ask, in the example at 10:25 when you are talking about the compound with a pi bond, why would the hydrogen not be taken from the end with the sigma bond as there are 3 there to play with? wouldn't that be another possible structure? And if that were the case, wouldn't they be more or less the same? I'm probably missing some key concept here so sorry in advanced haha.
Hi Henry, I'm going to assume you are speaking of the molecule in the middle of the page, with just one pi bond. Its called propene. You are right that there are more hydrogens on the left carbon, the methyl. And if we remove one of these, we get a resonance stabilized allylic anion. However, this anion that is formed would not be on an sp2 hybridized carbon that is part of a pi bond. That is what I was trying to show in this part of the video, the effect of hybridization on pka. Thus, I used an alkane, alkene and alkyne to show that effect. If we needed to, I could discuss the pka of the hydrogens in the methyl group on the propene in a separate part of the video. Hope that helps, its hard to reply without being able to make drawings!
Thank you
Nice video sir
In the first example you used electronegativty to decide which conjugate base is more stable. Oxygen was because of electronegativty. But the second problem you use atom size and sulfur was more stable? Carbon and Sulfur only differ in electronegativties by 0.03. So why were two different paths used?
That is a good question. When we compare two atoms in the same period, like , going across the periodic table, electronegativity is key. The more electronegative atom is more stable with the negative charge. When we compare atoms in the same group, like O and S, then size is the deciding factor. The larger atom, in this case S because its in the 3rd period, is more polariziable, meaning it is more stable with a negative charge than the O. I like to think of it like a classroom. If I am teaching a class with only 6 students in it, it will be really noticeable when one student brings her sister to observe class. But if my class has over 100 students in it, I may not even notice when one student brings her sister to observe. The same thing happens when atoms have more electrons. One more isn't much of a big deal.
@@RyanJeskeOrganicChem What if thr atoms are in different groups and different periods?
@@nadinethegamer2382 I think you are referring to the comparison of say, N-H and S-H? In that case, one can't use the "Atom" in the ARIO concept to predict acidity. A pka table would be your best bet. I'm not sure about your professor's opinion on this, but in my class, I wouldn't ask students to make that comparison without also providing them with an appropriate pka table. Let me know if that doesn't answer your question.
Thank you so much
Good video but s and p orbitals are all in the same shell so they all have the same energy which means they are all the same distance from the nucleus
Hi Brad, that's true, thanks for the comment. I should have been more specific about talking about sp vs sp2 vs sp3 orbitals (and not simply s vs p), and the shape of the sp orbital is more "round" or "squished" vs the elongated shape of an sp3 orbital. This more compact shape allows the electron density to reside closer to the nucleus, which offers a bit more stability compared to an electron pair in an sp2 or sp3 orbital.
top lad
can this also be used for aromatic compounds?
Hi ranya x, I'm not sure what you mean by used for aromatic compounds. If you mean to discuss the acidity of phenol, then, yes, you can use the ARIO concept for that. As for your other comment, an electron withdrawing group typically has electronegative atoms, likely a carbonyl. This type of group would pull electrons towards itself. A electron donating group typically has a lone pair that can be used to "donate" electrons. These groups are often -OH or -OCH3. If you have an organic textbook, you can probably find a good description of electron withdrawing vs electron donating in the "aromatic reactions" chapter. In our book, by David Klein, that is chapter 19. Also, this video may help to explain the concept. ua-cam.com/video/eXoi0Q4k1O8/v-deo.html
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