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Yeah bro thank you very much I have exam tomorrow in this lecture and search in many TH-cam videos about it But still confused Your video come in right time
My doubts were clarified in the first minutes. But this video is so good that I watched until the final. The examples are well explained. Thanks for the video ❤
For anyone confused about 24:01 , this is why I think it's equitorial and not axial. CH3 and OH, according to the structure needs to be CIS. This means both have to be pointing up. Axial and equitorial bonds alternate up and down, shown in 1:17. If you go in order, axial bonds in carbon 1 should go UP. Carbon 2 axial bonds should go DOWN. Carbon 3 axial bonds should go UP. And carbon 4 axial bonds need to go DOWN. That's why the OH group cant be axial up because any axial group on the carbon 4 needs to be DOWN. Hence, you're forced to put the OH group in an equitorial up position since if you go in order (again), Carbon 1 equitorial bonds should go DOWN, Carbon 2 equitorial bonnds should go UP, carbon 3 equitorial bonds go DOWN, and carbon 4 equitorial bonds go UP. CH3 and OH need to be CIS and the only way to do that is if you do an axial up CH3 and an equitorial up OH.
At 16:20, when you're doing a chair flip from form B to form A, you said that you move the carbons in a couterclockwise direction, but is it necessary to reverse the direction of the carbons, or just move them back one place? By the way, your videos are amazing, and I use them for practically every subject in school! Thank you so much!
Wow! Just have to say like everyone else that this video was super helpful and clear! Literally game-changing to understanding such an abstract and visual concept!
At 23:05 or even for the previous question, why can't both the groups be in the equitorial plane ? Why one needs to be in the axial and the other in the equitorial plane Having both the groups in the equitorial plane will make the compound more stable by avoiding 1-3 repulsion
@@Christophero Because they're cis, they either both have to be up or both have to be down. They're also right next to each other in positions 1 and 2. The way the chair is set up you can't have axial and equatorial bonds next to each other in the same direction. So, they alternate directions. Therefore, the set-up with the lowest energy is one with the bulkier group on the equatorial bond and the smaller group on the axial bond. This way they are right next to each other and still pointing in the same direction. Does that make sense?
This actually comes down to the geometry of the carbon atoms. Notice they are all sp3 hybridized - if they were to be in the same plane, the ideal bond angle for sp3 hybridized atoms (109.5) would be deviated from which would further destabilize the molecule.
Could someone explain to me @21:03. Why the second option has the carbon number starts 1 carbon shifted clockwise rather than at the same place as the first possible chair conformation? Thank you so much!
Hi! Since CH3 is a wedge it needs to be in the upward position regardless of whether it is in the axial or equatorial... in the first diagram he drew CH3 in axial up, which leaves the remaining position to be equatorial down... because of that, for the second diagram, he shifted the numbers clockwise because the next position has equatorial up open so that could be occupied then... I hope this helps
Reading the textbook sent me on a downward spiral lol. 🌀, thanks for your very clear and concise explanation. Sometimes the textbook helps but like sometimes not at all. 😂
Thank you so much for your explanation sir! And I have a doubt @24:02 Why do we put the Hydroxyl group in the equatorial up position but not in the axial up position sir?
Im confused on how you know if the substituents is either axial or equatorial and if they are up or down on on the chair. I understand the ring flip and which one would be more stable but I just dont understand the inital drawing process, someone plzzzzzzzzzz help
For anyone confused about 24:01 , this is why I think it's equitorial and not axial. CH3 and OH, according to the structure needs to be CIS. This means both have to be pointing up. Axial and equitorial bonds alternate up and down, shown in 1:17. If you go in order, axial bonds in carbon 1 should go UP. Carbon 2 axial bonds should go DOWN. Carbon 3 axial bonds should go UP. And carbon 4 axial bonds need to go DOWN. That's why the OH group cant be axial up because any axial group on the carbon 4 needs to be DOWN. Hence, you're forced to put the OH group in an equitorial up position since if you go in order (again), Carbon 1 equitorial bonds should go DOWN, Carbon 2 equitorial bonnds should go UP, carbon 3 equitorial bonds go DOWN, and carbon 4 equitorial bonds go UP. CH3 and OH need to be CIS and the only way to do that is if you do an axial up CH3 and an equitorial up OH.
At 24:40, if there are 3 of them or more, how will I know if the third substituent will be axial or equatorial? I anticipated the OH to be axial, but it seems to be equatorial. Someone please answer me
remember that the wedge is always upward while the dashed line is always downward regardless of the axial or equatorial position... OH is equatorial because in position 4 the only upward position was equatorial
19:55 why we don’t put the br axially down but equatorial? Is it bc we supposed to do a 90 angle with the bond? But without converting it into chair version how we supposed to know if it’s axially or equatorial?
I'm probably a bit late with the answer but.... as you already understand each bond in the cycle-hexane structure has two possible bonds, axial and equatorial. based on the shape of the chair we know which way the axial bond would point (the chair makes a kind of arrow in that direction of the axial bond). The equatorial bond always points the opposite direction to the axial (par example if the axial bond points up then the equatorial would point down). Hence if we know if a bond has to point down, we'd choose whichever bond (axial or equatorial) that is pointing down at that particular carbon.
hey, a question: in the last example where you draw isomers - you can choose between equatorial up or down and axis up or down positions, right? and that still counts as correct?
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Apart from other tutors that explain online, this organic chemistry tutor is the best one who has a fair understanding of all other science courses as well.
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I have exam tomorrow in this lecture and search in many TH-cam videos about it
But still confused
Your video come in right time
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For anyone confused about 24:01 , this is why I think it's equitorial and not axial. CH3 and OH, according to the structure needs to be CIS. This means both have to be pointing up. Axial and equitorial bonds alternate up and down, shown in 1:17. If you go in order, axial bonds in carbon 1 should go UP. Carbon 2 axial bonds should go DOWN. Carbon 3 axial bonds should go UP. And carbon 4 axial bonds need to go DOWN. That's why the OH group cant be axial up because any axial group on the carbon 4 needs to be DOWN. Hence, you're forced to put the OH group in an equitorial up position since if you go in order (again), Carbon 1 equitorial bonds should go DOWN, Carbon 2 equitorial bonnds should go UP, carbon 3 equitorial bonds go DOWN, and carbon 4 equitorial bonds go UP. CH3 and OH need to be CIS and the only way to do that is if you do an axial up CH3 and an equitorial up OH.
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At 16:20, when you're doing a chair flip from form B to form A, you said that you move the carbons in a couterclockwise direction, but is it necessary to reverse the direction of the carbons, or just move them back one place? By the way, your videos are amazing, and I use them for practically every subject in school! Thank you so much!
Thanks
Thanks I was super confused about chair conformation and this just helped me understand it omg
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He didn’t say it in this clip but the positive slope clockwise and negative slope counterclockwise solidified this for me and got me my A
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Why do examples 2 amd 3 from 22:40 onward Not have multiple possible conformations?
19:17 can we take the Bromine atom on axial down position as the methyl is on up so we still get a trans isomer
At 23:05 or even for the previous question, why can't both the groups be in the equitorial plane ?
Why one needs to be in the axial and the other in the equitorial plane
Having both the groups in the equitorial plane will make the compound more stable by avoiding 1-3 repulsion
I have this question too, we’re you able to find an answer?
@@Christophero Because they're cis, they either both have to be up or both have to be down. They're also right next to each other in positions 1 and 2. The way the chair is set up you can't have axial and equatorial bonds next to each other in the same direction. So, they alternate directions. Therefore, the set-up with the lowest energy is one with the bulkier group on the equatorial bond and the smaller group on the axial bond. This way they are right next to each other and still pointing in the same direction. Does that make sense?
This actually comes down to the geometry of the carbon atoms. Notice they are all sp3 hybridized - if they were to be in the same plane, the ideal bond angle for sp3 hybridized atoms (109.5) would be deviated from which would further destabilize the molecule.
@@leila8397 thank you.
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17:25 the second demo, second structure, I think the up isopropyl should be shooting toward right
Could someone explain to me @21:03. Why the second option has the carbon number starts 1 carbon shifted clockwise rather than at the same place as the first possible chair conformation? Thank you so much!
Hi! Since CH3 is a wedge it needs to be in the upward position regardless of whether it is in the axial or equatorial... in the first diagram he drew CH3 in axial up, which leaves the remaining position to be equatorial down... because of that, for the second diagram, he shifted the numbers clockwise because the next position has equatorial up open so that could be occupied then... I hope this helps
@@eshalshoaib5923 Thank you so much! That makes sense
on 15:00, the ring flip of the first molecule, was the labelling of axial and equatorial right?
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Reading the textbook sent me on a downward spiral lol. 🌀, thanks for your very clear and concise explanation. Sometimes the textbook helps but like sometimes not at all. 😂
salvando minha prova pq eu não achei nenhum canal BR que explicasse tão bem assim S2
Thank you so much for your explanation sir!
And I have a doubt @24:02
Why do we put the Hydroxyl group in the equatorial up position but not in the axial up position sir?
were you able to find out? I have the same question!
i think there might be steric hindrance between the methyl group and the hydroxyl group if you put it in the axial up position
Im confused on how you know if the substituents is either axial or equatorial and if they are up or down on on the chair. I understand the ring flip and which one would be more stable but I just dont understand the inital drawing process, someone plzzzzzzzzzz help
Hi @Mohamed Dahcheh, this video helped me understand the placement of substituents! th-cam.com/video/4xNQwohSRbo/w-d-xo.html
Good luck!
@@imanifreefun thank you, i had the same question but I understood it bc of this video series :)
Why is in the last example the oh in equatorial and not in axial? Because Br is in equatorial so the next one should be in axial???
i want to ask this ques too...
specifically looked in the comments for this question.. isnt it suppose to be axial since Br was equatorial
For anyone confused about 24:01 , this is why I think it's equitorial and not axial. CH3 and OH, according to the structure needs to be CIS. This means both have to be pointing up. Axial and equitorial bonds alternate up and down, shown in 1:17. If you go in order, axial bonds in carbon 1 should go UP. Carbon 2 axial bonds should go DOWN. Carbon 3 axial bonds should go UP. And carbon 4 axial bonds need to go DOWN. That's why the OH group cant be axial up because any axial group on the carbon 4 needs to be DOWN. Hence, you're forced to put the OH group in an equitorial up position since if you go in order (again), Carbon 1 equitorial bonds should go DOWN, Carbon 2 equitorial bonnds should go UP, carbon 3 equitorial bonds go DOWN, and carbon 4 equitorial bonds go UP. CH3 and OH need to be CIS and the only way to do that is if you do an axial up CH3 and an equitorial up OH.
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When dealing with 2 substituents, do we have to put them in opposite positions, like axial and equatorial?
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At 16:30 you drew the equatorial position of the tert-butyl toward the left direction but should it not be toward the right direction?
At 19:35 why was CH3 on Carbon 1 and Br on Carbon 3? I thought Br needs to be prioritized over CH3?
At 24:40, if there are 3 of them or more, how will I know if the third substituent will be axial or equatorial? I anticipated the OH to be axial, but it seems to be equatorial. Someone please answer me
remember that the wedge is always upward while the dashed line is always downward regardless of the axial or equatorial position... OH is equatorial because in position 4 the only upward position was equatorial
@@eshalshoaib5923upward position can be axial too so why equitorial please tell
24:15 for the last example, i did understand that OH has to be upward, but why now axial?
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@@aideno2597 Yep.
19:55 why we don’t put the br axially down but equatorial? Is it bc we supposed to do a 90 angle with the bond? But without converting it into chair version how we supposed to know if it’s axially or equatorial?
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at 22:52 how do you know when do draw the atoms in equatorial position or axial position based on the wedge and the dash? im confused
at the very last part... does it matter is it going axial or equatorial, as long as it follows the wedge pattern?
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little knit picky but I believe that the equatorial bonds in the middle are suppose to be a parallel with the right and left parallel lines.
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How do we judge axial or equatorial positions based on wedge or dash?
I'm probably a bit late with the answer but.... as you already understand each bond in the cycle-hexane structure has two possible bonds, axial and equatorial. based on the shape of the chair we know which way the axial bond would point (the chair makes a kind of arrow in that direction of the axial bond). The equatorial bond always points the opposite direction to the axial (par example if the axial bond points up then the equatorial would point down). Hence if we know if a bond has to point down, we'd choose whichever bond (axial or equatorial) that is pointing down at that particular carbon.
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hey, a question: in the last example where you draw isomers - you can choose between equatorial up or down and axis up or down positions, right? and that still counts as correct?
wondering the same thing! did you find out?
@@alyssabasant5651 still counts as correct ;)
I could be wrong, but I do not think you are slanting your equatorial positions in the right direction in the middle of the chair molecule
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