Thanks for the video, it's really helpful! I have two questions though: 1. Why do you count the positive charges as conjugated? Would you not only count them if there are atoms or compounds near-by that have lone pairs and could put them where the positive charges are? When only seeing the positive charge I learned it isn't counted, now I'm confused. 2. I found a structure where it says it's not aromatic but I would count double single double single there. It's a ring of five carbons with double bonds on the two lower sides (like around 4:12 when the lone pair is on top). On top there is a double bond and then an oxygen with two free pairs of electrons. Would you consider this structure aromatic or non-aromatic and why? Really hope you can help me with that one. If anyone else here is able to answer the questions feel free to do so!
Thank you for helping me understand but I only have one question .why aren’t you counting the +charge at 9:15 to make it 8 pi electrons resulting to anti aromatic?
Hi... thank you for watching! A +charge on a carbon indicates that it lost its 4th valence electron and therefore now only has 3 bonds (bond to H not shown) using its 3 remaining valence electrons, and no other electrons, so think of the +charged carbon as "void" of pi electrons (so, no electrons to count there), remembering that pi electrons are either lone pairs or the 2 electrons in a double bond. But we do consider the +charge when determining conjugation, because pi electrons can jump to the +charged carbon since there is a "void" of electrons, like there is a "vacancy" for pi electrons to jump into. It also helps to think about a carbon with a -charge; a negatively charged carbon has its 4th valence electron and gains a fifth electron that pairs up with the 4th to make a lone pair [plus the 3 bonds as mentioned above]; that's why we count the -charges as pi electrons (lone pairs). I hope this helps further understand aromaticity and why +charges are included but not counted as pi electrons.
Thank You for the feedback! I will make sure to crank it up a bit on all future videos. I was always afraid I was making it too loud; I appreciate the constructive feedback!
True... cyclooctatetraene predominantly exists as tub-shape due to its irregular bond lengths, however, I intended to show its planar transition state between its tub-shapes which is antiaromatic. I showed its antiaromatic state because that's how they usually show it on DAT or MCAT. If they were to show the 3D tub-shape, then indeed it would be nonaromatic. Thank You for watching!
Impressive explanation mam
so helpful! thank you amreen. there truly is no better trainer thank you!
Thank you very much. You clear all my doubts in a very easiest way.thanks a lot❤❤
I am so glad this helped make it easier! Thank You for watching!
Thank you for yet another helpful video!
very clear thanks
Thanks for the video, it's really helpful! I have two questions though:
1. Why do you count the positive charges as conjugated? Would you not only count them if there are atoms or compounds near-by that have lone pairs and could put them where the positive charges are? When only seeing the positive charge I learned it isn't counted, now I'm confused.
2. I found a structure where it says it's not aromatic but I would count double single double single there. It's a ring of five carbons with double bonds on the two lower sides (like around 4:12 when the lone pair is on top). On top there is a double bond and then an oxygen with two free pairs of electrons. Would you consider this structure aromatic or non-aromatic and why? Really hope you can help me with that one.
If anyone else here is able to answer the questions feel free to do so!
omg.... you made my exact issues with aromatic molecules so much more clear... thanks
I’m glad it cleared aromaticity up for you!
Amazingly simplified! Couldn't explain it better, thank you!!
Thank You!
Thank you for helping me understand but I only have one question .why aren’t you counting the +charge at 9:15 to make it 8 pi electrons resulting to anti aromatic?
Hi... thank you for watching! A +charge on a carbon indicates that it lost its 4th valence electron and therefore now only has 3 bonds (bond to H not shown) using its 3 remaining valence electrons, and no other electrons, so think of the +charged carbon as "void" of pi electrons (so, no electrons to count there), remembering that pi electrons are either lone pairs or the 2 electrons in a double bond. But we do consider the +charge when determining conjugation, because pi electrons can jump to the +charged carbon since there is a "void" of electrons, like there is a "vacancy" for pi electrons to jump into. It also helps to think about a carbon with a -charge; a negatively charged carbon has its 4th valence electron and gains a fifth electron that pairs up with the 4th to make a lone pair [plus the 3 bonds as mentioned above]; that's why we count the -charges as pi electrons (lone pairs). I hope this helps further understand aromaticity and why +charges are included but not counted as pi electrons.
Good question... btw... it's a common question among my students, and hopefully this helps others better understand aromaticity!
Hello, does the "including in counting as double bonds, but not in counting as pi electrons" apply to both positive and negative charges?
It help me so much thank u
Thank u for saving my time
The best video ever ❤❤
I am glad this helped!
you so so amazing! thank you so much for your help :)
I am glad this helped! Thank You for watching!
This is so helpful!
I am glad this was helpful!
God bless that’s all I gotta say
lol... thank you so much... glad this helped!
❤❤. Pls continue making it. Pls make a trick for gmp of hydrocarbons.
I will work on a trick for that. Thank You for watching!
You should increase the volume of the sound
Thank You for the feedback! I will make sure to crank it up a bit on all future videos. I was always afraid I was making it too loud; I appreciate the constructive feedback!
Well cyclooctatetraen isn’t planar in the first place
True... cyclooctatetraene predominantly exists as tub-shape due to its irregular bond lengths, however, I intended to show its planar transition state between its tub-shapes which is antiaromatic. I showed its antiaromatic state because that's how they usually show it on DAT or MCAT. If they were to show the 3D tub-shape, then indeed it would be nonaromatic. Thank You for watching!