From what I’ve understood, wouldn’t the second to last molecule be non-aromatic since a molecule would rather remain non-aromatic rather than switch to anti-aromatic ?
Very helpful.... But sir we can also consider that delocalised lone pair are sp2 and localised lone pair sp3 and stays in hybrid orbital... So whenever we see any lone pair participating in resonance we can say it is sp2..
I find that It's unnecessary to have it subtracted by 2 and find the n value. Simply check if the pi electron value is 4n or 4n+2. If it's the former then it's Antiaromatic, if it's latter then Aromatic. But very very helpful. Thanks!
Great video, thanks a bunch. Quick question. My prof heavily emphasized planarity in his lecture on this topic. Any comments on planarity and it’s importance?
Hi Dominic, yes planarity is very important and required for aromaticity. The reason planar ring systems are so vital is that they allow for p-orbital overlap which leads to proper conjugation of pi electrons. When a ring fails to meet planar standards, it is classified as non-aromatic (not anti). An example of this is 1,3,5,7-Cyclooctatetraene. One would normally expect anti, but it is non-planar.
I find your videos very helpful, but your intro music is very loud and hard on the ears especially with headphones on. Can you please consider changing it for future videos? Thanks for all your effort man!
The lone pair is in resonance with the pi bonds. In order to participate in the delocalization of electrons, the nitrogen must keep a p-orbital for overlap purposes and therefore can only be sp2.
The lone pair is in resonance with the pi bonds. In order to participate in the delocalization of electrons, the nitrogen must keep a p-orbital for overlap purposes and therefore can only be sp2.
The AMOUNT OF CLARITY is insane!!!!! Thank you so much ❤
After watching several videos without understanding,a planner compound, you have finally made it for me❤
Wow after watching a bunch of videos, your video made the most sense. Thank you!!
wow thank you!! After the whole semester this finally makes sense before my final!!
Same situation
I was not able to understand what's the hybridisation of the charged carbon atoms but now I do thanks a lot
Best video on aromaticity. Made me understand it 100%.
highly recommend this video!!! I can do my homework problems now. Thank you
at 12:23 it will be 3/2
👍🏼
5th compound n = 3/2
You are amazing !!! Final is on monday and you made this 10000 times easier. Thank u
I can't explain how good your lectures are.❤❤
Your explanation of the pi bond in the lone pairs 1 C away from a pi orbital just saved my exam grade lol
Wonderfully explained! Cleared my concepts about aromaticity👌
thank you so much, i understand everything now after watching this
Thank you! I was super confused until I watched your videos.
You are literally one of the best organic chem teachers I've known
Thank you for the high compliment!
very very amazing and clear explanation
From what I’ve understood, wouldn’t the second to last molecule be non-aromatic since a molecule would rather remain non-aromatic rather than switch to anti-aromatic ?
that’s what i think too lol
Very helpful.... But sir we can also consider that delocalised lone pair are sp2 and localised lone pair sp3 and stays in hybrid orbital... So whenever we see any lone pair participating in resonance we can say it is sp2..
Thank you so much for complete detailed information
Big thanks for you, Sir.. Your videos help me a lot, especially for my lecture.. Once again thank you so much. :)
You are most welcome
I find that It's unnecessary to have it subtracted by 2 and find the n value. Simply check if the pi electron value is 4n or 4n+2. If it's the former then it's Antiaromatic, if it's latter then Aromatic. But very very helpful. Thanks!
it was really helpful .Keep up ur momentum bro.
Thanks!
Great video!! on point❤
thanks so much! Hope you guys are able to push out more Orgo 2 videos haha
please how do i easily know that a ring structure is going to be conjugated?
Thank youuuuu ❤ I’m finally cleared
great explanation !!
thank you very much it was really helpful and clear
thanks a lot for an easy explanation
Great video, thanks a bunch. Quick question. My prof heavily emphasized planarity in his lecture on this topic. Any comments on planarity and it’s importance?
Hi Dominic, yes planarity is very important and required for aromaticity. The reason planar ring systems are so vital is that they allow for p-orbital overlap which leads to proper conjugation of pi electrons. When a ring fails to meet planar standards, it is classified as non-aromatic (not anti). An example of this is 1,3,5,7-Cyclooctatetraene. One would normally expect anti, but it is non-planar.
How know that aromatic is planar?
In furan how to jutch oxygen is sp2 hybridised ????
very helpful thank you !
BEST VIDEO !
thank you for this!
Thanks very u are d best
We need part two on the,same topic with different structures. You are a GOD
Thank you
Thanks 🎉
goated channel
Organic is bussin
thank youuuu so much
You're welcome!
Thank you!
Nice video
Thanks
Makes Sense
Good one
You are using the same 4n + 2 formula for both aromatic and anti aromatic
4n +2 = Pi electrons is the formula for determining aromaticity (aromatic or anti)
AMAZING
Good am now getting something
Why pyridine is marked not aromatic???
Pyridine is aromatic, it was also marked as such in the video!
I find your videos very helpful, but your intro music is very loud and hard on the ears especially with headphones on. Can you please consider changing it for future videos? Thanks for all your effort man!
BLESS
How the nitrogen atom is SP2 hybridized in last compound.
The lone pair is in resonance with the pi bonds. In order to participate in the delocalization of electrons, the nitrogen must keep a p-orbital for overlap purposes and therefore can only be sp2.
Good
4n+2=8 4n=6 6/4=n n=1.5
Value of n in thio pyrane not correct
i love u man
good vid
شكرآ
4n=6 n=3/2
Math was done wrong at 12:33
awesome
In last question how nitrogen is sp2 hybridised here ??plz clear confusion
The lone pair is in resonance with the pi bonds. In order to participate in the delocalization of electrons, the nitrogen must keep a p-orbital for overlap purposes and therefore can only be sp2.
Best
algebra dude!
you're great
Thanks for this 🙏
No problem 👍