how can you know if the sp2 electron pair is in the plane of the ring

[navneetdh]

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in this case the second figure has a nitrogen where its sp2 electron pair is not in the plane of the ring

But how can you tell whether or not the electron pair will be in the plane of the ring or not? I think this is very important also to determine aromaticity of the ring---sometimes when electrons are added or not added for Huckels rule.

Please explain this.

Thank you --i appreciate all your help sincerely.

 

[stiney44]

I believe it depends on if the nitrogen is part of a double bond or not. For example, pyrrole's electrons are in plane with the ring because it is not part of a double bond. In the second and third example the electrons wouldn't count, and I believe all three are aromatic. Someone correct me if I'm wrong.

 

[navneetdh]

I believe it depends on if the nitrogen is part of a double bond or not. For example, pyrrole's electrons are in plane with the ring because it is not part of a double bond. In the second and third example the electrons wouldn't count, and I believe all three are aromatic. Someone correct me if I'm wrong.

no i dont think that is correct--because both for the 1st and the 3rd the electrons ARE in the plane of the ring (according to achiever test 3). Its only the 2nd where they are not.

 

[yumi43]

in this case the second figure has a nitrogen where its sp2 electron pair is not in the plane of the ring

But how can you tell whether or not the electron pair will be in the plane of the ring or not? I think this is very important also to determine aromaticity of the ring---sometimes when electrons are added or not added for Huckels rule.

Please explain this.

Thank you --i appreciate all your help sincerely.

i think if the electron pair is outside the ring (like the 2nd example) you don't include those when you try to figure out aromaticity. if the electrons are inside the ring, like the 1st example, then they count as pi electrons that you'd wanna include. look through examples in any other book to verify this. i believe it's correct.

 

[dennisDDS]

Chad said this and I think it really helped

If the N is doubled bonded you don't count the lone pair

If the N is not double bonded you count the lone pair

hope this helps

 

[solstice]

Chad said this and I think it really helped

If the N is doubled bonded you don't count the lone pair

If the N is not double bonded you count the lone pair

hope this helps

Which Chad's Video talks about this. I need to review it; since I have problem with these too.

 

[dennisDDS]

Which Chad's Video talks about this. I need to review it; since I have problem with these too.

the condensed video one..

http://www.coursesaver.com/showthread.php?t=2791

 

[navneetdh]

Chad said this and I think it really helped

If the N is doubled bonded you don't count the lone pair

If the N is not double bonded you count the lone pair

hope this helps

yes but achiever answers didnt agree-----i dont know if the rule you stated is only for countic electrons in huckels rule---but achiever question was are the sp2 electrons in the PLANE OF THE RING----and teh answer was that 1 and 3 have their electrons in the plane of the ring but 2 does not---and I have no clue why.

 

[dennisDDS]

yes but achiever answers didnt agree-----i dont know if the rule you stated is only for countic electrons in huckels rule---but achiever question was are the sp2 electrons in the PLANE OF THE RING----and teh answer was that 1 and 3 have their electrons in the plane of the ring but 2 does not---and I have no clue why.

Wait so the first one you have 2 pi bonds and the lone pair electrons so
that gives you 6 electrons total.

4(1) +2 = 6 so aromatic

And for the third one you have 3 pi bonds so 6 electrons total..

so 4(1) + 2 = 6 so aromatic once again..

as you can see there are three substituents bonded to the n in the third one so it is in fact sp^2 hybrid.

 

[navneetdh]

Wait so the first one you have 2 pi bonds and the lone pair electrons so
that gives you 6 electrons total.

4(1) +2 = 6 so aromatic

And for the third one you have 3 pi bonds so 6 electrons total..

so 4(1) + 2 = 6 so aromatic once again..

as you can see there are three substituents bonded to the n in the third one so it is in fact sp^2 if the lone pair was counted in number three it would be sp3 and not aromtic so the answer must be wrong. because number three is in fact aromatic..

Sorry to keep contradicting you---but I dont think achiever was talking about aromatic or not. That was just my second question.

Achiever gave these 3 figures and asked which one IS in the PLANE of the ring. And the answer was 1 and 3. The electrons of 2 are not in the plane of the ring.

1. Now i have trouble with know ---how do you look at that and know if they are or are not in the plane of the ring.

2. Also according to the rule you stated---if N is double bonded you count the electrons---well then in the 2nd figure N is double bonded and you are suppose to count the electrons---so 2nd will be aromatic too right?

 

[dennisDDS]

Sorry to keep contradicting you---but I dont think achiever was talking about aromatic or not. That was just my second question.

Achiever gave these 3 figures and asked which one IS in the PLANE of the ring. And the answer was 1 and 3. The electrons of 2 are not in the plane of the ring.

1. Now i have trouble with know ---how do you look at that and know if they are or are not in the plane of the ring.

2. Also according to the rule you stated---if N is double bonded you count the electrons---well then in the 2nd figure N is double bonded and you are suppose to count the electrons---so 2nd will be aromatic too right?

ohh ok.. sorry.. the last thing I wrote didn't come out right so i deleted it..
yeah I dont know why the 2nd one would be out of plane and the 3rd one would be in plane that really doesnt make sense..

but as far as the second one you really have to start worrying about is the molecule planar which is key in other words is it flat? a 12 carbon ring is really huge so I doubt that it could take a flat shape.. so i would say no even thought it follows huckels rule..

 

[dennisDDS]

sorry one more thing

if N is double bonded you DONT count the lone pair..
If N is NOT double bonded you COUNT the lone pair..

Hopefully I make sense.. gL

 

[prsndwg]

sorry one more thing

if N is double bonded you DONT count the lone pair..
If N is NOT double bonded you COUNT the lone pair..

Hopefully I make sense.. gL

this way the middle one will also be aromatic!

 

[prsndwg]

Navneetdh, how did you do on the TS on that last achiever.. it sure kick my booty majorly!!

 

[jorasic]

Hey this thread is a mess, but I'll try to contribute!

I think you're confused or the achiever is unclear. Being part of the plane of the ring, it should refer to being part of the planar configuration of the conjugated ring? There are 2 places a lone pair like that can occupy. The plane of the ring, or in the p-orbital. pi bond overlap.
The lone pair is part of the ring and NOT part of the pi-bond/ p-orbital overlap.

Look at figure 2.
No formal charges, so no hyrdogen on it. The hybridization is sp2.
All sp2 hybridization will take the 1 s orbital and 3 p orbitals and translare them into 4 oribtals. As a result, there will be 3 sp2 orbitals of course, and the 4th oribtal will be the left over p(z) orbital that WASN'T hybridized. So its just there hoping to be filled.

p> _ _ _
s > _

turns into

sp2> _ _ _ + _

The C-N and C-N bonds will occupy the first 2 sp2 orbitals.
The lone pair will fill the 3rd sp2 orbital and there's nothing in the p-oribtal; its empty. So it won't contribute to the pi electon count.
Only if all 3 sp2 orbitals are filled can your lone pair be put in the extra p-orbital of an sp2 atom. So say it had another lone pair on that. Then that will fill the open p-orbital since all 3 sp2 orbitals are already filled by the C-N, C-N, and first lone pair. So in figure 2, the lone pair IS PART OF THE PLANAR ring, but not part of the pi bond overlapping. I think the Achiever's explanations might be unclear? You have 10 pi electrons which follows 4n+2.

Usually inside lone pair will be part of the ring, but don't go by that.
Don't go by the double bond to Nitrogen thing either. Use my logic, its the truth 🙂 Or look it up, cus this is how textbooks explain it.

 

[navneetdh]

Hey this thread is a mess, but I'll try to contribute!

I think you're confused or the achiever is unclear. Being part of the plane of the ring, it should refer to being part of the planar configuration of the conjugated ring? There are 2 places a lone pair like that can occupy. The plane of the ring, or in the p-orbital. pi bond overlap.
The lone pair is part of the ring and NOT part of the pi-bond/ p-orbital overlap.

Look at figure 2.
No formal charges, so no hyrdogen on it. The hybridization is sp2.
All sp2 hybridization will take the 1 s orbital and 3 p orbitals and translare them into 4 oribtals. As a result, there will be 3 sp2 orbitals of course, and the 4th oribtal will be the left over p(z) orbital that WASN'T hybridized. So its just there hoping to be filled.

p> _ _ _
s > _

turns into

sp2> _ _ _ + _

The C-N and C-N bonds will occupy the first 2 sp2 orbitals.
The lone pair will fill the 3rd sp2 orbital and there's nothing in the p-oribtal; its empty. So it won't contribute to the pi electon count.
Only if all 3 sp2 orbitals are filled can your lone pair be put in the extra p-orbital of an sp2 atom. So say it had another lone pair on that. Then that will fill the open p-orbital since all 3 sp2 orbitals are already filled by the C-N, C-N, and first lone pair. So in figure 2, the lone pair IS PART OF THE PLANAR ring, but not part of the pi bond overlapping. I think the Achiever's explanations might be unclear? You have 10 pi electrons which follows 4n+2.

Usually inside lone pair will be part of the ring, but don't go by that.
Don't go by the double bond to Nitrogen thing either. Use my logic, its the truth 🙂 Or look it up, cus this is how textbooks explain it.

ok thank you--that is definitely making more sense now pheww!!
👍🙂

 

[SaraForDental]

So confusing.. Naveendth on the first post you said the second figure says the electrons are NOT PART OF THE RING and jorsaic is saying it IS PART OF THE RING and in the end you agreed.. so is the achiever wrong? Im so lost..🙁

 

[navneetdh]

So confusing.. Naveendth on the first post you said the second figure says the electrons are NOT PART OF THE RING and jorsaic is saying it IS PART OF THE RING and in the end you agreed.. so is the achiever wrong? Im so lost..🙁

yeah the achiever explanation did not make much sense at all-----what jorsaic said--I can picture it--so I will stick to it----rest I hope I do not get a problem as complex as this

Also--the question in the achiever wasnt asking for aromaticity----it just wanted to know if electrons are in plane of the ring or not.

Now---if the nitrogen is sp2 hybridized then the electrons should be part of the ring since the electronic geometry is trigonal planar. This is the case for both the 2nd and the 3rd picture.

In the first picture nitrogen is sp3 and therefore has tetrahedral electronic geometry---so the electron pair would preferentially be above the rest and be perpendicular to the plane of teh ring.

I think achiever must be wrong---the answer should have been that 1=not in plane of ring
2=in the plane of ring
3=in the plane of the ring.

Also from what I understand---i may be wrong, plz correct me if I am----that the electrons that are perpendicular to the plane of the ring are the ones that actually count for Huckels rule. RIght? so since 2 and 3 have electron pair in the plane of the ring---NOT perpendicular---they wont count for huckels rule. And so without those electrons the system is aromatic for both 2 and 3.

For 1 the electrons are PERPENDICULAR to the plane of the ring--remember N is tetrahedral electronic geometry there (sp3)---so you WILL count them for Huckels rule---the system become aromatic if I include those electrons.

This gives 1,2,3 all aromatic.

I hope I understood this all properly and explained it to you fine. I hope this helps.