Trouble with counting Huckle's number of e-

[Little Etoile]

---

Members don't see this ad.

Can someone explain how you count the number of delocalized electrons in a ring? I can count the double bond pi bond electrons just fine, but I get confused when there is a lone pair (or two) on a non-carbon ring atom (e.g. O or N). It seems like sometimes they are counted toward the 2n +2 total, sometimes they are not. How do I know if the total is 2n +2?

As always, thanks in advance.

 

[sleepy425]

ok, here's how I teach it in my TPR class. This isn't how I would teach it in a real ochem class, but for the mcat this is an easy way to count huckel numbers:

Three steps:

1. Count every double bond in the ring as 2 electrons

2. If an atom has more than one lone pair, count exactly 1 of those lone pairs.

3. If an atom has only one lone pair and...
a) it is double bonded to another atom, don't count the lone pair
b) it is only single bonded to other atoms, count the lone pair

So then I give a few examples (I'm just gonna use the name, google each one and you can see the structure):

Pyridine:
three double bonds: 6 electrons
the nitrogen has a lone pair, but the nitrogen is also part of a double bond so don't count the lone pair: 0 electrons
total: 6 electrons and fulfills all the other conditions for aromaticity too

Furan:
two double bonds: 4 electrons
the oxygen has two lone pairs so count only one: 2 electrons
total: 6 electrons and fulfills all the other conditions for aromaticity too

Pyrrole:
two double bonds: 4 electrons
the nitrogen has one lone pair and the nitrogen is not double bonded to anything. that means you count the lone pair: 2 electrons
total: 6 electrons and fulfills all the other conditions for aromaticity too

I hope that helps. The real explanation is that you only count lone pairs that are perpendicular to the plane of the ring, the ones that are in p orbitals. if you have a nitrogen double bonded to something, the electrons in the double bond are perpendicular to the plane of the ring (they have to be otherwise they wouldn't be able to form the double bond). that means that the lone pair can't be perpendicular to the plane of the ring because it would occupy the same space as the electrons in the double bond. that's all there is to it. for an oxygen like in furan, one electron pair can be perpendicular to the plane, but the other can't be because it would occupy the same space.

 

[Soccerdoc11]

hey sleepy can you expand on this on when to see if the molecule will be basic or not (if the lone pair doesn't contribute to the ring then it can donate and be basic)... I am a little rusty on it, I'm sure you can explain it.. Thanks a ton..

 

[sleepy425]

hey sleepy can you expand on this on when to see if the molecule will be basic or not (if the lone pair doesn't contribute to the ring then it can donate and be basic)... I am a little rusty on it, I'm sure you can explain it.. Thanks a ton..

that's a really excellent point. i think you've got it down. it actually goes both ways.

so it's really bad to disrupt aromaticity. if you protonate the molecule by donating an electron pair that is contributing to aromaticity, it will now be part of a sigma bond and cannot contribute to aromaticity anymore. this means you've lost aromaticity, which is energetically unfavorable. so that means that the aromatic molecule won't be very basic because it won't be willing to donate an electron pair.

but in the case of pyridine, the electron pair on the nitrogen does not contribute to aromaticity so it can be protonated somewhat easily. if you remember from ochem, there's a reagent called PCC, which is used to oxidize primary alcohols to aldehydes. the P stands for pyridinium, which is the protonated form of pyridine, so that's just an example to show you that pyridinium does exist and thus pyridine is decently basic.

ok, I said it goes both ways. look up cyclopentadiene. cyclopentadiene is a nonaromatic molecule (not antiaromatic, nonaromatic because it's not fully conjugated). cyclopentadiene is much much more acidic than, say, 1,4-pentadiene (which is the same molecule except not cyclic). Why? Because if you pull off a proton from cyclopentadiene, it becomes aromatic. If you pull off a proton from 1,4-pentadiene, it's not aromatic so it's like pulling a proton off of an alkane (maybe a little easier, but still pretty hard, like pKa in the 40s or upper 30s). The pKa of cyclopentadiene is like 15, which means that it sooo much more acidic. So by pulliing a proton off, it restores aromaticity which is very favorable, so deprotonation occurs more readily so the pka is lower (more acidic).

 

[Soccerdoc11]

And it will then have 6 pi electrons making it aromatic (according to your rules) great thanks for clearing that up..👍

 

[Farcus]

holy crap great great explanation, this'll come in handy when I take ochem2, but another question I been baffled on what they mean conjugated... I mean do they mean like every other one? like double,single,double bond? Again great help on the lone pair thing, would this work for all compound or asking to see if its aromatic or not or just certain ones?

 

[prettyslick]

Can someone explain how you count the number of delocalized electrons in a ring? I can count the double bond pi bond electrons just fine, but I get confused when there is a lone pair (or two) on a non-carbon ring atom (e.g. O or N). It seems like sometimes they are counted toward the 2n +2 total, sometimes they are not. How do I know if the total is 2n +2?

As always, thanks in advance.

2n + 2????

I thought Huckel's rule was 4n + 2????

 

[sleepy425]

holy crap great great explanation, this'll come in handy when I take ochem2, but another question I been baffled on what they mean conjugated... I mean do they mean like every other one? like double,single,double bond? Again great help on the lone pair thing, would this work for all compound or asking to see if its aromatic or not or just certain ones?

well, i guess best explanation of conjugation is that all atoms in the ring must be sp2 hybridized (or sp hybridized i guess, but that usually doesn't apply in a ring b/c triple bonds are never seen in small rings because of the strain you'd create). the problem with this explanation is that a lot of times, it's not always apparent what the hybridization is. for example, let's take my example of pyridine and then pyrrole (again, if you could look them up on wikipedia it'd probably help). for pyridine, it's very obvious that all the atoms are sp2 hybridized.

for pyrrole, it's not so obvious. the carbons are clearly sp2 hybridized. the nitrogen would normally be considered sp3 hybridized, right? it's got three single bonds and one lone pair, so that's normally sp3. if you had ammonia or something, you would say it's sp3 hybridized. but for pyrrole it's not. the reason it's not is because you can draw a resonance contributor into the ring. so if you look at a resonance structure of it, you can draw at least two that have a double bond to the nitrogen and two single bonds to the nitrogen. so that'd be sp2.

i'm not that good at explaining it, but i hope you get the idea.

also, to answer your other question, the approach to determining whether or not lone pairs contribute to huckel's rule applies to all aromatic compounds.

so when checking for aromaticity, you go through the following checklist:
1. is it cyclic? determine like in first grade
a) if yes, move to step 2
b) if no, nonaromatic
2. is it fully conjugated? determine as said above
a) if yes, move to step 3
b) if no, nonaromatic
3. does it obey huckel's rule? determine as I described earlier
a) if yes, it's aromatic
b) if no, nonaromatic
1) check for antiaromaticity-does it obey the 4n rule?

Technically there's a fourth check: is it planar? Generally, for small rings, they will always be planar. Unless they explicitly tell you that a molecule is nonplanar, assume planarity.

sorry i couldn't be more help tho with the conjugated thing

 

[sleepy425]

2n + 2????

I thought Huckel's rule was 4n + 2????

Oh, it's 4n+2. I think that was just a typo. I mean, people don't really think of 4n+2 when they do it. They think of 2, 6, 10. and for antiaromatic they think of 4, 8. but you're absolutely right, 4n+2

 

[prettyslick]

Oh, it's 4n+2. I think that was just a typo. I mean, people don't really think of 4n+2 when they do it. They think of 2, 6, 10. and for antiaromatic they think of 4, 8. but you're absolutely right, 4n+2

I figured it was probably a typo but it was bothering me so I had to clarify!! haha

no confusion now...

I think OP was confusing 2n+2 rule for number of Hydrogens in alkanes where n = # of carbons ex.) C6H14 (2n+2 rule)

 

[Little Etoile]

2n + 2????

I thought Huckel's rule was 4n + 2????

Oops, sorry. Yeah. Totally a typo!