Aromatic, Non-Aromatic, or Antiaromatic? Some Practice Problems
Last updated: October 16th, 2022 |
Aromatic vs Antiaromatic vs Non Aromatic Practice Exercises
Our last post in this series on aromaticity went through the 4 conditions a molecule must fulfill in order to be aromatic.
- First, it must be cyclic
- Second, every atom around the ring must have an available p-orbital
- Third, the number of electrons in the pi system must be 2, 6, 10, 14, 18, or a higher number in the set that increases from 18 in increments of 4 (22, 26, 30…. etc). We usually abbreviate this as [4n+2] pi electrons.
- Fourth, the molecule must be flat.
In that post we tried to explain what each of those rules meant – so if any of these individual items seem unclear to you, it might be a good idea to go back to that post.
In this post,we’re going to apply these 4 rules with some practice examples.
Table of Contents
- A System For Determining If Molecules Are Aromatic: Build A Table
- The Benzene Anion
- The Cyclopentadiene Cation
- Pyrrole Conjugate Acid
- Cyclobutene Di-Anion
- Pyrylium Ion
- Summary: Aromaticity Practice Exercises
1. A System For Determining If Molecules Are Aromatic: Build A Table
Here’s my tip for those of you getting started on trying to answer the question “Is this aromatic?” . Make a table.
Tables are great for organizing information. They also serve as a built-in checklist.
For instance, here’s an image from a tutoring session I did with a student where we determined whether or not several molecules were aromatic:
You’ll notice that my checklist doesn’t usually bother having a column for “flatness” because it’s generally assumed. (You need to know the few exceptions that come up – we covered that last time).
Today, we’re going to look at the following 13 molecules and try to determine if they are aromatic or not:
Let’s get started!
The easiest example to start with is benzene, and it demonstrates how to use the table. It’s cyclic, conjugated, has 3 pi bonds and those pi bonds are all in the pi system. It has zero lone pairs that contribute to aromaticity. Therefore it has (3 × 2) + 0 = 6 pi electrons. We’re assuming it’s flat (it is). Checking off all the boxes, we can say that it’s aromatic.
3. The Benzene Anion
If you remove a proton from benzene, you get the benzene anion. Like benzene, it’s cyclic, conjugated, had 3 pi bonds and those pi bonds are all in the pi system. It’s tempting to look at that lone pair and to think that it might contribute to the pi system as well, giving 8 pi electrons total. However, that lone pair is in the plane of the molecule (along with the other C-H bonds) and thus can’t overlap with the p-orbitals. Therefore, for the purposes of determining aromaticity, we can ignore the lone pair.
4. The Cyclopentadiene Cation
The cyclopentadiene cation below is cyclic and conjugated (that positive charge represents a carbocation with an empty p-orbital). There are two pi bonds and zero lone pairs to contribute to the pi system. This gives us 4 total pi electrons, which is not a Hückel number. Therefore it isn’t aromatic.* [Note 1]
Pyrrole is cyclic and conjugated (that lone pair on nitrogen can contribute to the pi-system). There are two pi bonds and one lone pair of electrons that contribute to the pi system. This gives us 6 total pi electrons, which is a Huckel number (i.e. satisfies 4n+2). Therefore it’s aromatic.
6. Pyrrole Conjugate Acid
If pyrrole is protonated on the nitrogen, the lone pair can no longer participate in the pi-system. Therefore the molecule drawn below is not aromatic.
Pyridine is cyclic, conjugated, and has three pi bonds. It’s a lot like the benzene anion in that the lone pair looks like it might contribute to the pi system, but in fact is in the plane of the ring (along with the C-H bonds) and thus cannot contribute (the nitrogen is already contributing a p-orbital towards the pi-system – note that it’s drawn as participating in a double bond with an adjacent carbon). Therefore we can ignore the lone pair for the purposes of aromaticity and there is a total of six pi electrons, which is a Huckel number and the molecule is aromatic.
Thiophene, the sulfur analog of furan, is cyclic and conjugated all the way around the ring. It has two pi bonds. What’s interesting about thiophene (and furan) is that although there is an atom bearing two lone pairs in the ring, we can only count one of those lone pairs toward the pi-system. [Each atom can contribute a maximum of one orbital and two electrons towards aromaticity]. The other lone pair is in the plane of the ring, much like the lone pair on the nitrogen of pyridine, above. Therefore thiophene has six pi electrons total, which is a Huckel number, and thiophene is aromatic.
Cyclobutadiene is cyclic and conjugated. There are two pi bonds and zero contributing lone pairs. Two pi bonds gives us a total of 4 pi electrons, which is not a Huckel number. Therefore it is not aromatic, just like the cyclopentadienyl cation, above. [Note 1]
10. Cyclobutene Di-Anion
Cyclobutadiene is not aromatic. If we somehow pump two electrons into cyclobutadiene, however, then the situation changes considerably. The cyclobutadiene dianion is cyclic and conjugated. It has a single pi bond, and now two carbons bearing lone pairs which can contribute to the pi system, giving us a total of six pi electrons. This is a Huckel number and thus the cyclobutadiene di-anion is aromatic!
Naphthalene has two rings and is thus a bicyclic compound. It is conjugated around the perimeter of the rings and there are a total of 5 pi bonds and zero lone pairs, giving us 10 pi electrons total. 10 is a Huckel number (satisfying 4n+2 for n=2) and naphthalene is an aromatic molecule.
12. Pyrylium Ion
The pyrylium ion is a six-membered ring that, like benzene, has three pi bonds. It also bears a lone pair on the oxygen. Like pyridine and the benzene anion, this lone pair is actually in an orbital at right angles to the pi system so it does not in fact count towards aromaticity. Therefore the molecule has only 6 electrons in the pi system and is in fact aromatic.
Indole is a bicylic molecule that looks like a molecule of benzene fused to a molecule of pyrrole. It is conjugated; every atom around the perimeter of the two rings participates in the pi system. It has 4 pi bonds and a single pair of electrons on the nitrogen that participates in the pi system, giving 10 pi electrons in total. This is a Huckel number and indole is in fact aromatic.
Azulene is another example of a bicyclic molecule, like naphthalene and indole, above. It’s conjugated all around the perimeter of the pi system. There are five total pi bonds and zero lone pair electrons, giving a total of 10 pi electrons, which is a Huckel number. And azulene is, in fact, aromatic.
If it’s hard to visualize azulene as aromatic, it might be helpful to draw a resonance form where both rings look like they have aromatic components. For example, in the resonance form below, you can think of azulene as being composed of the cycloheptatriene cation (aromatic) fused to the cyclopentadiene anion (also aromatic).
Note that one ring bears a negative charge and the other one bears a positive charge. It turns out that this resonance form has a significant contribution to the hybrid, since azulene itself has a substantial dipole moment (1.08 D). Naphthalene, in contrast, completely lacks a dipole moment.
15. Summary: Aromaticity Practice Exercises
The whole point of using a table like the one above is to organize your thoughts. Like a set of training wheels, if you do problems like this often enough, you’ll start to find that you don’t need it.
The most common part where students get tripped up is in counting the number of pi electrons. Just remember that each atom can at most contribute one p-orbital towards the pi system, and that the p-orbital can have (at most) two pi electrons.
Are there any other examples you’re not sure about that we should apply this system towards? Leave a comment!
Note 1. there are two molecules in particular in the list above which will deserve further discussion. The cyclopentadiene cation and cyclobutadiene were each described, above, as “not aromatic”, which is technically true.
Just as we saw that benzene is unusually stable compared to the theoretical “cyclohexatriene” , cyclobutadiene and the cyclopentadiene cation are interesting in that they are unusually unstable. This is a property called “antiaromaticity”, which we will describe in some detail in the next post.
Antiaromaticity and Antiaromatic Compounds
Thanks to Matt Struble for assistance with this post.
38 thoughts on “Aromatic, Non-Aromatic, or Antiaromatic? Some Practice Problems”
Just a typo- “fused to the cyclopentadiene cation” kindly change it to anion!
Otherwise very helpful post!!!
Fixed. Thanks for telling me!
In furan,are the the two C-O bonds at right angles because they are covalent bond formed involving py and pz orbitals of oxygen.
Also does the one lone pair of oxygen which is counted in pi electrons shows resonance because it is in px orbital which is perpendicular to py and pz and the molecular plane,while the other lone pair in s orbital does not involve in resonance because it lies in the molecular plane?
The old enough paper
Aromatic molecular zones and fragments, L. Tarko, ARKIVOC, 2008, Part xi, p. 24 – 45
offers a very different answer for title question.
Above #1, #3 and #4 conditions, actually based on Huckel model, are long time ago obsolete.
That’s interesting. Thank you Tarko.
Loved this one. Helped a lot in understanding aromaticity. Still having problems in multiple ring questions what if one ring is aromatic and the other is antiaromatic?
Cyclobutadiene dianion has 6 pi electrons. However, there is no conjugation covering ALL the atoms. Is this not a violation of Huckel rule?
Sure there is. All four carbons are conjugated through their p orbitals. Two of them are half-filled, and two of them are completely full.
Sooooo helpful and clear. Thanks a million!
I have probably appreciated you before as well. I love you so much Sir James. You make organic chemistry so so so so so much easy. I thoroughly enjoy your lessons. God bless you more and more. Love, from Mumbai, India.
Thank you Samina – glad you find it helpful!
Regarding naphthalene, I was confused as to why it’s considered an aromatic compound? Isn’t there strain due to the repulsion of hydrogen atoms on the shared carbon atoms, thus rendering the molecule not flat? I understand that if a bridge was present, it makes the molecular compound flat again, thus fulfilling a criteria of aromaticity. Sorry, just a bit confused.
There aren’t hydrogen atoms on the shared carbon atoms in naphthalene – if you count the bonds, each carbon has 4.
Hi James, from down under in Oz. Thank you very much for your clear and concise explanations. My memory is not good, so understanding the mechanisms and the logic behind the chemistry really helps and saves taxing the memory.
I used your table above to answer questions like this and it was very helpful. However, one problem that I can’t find that they gave us was using thiophene C4H4S which you have in your example above which is aromatic. However, the question asked for the thiophene radical C4HS+ if that was aromatic? My thoughts were probably not because it would destabilize the aromaticity and even though it is five electrons close to Huckel’s number it is not a Huckel number.
Hi Peter – The thiophene radical will not be aromatic. It will be a radical cation. Thiophene has six pi electrons. Remove an electron from this and you have five electrons.
The same is true for the benzene radical (not to be confused with the phenyl or benzyl radical!)
Hope this helps! James
Cyclohexenyl dianion. … whether it is aromatic or not??? The two anions are in 1St and 4th carbon.. double bond is between 2nd and 3rd carbon….
Are any of the carbons in the ring sp3 hybridized? If so, then it’s not aromatic….
Hey James, for the Pyrrole conj. acid example, I was wondering how that isn’t conjugated. the reason why there is a positive charge to nitrogen is that it has 4 electrons when it needs 5 for the valence (formal charge)? My question is wouldn’t that positive charge be delocalized by moving a double bond adjacent to it, hence being conjugated??
There is no way that the formal charge on nitrogen can be removed through pushing the other double bonds around, since the nitrogen has no available p orbitals that can be conjugated with the pi-system. Nitrogen in this case has all 8 of its electrons bound up in single bonds, and we know that resonance forms never involve breaking single bonds, right?
Is the conjugate base of Pyrrole aromatic? It does seem to have 6 pi electrons but I read somewhere that the Pyrrole anion is unstable?
Yes it is. The N-H bond is in the plane of the ring along with the C-H bonds, and the lone pair liberated by deprotonating nitrogen does not affect aromaticity since it is at right angles to the pi system.
How do we determine aromaticity of bicyclo compounds high , one ring is aromatic (benzene ) and the other ring is a non planar cyclohexene ring ?
I am wondering about compounds that contain an aromatic system. Would the entire molecule be aromatic? Eg benzaldehyde
Just the benzene ring. The aldehyde is not considered part of the aromatic system.
cyclic C3H3 Free radical->Aromatic or non-Aromatic?
Odd number of pi electrons –> not aromatic
cycloheptatrienyl anion…non-aromatic or antiaromatic?
Count the pi electrons and that should tell you!
I can’t understand what is aromatic and also not give any example in simple way
Is toluene aromatic because its ring meets the rules, or does the methyl substituent not being planar make it non-aromatic?
Because the ring meets the rules. : – )
The methyl group has no impact on the aromaticity. Aromaticity is determined by the pi-electrons (i.e. the electrons in the double bonds).
why is ”Pyrrole Conjugate Acid” not Aromatic? I assume, it has an empty p-orbital. And with this empty p-orbital, it is possible to build a conjugated phi orbital system with the rest existing phi-bonds.
Thanks in advance!
It’s not conjugated all the way around the ring. If you look at the nitrogen you’ll see that it is sp3 hybridized (attached to four sigma bonds). If it had an empty p orbital that would give it 5 orbitals which violates the octet rule!
What will be the aromaticity of a compound having both an aromatic and an antiaromatic ring? Eg- phenyl cyclobutadiene
The molecule does not exist.
However if you look at azulene, it’s possible to push the arrows such that you get a cyclopentadienyl cation and a heptatrienyl anion, each of which are (in theory) anti aromatic. However based on the dipole moment the contribution of this resonance form is zero
I found a source saying cyclobutan-1,2,3-trione is aromatic .Is it correct? If yes please explain how?