The other night a student came to me with a question about aromaticity.

“There’s one thing I don’t get”, she said. “They say a molecule has to have 4n+2 electrons. *How do you find n*?”

Teachable moment!

“n” is **not** a characteristic of the molecule! Let me explain.

In order for a molecule to be aromatic, it has to have the following characteristics:

- It must be cyclic
- It must be conjugated (i.e. all atoms around the ring must be able to participate in π-bonding through resonance)
- It must be flat

And, it must have a certain number of π-electrons. This is known as Huckel’s rule. The number of π electrons must equal one of the numbers in this series:

2, 6, 10, 14, 18….and so on. For example, we can find aromatic molecules with 2 pi electrons, 6 pi electrons, 10 pi electrons, 14 pi electrons, 18 pi electrons, and so on. But we have never found aromatic molecules with 0, 1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16, 17 (and so on) pi electrons. Those numbers are not in the series.

To reprise: the number of pi electrons in an aromatic molecule will always be found in the series** [2, 6, 10, 14, 18 …and so on] **

However, there has to be a better way of expressing it than [2, 6, 10, 14, 18… and so on”]

There is! This is where we use algebra. **this is where n is going to come in – we are going to use math (algebra) to replace “2, 6, 10, 14, 18… and so on” with a condensed formula. **

2, 6, 10, 14, 18… is an example of an algebraic series.

Algebraically, you can express this with the formula (4n +2), where n is a natural number (0, 1, 2, 3…)

Let’s plug in different values of n (we will put **n** in **bold****)**

For the formula 4n + 2

For **n = 0**, we get (4 ×** 0** + 2) = **2**

For **n = 1**, we get (4 × **1** + 2) = **6**

For **n = 2**, we get (4 × **2** + 2) = **10**

For **n = 3**, we get (4 ×** 3** + 2) = **14**

For **n = 4**, we get (4 ×** 4** + 2) = **18**

We can keep going, but do you get the idea?

By using the formula [4n +2], we are expressing the same idea as [2, 6, 10, 14, 18… and so on] but it is a **lot** more precise.

**So “n” comes from algebra, NOT from chemistry. **

Molecules that have the 3 characteristics listed above (cyclic, conjugated, flat) **and **have this number of π electrons [4n +2] will be aromatic. The letter “n” is **not **a characteristic of the molecule!

[Thanks to commenters Shawn and Egon Willenhagen for suggested edits and to Jamey for a correction]

### Related Posts:

Tagged as:
4n+2,
aromaticity,
benzene,
huckel's rule,
pyridine

{ 41 comments… read them below or add one }

I find some students dissatisfied with the idea that n doesn’t actually refer to anything in the molecule. I don’t know what else to tell them. It seems like they can still apply the rule though.

Well, it’s not so much that it doesn’t refer to anything… the original description is just missing an implied “…where n can be any natural number”

Thanks for mentioning that. I’m going to put that in there explicitly.

I think ,there’s no need of (4n+2),

Learn only this series 2,6,10,14,..etc am i right?

(4n + 2) is shorthand for the series that you mention.

n is a natural number. And how can v find that

When the n will be equal to 0,1,2,3..

You can work out the number of pi electrons from the molecule (x)

Then working backwards using Huckel’s Rule –> (4n+2) = x

(x-2)/4 = n

If n is a natural number (0,1,2,3…) and the structure is a planar monocycle with an uninterupted cyclic pi-cloud then the molecule is aromatic.

I find this a little bit confusing. How do i get that the molecule is flat? I always thought that the fact that the aromatic molecule is planar is caused by aromaticity itself (or even distribution of electrons in the rings above and below aromatic core). From this article it seems to me like the planarity is something like prerequisite (i think of prerequisite like that it must be cyclic and conjugated). I’m not sure if i explained my problem properly but if you got some of my idea, can you please explain it?

Good question. It’s generally a good assumption that if the molecule fits the first three criteria, the molecule is flat, unless there is some steric impediment to it doing so. One notable exception is one of the isomers of [10]-annulene (see figure 3 in this page) which cannot lie flat due to steric clash between the hydrogens.

Thanks you!

Thanx for this post; there is a lot of confusion about the Hueckel rule. I personally like azulene more for the 10-electron example, as each individual ring does not conform the Hueckel rule, while your example still has to separate rings with each 6 π electrons. Another aspect of this is that ‘aromatic’ bonds, those in the delocalized bond system, have a bond length somewhere in between that of a single and of a double bond.

Azulene is a much better example. Good idea Egon.

Huckel model is obsolete. See

ARKIVOC, 2008, Part xi, p. 24 – 45

According to above comments the very aromatic hydrocarbon pyrene is anti-aromatic because it includes 16 pi electrons?

According to above James’ amusing comments regarding ‘algebraic sens of n’ the aromatic hydrocarbon pyrene is anti-aromatic because it includes 16 pi electrons! Also, the non-existent hexazine should be quite stable because his Huckel aromaticity!

Huckel model became obsolete many years ago, see, for instance, ARKIVOC, 2008, Part xi, p. 24-45

Yes, it’s worth noting that the Huckel model starts to fall apart for even moderately complex polyaromatic hydrocarbons, so thank you for pointing that out. However, as far as the most common examples students will encounter in an introductory class (i.e. the intended audience of this site), the Huckel model is fine. By all means learn the exceptions once you’ve mastered the basics.

I have confusion that Benzenoids obey Huckel’s rule where non benzenoids not but we call them aromatic compounds?

Dear Adil! The term aromatic is an old name used for those compounds having pleasant odor, although they are not in the category what we call as “Aromaticity”.

It should read:

For n = 0, we get (4 × 0 + 2) = 2

For n = 1, we get (4 × 1 + 2) = 6

For n = 2, we get (4 × 2 + 2) = 10

For n = 3, we get (4 × 3 + 2) = 14

Thanks, Jamey, that confused me, too.

thanks i had the same problem

Very helpful. Thanks

I get it!!!! You have no idea what this means… Thank you!!!

I have little confusion … n is not the characteristic of a molecule so what is it???

all have same value of n’ or every molecule has different n’ value how we can find the value of ‘n???

n, is the number of ring or rings

NO it absolutely is not!!

No, its the number of electrons in the pi-bonds. Take benzene, for example, we construct this as alternating pi bonds in a six membered ring. If the pi, bonds are alternating, then we know that there must be half the number of double bonds as there are single (3 in this case). So, we find that that the number of electrons in the pi bonds are 3×2 (because there are two electrons in every double bond), so we see that the 6 electrons in the pi bonds, satisfies the 4n+2 rule. Another way to put the 4n+2 rule is that if you set 4n+2 equal to the number of electrons in the pi bond and solve for n, you will find that n will be a whole number. Therefore n must be a whole number that satisfies this equation 4n+2=x, where x = the number of electrons in the pi bonds. The reason for this is probably related to quantum mechanics, since n must be a whole number, it must be a quantized value of some kind…that’s just a guess though.

Hi I’m confused on the last structure: I don’t see how the last dicyclic structure has 10 pi electrons. At the tertiary carbons that the cycloheptane and cyclopentane are connected to: aren’t there no space for electrons? Or they don’t have Hydrogens, which I’ve been counting as electrons for the previous examples.

Sophia !the last structre has 10 pi bond because it contain 5 double bond and each double bond contain 2 pair of electron so they become 10 pi electron.

You would think it would contain 4 electrons (since a double bond implies that there are double the amount of the number of electrons in a single bond), but its important to note that that the 4n+2 requirement isn’t dependent on the electrons of the double bond, but of the pi bond…there is a difference. A double bond = 1 pi bond + 1 sigma bond. The pi bond refers to sideways orbital overlap of one of the 3 p orbitals, resulting in a shorter bond.

We are taught that a p orbitals house 6 electrons, but this is wrong. the p sublevel houses 6 electrons. It also has 3 individual p orbitals in it, each containing 2 electrons. If you’re curious as to why, look up the Zeeman effect. So basically because its only one p orbital for each carbon in the bond, there are 2 electrons total in the pi bond, but four in the double.

Sir ,, i am still confused why annulene is not aromatic,,,,how steric hinderence affect it????

Is a compound having more than one ring aromatic even if a single ring is aromatic? For eg. if benzene is linked to another five-membered ring (with no pi-bonds) to form a bicyclic compound, can it be called aromatic?

Yes, what you’re describing is simply a substituted benzene which happens to have substituents which form a ring. Doesn’t change aromaticity.

What if you have a molecule that is made from taking 1 benzene and putting 6 benzene rings around it?

You get a cyclohexane ring in the center. Does this benzene bonded to 6 other benzene rings change the aromaticity or is the molecule itself still aromatic even though the ring on the inside of the molecule(cyclohexane) is not?

and why can’t 2 aromatic rings bonded to each other keep their pi bond and form an alkyne? I mean surely there are cases of cycloalkynes.

Hey James, I have 2 questions about ortho meta para directors and i was wondering if you would be able to answer them.

1. If you have HSO3 on a ring and add NaOH I believe the OH replaces the HSO3. Would you need to add NaOH and heat, and then H3O and H2O as the total step to accomplish this? or just NaOH and heat?

2. If you have two ortho/para directors on a ring and one is stronger than the other, say NH2 and CH3, and all other spots are open and you are adding Br2 in say acetic acid, would the Br2 only add to the NH2 directed postions? Or would it go to every spot on the ring?

Thank you ! it was very helpful , however I wanted to know while counting the pi bonds , do the lone pairs of electron play any role ?

It depends whether or not they are aligned with the other electrons of the pi system.

Thank you for this great explanation! I was also confused.

Thank you for this, life saver!

l understand that for a compound to be aromatic using Huckel’s rule,first you should know the number of pi electrons in the compound and if you substitute any integer value into the formula and the result amounts the number of pi electrons in the compound then that integer substituted becomes the value of n for that compound.

This was so helpful, so clear and consice and really helped. Every time I’m stuck on a concept in organic chem, I come to this website, it’s been an absolute life saver. Thank you!