[confusedstud]

For a molecule to be aromatic it must be planar, cyclic system of conjugation and a total of 4n+2 pi electrons.

Does a conjugated system need to have its pi electrons to be delocalized? So it not only must have alternating single and double bonds but it must also be able to delocalize its electrons.

So cyclopentadiene would not be considered conjugated despite its alternating single and double bond? Hence it is neither aromatic or antiaromatic?

[cseil]

Draw the cyclopentadiene and write the hybridisation of every carbon in the molecule.
Now again think about the 3 rules you wrote: it must be planar, with a cyclic system of conjugation and a total of 4n+2 pi electrons.

Do the same thing but with cation cyclopentadienyl and anion cyclopentadienyl.

[confusedstud]

Draw the cyclopentadiene and write the hybridisation of every carbon in the molecule.
Now again think about the 3 rules you wrote: it must be planar, with a cyclic system of conjugation and a total of 4n+2 pi electrons.

Do the same thing but with cation cyclopentadienyl and anion cyclopentadienyl.

The cation and anion do exhibit conjugation as all the carbons involved are sp2. But am I right that the cyclopentadiene is not a conjugated systeM?

[Dan]

The cation and anion do exhibit conjugation as all the carbons involved are sp2. But am I right that the cyclopentadiene is not a conjugated systeM?

For aromaticity, every atom in the ring system must contribute to the delocalisation. If this is not true, then the molecule is non-aromatic.

Conjugation is more general, and applies to linear systems as well. The alkenes in cyclopentadiene are conjugated in the same sense as the alkenes in (linear) butadiene. All aromatic compounds have conjugation, but not all conjugated systems are aromatic. For aromaticity, you need an uninterrupted ring of conjugation (ie an uninterrupted ring of overlapping p-orbitals).

[confusedstud]

The cation and anion do exhibit conjugation as all the carbons involved are sp2. But am I right that the cyclopentadiene is not a conjugated systeM?

For aromaticity, every atom in the ring system must contribute to the delocalisation. If this is not true, then the molecule is non-aromatic.

Conjugation is more general, and applies to linear systems as well. The alkenes in cyclopentadiene are conjugated in the same sense as the alkenes in (linear) butadiene. All aromatic compounds have conjugation, but not all conjugated systems are aromatic. For aromaticity, you need an uninterrupted ring of conjugation (ie an uninterrupted ring of overlapping p-orbitals).

In butadiene I can see how the electrons can delocalize. But for cyclopentadiene I don't really see how the electrons can delocalize because of the sp3 carbon though. How would it delocalize?

[Dan]

In butadiene I can see how the electrons can delocalize. But for cyclopentadiene I don't really see how the electrons can delocalize because of the sp3 carbon though. How would it delocalize?

I'm struggling to see why you are happy with butadiene but not cyclopentadiene. If the pi bonds are adjacent to each other, they are conjugated (barring any conformational effects that prevent them being coplanar).

Ok, so think about butadiene - no problems there.

Now think about 2,4-hexadiene. It's exactly like butadiene, just with a methyl group at each end. An sp³ atom at each end of the conjugated diene.

Now cyclopentadiene, that's just like 2,4-hexadiene - an sp³ atom at each end of the conjugated diene.

[confusedstud]

Hmm how would 2,4-hexadiene have resonance? Let me draw out the resonance structures to illustrate my misconception.

https://imgur.com/mBOfNpy

I thought that this was impossible because the third and sixth carbon from the left would now have 10 bonding electrons around it. The same problem occurs for the cyclopentadiene.

The only way I can think of is this https://imgur.com/07s6zfe which can occur for the cyclopentadiene as well.

[Dan]

Hmm how would 2,4-hexadiene have resonance? Let me draw out the resonance structures to illustrate my misconception.

https://imgur.com/mBOfNpy

I thought that this was impossible because the third and sixth carbon from the left would now have 10 bonding electrons around it. The same problem occurs for the cyclopentadiene.

Yes, that is impossible, but it is not a requirement of conjugation. See: definition of conjugation.

The only way I can think of is this https://imgur.com/07s6zfe which can occur for the cyclopentadiene as well.

Exactly, this is a representation of conjugation. The same can be drawn for butadiene (n.b. I don't mean cyclobutadiene).

[confusedstud]

Oh i see. So would it be correct to say that as long as there is a alternating single and double bond there would be conjugation?

But for a molecule to be considered aromatic, it has to conjugate completely throughout the ring structure?

[Dan]

Oh i see. So would it be correct to say that as long as there is a alternating single and double bond there would be conjugation?

Usually yes, but there can be conformational effects that force the pi orbitals out of coplanarity, breaking the conjugation.

But for a molecule to be considered aromatic, it has to conjugate completely throughout the ring structure?

Yes, this is one of the basic requirements of Huckel aromaticity: https://en.wikipedia.org/wiki/H%C3%BCckel%27s_rule