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Does benzene have both isomers and resonance structures?

I know that it has resonance structures because you can manipulate the position of the double bonds but what about isomers?

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Well, it depends on what kinds of isomers you have in mind. It certainly has constitutional isomers: For example, the sum formula $\ce{C6H6}$ does describe benzene as well as hex-1,5-en-3-in $\ce{H2C=CH-C#C-CH=CH2}$. There are also a lot ring structures that are isomers of benzene:

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As far as I know there aren't any configurational isomers of benzene since a six-membered ring doesn't allow for trans double bonds (this would be sterically impossible).

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  • $\begingroup$ On a sidenote, Bicyclo[2.2.0]hexa-2,5-diene is also known as Dewar benzene, prismane as Ladenburg benzene. The smallest stable (at RT) ring with a trans double bond is indeed trans-cyclooctene. $\endgroup$ – Klaus-Dieter Warzecha Mar 4 '14 at 11:08
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As you note, benzene has resonance structures, but it does not have isomers. All of the bond lengths in benzene are identical. That is why 1,2-dideuteriobenzene or 1,2-dimethylbenzene (o-xylene) both exist as single compounds. Here's an interesting way to look at it. Look at the energy diagram below on the left. If benzene had 2 isomers, that would be them as the two stable wells at the bottom of the energy diagram, the "resonance delocalized" version of benzene is shown as the transition state involved in the interconversion of these two "isomers". Now look at the energy diagram below on the right. This is the real case. It turns out that the "transition state" is more stable than either of the 2 "isomers"! It could be said that there is a negative activation energy for the formation of delocalized benzene from either of the 2 localized "isomers". The negative activation energy would correspond to the resonance stabilization of benzene compared to either of the localized "isomers". Hence, benzene exists as a fully delocalized structure with identical bond lengths at the bottom of the potential well.

enter image description here

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