Why is the C_6H_8 molecule different from benzene with respect to resonance?

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Aug 9, 2015

Benzene has two important equivalent resonance contributors, while any other ${\text{C"_6"H}}_{8}$ isomers have only one.

Explanation:

There are many molecules with the formula ${\text{C"_6"H}}_{8}$, but I suspect that you are thinking of hexa-1,3,5-triene or cyclohexa-1,3-diene

Benzene

Benzene has two equivalent resonance contributors, neither of which involves the separation of charge.

Hexa-1,3,5-triene

Hexa-1,3,5-triene has only one important resonance contributor.

(from www.chemsynthesis.com)

Any other contributor we might draw has a separation of charge. For example,

${\text{CH"_2"=CH-CH=CH-CH=CH"_2 ⟷ stackrel(-)("C")"H"_2"-"stackrel(+)("C")"H-CH=CH-CH=CH}}_{2}$

Cyclohexa-1,3-diene

Cyclohexa-1,3-diene is an "almost"-benzene.

But it is still a "linear" diene.

Like hexa-1,3,5-triene, it has only one important resonance contributor, with any other contributors having separation of charge.

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