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

1 Answer
Aug 9, 2015

Answer:

Benzene has two important equivalent resonance contributors, while any other #"C"_6"H"_8# isomers have only one.

Explanation:

There are many molecules with the formula #"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.

Benzene
(from chemistry.stackexchange.com)

Hexa-1,3,5-triene

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

Hexatriene

(from www.chemsynthesis.com)

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

#"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.

www.sigmaaldrich.com

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.