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exaplanation of cyclobutadiene resonance


rakuenso

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To explain why cyclobutadiene is not aromatic requires one to talk about molecular orbitals, unfortunately, since resonance, conjugation and aromaticity are based on molecular orbital theory.

 

Basically, the four p-orbitals on each of the four carbon atoms will form four Pi orbitals, a bonding orbital, two non-bonding orbitals, and an antibonding orbital. Since each p-orbital contributes one electron to the Pi system, there will be four electrons: the first two fill the bonding orbital, so the other two go into the non-bonding orbital. However, since the two non-bonding electrons will go into opposite orbitals, this makes cyclobutadiene a highly-unstable diradical species (hence, cyclobutadiene is known as an anti-aromatic molecule, a molecule which becomes highly unstable when it forms a cyclic conjugated Pi system).

 

Since the formation of this conjugated Pi system is so unfavorable, cyclobutadiene will adopt a conformation which prevents the two adjacenct pi bonds from becomming conjugated. Since conjugation is essentially resonance, cyclobutadiene will not have a resonance form.

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ok, another unrelated question but didn't want to start up another thread in fear of spam.

 

It's about hybridisation:

 

Take for example CH3+, carbon here would be sp2 hybridized correct? Since it appears that it would sit in a planar, and thus has an empty p orbital. OH-'s Oxygen would be sp3 hybridized since it would have 3 lone pairs and one sigma bond, and it would resemble a tetrahedral shape.

 

So when CH3(+) + OH(-) -> CH4O, you end up with carbon's empty p orbital overlapping Oxygen's sp3 orbital and a sigma bond is formed. How does this work? The net result is that both Carbon and Oxygen becomes sp3 hybridized.

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