Why is the hybridization of oxygen in the seven membered ring pictured below $sp^3$, while the oxygen has a hybridization of $sp^2$ in the 5 membered ring? This also may be an error in the answer key as well.
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2$\begingroup$ Count the number of pi electrons - what would happen if that oxygen were to be sp2 instead of sp3? $\endgroup$– orthocresolApr 29, 2017 at 21:58
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2$\begingroup$ Honestly, this is a bad question. Are we assuming the ring is flat? If it's puckered, you could make a case for $sp^{2}$. If it's flat, $sp^{3}$ is probably better... $\endgroup$– ZheApr 29, 2017 at 22:06
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1 Answer
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It relates to the Hückel rule for aromatics. With the electron pair of the oxygen participating in the $\pi$ system of the oxazol (5-membered ring) we get 6 $\pi$-$\ce{e-}$. Doing the same for the oxazepin would get 8 $\pi$-$\ce{e-}$.
Hückels rule states that planar, circular $\pi$ systems with $4n+2$ electrons are aromatic, resulting in a lower overall energy if the system is conjugated. (with $n$ being a positive integer or zero) The other statement that Hückels rule does is that $\pi$ systems with $4n$ electrons are anti-aromatic, which avoid conjugation as it would increase the overall energy of the system.
Therefore it is favoured for your oxazol to conjugate, which results in a sp$^2$-hybridization for the oyxgen, while the oyxgen in the oxazepin avoids conjugation, resulting in a sp$^3$-hybridization.
The Hückel rule in turn can be quantized by a Hückel method computation or graphically utilizing a Frost-Musulin-diagram.
