0
$\begingroup$

From the Chem LibreTexts article on Aromaticity and the Huckel $4n + 2$ Rule:

So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain one or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are $\mathrm{sp^2}$ hybridized. But is the oxygen atom $\mathrm{sp^2}$ hybridized? The oxygen has at least one lone electron pair and is attached to an $\mathrm{sp^2}$ hybridized atom, so it is $\mathrm{sp^2}$ hybridized as well.

enter image description here

Is this a general pattern - if an adjacent atom is $\mathrm{sp^2}$-hybridized, and if there's at least one electron pair, then our atom it $\mathrm{sp^2}$-hybridized as well?

Why can't oxygen in furan be $\mathrm{sp}$-hybridized? Is this only because the overall structure will not be consistent, because sp-hybridization will make the $\ce{C-O-C}$ line straight, and thus the $\ce{O-C-C}$ angles would have to be lesser than the carbons' $\mathrm{sp^2}$-hybridization would allow?

$\endgroup$
8
$\begingroup$

Is this only because the overall structure will not be consistent, because sp-hybridization will make the COC line straight, and thus the OCC angles would have to be lesser than the carbons' sp2-hybridization would allow?

Yes, this is the correct answer. A very common misconception is that hybridisation can be used to predict the geometry, or that hybridisation somehow involves an energy cost associated with 'promoting' electrons into the hybrid orbitals. This is entirely wrong.

Hybridisation is always determined by geometry. You can only assign hybridisation states to an atom if you already know its geometry, based on some experimental or theoretical evidence. The geometry of the oxygen in furan is trigonal planar and therefore the hybridisation must be $sp^2$.

$\endgroup$
4
$\begingroup$

I think bon's answer is excellent, but I wanted to comment about one piece of your question:

Is this a general pattern - if an adjacent atom is $\mathrm{sp^2}$-hybridized, and if there's at least one electron pair, then our atom it $\mathrm{sp^2}$-hybridized as well?

Whenever students ask "general pattern" kinds of questions, I hesitate, because chemists love to find counter-examples to general patterns.

The point of Hückel's rule (article) is that even heteroatoms can participate in aromatic rings. The specific rule is that if you have an $\mathrm{sp^2}$ conjugated system, the lone pair will be involved if it makes the system more stable. In this case, conferring Hückel $4n+2$ aromaticity.

Consider furan - in the Lewis diagram, you have two lone pairs on the oxygen atom. If we count electrons from the carbon atoms, we have 4 (one per carbon). So adding two electrons from one of the lone pairs will give 6 = 4(1) + 2, so Hückel aromatic.

If we look at thiazole:

enter image description here

Now there are two heteroatoms. The sulfur will act like the oxygen in furan (i.e., as a two-electron donor). So how many electrons does the nitrogen contribute? Do you think the nitrogen is $\mathrm{sp^2}$? Yes, clearly this is also aromatic.

$\endgroup$
0
$\begingroup$

Hybridization is driven by geometry, so ask whether a geometry that drives $sp$ hybridization on the oxygen, including a bond angle of 180°, is plausible in this small ring. I'm guessing you might find such a geometry would be a bit destabilized by angle strain.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.