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I’ve found 4H‐1,4‐oxazocine in my chemistry book and I’m perplex as it states as aromatic:

4H‐1,4‐oxazocine

I can count six electrons on p orbitals which could satisfy Hückel’s $n = 1,$ but the oxygen is sp3-hybridized. Even if we’d consider one lone pair as one sp2-bonding, the second lone pair would add 2 electrons to the p system, counting a total of 8 electrons, which doesn’t satisfy Hückel’s rule.

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    $\begingroup$ So what about the nitrogen lone pair? $\endgroup$
    – Waylander
    Jan 13, 2021 at 17:41
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    $\begingroup$ @Waylander so basically I could count n=10 considering O as $sp^2$ with a lone pair on the pi system? $\endgroup$
    – Pseuronimo Stilton
    Jan 13, 2021 at 17:44
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    $\begingroup$ Didn't we do the same in furan? Didn't we do the same in pyrrole? Well, now put the two together. $\endgroup$
    – Ivan Neretin
    Jan 13, 2021 at 18:04
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    $\begingroup$ @IvanNeretin Thanks! I actually had this preconception I’ve read somewhere that n=number of aromatic rings, thus in my reasoning this compound couldn’t have more than 6 e^- in its pi system. $\endgroup$
    – Pseuronimo Stilton
    Jan 13, 2021 at 18:13
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    $\begingroup$ @pseuronimo n is not the number of aromatic rings, nor could it be since the 4n+2 rule is really accurate only for a single ring. Rather it's an optimal value that depends on ring size. Typically n=0 for a three-atom ring, 1 for 4-7 atoms, 2 for 8-9 atoms. Aromaticity is usually insignificant for larger rings than that. $\endgroup$
    – Oscar Lanzi
    Jan 13, 2021 at 19:49

1 Answer 1

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According to Huckel's rule four criteria for aromaticity

The molecule needs to be (1) planar, (2) cyclic, (3) fully conjugated, and has (4) 4n+2 electrons.

Your molecule does not seem to be planar (b/c of N and O atoms) but assuming that it has a nearly planar conformation it satisfies all other criteria. You only need to count one of oxygen's lone pair and the only nitrogen lone pair:

$10=4*2+2$

So it could be aromatic (if it is nearly planar)

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