I'm not able to understand how the following compounds are aromatic.
When should the lone pairs on heteroatoms be taken into consideration when counting the number of π electrons?
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asked Oct 2 '13 at 13:30
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Only count the lone pairs/ pi-bonds/ groups which are participating in conjugation and ignore them in all other cases.
For example, in compound 2 (thiophene), there are two lone pairs on sulfur.
One lone pair (brown) is in a p-orbital, and hence participates in conjugation with the two π-bonds. The other lone pair (blue) is pointing outwards from the ring in an $\mathrm{sp^2}$ orbital. This lone pair is orthogonal, or perpendicular, to the π-system and hence cannot take part in conjugation. In total there are six π electrons, and the compound is therefore aromatic.
See also: Conjugated system on Wikipedia.
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- The Nitrogen atom is already an $\mathrm{sp^2}$ hybridized atom, so its lone pair does not take part in conjugation
- The sulfur atom is $\mathrm{sp^3}$ hybridized, it has two lone pairs, so only one of its lone pair will take part in conjugation for it to be $\mathrm{sp^2}$ hybridized atom
- $\ce{N}$ bonded to $\ce{H}$ is $\mathrm{sp^3}$ hybridized and has its lone pair taking part in conjugation, thus its lone pair counted as 2 pi electrons. the lone pair of the $\mathrm{sp^2}$ hybridized $\ce{N}$ atom is not pi electrons
- Lone pair on $\ce{S}$: one pair is pi electrons, lone pair on $\ce{N}$: not pi electrons
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