5
$\begingroup$

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?

Heterocycles: isoquinoline, thiophene, imidazole, thiazole

| improve this question | | | | |
$\endgroup$
11
$\begingroup$

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.

Lone pairs on sulfur in thiophene

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.

| improve this answer | | | | |
$\endgroup$
0
$\begingroup$
  1. The Nitrogen atom is already an $\mathrm{sp^2}$ hybridized atom, so its lone pair does not take part in conjugation
  2. 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
  3. $\ce{N}$ 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
  4. Lone pair on $\ce{S}$: one pair is pi electrons, lone pair on $\ce{N}$: not pi electrons
| improve this answer | | | | |
New contributor
Chm is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE! To help you become acquainted with the site, I would just like to suggest if you haven't already, please take a minute to look over the help center and tour page to better understand our guidelines and question policies. You will find StackExchange (SE) to be a different kind of Q&A website from the mainstream. To learn more visit Meta.Chemistry, Meta.SE, or chat $\endgroup$ – Zenix yesterday

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.