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In Clayden, page 735, Aromatic Heterocycles 1:Reactions:

The other simple five-membered heterocycles are furan, with an oxygen atom instead of nitrogen, and thiophene, with a sulfur atom. They also undergo electrophilic aromatic substitution very readily, although not so readily as pyrrole. Nitrogen is the most powerful electron donor of the three, oxygen the next, and sulfur the least. Thiophene is very similar to benzene in reactivity.

This is in the context of electrophilic aromatic substitution.

My question is, why is nitrogen the most powerful electron donor? Sulphur is the least electronegative, so will it not be less reluctant to donate electrons?

Or are there some resonance/aromatic or other effects I'm failing to see in pyrrole, furan, and thiophene?

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  • $\begingroup$ Sulphur doesn't like bond order higher than one. One could think Se would be even better, but it's likely the opposite. $\endgroup$
    – Mithoron
    Commented Jan 31 at 20:12
  • $\begingroup$ I remember this on heterocyclic chemistry class. It is obvious that oxygen would donate less than nitrogen due to electronegativity. Sulphur however, are on the 3rd period. This allow it to contained more than 8 electrons. Basically, because you're allowed to draw thiophene with $=S=$ bond in the resonance structure, making Sulphur the least donating of the three. $\endgroup$
    – Tensor
    Commented Feb 1 at 3:10
  • $\begingroup$ I once heard an explanation related to larger atoms having weaker resonance because they are geometrically less able to. I wonder if this is correct $\endgroup$
    – Gaurav Sai Maddipati
    Commented Feb 1 at 5:36
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    $\begingroup$ Sulfur is not a good donor to carbon pi systems due to its larger size. Sulfur's larger 3p orbital (you can imagine a lone pair there) does not overlap efficiently with the pi system of the ring. $\endgroup$
    – lou
    Commented Feb 1 at 21:40
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    $\begingroup$ It can't be simplified to comment length, but essentially, most of the times longer bonds mean weaker bonds. This is usually due to more nodes in the eave function and less efficient overlap. Most importantly though, since it came up, there are no d orbitals in binding for sulfur. There's not more than 8 electrons allowed. And I have no clue what geometrically stable could mean in this context. $\endgroup$
    – Martin - マーチン ♦
    Commented Feb 2 at 0:57

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Nitrogen is able to donate its electrons to the aromatic pi-system more readily than oxygen and sulfur because its p-orbital (which contains the lone pair) is energetically closer to carbon (i.e., only slightly lower in energy than carbon's p-orbitals). Oxygen's p-orbitals are lower in energy still, meaning that overlap is weakened, and sulfur's p-orbitals are much higher in energy and larger in size making overlap much weaker.

You can think about it in terms of which atom's p-orbitals are most similar to those on carbon, since the aromatic system is mostly made of carbon p-orbitals.

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