When pyrrole undergoes electrophilic aromatic substitution, at which position does substitution occur?

When pyrrole undergoes electrophilic aromatic substitution, at which position does substitution occur?

May 4, 2016

Suppose we used ${\text{Br}}_{2}$ as an electrophile. Then, suppose we examine the intermediate where one $\text{Br}$ has already attached on either carbon-2 or carbon-3 as follows:

The difference is fairly clear: electrophilic aromatic substitution (EAS) occurring at carbon-2 gives one more resonance structure than on carbon-3, so that intermediate is more stable.

Thus, EAS more readily occurs on carbon-2 starting with pure pyrrole.

The mechanism is fairly simple; just get an electrophile, then pyrrole can slowly act as a nucleophile. Afterwards, it quickly restores its aromaticity.

Overall, the carbon position depends on the stability of the resonance structures for the intermediate that forms. In this case it worked out as carbon-2 for pure pyrrole.

BUT WAIT! WE'RE NOT DONE.

If both carbon-2 positions are occupied, then carbon-3 will be the reactive position.

The same chemistry will happen with furan and thiophene as well.

RELATIVE REACTIVITIES IN ELECTROPHILIC AROMATIC SUBSTITUTION

NOTE: In terms of reactivity, it goes:

$\text{pyrrole" > "furan" > "thiophene" > "benzene}$

Oxygen, being more electronegative than nitrogen, distributes more negative charge density upon itself and less upon the ring, thus stabilizing the carbocation intermediate less, making furan less reactive towards EAS than pyrrole.

Sulfur's $3 {p}_{z}$ orbital overlaps less effectively with carbon's $2 {p}_{z}$ orbitals, thereby sharing electron density more poorly than furan's oxygen does, stabilizing the carbocation intermediate less, making thiophene less reactive towards EAS than furan.

And pyrrole, furan, and thiophene are all more reactive than benzene with EAS because the lone pair on the heteroatom can donate electron density into the ring by resonance, thus stabilizing the carbocation intermediate more effectively.