# Why is pyrrole more reactive than pyridine and benzene for an electrophilic substitution reaction?

According to me, pyridine should be the most reactive as it has got an extra pair of electrons on the nitrogen atom which is further localised and readily available for electrophile.

From Peter Sykes:

Pyridine (62), like benzene, has six at electrons (one being supplied bynitrogen) in delocalised it orbitals but, unlike benzene, the orbitals will be deformed by being attracted towards the nitrogen atom because of the latter's being more electronegative than carbon. This is reflected in the dipole of pyridine, which has the negative end on N and the positive end on the nucleus: Pyridine is thus referred to as a $\pi$-deficient heterocycle and, by analogywith a benzene ring that carries an electron-withdrawing substituent, e.g.$\ce{NO2}$ (p. 6.22), one would expect it to be deactivated towards electrophilic attack. Substitution takes place, with difficulty, at the 3-position because this leads to the most stable Wheland intermediate (63); the intermediates for 2- and 4-attack (64 and 65, respectively) each has a canonical state in which the charge is located on divalent N a highly unstable,i.e. high energy, state. There are certain formal analogies here to m- attack or p. 6.24), byt pyridine is very much more difficult to substi Thus nitration, chlorination, bromination and FriedelCraf really be made to take place usefully, and sulphonation on ing with oleum for 24 hours at 230°C, with an $\ce{Hg^2+}$ catalyst attack is due partly to the fact that pyridine has an available nitrogen, and can thus protonate (66), or interact with an electrophile (67):

The positive charge will clearly further destabilise any of the $\sigma$-com plexes for electrophilic substitution, as did a substituent such as $\ce{-NR3^+}$, on the benzene nucleus (p. 6.24), but the destabilisation will be much more marked than with $\ce{-NR3^+}$; as the charge is now on an atom of the ring itself and not merely on a substituent.
Pyrrole (68) also has 6 electrons in delocalised orbitals, but here the nitrogen atom has to contribute two electrons to make up the six (thus be coming essentially non-basic in the process, cf. p. 3.26), and the dipole of pyrrole is found to be in the opposite direction to that of pyridine, i.e. with the positive end on nitrogen and the negative end on the nucleus: Pyrrole is thus referred to as a π- excessive heterocycle and behaves rather like a reactive benzene derivative, e.g. aniline (p. 6.25), undergoing very ready electrophilic attack. This may be complicated by the fact that in strongly acid solution protonation (69) is forced even on the weakly basic pyrrole (it takes place on the 2-carbon atom rather than on N. cf. p. 3.26)

TLDR

• Pyridine is a $\pi$-deficient heterocycle owing to nitrogen's more electronegative nature than carbon. Pyrrole on the other hand is $\pi$ excessive since nitrogen donates its lone pair to the ring.
• The non-delocalized lone pair on pyridine tends to interact with the electrophile/$\ce{H+}$. This further destabilizes any of the $\sigma$ complexes formed in electrophilic aromatic substitution (EAS).
• Well, not really. You didn't rewrite anything. And it's not like we don't know how pyridine looks. If you need pictures badly, you can use some from Wikipedia (with proper attribution). – Mithoron Mar 26 '18 at 21:03

This can be explained by presence of the aromatic lone pair owing to nitrogen in pyrrole makes the aromatic system more electron rich, and you can get a hint of it by looking at the electron pushing mechanism.

The electron pushing from pyridine fails,

Initially, it was due to the lone pair of electron is fixed at an orthogonal position away from the aromatic ring.

Secondly, is that particular intermediate cannot be obtained because of structural geometry problems, (=N=)+ should be linear and cannot be formed while maintaining a ring.

One way to awaken electrophilic substitution in pyridine is to N-oxidize it. See here, about halfway down the page, for an example with nitration (predominantly opposite the N-oxide function), and this answer for an explanation of what happens to the electrons as a result of this oxidation. In effect N-oxidizing pyridine works by making that molecule more like ... pyrrole.