91Ó°ÊÓ

Imidazole isthe heterocyclic ring of histidine. It is an aromatic compound because it is cyclic and planar. Every atom in the ring has a p orbital, and the pi cloud contains three pairs of pi electrons.The lone-pair electrons on N-1 are part of the pi cloud because they are in a piorbital, whereas the lone-pair electrons on N-3 are not part of the pi cloud because they are in a sp²orbital that is perpendicular to the pi orbitals. Imidazole also undergoes hydrogen bonding.

There are three possible sites for electrophilic substitution in imidazole i.e. C-2, C-4 and C-5.

Substitution at C-2:

It forms an intermediate with three resonance contributors. All the atoms in one resonance contributor have complete octet and positive charge on N and the other two have carbon and nitrogen with incomplete octet.

Substitution at C-4:

It forms an intermediate with two resonance contributors. All the atoms in one resonance contributor have complete octet and positive charge on N: one contributor has a carbon with an incomplete octet.

Substitution at C-5:

It forms an intermediate with three resonance contributors. All the atoms in one resonance contributor have complete octet and positive charge on N and other two contributors have a carbon with an incomplete octet.

To determine the major product of the reaction, we have to compare the relative stabilities of carbocations formed in the first step of the reaction.

The substitution at C-5 forms the most stable intermediate because a positively charged carbon with an incomplete octet is more stable than the positively charged nitrogen with an incomplete octet (Substitution at C-2).

Whereas substitution at C-4 forms the least stable intermediate because it has only two resonance contributing structures as compare to C-2 and C-5, which are having three contributors as shown above in step 2.

Thus, the major product of the reaction is 5-bromo-N-methyl Imidazole as shown in the figure given below: