Why does imidazolium ring contain a positive charge?

[sameeralord]

Just interested how this positive charge occurs. Thank you

 

[ChrisW]

Draw the cyclic structure, then draw in double bonds where necessary to keep the octet rule for carbon. Should end up with 4 bonds to one of the Nitrogens, which gives a + charge. Now if you draw the resonance forms you'll see that the double bonds (and + charge) can bounce around throughout the ring system, so it's easier to just draw the whole ring system as positively charged.

 

[chemisttree]

Actually, your diagram needs a bit of revision (or definition). In the example you provide, R1 and R3 cannot be equal to H. If this were the case, you would have a substituted imidazole molecule. Deprotonation would yield (R1 or R3 = nothing) a substituted imidazolium species whereas alkylation (or quaternization) of one of those species would yield one as well. Keep in mind that R1 and R3 cannot both be quaternized, only one (in your example). Therefore, either R1 or R3 must be equal to H (they can both be equal to H as well) but they both can't be an alkyl group.

Edit: Bolded item is wrong. It should read, "either R1 or R3 must be equal to 'nothing' and either one or the other may be equal to an alkyl. Both may also be equal to 'nothing' ".