In an Imidazole ring the NH acts as a better nucleophile as electron density is more on the nitrogen as H in NH  is more acidic than H of CH. Hence,  the alkylation occurs at N preferably.

 Prof. Deshmukh

Thats fine, but my quarry is in presence of base both protons will go or any one  specifically leaves

IN IMADAZOLE the NH group is more nucleoohile than CH . lone pair of NH more active. than the alkylation occures on NH group. 

It depends on the base you will like to use. If you use a base strong enough, both NH and CH can ionise; however, N bears better the negative charge than C: the first H to go will be the one on N. 

Yes it strongly depends on the base  you use. If you want to make a C-N coupling over the N, I would recomend you to use a mild base such as carbonate, phosphate or even the stronger tert-butoxide.

N-alkylation occurs with milder bases than C-alkylation. Nitrogen is a better electron donor than carbon and thus is more nucleophilic.

The NH of imidazole is more nucleophilic than carbon between the two nitrogen 

Imidazole is an N-nucleophile. Depending on the amount of alkylating agent, its reactivity and additional base added (imidazole is a base by itself) you can get a wild mixture of products. You will first get alkylation on the first nitrogen, then on the second, and only if you have excess alkylating agent and additional strong base you will get a C-alkylation.

Of course, NH is more good nucleophile than carbon between two nitrogen s.

Actually, the other nitrogen (not the NH) of an imidazole is the better electrophile.  It has a lone pair of electrons available where the NH needs to deprotonate first to act as a nucleophile.  Minor structural and electronic differences can greatly infulence the selection of which nitrogen is more likely to act as the better nucleophile.  There still is a competition even then.

William is correct - in neutral imidazole, the unsubstituted nitrogen is the nucleophile.  For deprotonated imidazole, the negative charge density is mostly on the nitrogens, so either one will react with an electrophile.  If you want to deprotonate the carbon, you first have to alkylate both nitrogens, forming an imidazolium salt.  The pKa of the methine H is then about 24, and you can deprotonate it and alkylate it in reasonable yield.

http://pubs.rsc.org/en/Content/ArticleLanding/1995/C3/c39950001267#!divAbstract

Hi Nitika!

As I see, some proficient collegues pointed that PYRIDINE-type nitrogen is most active nucleophile (in absence of base!). PYRROLE-type nitrogen is not active becase not have lone pair for nucleophilic attack, but it will be more active in presence of deprotonation base (charge carrier is more active in common over lone pair).

In reality, because of rapid proton exchange in non-substituted (symmetrically substituted) imidazole both atoms are equal at near RT and above. Imidazole with C4 (C5)-substituent(s) show some preference against one of nitrogen atoms.

From your question it is not clear wich electrophile was used. Alkylation and acylation-type reagents will produces  various products by subsequent substitution (in term of position).

In an Imidazole ring acts as a better nucleophileis  NH as the electron density on nitrogen is more then carbon.