What is the product formed from a reaction between pyrrole, with i)n-BuLi and then ii) MeI

Question

i know that strong bases such as n-BuLi can deprotonate N on the pyrrole ring, but i'm not to sure what the product formed will be upon addition of MeI....

any help would be appreciated. if you need any more detail than this let me know

Answer 1

well i think you i)n-buli is BULL-i

Answer 2

OK, let's say you're right about the first reaction: acidic H on the pyrrole N, basic/nucleophilic alkyl anion gabs the proton.  Push arrows from the carbanion to the H, from the N-H bond back onto the N.  What do you get?  Should be the conjugate base of the pyrrole, with two lone pairs and the formal negative charge on the N atom, right?  (Plus butane.)

So, then react that species with MeI.  What do you know about MeI?  Is the C-I bond polar or non-polar?  Is the molecule nucleophilic or electrophilic, and at which atom will the reactivity normally occur?  Is iodide a good leaving group or a poor leaving group?  What properties should the pyrrole anion have?  Nucleophile or electrophile, and at which atom?

Draw arrows from the N to the MeI C, from the C-I bond to the I.  Displace [I]– and form a new N-C bond in an SN2 reaction.  Your product is N-methylpyrrole.

You should probably leaf through some chapters on heterocyclic chemistry to convince yourself that you won't actually activate the alpha-carbon (next to the N), because those C-H bonds are also very acidic: lots of resonance forms in the anion, obviously.  There is a lot of chemistry on pyrroles and indoles where the type of reactivity you're descrbing substitutes as the carbon next to the N, not at the nitrogen.