[NewtoAtoms]

Can anyone help me with this.... what reaction should I be looking at???

[macman104]

Well, what do you see as possible reaction sites?  What are your nucleophiles, what are you electrophiles?

[lutesium]

AFAIK this Reaction would depend on the solvent!
If CH3CN is used as the Solvent the answer would be 4 if the Reaction is done under another Solvent the answer would be 3!

[macman104]

I respectfully disagree.  The question clearly states excess CH₃I as the reagent.  Excess being the key word.  To me that says "go until you can't anymore".  I could see a second step of this reaction if there were another reagent added, but that's not part of the question

[NewtoAtoms]

Hello Macman,

I am a single mother learning at home by myself usually studying from 9 - 4 am, so I don't have a teacher to bounce my ideas off of.  Therefore I am so grateful for your help, because sometimes I am truly stuck. 

In this reaction:

                 H
                 l
(CH3)2CHCHN:                                                                  CH3I
             l   l
         CH3  H

This is the lewis base and                                         This is the acid and the electrophile
the nucleophile

Therefore in the first part of the reaction I would find:

                H
(CH3)2CHCHN⁺I^-CH3
             l   H
          CH3

However then I am stuck.. would there be a leaving group (ie. HI) to leave another bonding site on the N??

[NewtoAtoms]

Let me continue..

                 H
                 l
(CH3)2CHCHN.   + HI               CH3I
                     l
                CH3
This is still a lewis base and a nucelophile            Lewis acid, electrophile


THE RESULT

                 H
                 l
(CH3)2CHCHN²⁺I^-CH3
                 l
                 CH3
                               
                 
(CH3)2CHCHN-(CH3)2
                 
COMPLETE???

Therefore the correct answer would be 3

Thank you for your time.

Newtoatoms

[macman104]

You're on the right track.  You identified your nucleophile and electrophile and reacted them, good.  But you've stopped, do you feel that the nitrogen can no longer act as a nucleophile, why?

[NewtoAtoms]

Hello Macman

Well I stopped because there were not more hydrogens left, which combined with I as a leaving group. That was my rational.
However, I do realize that it still has two lone electrons right?!?
That means that technically we continue the reaction attaching another 1 to 2 methyl's ?

If the N could continue to act as a nucleophile, then #4 would be correct at:

                 CH3
                 l
(CH3)2CHCHN⁺-CH3  I^-
                 l
                 CH3

[macman104]

Indeed, it can, and this is the basis of the hofmann elimination/degradation reaction.

http://en.wikipedia.org/wiki/Hofmann_elimination