Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: NewtoAtoms on March 22, 2009, 03:41:45 AM
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Can anyone help me with this.... what reaction should I be looking at???
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Well, what do you see as possible reaction sites? What are your nucleophiles, what are you electrophiles?
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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!
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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
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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??
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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
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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?
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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
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Indeed, it can, and this is the basis of the hofmann elimination/degradation reaction.
http://en.wikipedia.org/wiki/Hofmann_elimination