hoffman elimination?

[Ramoray]

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ok so i thought i was good at ochem but i cannot feel bad as i just looked in my ochem text and online and still could not figure out hte answer to a quesiton on form EG today. It was the hoffman elimnation of N ethyl methyl amine. and it asked for how many alkene products. Now in my book and what i thougth on the test was the only example in my book was o fa primary amine undergoing hoffman elimnation to form the least sumbstituted product. But nowhere can if ind if a secondary amine also undergoes hoffman elimnation?? so i picked it would not and hence give 0 alkene products. but **** that is when iknow it is not something i could have studies if i am sitting here with the whol einternet and ochem text and STILL cannot find the answer! ok anyone? thnx

 

[fun8stuff]

hoffman elimination can occur on amines other than primary ones. it is an E2 rxn that converts an amine into an alkene. it forms least substiuted double bond. maybe 4 products would be formed? .....

 

[Ramoray]

so what is the answer? you are right i thought it said the hoffman rearangment, and that is what i meant to type, so was it just being tricky and trying to ocnfuse you with the different hoffmans? that is why i put 0 as i was like this doenst undergo rearrangment, but then the hoffman and double bonds terms got in my head since it asked it in the question so i was wondering. so anyone else remember this one and what is the freakin answer?

 

[spumoni620]

hey i happened to have that ? too on form as. if i remember right it was N-ethylmethanamine or something. i put one b/c the way i drew it there was one hoffman prod possible (elimination forming the least substitutued alkene).

but heck i am by no means confident about that ans...haven't even taken 2nd sem ochem yet so who knows if i was right?????

 

[jasonparks]

Here's what I think. Only a quaternary ammonium ion will undergo a Hoffman elimination. So I think the answer is zero. Of course this isn't the answer I selected. I could only remember that a Hoffman forms the least stable alkene (anti-Zaitsev), and that two equally stable alkenes could be formed from the alkene shown. So I picked 2.

 

[Ramoray]

but i think indio said it right.. wasnt your questoin the Hoffman Rearrangment? not eliminatoin. i just wrote that wrong. So was this tricky or hwat. i know its just a question so i dont care i just think that is tricky to **** u over with the name hoffman!

 

[Mudd]

Jasonparks is correct that Hoffman elimination only occurs if the amine is quaternary and cationic.

However, if the question was on Hoffman rearrangement, then the original compound must have been an amide. In Hoffman reaarrangement, an amide loses the CO to become an amine.

There must be more to that question.

 

[closertofine]

This is what I remember (and I have a pretty good visual memory): it was a discrete question, no passage...for me it was the first in a group of discrete questions....so there was no more to it than was given in the question.

And (again, only what I remember), it was Hoffmann elimination, not rearrangement, of N-ethylmethylamine. Not having taken Kaplan or memorized any crazy details, I had NO idea what Hoffmann meant...though I guess now it sounds a little familiar from O-chem several years ago.

Based just on the elimination part, I guessed there would be 2 possible products...but now that you guys explain what Hoffmann is, I guess I was wrong...it must be zero, right? Then again, I could always overthink these questions...like if you MADE the amine quaternary, you could do the elimination then...but I guess that's not what they meant!

Anyway, I need to stop thinking about these questions...there's nothing else I can do about my score, unfortunately...!

 

[Ramoray]

i am 100 percnet sure it was rearrangment i rmeember thinkign WTF can u do an arrangment on here..

 

[jasonparks]

It is possible that the question assumes exhaustive methylation, which is often a preliminary step in a Hoffman elimination. But I tend to read too much into the questions, and they clearly are not making this assumption.