OCHEM: Hoffman Rearrangement?

[informatique]

I've been working on the problems in OChem Odyssey and there is one problem where the entire carbonyl group is eliminated to yield a primary amine and there is another problem where only the oxygen is eliminated.

Phenyl - CH2 - CH2 - NH2 (entire carbonyl group eliminated)

vs.

Phenyl - CH2 - CH2 - CH2 - NH2 (only oxygen is eliminated)

Can someone please explain the hoffman rearrangment to me? I definitely do not want to get this one wrong on the exam.

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[tncekm]

I've been working on the problems in OChem Odyssey and there is one problem where the entire carbonyl group is eliminated to yield a primary amine and there is another problem where only the oxygen is eliminated.

Phenyl - CH2 - CH2 - NH2 (entire carbonyl group eliminated)

vs.

Phenyl - CH2 - CH2 - CH2 - NH2 (only oxygen is eliminated)

Can someone please explain the hoffman rearrangment to me? I definitely do not want to get this one wrong on the exam.

I"m pretty sure the hoffman rearrangement always loses a carbon (entire carbonyl group) via carbon dioxide. Maybe the other problem is a typo?

 

[DRHOYA]

I'm also working with the Odyssy. I'm finding it hard to explain the Hoffman rearrangment though with the rxn's you posted. Do you know what question #'s they were from what chapter? I know that in Elimination Rxn's, the Hofmann product is the least substitued alkene. So, depending on the rxn conditions, you will have a "hofmann rearrangement". Hope that helps.

 

[tncekm]

I'm also working with the Odyssy. I'm finding it hard to explain the Hoffman rearrangment though with the rxn's you posted. Do you know what question #'s they were from what chapter? I know that in Elimination Rxn's, the Hofmann product is the least substitued alkene. So, depending on the rxn conditions, you will have a "hofmann rearrangement". Hope that helps.

No, the hoffman product as a result of exhaustive alkylation of an amine is different from the hoffman rearrangement. The hoffman rearrangement is a rather bazaar reaction where a primary amide loses a carbonyl group (to carbon dioxide) and is converted to a amine that is short one carbon (the carbon from the carbonyl group).

So:

Amide ---(hoffman rearrangement)---> Amine + CO2

 

[DRHOYA]

No, the hoffman product as a result of exhaustive alkylation of an amine is different from the hoffman rearrangement. The hoffman rearrangement is a rather bazaar reaction where a primary amide loses a carbonyl group (to carbon dioxide) and is converted to a amine that is short one carbon (the carbon from the carbonyl group).

So:

Amide ---(hoffman rearrangement)---> Amine + CO2

Ohh ok. So, what are the special rxn conditions that would cause this rare rxn?

 

[tncekm]

Ohh ok. So, what are the special rxn conditions that would cause this rare rxn?

I just remember it b/c its a primary amide w/ strange reagents (Br2 and a base). Whenever I see that, I'm like "oh, i don't know how to move the electrons around so it must be the Hoffman rearrangement" Just google it if you're interested in the details.

 

[DRHOYA]

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

Yeah you are right, it is characteristic when you have Br2 and a strong base (such as NaOH). Also, its characteristic to "lose a carbon". Thanks man!, b/c now I know the difference between the hofmann elimination and this, because there is a big difference.

 

[tncekm]

Yes sir. I hope that helps you out on your big day Its a tricky distinction, but once you know it, its an easy way to pickup some points that the next guy didn't

 

[doc3232]

You can also use BrO-
The only thing is that I know you cannot do the Hof Rearrangement without losing both the C and the O. The rearrangement has the transition state of N=C=O so there is no way that I can see for that carbon to get back in.
If anyone does please explain.

 

[pistolpete007]

i know hoffman elimination is completely diff than arrangment.....and i know the elimination obviously has to do with elimination but what are the specifics of it again? it makes the least subs. alkene but what kind of substrate and reagents are we looking at?

 

[tncekm]

Okay, so you've got a secondary amine. You will use an organometallic reagent to "exhaustively" alkylate. I.e. R-NH2 --> R-N(CH3)3+. The positively charged ammonium ion is a great leaving group, so it leaves, and the hoffman elimination product is the major product with the zaitsef being the minor.

 

[TexasDent]

I've been working on the problems in OChem Odyssey and there is one problem where the entire carbonyl group is eliminated to yield a primary amine and there is another problem where only the oxygen is eliminated.

Phenyl - CH2 - CH2 - NH2 (entire carbonyl group eliminated)

vs.

Phenyl - CH2 - CH2 - CH2 - NH2 (only oxygen is eliminated)

Can someone please explain the hoffman rearrangment to me? I definitely do not want to get this one wrong on the exam.

The first one is probably the hoffman rearrangement (Br2 and strong base i.e. NaOH) This always cuts out the C=O to make it one less carbon.

The second one is probably (H2NNH2 and strong base i.e. KOH) This always just cuts the =O bonds and replaces with two hydrogens.

Correct?

 

[TexasDent]

Actually I just realized, the second one is more likely LiAlH4 which is a reducing agent and will reduce an amide to an amine just cutting the =O bonds and replacing with 2 H's.

The reagents, (H2NNH2 and strong base) are used to reduce a ketone, not an amide.

Sorry about that.

 

[informatique]

Actually I just realized, the second one is more likely LiAlH4 which is a reducing agent and will reduce an amide to an amine just cutting the =O bonds and replacing with 2 H's.

The reagents, (H2NNH2 and strong base) are used to reduce a ketone, not an amide.

Sorry about that.

Yeah, I went back and redid the problems in that section and the second one was LiAlH4. I had mistaken that problem for being Br2/OH-