Hoffman Elimination question

[SaintJude]

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Question:

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I understand what the products of a typical Hoffmann elimination question would be.

Usually the least substituted alkene will form and a separate tertiary amine. Here the amine is attached to the alkene. Why? Is it because the nitrogen is flanked by 2 carbons?

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[SaintJude]

Haha--I know that people don't like orgo...but I didn't realize the contempt was so great...no responses 🙁

 

[MT Headed]

Haha--I know that people don't like orgo...but I didn't realize the contempt was so great...no responses 🙁

I just want to go on record as saying I'm sooo glad I don't need to know this junk anymore. I sold all my books so I tutor here from memory and wikipedia, and I just can't get excited enough to even google this type of problem anymore.

Good luck!

 

[theseeker4]

Question:

View attachment 18521

I understand what the products of a typical Hoffmann elimination question would be.

Usually the least substituted alkene will form and a separate tertiary amine. Here the amine is attached to the alkene. Why? Is it because the nitrogen is flanked by 2 carbons?

You are right, it is in a ring, and the reaction causes only a single bond to break, meaning the second bond remains attached. Think of that second bond, the one that didn't break, as the R group on the amine. That group doesn't become separated from the amine, which is the same as in this case. The R group just happens to be the opposite end of the alkene.

 

[lkthlttr]

This is where a cursory understanding of the mechanism will get you a long way. You should remember that in the Hofmann elimination that the leaving group is a positively charged amino group (usually following methylation with methyl iodide as is the case in this question). However as opposed to the instance you are referring to in which there are two possible protons can be removed by a base, and the less sterically hindered one is removed preferentially, if you try to draw out this reaction it should be clear that due to symmetry there is only one choice of proton which will result in elimination of the positively charged ammonium group. This will result in a neutral amine product, and like theseeker4 says, because it was in a ring, the other carbon is still attached to the amine.

 

[ljc]

To help myself understand -- at this point, you could run a 2nd equivalent of Hoffman Elimination to cleave the amine group, correct? Resulting in another double bond, and a separate N(CH3)3 molecule?

 

[SaintJude]

Yes.

 

[chiddler]

question: how is the hydrogen removed from the amine?

 

[SaintJude]

In the Hoffmann elimination, an amino group is converted to a quaternary ammonia salt through several alklylations (aka "exhaustive methylation") --see step 1 of the equation I posted.

The resulting quartenary ammonium salt is a really good leaving group.

 

[chiddler]

In the Hoffmann elimination, an amino group is converted to a quaternary ammonia salt through several alklylations (aka "exhaustive methylation") --see step 1 of the equation I posted.

The resulting quartenary ammonium salt is a really good leaving group.

Yes i understand this. But! The nitrogen initially has an H. Then its gone. What happens to it?

 

[lkthlttr]

After the first alkylation reaction, you have a quaternary nitrogen (a nitrogen attached to three carbons and one hydrogen). The pKa of an ammonium group is ~10, so that hydrogen will dissociate relatively easily.

 

[chiddler]

After the first alkylation reaction, you have a quaternary nitrogen (a nitrogen attached to three carbons and one hydrogen). The pKa of an ammonium group is ~10, so that hydrogen will dissociate relatively easily.

oh!

thanks :-3