Why Hoffman E produces least substituted alkene??

[johnwandering]

Im wondering why this is.

The OH- is a fairly small base, so why does this reaction Always favor the least substituted??

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[chiddler]

http://www.organic-chemistry.org/namedreactions/hofmann-rule.shtm

 

[johnwandering]

Thanks, id figured.

But what is it about Hoffman that causes the least subsituted alkene to be created?
Whats different about it that makes the thermodynamic product nonexistent?

Most other E2 reactions with OH- at low T produce the most substituted.

 

[MedPR]

Thanks, id figured.

But what is it about Hoffman that causes the least subsituted alkene to be created?
Whats different about it that makes the thermodynamic product nonexistent?

Most other E2 reactions with OH- at low T produce the most substituted.

Hoffman occurs because the base is too bulky to attack and add to the most substituted carbon.

 

[chiddler]

i don't know and i doubt that is required knowledge. we don't even know the mechanism of what the silver oxide does.

 

[MedPR]

i don't know and i doubt that is required knowledge. we don't even know the mechanism of what the silver oxide does.

Knowing that bulky bases result in Hoffman and not Zaitsev is definitely required knowledge.

 

[MedPR]

Well look at me.. I didn't even read the entire OP.

What reaction specifically are you talking about where OH- results in Hoffman?

 

[chiddler]

Well look at me.. I didn't even read the entire OP.

What reaction specifically are you talking about where OH- results in Hoffman?

good point...

wait. there's an OH. oh! it's an elimination!

(i mean i didn't realize it was a traditional elimination reaction)

 

[MedPR]

I guess I'm missing something... There's only one leaving group and the final product is the most substituted alkene. The only other place to put a double bond is on one of the R groups which would make a mono-substituted alkene vs the di-substituted that is created above.

 

[johnwandering]

Oh, so OH- is a strongly hindered base?
(yes the base for Hofmann Elimination is OH- (Ag2O + H2O---> OH- and Ag+)

So for every E2 reaction with OH-, we choose the least substituted Alkene?

 

[MedPR]

Oh, so OH- is a strongly hindered base?
(yes the base for Hofmann Elimination is OH- (Ag2O + H2O---> OH- and Ag+)

So for every E2 reaction with OH-, we choose the least substituted Alkene?

No, I didn't read the part about OH until later. See the above post and let me know if that makes sense.

 

[johnwandering]

No, I didn't read the part about OH until later. See the above post and let me know if that makes sense.

Thanks, but there are usually more than one possible alkene. And the product is always the least substituted alkene for some reason.

http://www.organic-chemistry.org/namedreactions/Hofman1.gif

I understand that there are more hydrogens/a less hindered target in the end carbon, but I was under the impression that OH- was small enough to produce a thermodynamic product...

 

[chiddler]

.

 

[MedPR]

Thanks, but there are usually more than one possible alkene. And the product is always the least substituted alkene for some reason.

http://www.organic-chemistry.org/namedreactions/Hofman1.gif

I understand that there are more hydrogens/a less hindered target in the end carbon, but I was under the impression that OH- was small enough to produce a thermodynamic product...

I thought OH- was small enough as well. I have no idea.

 

[milski]

We just did that today in class. It's an E2, which means that it has to happen in one step and the leaving group and the leaving H have to be antiperiplanar. When you do the newman projections for the case where the most substituted and the least substituted C get the double bond, you should see that E2 happening on the more substituted side requires the leaving group and the rest of the alkyl chain to be in gauche configuration. Since both the leaving group N(CH3)3 and the alkyl chain are really bulky, that does not happen and E2 proceeds at the less substituted C.

BTW: The CH3I in the first step reacts with NH2 and results in N(CH3)3. I don't remember what the deal with the AgO2/H2O/heat was but I think AgO2/H2O lead to primary alcohol and then the heat made the elimination happen. But that would imply OH leaving too, so it does not sound right. Notes from class are not posted yet, so if you want the details, I can come back in a day or two.

 

[johnwandering]

No I think that about solves it.
Thanks man