Organic Question

[poc91nc]

Advertisement - Members don't see this ad

Take methylenecyclohexane. If we were to treat this compound in the presence of HBr and organic peroxide...we would get anti-markovnikov addition right? The initial radical is generated from the cleavage of the organic peroxide...which propagates the formation of a radical bromine. The final reason for anti-markovnikov addition is that addition of the bromine radical to the carbon with more H's gives a tertiary radical and more stability.

For the NBS reaction, a bromine radical is also generated. Why does the SAME bromine radical abstract an allylic hydrogen and not attack the double bond as in the HBr/ROOR reaction? The allylic radical is the most stable....therefore it is most likely to be formed...I know the argument for why allylic radicals are formed over other radicals (including tertiary). HOWEVER....why does that line of reasoning not apply to the HBr/ROOR reaction?

 

[skyisblue]

NBS Bromination of an alkene is always at the allylic position b/c of the stability of the allylic radical due to resonance so you're right about that.

For HBr/Peroxide, are you talking about like a hydroboration reaction of an alkene? If that is the case, then that mechanism is entirely different than that of NBS Bromination of an alkene.

 

[poc91nc]

That's the question...why is the mechanism different??? Why does the SAME Bromine radical not abstract an allylic hydrogen (in the HBr/ROOR reaction) to generate a product with substitution at the allylic position (preceeded by an allylic radical intermediate)? Allylic radicals are the most stable. Instead we get the anti-markovnikov product that is preceeded by a radical intermediate formed from the homolytic cleavage of a pi bond. However, the justification of the anti-markovnikov product (for example try the reaction with methlylenecyclohexene, 1-propene) is that BECAUSE it is preceeded by a more stable radical intermediate...in the case of methylenecyclohexene we get a tertiary radical intermediate...1-propene we get a secondary. If, for both of the previously mentioned compounds, we added the bromine radical in a markovnikov fashion...we generate radical intermediates of LESS stability. Therfore, anti-markovnikov addition is justified by stability. Nonetheless, the SAME type of stability argument is applied to the NBS reaction that yields an allylic halide. Allylic radicals are more stable..therfore we get substitution at the allylic position.

Hence the question...why does the SAME bromine radical react one way in HBr/ROOR to give one type of product....yet another way in the NBS reaction to give allylic substitution?

 

[poc91nc]

Another way of stating the question is....why don't we "GO FOR BROKE" in terms of stability and generate an allylic radical in the HBr/ROOR reaction? What is it that prevents us from generating an allylic radical and hence a product with allylic substitution in the HBr/ROOR reaction?

 

[skyisblue]

I suspect that it's simply due to reaction conditions.

The way I've studied it is that Alkene + NBS = bromination at allylic position due to stability via resonance structures.

http://members.aol.com/logan20/alkenes.html

Here's a good link to the mechanism for peroxides and bromination of alkenes.

 

[MSTPbound]

Another way of stating the question is....why don't we "GO FOR BROKE" in terms of stability and generate an allylic radical in the HBr/ROOR reaction? What is it that prevents us from generating an allylic radical and hence a product with allylic substitution in the HBr/ROOR reaction?

It's a good question. If I didn't have to get back to MCAT prep, I would try to work it out, but here's my guess; if you compare the differences in enthalpy changes for the intermediates and products for both possible mechanisms in both reactions, you would probably get precisely the answer to your question. I'm certain that differences in the mechanism, like the fact that bromine is kept in low concentration for the NBS reaction since it gets regenerated, alter the kinetics and thermodynamics of each process to result in the product distribution you see.

If you don't get a more definite answer in a week, I'll stop back in this forum and try to mess around with it. If you DO get a more solid answer (or come up with it yourself!), I'll be back to read it for my own edification!🙂

Take Care,

MSTPbound

 

[MSTPbound]

Ok, here we go.

So it's more or less what I suspected, and perhaps not as complicated to understand as I thought. It does come down to reaction kinetics and thermodynamics (doesn't everything?).🙂

Why does the SAME bromine radical abstract an allylic hydrogen and not attack the double bond as in the HBr/ROOR reaction? The allylic radical is the most stable....therefore it is most likely to be formed...I know the argument for why allylic radicals are formed over other radicals (including tertiary). HOWEVER....why does that line of reasoning not apply to the HBr/ROOR reaction?

Like everyone has been saying, it's in the mechanism. The same line of reasoning "can" be applied to the HBR/ROOR reaction and, in fact, it DOES take place… but the peroxide-catalyzed reaction is much FASTER.

If just a little bit of peroxide is present, you get a mixture of products; but an appreciable amount renders the peroxide reaction so much faster that only the anti-Markovnikov product is observed. This is called the “peroxide effect” and only occurs with HBr addition to alkenes.

That’s why you won’t get the effect with HCl; because the radical reaction is strongly endothermic, so the Markovnikov product is observed, even if peroxide is present. It's also why this doesn't become an issue in the NBS reaction; you down't have a catalyst like peroxide throwing a competing, faster paced reaction into the mechanism.

Hope this helps. I re-read my previous post and was afraid it was a little ambiguous.

Cheers,

MSTPbound

 

[poc91nc]

With all due respect, I appreciate the insight, but your explanation of the organic peroxide mechanism does not explain my orignal question. I know how the HBr/ROOR mechanism works.

"Like everyone has been saying, it's in the mechanism. The same line of reasoning "can" be applied to the HBR/ROOR reaction and, in fact, it DOES take place… but the peroxide-catalyzed reaction is much FASTER."

In the case of HBr/ROOR, the peroxide is cleaved homolytically which in turn propagates the formation of a bromine radical from HBr. That bromine radical attacks the double bond to give a carbon radical. I have made it clear, in the previous posts, as to why the final compound is anti-markovnikov.

Insert thought experiment with any compound with an allylic hydrogen:

Imagine the generation of a bromine radical in the standard fashion as dictated by the peroxide mechanism. Let us use the the compound methylenecyclohexane....

Choice 1: bromine radical cleaves the double bond by homolytic cleavage and the bromine adds to the carbon with more H's. RESULT: A tertiary radical.

Choice 2: bromine radical cleaves an allylic hydrogen to give an allylic radical.
Result: Allylic radical

Now...which is more stable? tertiary radical? or allylic radical? The allylic radical is more stable...

Right off the bat...we can't say anything about the activation energy/kinetics (unless you have tabulated values readily accessible)....only the start and final products. We only have enthalpic generalizations (stability) which imply nothing about kinetics. In the above case...a tertiary radical is formed to generate the anti-markovnikov product we are so used to. That is the paradox that must be resolved. You imply that it DOES take place....but don't you see...for stability reasons...if it does take place, we would get a totally different product in high amounts. Yet, when this reaction is done...you get almost exclusive anti-markovnikov addition to the double bond.

Hope this makes the question a bit clearer...sorry I'm not the most articulate person.

 

[poc91nc]

I am only concerned with the reaction involving the bromine radical and carbon compound. Anything preceeding the formation of the radical gives the same result (regardless of whether you use NBS or ROOR)....the bromine radical. That is the reference point. Starting from the reference point...if we agree that allylic radical formation is preferred over a tertiary radical...we must be forced to swallow the poison and somehow justify a product that does not seem to exist.

You may be correct in assuming the kinetcs are faster since the enthalpic generalizations of both radicals make no implication regarding kinetics. But...where does the mechanistic justification lie? You have to make some argument for stability of the activation complex. Even still.....you cannot explain the absence of allylic substitution in HBr/ROOR.