Radical bromination rxn

[chiddler]

Major product from the radical addition of Br2 to 1-butene would be?

A. 4-bromo-1-butene
B. 3-bromo-1-butene
C. 2-bromobutane
D. 1-bromobutane

Answer: D.

I'd like to add that the passage provides this info: "When reacting with alkenes, bromine radicals tend to add to the least substituted carbon of the double bond, the reverse of the case with alkanes".

So it's easy to see what the answer is within the context of the passage.

Outside this context, I don't understand especially given this.

The specific question is this: How is it that the answer to the question I put is that while I see a similar reaction with a completely different product?

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[syoung]

resonance? when you remove the H on the allylic C (carbon next to double bond) it can resonance stabilize with the double bond. I found that resonance is almost always the best answer for most ochem rxns

 

[syoung]

what is the solvent? H2O is a diff reaction than with CH2Cl2 right?

 

[chiddler]

resonance? when you remove the H on the allylic C (carbon next to double bond) it can resonance stabilize with the double bond. I found that resonance is almost always the best answer for most ochem rxns

can you please explain what you mean? I know there is resonance but i don't see how that explains how the two reactions I wrote are different.

what is the solvent? H2O is a diff reaction than with CH2Cl2 right?

No, alkenes are not miscible in water.

 

[syoung]

Sorry my bad on the Water thing.
FIrst of all
your question asks about 1 butene, the picture shows 1 propene
Second, the first two answers are wrong, because they say butene still and that can't happen when you add Br2 to an alkene, the double bond WILL come off.
Third the passage stated that the Br2 will add via radical rxn TO the least substituted, which will be the left side --> H2-C=CHCH2CH3
Normally if you have a lot of Br2, you will get 1,2 dibromo butane (both on the first carbon) but in this case the answer lists only 1 Br is added.

 

[syoung]

http://www.chemgapedia.de/vsengine/...le/radi_brom_allyl/radi_brom_allyl.vscml.html

But to answer your question, the difference is in the conc of the Halogen. Low = formation of allylic radical (double bond exists) while the high = formation of vicinal halogens (no double bond lefT)

 

[chiddler]

Sorry my bad on the Water thing.
FIrst of all
your question asks about 1 butene, the picture shows 1 propene
Second, the first two answers are wrong, because they say butene still and that can't happen when you add Br2 to an alkene, the double bond WILL come off.
Third the passage stated that the Br2 will add via radical rxn TO the least substituted, which will be the left side --> H2-C=CHCH2CH3
Normally if you have a lot of Br2, you will get 1,2 dibromo butane (both on the first carbon) but in this case the answer lists only 1 Br is added.

Yes my question does ask about 1 butene. Yes the picture does show propene. My question is what is changing in these two reactions that causes such a huge difference in products?

Like I wrote, the answer to the actual passage question was pretty clear. I don't need help with that.

To help clarify, I edited in my specific question a little bit late, but that is what I need help with. Not the passage question.

 

[chiddler]

http://www.chemgapedia.de/vsengine/...le/radi_brom_allyl/radi_brom_allyl.vscml.html

But to answer your question, the difference is in the conc of the Halogen. Low = formation of allylic radical (double bond exists) while the high = formation of vicinal halogens (no double bond lefT)

nice find. then this is a reaction specific enough that some info must be given to us anyway.

thanks for the helpful responses.

 

[Class1P]

At first I was like, "huh?" then I reread the problem and realized it was an Alkene and that the reaction I visualize happening would be the first carbon in the double bond using it's shared electron to attack the Br2 which would be more likely because it would leave C2 as a secondary carbon with a single extra electron and no net charge. The radical is better suited for the secondary carbon because of the electron cloud that can balance it out.