When to use HBr

[911dds]

This might be the dumbest and easiest question thus far on SDN, nevertheless here I go:
HBr reagent rxn are so simple, but for some reason, I've got some things messed up in my head. Some HBr reagent rxn place the Br at the more Substituted carbon (Mark.) and in other rxn do not. When is HBr Mark and when is it not? Is there a rule of thumb I'm missing such as: when using reactant X, HBr is Mark. thanks for your help.

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[joonkimdds]

I am feeling really high so my answer might be edited when I come back after 10 hrs...but I think it's with presence of ROOR that causes anti M's rule.

 

[Svart Aske]

I have a related question: I know that addition of Br using peroxide initiator (ROOR) is anti-Markovnikov, but I'm not so sure about UV light. If we're using HBr and UV on an alkene, will the Br add to one of the C's in the double bond or will it add to an adjacent carbon because of resonance?

 

[Sublimation]

I have a related question: I know that addition of Br using peroxide initiator (ROOR) is anti-Markovnikov, but I'm not so sure about UV light. If we're using HBr and UV on an alkene, will the Br add to one of the C's in the double bond or will it add to an adjacent carbon because of resonance?

Good Q. Okay this is how such a situation will react. Now if u have HBr and UV light u will form a radical of Br and H. Now what happens is the, Br will add to the alkene>>one electron from the Pi bond goes to the most sub. carbon(cuz its most stable there) and the other will bond with the Br. [remember that you will get a mixture of both but the major product will be present more then the other] now we have a radical that of the former alkene which is a radical alkane, now this radical will attack the HBr pulling the H off and forming a Br. radical. Remember the light is the initiation, and the propagation takes it from there its a chain reaction. Then termination steps will end it. Hope that helps. Remeber HBr with no peroxide or light will form a mark. halide.

 

[911dds]

Remeber HBr with no peroxide or light will form a mark. halide.

Ok..that was what I was looking for! Thanks man.
How about Br2? They always form the most substituted halide when reacted with light...right? also, what is NBS with peroxide? Do they always form the less substituted Br?

Thanks for the feedback.

 

[joonkimdds]

For the free radical question, don't forget to puts Br on a stable carbon. BUT! if it's chlorination, Cl does not care whether C is stable or not. It makes a big difference and you will often encounter a question asking about chlorination and how many stereoisomers it will make...etc.


by the way, i don't remember seeing Br2 with light.

 

[doc3232]

If a radical is involved then it must go most substitued carbon. This leaves Bromine with the least substitued one.

NBS: Br goes on allylic position

 

[joonkimdds]

If a radical is involved then it must go most substitued carbon. This leaves Bromine with the least substitued one.

NBS: Br goes on allylic position

shouldn't it go to the most substituted carbon? I thought Br attaches to the most stable one.

 

[911dds]

I think Br2 reagents reacts like the HBr. If it reacts with light or ROOR group, it attaches to the least substituted...somebody correct me if I'm wrong.

 

[doc3232]

shouldn't it go to the most substituted carbon? I thought Br attaches to the most stable one.

When you have radicals, the radical takes the most substituted.

 

[Sublimation]

I think Br2 reagents reacts like the HBr. If it reacts with light or ROOR group, it attaches to the least substituted...somebody correct me if I'm wrong.

As in any reaction the variables play a major role. Now if you add a dihalide to a alkene. You will get a anti addition. There is no markovnikov or antimarkovnikov. You form a racemic mixture. because it can attack from the top or bottom of the Pi bond. Now if you take a alkane and a dihalid, and you add some every i.e. light<<<<initiationstep] will undergo afree radical reaction. chlorination and bromination are 2 examples, and remeber the propagation steps. this will also for a markovnikov product. However beware that the H that is attached to a C that has more S characteristic will be harder to pull off SP>SP2>SP3.