Free Radical Halogenation in Alkenes

[justadream]

Let's say you have 1-butene. If you do FRH, why is the product (according to TBR*):

1-bromobutane

Why does the Br add to the alkene (as opposed to one of the normal sp3 carbons)?
Also, why does FRH in alkenes favor adding the halide to the the less substituted side?

*I can't seem to find the page at this moment but this is from what I remember

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[sillyjoe]

Let's say you have 1-butene. If you do FRH, why is the product (according to TBR*):

1-bromobutane

Why does the Br add to the alkene (as opposed to one of the normal sp3 carbons)?
Also, why does FRH in alkenes favor adding the halide to the the less substituted side?

*I can't seem to find the page at this moment but this is from what I remember

This is interesting. I don't remember them discussing free radical halogenation of alkenes. Do you at least remember if it was book 1 or book 2? I only remember a discussion of free radical halogenation of alkanes and the different selectivity depending on the reaction rate which forms a kinetic and thermodynamic product. For example, Cl free radical halogenation of an alkane is much faster and therefore less selective than Br free radical halogenation of an alkane)

 

[justadream]

I was wrong - it's not in TBR.

It's actually from TPR SW Organic Chem Passage 6 #4.

"The major product from the radical addition of bromine to 1-butene would be:"

Answer: 1-bromobutane

"The addition of Br radicals to alkenes gives a predominance of the less substituted product"

So now that I think about it, maybe this it not referring to Free-Radical Halogenation of alkanes but instead to Free radical addition of HBr to alkenes in which you get the anti-Mark product (which I thought was not a testable topic?)

 

[sillyjoe]

I was wrong - it's not in TBR.

It's actually from TPR SW Organic Chem Passage 6 #4.

"The major product from the radical addition of bromine to 1-butene would be:"

Answer: 1-bromobutane

"The addition of Br radicals to alkenes gives a predominance of the less substituted product"

So now that I think about it, maybe this it not referring to Free-Radical Halogenation of alkanes but instead to Free radical addition of HBr to alkenes in which you get the anti-Mark product (which I thought was not a testable topic?)

I am just using TBR for Organic and I don't remember that being discussed. Sorry I can't be of more help!

 

[y123]

Found this:

Should the answer be Br2+1-butene --> 1,2-dibromobutane?

 

[DrknoSDN]

this it not referring to Free-Radical Halogenation of alkanes but instead to Free radical addition of HBr to alkenes in which you get the anti-Mark product (which I thought was not a testable topic?)

Yes... Details Here

Specific Alkene reactions were on prior versions of the AAMC organic topic outline but are no longer listed on the most recent version. Just remember that anything is testable. AAMC puts experimental questions into actual tests that are not weighted into your final score. This information can be found on the official website.

 

[amy_k]

If it is radical addition of Br2 it would give dibromobutane. You could read this as the addition of "a bromine radical" i.e. Br* to 1-butene which would give 1-bromobutane, it is not very clear.

 

[BerkReviewTeach]

Let's say you have 1-butene. If you do FRH, why is the product (according to TBR*):

1-bromobutane

Why does the Br add to the alkene (as opposed to one of the normal sp3 carbons)?
Also, why does FRH in alkenes favor adding the halide to the the less substituted side?

*I can't seem to find the page at this moment but this is from what I remember

TBR removed that reaction from their books about ten years ago, because it is NOT tested on the MCAT. Alkene reactions were removed starting in 2004. I think you may be mistaking your source or you are using TBR books that are over ten years old.

Either way, you are wasting precious study time with subjects that will not appear on your exam.

 

[justadream]

TBR removed that reaction from their books about ten years ago, because it is NOT tested on the MCAT. Alkene reactions were removed starting in 2004. I think you may be mistaking your source or you are using TBR books that are over ten years old.

Either way, you are wasting precious study time with subjects that will not appear on your exam.

Yes @BerkReviewTeach is correct. I went back and my source was TPRSW - not TBR.

 

[neuroresearchguy23]

I'm not sure about it being on the exam but I think the reason it goes to the less substituted side is because with radical reactions (along with peroxides) anti markovnikov (spelling might be wrong ... Sorry) is presented. Reasoning behind why it's anti markovnikov... Honestly I'm not 100% sure but I think that's out of the scope of the mcat. Anyone want to confirm or deny this?

 

[Cawolf]

@nsharma23

Yes, the Br radical is added first (from HBr) and adds to the more substituted carbon to form the more stable radical intermediate.

This radical then abstracts H radical from HBr to complete the addition.

I am relatively sure this is out of the scope of the MCAT.

ROOR --> 2 RO•
RO• + HBr --> ROH + Br•

CH2=CH2CH3 + Br• ---> CH2BrC•H2CH3 + HBr

CH2BrC•H2CH3 + HBr --> CH2BrCH3CH3 + Br•