How do you predict when Free-radical-halogenation is Markovnikov or Anti...?

[kevini200]

In Chad's videos, he presents the reaction between tert-butyl and Br2 in the presence of heat and light; the bromine is added to the most substituted carbon (the tertiary carbon)--- (Markovnikov).

I'm reviewing a question from DAT Achiever (Test 1 Question 82) that I got wrong, and it states in the solution that "Free radical addition is anti-Markovnikov where bromine radical selectively attacks the less substituted carbon to produce the most stable form of tertiary alkyl radical."

Now, in Chad's video, the bromine radical is formed from Br2, where in DAT Achiever it is formed from HBr. Does that matter?

Can anyone help me shed some light on this issue?

---

Members don't see this ad.

 

[UBCStudent128]

I can't quite remember what the reagent was, it should be written in the question in DAT Achiever tho.

HBr in presence of "XXXX" creates anti. It's just one of the exceptions you need to remember.

 

[kevini200]

I can't quite remember what the reagent was, it should be written in the question in DAT Achiever tho.

HBr in presence of "XXXX" creates anti. It's just one of the exceptions you need to remember.

OK, do you think its peroxide (ROOR)? Achiever forgot to list that; they just had HBr,heat and light...

 

[wired202808]

OK, do you think its peroxide (ROOR)? Achiever forgot to list that; they just had HBr,heat and light...

Yes HBR, hv and HBR, ROOR is the same thing and goes anti-markovnikov

 

[UBCStudent128]

OK, do you think its peroxide (ROOR)? Achiever forgot to list that; they just had HBr,heat and light...

Yea I think it is.
They should list that though.....

 

[kevini200]

Yes HBR, hv and HBR, ROOR is the same thing and goes anti-markovnikov

OK, so if a question (in the future) has Br2 heat and light, it would go Markovnikov, though- right?

 

[wired202808]

OK, so if a question (in the future) has Br2 heat and light, it would go Markovnikov, though- right?

Yes, just remember where to stick the Br, that also varies on whether the problem is using the starting material as a cyclohexane vs. a benzene.

 

[Golfguy]

OK, so if a question (in the future) has Br2 heat and light, it would go Markovnikov, though- right?

Are you talking about addition to an alkane? If so, a Br will substitute in the most stable radical position. It depends on stability of the position and probability (how many H's can be substituted for (with Cl mostly)).

With an alkene. It adds anti-markovnikov to the carbon with the most H's.

You have to specify what the substrate is. T-butyl is a group....

 

[wired202808]

Are you talking about addition to an alkane? If so, a Br will substitute in the most stable radical position. It depends on stability of the position and probability (how many H's can be substituted for (with Cl mostly)).

With an alkene. It adds anti-markovnikov to the carbon with the most H's.

You have to specify what the substrate is. T-butyl is a group....

WRONG! Br2, hv DOES NOT react with an alkene. Only Br2, CCl4 and other Br2, H2O reagents react with alkenes and the product would be Markovnikov and the addition is Anti. (Its not anti-Markovnikov) Please be careful when spreading out mis-information.

 

[Golfguy]

WRONG! Br2, hv DOES NOT react with an alkene. Only Br2, CCl4 and other Br2, H2O reagents react with alkenes and the product would be Markovnikov and the addition is Anti. (Its not anti-Markovnikov) Please be careful when spreading out mis-information.

How about you stop trolling!

Yes, HBr and ROOR (or hv) add anti-markovinikov, however the OPs question is talking about alkanes. BTW, I never said Br2, ect adds to an alkene. I said Br, and didn't specify the source as HBr, so please stop putting words in my mouth.

 

[wired202808]

How about you stop trolling!

Yes, HBr and ROOR (or hv) add anti-markovinikov, however the OPs question is talking about alkanes. BTW, I never said Br2, ect adds to an alkene. I said Br, and didn't specify the source as HBr, so please stop putting words in my mouth.

You responded to a thread about Br2... which is very different from HBR. Im not trolling Im reading and trying to give the proper information out. instead of making up stuff.

 

[Bis-GMA111]

hm... in my notes i have the 2-methylpropane reacting with Br2--> (hv OR ROOR) resulting in the most substituted alkyl halide...am i missing something here?

 

[Bis-GMA111]

CLARIFICATION:


it's methylenecyclohexane reacted with HBr and hv..which is free radical addition..different from halogenation i believe

 

[wired202808]

CLARIFICATION:


it's methylenecyclohexane reacted with HBr and hv..which is free radical addition..different from halogenation i believe

yes because as I mentioned up top, in terms of an Alkene HBR and UV and HBR/ROOR are essentially two of sides of the same coin (aka the same thing) hence you can reduce them via Anti-Markovnikov. Meanwhile you CANT use Br2 / CCl for Anti-Markovnikov; it would go Markovnikov and the Stereochemistry would be Anti.

 

[jk595090]

If it's Br2 and there is an alkene, doesn't it HAVE to go to the allylic carbon? There are 3 different mechanisms. Hv and really high heat that creates radicals with Br2 must go to the allylic carbon. With HBr or Br2 when it's NOT in peroxide it's markov and when it's IN peroxides it's antimarkovnikov addition.... So there's 3 mechanisms, maybe.... I'm confused by it. I know in allylic you don't get a major and minor product. You only get 1 product which is different than the others

 

[Moka11]

So to revive an old thread and just summarize quickly. Free radicals will always add anti-markov?
If not always, can someone please give 2 examples to differentiate?

Thanks!

The back and forth up top is confusing.

 

[J777]

The very short and simple answer to your question is yes. However, the radical reactions discussed above are still very important, but they do not deal with markovnikov or anti-mark for various reasons (discussed below). Understand the difference between markovikov addition (regioselectivity), syn/anti addition (stereoselectivity), and general selectivity (will the rxn yield the more or less substituted product).

There are several different radical reactions to be aware of.

1.) HBr with ROOR (or HV or heat) + an alkene = anti mark (as you had pointed out)
2.) NBS with ROOR (or HV or heat) + an alkene = allylic brominated product (with this mechanism, you can get a mixture of products if multiple resonance structures can be drawn for the allylic radical) but we don't consider markovnikov here since we are not adding across a double bond
3.) Br2 with ROOR (or HV or heat) + an alkene = bromine is added across the double bond in an anti fashion. Note that this is not anti-markovnikov. Because we are adding two bromines across the double bond, this reaction is not regioselective so we don't even think about mark or anti-mark. Note the difference between anti and anti-markovnikov
4.) Br2 with ROOR (or HV or heat) + ALKANE = most substituted brominated product, but we don't consider markovnikov here since we are not adding across a double bond
5.) Cl2 with ROOR (or HV or heat) + ALKANE = mixture of products, again not considering mark...

Reactions 4 and 5 yield different products because of thermodynamic considerations. Bromination is far more selective and so it only yields the MOST substituted product (for our purposes). Chlorination is an exothermic process and will give a mixture of products (i.e. the chlorine can be placed on any of the carbons of the product because it is not selective).

I hope that this helps you. It is an important concept so if there is still some confusion, plz say so.

 

[Moka11]

wow thanks!