destroyer #90orgo

[Jo07]

Advertisement - Members don't see this ad

What is the deal with it saying "HCL and ROOR simply add Mark." Then saying ROOR will only work with HBr?

 

[sacjumpman]

If you have a peroxide (ROOR) and HBr, then it does a little trick and adds Anti-Markiv.

As opposed to when you don't have the ROOR present then HBr will add Markovnikov.

So what he's pointing out is that HCl doesn't work with this little trick. Instead, it just adds as if the ROOR was not present: Markov.

I was glad he pointed this out, cause I had no idea HBr was the only thing you could add anti-mark. that way.

 

[musl85]

well, just learned something new too hehe👍

 

[fancymylotus]

If you have a peroxide (ROOR) and HBr, then it does a little trick and adds Anti-Markiv.

As opposed to when you don't have the ROOR present then HBr will add Markovnikov.

So what he's pointing out is that HCl doesn't work with this little trick. Instead, it just adds as if the ROOR was not present: Markov.

I was glad he pointed this out, cause I had no idea HBr was the only thing you could add anti-mark. that way.

You're sooooooooooooo smart

 

[Jo07]

Ahhh...I see. What would I do without you guys?

Perhaps you can also explain this to me:

Road map 2 (left bottom corner)

Br2, hv --> anti markov. So shouldn't it add to the least substituted carbon?

Road Map 3

Starting reactant with br2 and hv would add antimark
CH3
that compound is C-C-C-C-C ..but wouldn't it add to least substituted?
Br

I definately prefer a compound in the C-C-C form rather than in any other shape so it's hard for me to see it. Sorry if these questions are stupid. Orgo is not my strong point.

 

[sacjumpman]

Ahhh...I see. What would I do without you guys?

Perhaps you can also explain this to me:

Road map 2 (left bottom corner)

Br2, hv --> anti markov. So shouldn't it add to the least substituted carbon?

This is just your classic radical halogenation. It adds to the most substituted carbon, or in this case, the allylic carbon because of special stability. Think of stabilization of the radical intermediate.

Road Map 3

Starting reactant with br2 and hv would add antimark
CH3
that compound is C-C-C-C-C ..but wouldn't it add to least substituted?
Br

This is the same thing again, as up above. Br2 with light (hv) adds to the MOST substituted carbon, in this case the tertiary carbon.

Again, it wants the most stabilized radical transition state. Perhaps google, "radical halogenation" or find it in your book. This was likely the first reaction you learned in Ochem class.

I definately prefer a compound in the C-C-C form rather than in any other shape so it's hard for me to see it. Sorry if these questions are stupid. Orgo is not my strong point.

See above in bold.

 

[SeeBrick]

radicals, peroxides, UV light... all add HBr anti-Mark...

 

[Mstoothlady2012]

i thought Br2/hv prefers benzyllic hydrogen...

 

[Jo07]

My understanding is that markovnikov is when the most substituted carbon will become more substituted and the H will go to the carbon with the most Hs.

Antimark is when the halogen adds to the less substituted C.

So, in these free radical (antimark) it should add to the least substituted but rearranges to form the most stable product?

I'm sorry that this is redundant.. I'm looking in books and I'm not sure if my head is on right because this is confusing me.

 

[Mstoothlady2012]

well I think Br2/hv prefers benzyllic hydrogens if available like showed in road map # 2...if not then it follow markovnikov like showed in road map # 3