Anti-Mark Question

[BeaverLover]

Why did second anti-Mark rxn add H to #2 when both 1&2 equally substituted?

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[Carbohydrates]

You numbered the chain wrong. You have to number the chain from right to left, which mean your C3 suppose to be C1.

 

[DAT Destroyer]

Why did H get added to C2 instead of C3? Help!

This is poorly written.....confusing indeed. Assuming the author of the problem meant excess HBr/ ROOR.....it was simply 2 consecutive anti-Markovnikov additions. Anti-Markovnikov addition only works when HBr/ROOR is used. This reaction is very very important to know, and is used in organic synthetic procedures on a wide scale.

Hope this helps.

Dr. Romano

 

[BeaverLover]

This is poorly written.....confusing indeed. Assuming the author of the problem meant excess HBr/ ROOR.....it was simply 2 consecutive anti-Markovnikov additions. Anti-Markovnikov addition only works when HBr/ROOR is used. This reaction is very very important to know, and is used in organic synthetic procedures on a wide scale.

Hope this helps.

Dr. Romano

Sorry Dr. Romano, I thought I had it figured out but I'm still confused. I didn't understand the reasoning as to why the second anti-Markovnikov addition had added an H to #2 carbon when both 2+3 are equally substituted. I googled around and it seemed that they would produce a equal mixture of product if the carbons were equally substitued, but the carbons in the examples that I saw were both secondary carbons.

Based on the picture above I would have guessed the second anti-Mark reaction would have added the hydrogen to #3 to form a secondary carbocation and produce 1,2 dibromopropane since both carbons are equally substituted, whereas adding the hydrogen to #2 would be less favorable because of the formation of a primary carbocation (which I believe leads to the formation of the final product depicted above). How am I viewing this wrong? Wouldn't the primary carbocation undergo a hydride shift anyways to become 1,2 dibromopropane and not the answer above? I just don't seem how the final answer is correct

 

[Futuredentista62]

This goes through a radical intermediate, not a carbocation intermediate.
The intermediate formed also has the radical on the most substituted carbon because, in the mechanism, the bromine adds first, not the hydrogen as we're used to seeing.

 

[JTLresces]

Is the radical more stable on the C with Br because Br is highly ekectronegative to stabilize it?

 

[Futuredentista62]

The radical does not for on the carbon with the bromine. It's more stable there because it is on the secondary carbon, not the primary carbon.
Take a look at the mechanism. Hope it helps.

 

[BeaverLover]

The radical does not for on the carbon with the bromine. It's more stable there because it is on the secondary carbon, not the primary carbon.
Take a look at the mechanism. Hope it helps.
View attachment 194712

Interesting...didn't see the formation of a radical, thx.