Ochem RoadMap 3 Question

[NuckingFutz]

I've been studying the Destroyer roadmaps all day and I have most of it down

However, there is a couple things that are confusing me on the 3rd Road Map (probably due to brain fatigue).

From the starting point, going either left or right yields in the binding of Br to the most substituted carbon. However, at the very top to the left, we see the Br from the HBr acid binding to the LEAST substituted carbon. I'm probably missing something simple but I can't really think of the reason why at the moment...

At first I thought that because Br- is a weak base, it would be an E1 mechanism and thus bind to the most substituted carbon, but this isn't the case with the reaction having the HBr reagent.

Thanks

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[DRHOYA]

Very good question. The way I like to think of it is this. At the starting compound in the center, you react that with Br2 under hv. In this rxn, the Br will add to make the most stable, which here would be adding to the C with the CH3 group attached which is a tertiary carbon. In the rxn at the top of the page, the difference arises in 2 aspects. 1.) There is no longer a CH3 group attached to the cyclopentane, but a =CH2. 2.) The rxn this time is HBr/ROOR. In this rxn, b/c of the double bond and rxn conditions, the Br adds to the least substituted carbon in an elimination rxn. This is why you get CH2Br. Hope that helps.

 

[osimsDDS]

just to add something important that i actually learned yesterday...HBr/ROOR adds anti-mark while HCl/ROOR adds mark...

 

[NuckingFutz]

Thanks for the tip!

On another somewhat related note... how do you know when a group is added in either a syn or anti orientation? When it comes down to stereo chemistry, I always fail at choosing -__-

 

[DRHOYA]

just to add something important that i actually learned yesterday...HBr/ROOR adds anti-mark while HCl/ROOR adds mark...

Ohh, thats good to know. Does that have anything to do with the fact that Cl is less selective than Br, and will go to form most stable via mark??