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Ochem RoadMap 3 Question


Full Member
10+ Year Member
7+ Year Member
Nov 20, 2006
  1. Dental Student
    I've been studying the Destroyer roadmaps all day and I have most of it down :D

    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 :p


    Full Member
    10+ Year Member
    May 15, 2008
    1. Pre-Dental
      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.
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