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How to solve these type of Ochem problems?

  1. Jul 24, 2012 #1
    Addition of HCI to 3-methyl-l-pentene gives two products. One of these is 2-chloro-3-methylpentane. What is the other product?

    answer: 3-chloro-3-methylpentane

    Please help me, I have a test tomorrow and I'm unsure of how to do this type of problem. I started out by drawing the molecule, but I seem to have added HCl worng.
  2. jcsd
  3. Jul 25, 2012 #2
    The hydrogen atom and the chlorine atom will add on each side of the double bond.

    The basic steps are: the double bond attacks the hydrogen atom, which is added to the "less substituted" carbon center. The carbon on the other end of the double bond temporarily becomes a carbocation, which is then attacked by the chlorine ion.

    The "major" product (most commonly produced) should be 3-chloro-3-methylpentane. When you have a temporary carbocation, it is most stable on a tertiary carbon center.

    In this case, you initially have a carbocation on either the 1 or 2 carbon (2 is more stable than 1), and in both cases you will have a hydrogen shift, from either the 3 to the 2, or the 2 to the 1, which results in either a secondary or tertiary carbocation.

    This would all make much more sense if it could be drawn out. Look up hydrohalogenation reactions and find some with carbocation shifts.
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