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Elementary dichlorination rxn

  1. Nov 3, 2004 #1

    Cyclopentane is dichlorinated and I have to write down the structures of all the dichlorinated cyclopentanes formed. I took monochlorocyclopentane and monochlorinated it. The point is that I get confused with the stereochemistry here since the answer is different from the one I have computed. Supposing I take monochlorocyclopentane, I can get the (1,1), (1,2) and (1,3) dichloro derivatives. The (1,1) is achiralic so it poses no problem. That means I get the following products:

    1. 1,1-dichloro cyclopentane. I know that when the second chlorine is cis to the first one, we have a meso product. But when it is trans, are there two possible products (enantiomeric)? If yes, then a racemic mixture should be formed right?

    2. 1,3-dichloro product (same problem as part (1)).

    The correct answer is 5 products of which 2 are optically active. I would be very grateful if someone could give me a detailed description of this (diagrams would be appreciated but a description will do for I can figure out myself).

  2. jcsd
  3. Nov 3, 2004 #2
    It seems natural to me that the following products will be formed

    1. (meso)1,2-dichlorocyclopentane
    2. (rac)1,2-dichlorocyclopentane (this means the trans attack produces its enantiomer too...)
    3. 1,3-dichlorocyclopentane (this wouldn't be optically active since there is no chiral carbon in the first place)
    4. 1,1-dichlorocyclopentane

    Now I need help figuring out why should we call these 5 products (in principle the racemic mixture is a mixture so we get 4 products right?). But if we do consider the racemic mixture as being composed of two distinct products then 5 products are indeed formed and two--the enantiomers--are optically active. So I believe that part is figured out. Is my reasoning correct? Thanks anyway.

  4. Nov 5, 2004 #3
    I'm sorry but I gave the wrong answer: the correct answer is 7 isomers. Could you please help me?
  5. Nov 5, 2004 #4


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    I think you have all of the right compounds. I count 7 isomers if you think of a racemic mixture as a mix of two isomers (stereoisomers).

    Are you confused by the stereoisomer/diastereomer nomenclature?
  6. Nov 7, 2004 #5
    Thanks movies. I had trouble visualizing the 1,3-dichloro product. (Perhaps I still do...yeah I get confused sometimes). Regarding the earlier problem, I was seeing a plane of symmetry in 1,3-dichlorocyclopentane...

    Last edited: Nov 7, 2004
  7. Nov 7, 2004 #6


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    There are three possible stereoisomers for 1,3-dichlorocyclopentane. The meso one does have a plane of symmetry but the other two do not.
  8. Nov 7, 2004 #7
    Rember the total number number of steroisomer possible is 2**n where n is the number of sterogenic center contained with the compound. Beware that this not allways the case. If the compound is meso, it has one less steroisomer.
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