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    Reaction: but-2-ene + hydrogen chloride. Forms a racemic mixture of the stereoisomers of 2 chlorobutane.

    Consider the shape of the reactive intermediate involved in the mechanism of this reaction and explain how a racemic mixture of the stereoisomers of 2-chlorobutane is formed.


    I know that something is planar, and so can be attacked above or below the bond, but I don't know what?
    My answer "the C=C bond in but-2-ene is planar and can be attacked by hcl above or below the bond. there's an equal chance of each enantiomer forming."

    Wrong.

    I have no clue how to mention the carbocation. How do I know what is planar?
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    The racemate mixture has both enantiomers, but if you have one type by itself you can distingush if it's L or D by using polarised light.
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    Sorry, I changed the question
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    The planar part is the carbocation intermediate:

    CH3-CH2-C(+)H-CH3

    The carbocation, which I put as C(+), has a trigonal planar arrangement, and can be attacked from above or below with equal probability, so there's an equal probability of the resulting C-Cl bond pointing up or down. This leads to a racemic compound.

    Just for your interest, as you seemed to be asking this before you changed the question, you can see the separate enantiomers in a racemate with chiral HPLC (high performance liquid chromatograhy). I'm not sure how much you know about HPLC but the two enantiomers will interact differently with the chiral environment they're being passed through so you get two separate peaks in the chromatogram. You can also do this with 1H-NMR by adding something calling a chiral shift reagent into the NMR sample. This splits peaks for each enantiomer. If you understood nothing about this last paragraph, ignore it
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    (Original post by Kyri)
    The planar part is the carbocation intermediate:

    CH3-CH2-C(+)H-CH3

    The carbocation, which I put as C(+), has a trigonal planar arrangement, and can be attacked from above or below with equal probability, so there's an equal probability of the resulting C-Cl bond pointing up or down. This leads to a racemic compound.

    Just for your interest, as you seemed to be asking this before you changed the question, you can see the separate enantiomers in a racemate with chiral HPLC (high performance liquid chromatograhy). I'm not sure how much you know about HPLC but the two enantiomers will interact differently with the chiral environment they're being passed through so you get two separate peaks in the chromatogram. You can also do this with 1H-NMR by adding something calling a chiral shift reagent into the NMR sample. This splits peaks for each enantiomer. If you understood nothing about this last paragraph, ignore it
    Hey thanks

    So the C=C bond in but-2-ene is not planar? it is the carbocation? So how would I write that as an answer....?
    "The carbocation is planar and has equal prob. of being attacked by HCl above or below the bond". Is that OK?
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    The C=C group is planar, but this is not why you get a racemate. This is because the first step is addition of H+ and whatever carbon you add to you will already have a hydrogen on there so it won't be a chiral centre. The carbocation is also planar and the addition of Cl- will result in that carbon having four different group and thus be chiral. The direction of the attack of Cl- determines which enantiomer you have and seeing as there's equal chance of above and below you get a racemate.

    Your answer is ok but I would say "The carbocation is planar and has equal prob. of being attacked by Cl- above or below the plane". This is because H+ has already been added so you've only got Cl- attacking now. And you're not attacking a bond, but rather the carbocation centre itself.
 
 
 
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