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    hello!
    I'm having some problems with a simple substitution reaction on halogenoalcans.
    An exercise is asking which reactives you would chose to prepare the 2-iodobutane.
    The correct version proposed is 2-bromobutane and iodine anion.
    My problem : as the linking strength C-Br is stronger than C-I, is the reaction going to proceed anyway? with a small result then?
    isn't there any other reactive that one could chose to have more efficient result?
    many thanks in advance for your help
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    Thats a very good question, yes normally the Br- ion would displace the iodine, however there are ways to get around this using different conditions. Although you do not need to know this for A-level, if you dissolve NaI in acetone and add the bromide to the solution a precipitate forms... which is NaBr which is not soluble in acetone and so is removed from the reaction before it has time to displace the Iodine. It's call the finkelstein reaction and it works very well
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    Another suggestion is probably although the c- Br bond is stron, the bromine is however more electronegative than carbon and so polarise the bond.

    As the bond becomes polarised, it is weakened, and when iodide ions are present around such a compound, the bromide would 'like' to leave, as it is a better leaving group and therefore once it leaves, the iodide ions can come in and hence replace it.

    That is what i think though.
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    Thank you both!
    this Finkelstein reaction is great, many thanks EierVonSatan! (I'm not preparing for A-level but for an university oral exam so this is very good )
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    (Original post by Lounor)
    Thank you both!
    this Finkelstein reaction is great, many thanks EierVonSatan! (I'm not preparing for A-level but for an university oral exam so this is very good )
    Ah okay then...you're welcome
 
 
 
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