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Homework Help: Reaction between propanone and iodine

  1. Sep 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Does iodine take part in the rate determining step of reaction between propanone and iodine in acidic medium?

    2. Relevant equations

    3. The attempt at a solution
    For positive catalyst the rate determining step is the 1st step of the at least 2 steps of reactions after being catalysed(usually??).
    Since the reaction is order 1 with respect to iodine and also to hydrogen ions and 0th order to propanone, and the 1st step should be the reaction between the catalyst and 1 of the reactants, it should be between iodine and hydrogen ions because iodine controls the rate of the reaction and so it is 'rate determining' and take part in the r.d.s.

    Are the hydrogen ions catalyst of the reaction? If so how do they act as the catalyst?? I mean are there like substeps with high reaction rates or it's about something else like adsorption to ions??
    How would I know whether it takes part in the rate determining step or not?
    What should be the answer actually?
    I am highly uncertain about my assumptions in the above deduction, please correct me also!
    Last edited: Sep 20, 2012
  2. jcsd
  3. Sep 20, 2012 #2


    User Avatar
    Gold Member

    Clear up some misconceptions first.

    Catalyst - A reactant which speeds up/down the reaction rate by affecting the formation of intermediate product or altering the reaction mechanism, and is recovered at the end of reaction, so catalyst isn't technically a reactant in the overall reaction.

    The reaction of acetone with Iodine is Nucleophilic addition. In the 1st step, which is also the RDS, the pi-bond in C=O polarizes towards O atom, creating valency vacancy on C atom. Nucleophile attacks on the C atom and forms Tetrahedral intermediate.

    Now note that polarization of pi-bond depends upon the attacking electrophile, so Iodine is involved in R.D.S.

    In the second step, protons are attached to polarized O atoms, then you get your compound. It is a fast step, thus not serving relevance in rate law.
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