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Hydrogenation of 1-Octene to n-Octene

  1. Jan 19, 2005 #1
    1) how many moles of h2 gas should be taken up by 1 octene in the courseof this reaction?

    CH3-(CH2)5-CH=CH2 -> (H2PtCl6) -> NaBH4, C2H5OH, HCL -> CH3-(CH2)5-CH2-CH3

    Answer is 2 right???

    2) What volume does this correspond to at STP?

    How can i solve this part?

    3) how many moles of molecular hydrogen are consumed per mole of the precatalyst (H2PtCl6) in order to convert it enterily to platinum metal?

    Isnt this 2 also?? what is the diff between this and the moles of H2 gas for the question #1?

    4) How many moles of precataylst (H2PtCl6) are used in this reaction?

    Again is isn't this 2??

    My answers don't seem right and i am still a little dubios before submitting them...am i misunderstanding a simple concept here?

  2. jcsd
  3. Jan 20, 2005 #2


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    Gold Member

    First, don't be fooled by the number of strange formulae here, just look at the difference between the first and the second formulae; you'll see that two hydrogens are added to the second formula.

    Hydrogen gas is generated by the action of HCl to the alcoholic solution of NaBH4. Four moles of HCl is needed to generate 2 moles of hydrogen gas, which will saturate two moles of the alkene. Hexachloroplatinate acid is the catalyst in this reaction, since platinum (and palladium) has the ability to absorb hydrogen gas inside, and this will catalyze the hydrogenation. Generally, catalysts are used in 1-5% proportion, or even in 0,1%. This depends on the efficiency and turnover number of that catalyst.

    How many moles of 1-octene was used in the reaction? So, you can assume the mole number of hydrogen gas needed. Also, the mole numbers of NaBH4 or HCl are needed, since these are the generators.Then you'll be able to calculate the volume at STP.

    The precatalyst, hexachloroplatinate, seems to absorb two hydrogen ions already, making two hydride ions available to attack the carbocation, i.e., CH-CH2, as I indicated in boldface. In this respect, 0.5 moles seem to be enough in my opinion, since only one hydride and one hydrogen cation are needed to saturate the alkene.

    In your penultimate question, (#3), I think a redox reaction takes place, so platinum(IV) is reduced to platinum(0) with the action of hydrogen gas, so hydrogen will be oxidized to proton, taking six chlorine atoms to form HCl.

    [tex]Pt^{4+}+4e^-\longrightarrow Pt[/tex]
    [tex]2H_2\longrightarrow 4H^++4e^-[/tex]

    [tex]H_2PtCl_6 + 2H_2 \longrightarrow Pt + 6HCl[/tex]

    This will help you, but they are not available for you to directly submit them, You'll have to understand and simplify them in order to submit in your homework. Please be respectful to my effort and ask anything you didn't understand.
    Last edited: Jan 20, 2005
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