Organic is Yummy!

Chemistry Level 4

Take E t h y n e ( C X 2 H X 2 ) Ethyne(\ce{C2H2}) , and perform the following reactions sequentially.

Reactions

  • Keep it in red hot F e \ce{Fe} tube at 873 K 873K

  • Then add O X 3 / Z n \ce{O3/Zn}

  • Then add H X 2 / P d \ce{H2/Pd} (1 eqv)

  • Then add H X + / Δ \ce{H^{+}/}\Delta

  • Then add H X 2 / P d \ce{H2/Pd} [2 eqv.]

  • Then add H X + / Δ \ce{H^{+}/}\Delta

  • Then B r X 2 / C C l X 4 \ce{Br2/CCl4}

  • Then alcoholic K O H \ce{KOH} (1 eqv.)


Now input the answer as follows.

First of all check if it is alkane, alkene or alkyne.

Let

  • X=1 (For Alkane Product)

  • X=2 (For Alkene Product)

  • X=3 (For Alkyne Product)

And Let Y= number of carbons in the product.

And Z= Number of Halogens attached to it.

Find the value of ( X + Y + Z ) ( X Y + Y Z + Z X ) (X+Y+Z)^{(XY+YZ+ZX)}


Original.

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The answer is 390625.

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1 solution

Suhas Sheikh
May 29, 2018

Initially reductive ozonolysis yields glyoxal(3 moles)which then is reduced to an alcohol The alcohol(of 2 carbons) is subsequently reduced to an alkene via E1 pathway and the Pd reduction sets up conditions once more for electrophilic halogenation by the cyclic bromonium complex Finally alcoholic KOH causes an E2 reaction Allowing one to conclude X=2,Y=2,Z=1 And we're done

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