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anonymous

  • one year ago

Complete the 6 reactions and for each: a. draw the starting hydrocarbon for each reaction b. complete and balance the reaction c. draw the final structures showing how the compounds are bonded, and d. name the type of reaction 1) cis-1,2-diethylcyclobutane + O2 -> 2) 1-(1-methyl-propyl)cyclopent-1-ene + Cl2 -> 3) Heptane + Cl2 -> 4) E-1,3-dichloropropene + H2 -> 5) trans-2-methyl-3-hexene + H20 -> 6) (2E,4Z)-3-ehtyl-5-methylheptadiene + 2HCl -> I will post what I have done so far below.

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  1. anonymous
    • one year ago
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    |dw:1443380486090:dw||dw:1443380746565:dw||dw:1443380899627:dw||dw:1443381140894:dw||dw:1443381302490:dw| And I have no clue how to do the last one.

  2. Photon336
    • one year ago
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    |dw:1443410277118:dw| well considering your first reaction. the pi electrons in that double bond are mobile they can act as a nucleophile, so what happens is that they attack the chlorine (Cl2) and a chloronium ion intermediate is formed. notice i drew two triangles, well it could be the other way too, because this structure can technically be attacked by the nucleophile, in this case chlorine- from either side. this will give you the anti product. @elfqueen |dw:1443410506403:dw|

  3. Photon336
    • one year ago
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    I assume for this the following would happen. |dw:1443410649031:dw| or wait' maybe it could be this:

  4. Photon336
    • one year ago
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    |dw:1443411090991:dw| I believe that the above depiction is incorrect. If the mechanism went primarily as i first drew it, through a carbocation intermediate, that would imply that there would be some kind of shift VIA a hydrogen, called hydride shift, would shift to form a more stable carbocation intermediate that would react. but from the regiochemistry, the OH seems to like going on the carbon that's Beta to the one where the methyl group is. |dw:1443411341017:dw| I'm wondering though.. it's possible that this could happen through a radical mechanism. that would probably be more plausible @woodward

  5. anonymous
    • one year ago
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    The ones typed below are not a separate reaction. It's the same thing just without the structures drawn out. Sorry for the confusion.

  6. anonymous
    • one year ago
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    And as for the circle of "What's this?" If you look up at the question I posted, it represents trans or "opposite side."

  7. Photon336
    • one year ago
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    |dw:1443411754068:dw| @elfqueen my best guess is that number four is a hydrogenation reaction: If you use hydrogen/ palladium catalyst the hydrogens add both to one side. so the two chlorine groups will either be facing both out or in. but the you can't draw it that way for one reason, the carbons aren't chiral because there aren't four different groups attached to the carbons afterwards.

  8. Photon336
    • one year ago
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    I guess you can draw it the way you did to say hey the hydrogens were both added in back, by the catalyst.

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