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How many cis-teams isomers

  1. Sep 8, 2015 #1
    1. The problem statement, all variables and given/known data
    image.jpg
    I chose B, because since there are 3 C=C in the compound and each C=C constituting in both cis-trans isomers. But the correct answer is C, how come there are 8?

    Sorry the title should be "how many cis-trans isomers"
     
    Last edited: Sep 8, 2015
  2. jcsd
  3. Sep 8, 2015 #2

    DrClaude

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    Staff: Mentor

    So there are two possibilities at the first double bond, two possibilities at the second, and two possibilities at the third. How many possibilities in total?
     
  4. Sep 8, 2015 #3
    6?
     
  5. Sep 8, 2015 #4

    DrClaude

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    Staff: Mentor

    No. Think of it as a coin toss.
     
  6. Sep 8, 2015 #5
    Well.. In a coin toss, the probability of getting either a heads or a tail is 1/2. So if we toss three times, then the total probability of getting either a heads or a tail would be (1/2)^3 , am I right?
    But if we were to choose the number of ways, shouldn't it be..permutations?
     
  7. Sep 8, 2015 #6

    DrClaude

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    Staff: Mentor

    Right. But you should be looking at how many possible outcomes.

    You're so close that I'll give it away: you have to multiply the possibilities, so you get 23 = 8 possible outcomes:
    CCC CCT CTC TCC CTT TCT TTC TTT
     
  8. Sep 8, 2015 #7
    I see. Thanks!
     
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