Possible Outcomes for Cis-Trans Isomers

[Janiceleong26]

Homework Statement

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"

 

[DrClaude]

I chose B, because since there are 3 C=C in the compound and each C=C constituting in both cis-trans isomers.

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?

 

[Janiceleong26]

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?

6?

 

[DrClaude]

6?

No. Think of it as a coin toss.

 

[Janiceleong26]

No. Think of it as a coin toss.

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?

 

[DrClaude]

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?

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

 

[Janiceleong26]

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

I see. Thanks!