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Author Topic: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene  (Read 10699 times)

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KSUPaintballa

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Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« on: January 24, 2010, 09:31:37 AM »

 ;D Hey guys,

I need some help with this problem.  I need to draw (3E,5Z)-5-ethyl-3,5-nonadiene.  I know there's 2 double bonds on the 3rd and 5th carbon and an ethyl group on the 5th carbon.  I just dont know how to show the 3E,5Z. ???

Thanks!

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stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #1 on: January 24, 2010, 09:51:44 AM »

Forums rules says you need to show your attempt first.
What do E and Z mean for a double bond?
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KSUPaintballa

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #2 on: January 24, 2010, 10:08:01 AM »

Ok, this was my first attempt. Z configuration has high priority groups on the same side of the double bond, and an E configuration had the high priority groups on opposite sides of the double bond.  The only thing is, what priority groups are there at the double bonds?
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bromidewind

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #3 on: January 24, 2010, 01:19:42 PM »

Try building a model. It will help you to visualize the E and Z configurations. Priority groups (in this situation) are based on molecular weight (i.e., length). If you don't have a molecular model kit (which I highly recommend purchasing), you can model it on the computer with several freeware applications (ChemSketch is an excellent program).
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stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #4 on: January 24, 2010, 02:31:56 PM »

So higher priority is determined by the Cahn-Ingold-Prelog rules. In simplest terms, the higher the atomic weight, the higher the priority (thus O is higher than C, F is higher than N, so on). Then there's rules for single versus double bonds. You should read about them and then reattempt. Specifically look at the 5Z alkene. 
Your definition of E and Z is good. Just need to review your C-I-P rules.
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KSUPaintballa

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #5 on: January 24, 2010, 04:58:04 PM »

Ok, I think I got it.  I don't know where the E,Z configuration comes in though.  What makes it E at 3 and Z at 5?
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stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #6 on: January 25, 2010, 01:53:06 AM »

Yeah that looks better. Here's a bit on E and Z naming from wiki: http://en.wikipedia.org/wiki/Alkene
And see http://www.cem.msu.edu/~reusch/VirtTxtJml/sterisom.htm#isom2
It's E at 3 b/c you've got the two higher priority groups (C vs. H with respect to both alkene carbons) on opposite sides of the double bond. It's Z at 5 b/c you've got the two higher priority groups (CC vs C on one carbon and C vs. H on the other alkene carbon) on the same side of the double bond.
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bromidewind

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #7 on: January 25, 2010, 08:42:21 AM »

Just to state the fact, E stands for the German word entgangen, which means opposite. Z stands for zusammen, which means together. That helps me remember how the two work together. I also hate stereochemistry :( it's never really been one of my strong points.

Oh, and the reason that you would use the E/Z system rather than cis-trans is because the priority groups can be ambiguous. If you can name it cis/trans without a doubt, and in a way that would not be confusing to other chemists, then go for it. But if it seems like it might be confusing to other chemists out there, go for the E/Z system. It's always a nice fallback.

And correct me if I'm wrong, but double bonds essentially count as two carbons, correct? Not atomically or anything, just when you're adding up the carbons.
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stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #8 on: January 25, 2010, 08:59:24 AM »

Yeah a C=C would count as two carbons. That's what I was trying to indicate when I said it was CC vs C.
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