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Topic: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene  (Read 19766 times)

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Offline KSUPaintballa

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Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« on: January 24, 2010, 03:31:37 PM »
 ;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!


Offline stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #1 on: January 24, 2010, 03:51:44 PM »
Forums rules says you need to show your attempt first.
What do E and Z mean for a double bond?

Offline KSUPaintballa

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #2 on: January 24, 2010, 04:08:01 PM »
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?

Offline bromidewind

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #3 on: January 24, 2010, 07:19:42 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).

Offline stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #4 on: January 24, 2010, 08:31:56 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.

Offline KSUPaintballa

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #5 on: January 24, 2010, 10:58:04 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?

Offline stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #6 on: January 25, 2010, 07:53:06 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.

Offline bromidewind

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #7 on: January 25, 2010, 02:42:21 PM »
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.

Offline stewie griffin

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Re: Drawing (3E,5Z)-5-ethyl-3,5-nonadiene
« Reply #8 on: January 25, 2010, 02:59:24 PM »
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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