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I understand the principal of the Cis-Trans naming system (at least I think so), until I saw this example in my learning material. My teacher doesn't know either so I thought I would post it here.

This is the molecule: enter image description here

There are two ethyl groups both on the bottom side of the double bond, so it could be cis right? enter image description here

Apparently not, the book says this is the answer: enter image description here

Could someone please explain the reasoning behind this? This particular example is unusual because there are three ethyl groups attached to the double bond so I cannot determine which one should be the parent chain. The book doesn't explain.

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  • $\begingroup$ Going from right to left...at the three carbon there are two potential ways to go...both lead only to two more carbons in either direction. either one is correct. Don't think that because only one is numbered that is "the" correct parent chain. Just because one is written in condensed form (CH<sub>2</sub>CH<<sub>3</sub>) do not think that there is any difference between them. They are identical. Either number is is exactly the same constitutive name. (That is a terrible book) $\endgroup$ – DrAzulene May 17 '16 at 17:51
  • $\begingroup$ Also, I notice you mention that there are three ethyl groups on the double bond. You must number parent chains so that the olefin pi bond is in the parent chain...is that is so there is no way to incorporate all the carbons in a molecule into the parent. Furthermore, if the ethyl group on 3 were missing...it still would not be "trans"...this book is archaic and mistaken...organic chemists do not use cis and trans to describe olefins..those are RELATIVE terms. Organic chemists correctly use E and Z to designate alkene configuration. $\endgroup$ – DrAzulene May 17 '16 at 18:02
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In my previous (deleted) answer, I overlooked the fact that this molecule has indeed three identical substituents and not two. Therefore, I would consider Greg E's answer as correct.

Also, C3 has 2 identical substituents on the same sp2 carbon, so cis/trans is not possible.

Example from organic chemistry 6th edition Paula Yurkanis Bruice:

enter image description here

So, your book is wrong!

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  • $\begingroup$ Thank you very much for your answer, I will mark it correct because you took the time to include a source as such. $\endgroup$ – Klik Dec 13 '13 at 19:00
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Unless I'm very much mistaken, this molecule should not be labelled with any stereochemical descriptors, whether they be relative (cis/trans) or absolute (E/Z). The double bond has three identical ethyl substituents, therefore there is no stereoisomerism possible. By way of explanation, imagine if the ethyl group and the hydrogen on carbon #4 were interchanged: the resulting molecule would be completely identical to the one pictured. This is always the case when an alkene bears two identical substituents on the same carbon atom.

To correctly use cis/trans or E/Z nomenclature, it is necessary that there not be two identical groups on the same carbon of the alkene; if there are, the labels are meaningless, since interchanging the two groups results in an identical molecule.

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  • $\begingroup$ This is such an incredible relief to hear. I spent so much time trying to understand this since it didn't seem to make sense. Thank you very much for confirming this. $\endgroup$ – Klik Dec 13 '13 at 18:59
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Cis/trans nomenclature for aklenes is usable only for the simplest cases, so you can use it safely when there are two hydrogens on the double bond. If you want to be fully correct, use the E/Z, it's the only bulletproof way.

Cis/trans is also widely used, if you have e.g. chlorine on the double bond, which makes the E/Z nomenclature counter-intuitive (but still correct).

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  • $\begingroup$ So you're telling me that my text book is wrong and that there is no way to classify this specific example using the Cis/Trans system? $\endgroup$ – Klik Dec 13 '13 at 7:48
  • $\begingroup$ In your very example, it does not even matter, because the two substituents on "left" carbon are the same. To put it clear: goldbook.iupac.org/C01092.html $\endgroup$ – ssavec Dec 13 '13 at 8:58
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Any time there are two of the same substituent on a carbon, that carbon makes the alkene non-stereogenic. Consider any terminal alkene (they have two hydrogens atoms on the same terminal carbon). Then no unambiguous relationship can be made between those hydrogens and the groups on the other carbon.

What was circled in the (Khan Academy lol) video is a cis relationship, but the molecule as a whole is not stereogenic and no diastereomer of it exists.

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