[rleung]

Hey,

I am confused as to why the following two pairs of molecules are cis and trans isomers.  The answers in the solutions manual say they are cis and trans isomers.  

To me, the first pair seems like the same molecule.  Besides the double bond that is there, the rest of the bonds are only single ones, and it would not matter where the CH3 group is attached.

In the second one, the molecules seem like constitutional isomers since the double bond is in a different place relative to one of the bonds that links the two ring structures.

Thank you so much for your help.

Ryan

[rleung]

Here is the 2nd molecule

[rleung]

Whoops, I put that on upside down.  Here it is rightside up.

[arnyk]

When you have a double bond as in cis 2-butene and trans 2-butene the two CH3 at either ends cannot rotate about the double bond.  Thus, cis 2-butene has both CH3's on the same side (you can't rotate them from one side to the other) and trans 2-butene has the two CH3's on opposite sides.  They are geometric isomers.

[rleung]

Thanks for the help, but I was actually asking about problem e and problem g in my second scanned picture.  Each set are a cis-trans isomers, but I do not see how.  Thank you.

[arnyk]

Geometric isomers are when two structures have the exact name chemical formula AND IUPAC name, but are geometrically different because of their multiple bonds not allowing for rotation.  For example, in question e) if you straightened out that chain you would not be able to distinguish cis and trans.  However, look at how the carbons branch off the double bonds on either side.  How do they branch off on the first structure's double bond, and how is this different to the way they branch off the second structure?  They would both be C6H14, and called 3-methyl-3-pentene, but one is cis and the other is trans.  

[rleung]

Hmmm, I am still a bit confused.  Basically, to me, cis-3-methyl-3-pentene in problem e looks like trans-3-methyl-3-pentene turned upside down.  Now, I know that double bonds cannot be twisted, but single bonds can be twisted, which means that when cis-3-methyl-3-pentene is turned upside down, the single bonds to the right and left of the double bond are able to twist and turn as much as they like.  Therefore, I do not see how the two structures in problem e are cis and trans isomers.  To me, they look like the same molecule but only one is turned upside down relative to the other.  

What am I not getting?  Thanks so much for the help.

Ryan

[arnyk]

Are you sure they are the same?  Try fliping the second one without touching the double bond and the two carbons attached to either side.  Ignore the end carbons with single bonds, they can rotate as much as they want.  But focus on the middle C1-C2=C3-C4 structure.  The key lies in the way C1 and C4 branch off the C2=C3 double bond.  Remember that C1 and C4 cannot rotate, so however way they branch off, that is where they stay.

If you can get a hold on an organic chem building set, try and build the two molecules.  Being able to see it in 3D really helps.  You will also be able to understand why you can't physically rotate the molecules to make them the same.

[Winga]

About Q.(e), just figure out the cis & trans isomerism of each compound without any rotation, translation. Once you found out the isomerism of that compound, no matter you rotate it (and bond rotations), the isomerism will never be changed.

Have you heard about E & Z isomerism?

About Q.(g), they should be two different compounds (constitutional isomerism).

 

[rleung]

Thank you.

I kind of see how it is different.  In the first molecule, the C1 curves up from the double bond, where in the second molecule, the C1 curves down from the molecule.  Is that the key in identifying the two molecules as isomers?

As for problem g, the solutions manual says it is cis- and trans- isomers, but Winga said it was constitutional isomers.  I see a constitutional isomer myself, but I just want to check with you to make sure

[movies]

The easiest way to tell if you have a cis/trans isomer is to find the highest priority group on one 'end' of the double bond and then look at the other end of the double bond and see if there is an alkyl group on the same side (cis) or the opposite side (trans) of the double bond.  This should minimize any rotating that you have to do, although it is often helpful to try to rotate the structures around so that you can overlay the two and know for sure.

Problem (g) is a tough one because you can't overlay them the way I've described without some major redrawing.  Try applying the first method I described.

By the way, as Winga mentioned, E/Z notation is superior to cis/trans notation.  More on E/Z in this thread:
http://www.chemicalforums.com/index.php?board=3;action=display;threadid=1302

[rleung]

Thanks.

I understand problem e now, but I am still confused about g.  I have not learned E-Z yet, so cis- and trans- is all I have.  I am just beginning Orgo, so I am sorry for sounding stupid and not knowing all the correct terms for everything.  

For problem g, I thought that a double bond in a ring structure cannot have cis- and trans- isomers because of the ring strain?  

Thanks.

Ryan

[Winga]

We can also applied cis & trans (or E & Z) isomerism to the ring systems but not aromatics.

Yes, it's true that you can also assign the cis & trans (or E & Z) isomerism to Q.(g), but I think they should not be geometrical isomeric to each other.

[arnyk]

I think the point the book is trying to emphasize is how to distinguish cis from trans.

cis - same side
trans - opposite side.

That's probably what they want you to identify when looking at the structures.

[movies]

For problem g, I thought that a double bond in a ring structure cannot have cis- and trans- isomers because of the ring strain?  

True, ring strain does come into play at some point, but larger rings can accomodate trans alkenes.  The cut-off point is 8-membered rings (smaller trans-cycloalkenes haven't been isolated).

Do you see how the two structures in (g) are cis/trans isomers?  The way that they are drawn is misleading because it looks like the double bond is in a different place on the ring, but since there isn't another substituent on the ring it doesn't really matter, it's just a consequence of the way it is drawn.  Think of it this way: how would you name the two structures in (g)?