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

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

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.

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

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).

 

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

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.

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)?