Conceptual Orgo Question. Please Help!

[skyisblue]

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A cumulene is a compound with 3 adjacent double bonds. This is it's general formula: R2C=C=C=R2

The end carbons are sp2-hybridized and that all the substituents "R" all lie in the same plane. BUT how can cumulene then exhibit cis-trans isomerism??

 

[Chga]

i dont think you can have cis, trans isomerism

 

[skyisblue]

that's what I thought, but my textbook says that you can. Anyone know this?

 

[dentstd]

This is not normal "cis-trans" isomerism. The "cis-trans" isomers the book is trying to get you to see are: one where the two R2 groups are in the same plane, and one where those two groups are in perpendicular planes.

This is because the unique configuration of the carbon Pi bonds. The central Carbon has two Pi bonds. Those Pi bonds around it can take one of two forms.... on the X-plane, and on the Z-plane. In one configuration, the two Pi bonds are X-X, while in the other it is X-Z.

Hope that makes sense. Draw out the orbitals on a sheet of paper and it'll be easier to see.

 

[skyisblue]

I know what x y and z coordinates are, but x y and z planes are hard for me to picture. Do you know of a link?

 

[skyisblue]

the middle carbon is sp-hybridized so I'm not sure how the pi bonds can be in different planes.

 

[dentstd]

the Pi bonds are in different planes, not the sigma bonds. The sigma bonds are on one axis.

and by x-plane and z-plane, I meant thru the x-axis and z-axis. whoops.

 

[dentstd]

 

[skyisblue]

First off, how did you add a picture onto this thread?

Anyhow, from your picture it looks like one isomer is planar. Every atom is on the same plane.

The second isomer is like a chair with the middle carbon being the intersection between the seat and the back of the chair. So basically one R2 group is on the plane of the paper and the other is sticking up perpendicular out of the paper or sticking down perpendicular behind the paper. Is this what you're getting at?

That's so weird because the end carbons are sp2-hybridized so we have both R2 groups on the same plane with a vacant p-orbital perpendicular to the sp2-hybridized orbitals. The middle carbon is no doubt an sp-hybridized carbon with 2 vacant p-orbitals that make up the pi bonds and we know that sp is linear........


I have another finicky question posted on this forum on ammonia-solvated electrons. I would appreciate it if you can look at that too.....

 

[dentstd]

First off, how did you add a picture onto this thread?

Anyhow, from your picture it looks like one isomer is planar. Every atom is on the same plane.

The second isomer is like a chair with the middle carbon being the intersection between the seat and the back of the chair. So basically one R2 group is on the plane of the paper and the other is sticking up perpendicular out of the paper or sticking down perpendicular behind the paper. Is this what you're getting at?

That's so weird because the end carbons are sp2-hybridized so we have both R2 groups on the same plane with a vacant p-orbital perpendicular to the sp2-hybridized orbitals. The middle carbon is no doubt an sp-hybridized carbon with 2 vacant p-orbitals that make up the pi bonds and we know that sp is linear........


I have another finicky question posted on this forum on ammonia-solvated electrons. I would appreciate it if you can look at that too.....

Yep, that's the distinction between the two molecules. One's entirely planar. In the other, one R2 group is perpendicular to the other R2 group.

The central Carbon is still linear and sp hybridized in both. Just that the unique double bond-double bond combo allows for a rotation.

If you have more questions, ask away and someone may answer, at their leisure.

 

[skyisblue]

Mad props to you Dentstd. You'll get into UCLA for sure!!! Thanks for your help.