# Visualizing rotation about a double bond

So I'm having trouble visualizing rotation about a double bond without using a model kit. The 2nd image has a red dot where I rotated it to make the molecule look exactly like the second image with my head and then my model kit. This is also consistent with my answer key that says these molecules are identical.

Now, the next two molecules:

I visualize rotating the left molecule at the red dot carbon and in my head, it matches up with the right molecule. However, when I pull out my model kit, it does not rotate like that and does not match up. This is also consistent with my answer key that says these molecules are diastereomers. So my question is, are there any rules that I could use or a way to visualize that will help me realize if a bond can be rotated to match a molecule without using a model kit.

Also, when people talk about it is not possible to rotate a double bond without breaking it, what exactly are they talking about that makes it different from this?

• In your first example, you are not rotating about the double bond but rather the C3-C4 single bond. Both structures are identical. The pair of chlorides are diastereomers, Z and E, respectively. If in either one you change the geometry of the double bond, which is a structural change and not a rotation of a bond, then you will have a pair of enantiomers. To "rotate about a double bond" requires the breaking of a pi-bond, an unlikely event. – user55119 Jun 3 '19 at 2:29
• If your rotate about the C3-C4 bond, the atoms attached to C3 (the H and CH3O) would change from wedge to dash, not making the structures identical. I meant just rotating C4 (the red dot). – Lat Jun 3 '19 at 2:58
• Rotating just the red dot does mean anything. At most is an handle for rotation around the near bond, when you have a model in your hands. – Alchimista Jun 3 '19 at 10:11
• There is a difference between rotating the entire molecule along an axis (no change in conformation and energy) and rotating about a bond (leave one part of the molecule where it is and rotate the other part - change in conformation). – Karsten Theis Jun 3 '19 at 12:41
• The red dot atom is C3. C4 is the carbon bearing the CH3O group. C4 and C5 and everything attached to them stays as is in the Fischer projection. A 180 deg. rotation about the C3-C4 bond gives the two views you have shown in your structures. I hope this issue is clear now. Make a molecular model! – user55119 Jun 3 '19 at 16:11

Adding one more element to help you long-term: I think the underlying reason you are having a problem with this is that the visualization you are using is a little misleading.

You're selecting an atom with your red dot. You can't rotate around an atom. You can rotate around the bonds between the atoms, though, assuming that the constraints of that bond allow it.

So in the first example, there are two possible carbon-carbon bonds you can rotate around that would both correspond to that red dot. Your dot should be on the sigma bond, based on your diagram.

In the second, you have three carbon-carbon bonds connected to your red dot. The one you are rotating is the double bond, not either of the bonds in the ring. So the dot should instead be on that bond.

You can't freely rotate double bonds

Double bonds don't have free rotation. This is "built in" to the nature of the bonding in them: in effect you would have to break a bond to allow the molecule to rotate around a double bond.

In the first picture you drew, you didn't rotate around the double bond but actually rated around the single bond attached to the red dot. The second molecule has no single bond that can rotate (the single bond attached to the red dot is part of a 3-membered ring which locks to bond so it can't rotate). The "rotation" you show is not possible: the apparent result is a structurally different compound that can only be reached by breaking some bonds and not from free rotation.

The results you see with a model kit are right. And the rule you need to remember is that you cannot rotate around double bonds.

It is not possible to rotate a double bond without breaking the double bond . Below 300$$\ce{^o}$$ C, double bonds are configurationally stable; that is, cis stays cis and trans stays trans.$$\ce{^1}$$

In alkenes, there is no free rotation about the double bond, unlike in alkanes. Consequently, groups attached to a double bonded carbon cannot exchange places. Therefore, the cis and trans isomers are separable compounds because they cannot interconvert.

source :ORGANIC CHEMISTRY , PETER VOLLHARDT &NEIL SCHORE pg439

In the first molecule ,red dot is on carbon bonded to a double and a single bond .If rotated along an axis passing through the red dot same molecule is formed.

However,in the second molecule,rotation along the red dot is not possible since the bonds are locked.

Your molecular model kit is correct.