[thedy]

What does reference plane mean in cis-trans isomers?How I know,which of all possible planes,that has a molecule is reference plane.Is this reference plane a convention of scientists?I was trying to find on net some pictures of reference plane,but I found nothing.
Thanks

[Dan]

The appropriate reference plane of a double bond is perpendicular to that of the relevant s-bonds and passes through the double bond. For a ring it is the mean plane of the ring.

Source

[fledarmus]

I believe that what you are talking about is a convention that is particularly applicable to cis-trans isomers in a ring system. For a double bond, it is fairly easy to see what cis and trans means - cis has both substituents on the same side of the double bond, trans has the substituents on opposite sides of the double bond. For a ring system, which is three dimensional and bent, "same" and "opposite" are somewhat ambiguous. So by convention, you draw the ring system flat, contained in a reference plane, with substituents either above or below the plane. If both substituents are on the same side of the reference plane, they are considered cis, if they are on opposite sides, they are considered trans. This allows you to define cis or trans orientation for two substituents whether they are axial or equatorial, and whether they are on adjacent carbon atoms or all the way across the ring from each other.

Not a great picture, but I've attached a diagram of 2 different 1,4-dimethylcyclohexanes below so you can see what I am talking about. I hope this makes sense...

[thedy]

I see,thanks to you for answers.
One more question.How much is rigid molecule in cis-trans molecules?They can rotate a little bit?Or not?Cis-trans isomers are boundary possibilities?For example if I have cis isomer,does it mean that this cis isomer can rotate,but only a little bit and thus,cis isomer cannot re-form to trans isomer(ordinary conditions).So cis isomer can be skewed,but it cannot be so much,that can form trans isomer.That means,cis isomer cannot reach other plane of molecule,but it can be exactly at the border of trans isomer.
So,what do you say?
Thanks

[fledarmus]

If you are considering rotation around double bonds, you have to realize that the double bond will only form if the p orbitals from the two atoms in the double bond can overlap. They will have maximum overlap when they are parallel to each other, and zero overlap when they are perpindicular. To rotate around the double bond, you will be passing through a point where they are perpindicular, which means you must completely break the double bond. The barrier to rotation will be at least the bond strength of the pi bond which has to break. A carbon-carbon pi bond has a bond strength of around 70 kcal/mole, which is a lot of energy.

In short, no, under normal conditions you do not get rotation around double bonds.

[thedy]

 They will have maximum overlap when they are parallel to each other, and zero overlap when they are perpindicular.  

And what case,when they are close to be perpendicular,hard upon,but not totally.They are still cis isomers,or they goes to trans form?Or something between? :)Strange question,I know,but that I am.
Thanks,I hope,that this is my last question..

[sjb]

 They will have maximum overlap when they are parallel to each other, and zero overlap when they are perpindicular.  

And what case,when they are close to be perpendicular,hard upon,but not totally.They are still cis isomers,or they goes to trans form?Or something between? :)Strange question,I know,but that I am.
Thanks,I hope,that this is my last question..

I think you can define cis isomers as those having a dihedral angle of between 0 and 90°, and trans between 90° and 180° for the highest priority groups or similar. Perhaps look into atropomers?

[thedy]

I think you can define cis isomers as those having a dihedral angle of between 0 and 90°, and trans between 90° and 180° for the highest priority groups or similar. Perhaps look into atropomers?

Understood,thanks

[thedy]

Hi,what does it mean,when I say equatorial hydrogens in cyclohexane?Carbons in cyclohexane have sp3 hybridization,so that means,hydrogens cannot be equatorial,can be?These hydrogens are not in one plane with carbons.That my first question.But I have another.
When I have chair conformation of cyclohexane,C2,C3,C5,C6 are in one plane.So that means,C1 and C4 are above,and below this plane.And that means that hydrogens on C1 are above plane,and hydrogens in C4 are below plane,or not?I m confused with these planes.Another question.
In addition to this plane(C2,C3,C5,C6)we have also other planes.Between C1-C6 and C3-C4 and between C1-C2 and C4-C5,they are also parallel,and thus they lie in one plane.So,that means,it is relative which carbon and which hydrogen lies in which plane,and which hydorgen/carbon is above or below the plane.
I think,that for example C1 carbon in chair conformation is above the plane(for example)and thus all carbons attached to this carbon is also above the plane.But only if I mean with respect to C2-C3 and C5-C6 plane.If I mean plane with C1-C6 and C3-C4,then C2 is above the plane and thus hydrogens are too above the plane,and C5 is below the plane etc.....But how we can see,it doesn t hold,because  according a web sites equatorial hydrogens for example are in one plane,but according a me,they are above or below the plane due to carbon.Why?
thanks a lot for patient

[fledarmus]

I think you are trying to be too pedantic with your use of the word "equatorial". Since a cyclohexane ring isn't perfectly flat, you're right, the "equatorial" substituents can't lie in a perfect plane with the plane of the ring. The terms "axial" and "equatorial" when applied to a cyclohexane ring are simply useful conventions that describe the two possible orientations of substituents about the normal chair conformation of a cyclohexane ring. These two orientations are important, because they do control such things as reactivity and stability of various substituents and conformations.

So, by definition, given the following chair conformation of cyclohexane:



(from Wikipedia, Cyclohexane, http://en.wikipedia.org/wiki/Cyclohexane)

you can observe two different types of hydrogens (one designated by red balls, the other by blue balls). All of the red balls are exactly the same, and can be interchanged by rotating the molecules about various axes, and the C-H bonds designated by the red balls are parallel to an axis through the center of the cyclohexane ring. So those are called the axial substituents. All of the other hydrogens, the blue balls, although not truly perpindicular to the central axis of the cyclohexane ring, come reasonably close. They are also exactly the same, and can be interchanged by rotating the cyclohexane ring about various axes, and are called the equatorial substituents.

All of the axial substituents can be interchanged by simple rotations; all of the equatorial substituents can be interchanged by simple rotations.

An axial substituent cannot be interchanged with an equatorial substituent by simple rotation of the molecule. However, an unsubstituted cyclohexane molecule can "flip" it's chair conformation to the opposite chair conformation, a transformation that occurs rapidly at room temperature. When this transformation happens, all of the axial hydrogens are transformed into equatorial hydrogens and all of the equatorial hydrogens are transformed into axial hydrogens.

Make yourself a model of cyclohexane, with all of the hydrogens present, and play with it. Practice drawing it from various viewpoints, and practice interpreting different "flattened" 2-D pictures of it. This is just something that takes a little time to be able to see in your head, and it will only get more important as you go.