@JohnRennie Do you know stereo chemistry?

a bit ...

@JohnRennie Why will this compound not show Geometrical Isomerism?

My teacher told that the way to identify GI across a ring is to see that there should be at least two carbons of the ring with different groups attached to them.

There are two such carbons as we can clearly see.

One on the top.

and other on its right

Is that molecule chiral?

geometrical isomerism is different from chirality, ain't it?

Oops, yes, I was getting mixed up. GI is the cis/trans distinction

You can't have cis-trans isomerism in that molecule because the ring means you can't rotate about the double bond.

Ok

@JohnRennie Don't get this properly.

cis-trans isomerism is a rotation about a double bond. Yes?

rotation?

You get from the top molecule to the bottom one by rotating 180 degrees about the double bond

@JohnRennie Ok, got it.

And the ring in your molecule constrains the geometry. If you tried to rotate 180º about the double bond you physically couldn't do it.

@JohnRennie 8th doesn't show GI but 10th does, why?

Is there a picture?

The double bond in question is the CN double bond because you can't rotate any of the other double bonds. Yes?

It's gone quiet ... :-)

@JohnRennie yes

@Abcd OK, so if you rotate the CN double bond for molecule 8 does it change the molecule?

@JohnRennie Is that C atom sp2 hybridised?

I think it is.

@Abcd Yes

@JohnRennie How did you calculate it? I used steric number. (..slightly off topic but related to the cis trans isomerism)

A carbon atom at the end of a double bond is always sp2 hybridised

One of the p orbitals forms the pi bond, and that leaves one s and two p oribtals to form the sigma bonds

@JohnRennie sp2 has two p orbitals so one p should be left IMO...

@Abcd Yes. The s and two p orbitals hybrise to give sp2 and they form three sigma bonds. Then the p orbital left over forms the pi bond to make the double bond.

There's a good explanation here

@JohnRennie if we flip it and place it than OH would be on left side

right now it's on right side

@Abcd Yes, but have you actually changed the molecule? If you simply turn over the whole molecule without changing the double bond don't you get the same result?

Ok, yes.

Now what about 10th?

For molecule 10 if you simply turn over the whole molecule do you get the same result as rotating the CN double bond 180º?

@JohnRennie I don't understand what rotating the CN double bond by 180$^o$ means tbh.

Image you grab hold of the two ends of the double bond and you twist it so one end rotates 180° relative to the other.

That's basically what cis-trans isomerism is

@JohnRennie so one Carbon rotates 180 relative to other?

Well, in this case we have one C and one N, but yes the two ends rotate relative to each other

Single bonds, i.e. sigma bonds, can rotate (as I'm sure you know). With a double bond rotating it 90° breaks the pi bond so it costs energy. Then when we rotate another 90° the pi bond can form again. This means there is a barrier to rotation. But given enough energy we can make the bond rotate.

The 180° rotation flips between the cis and trans forms

@JohnRennie Okay, and I feel that in your diagram the double bond of the ring should be down not up. (after rotation)

right?

If you're rotating the CN bond 180° then either you leave the ring unchanged and rotate the NOH group, or you leave the NOH group unchanged and rotate the ring.

What I did was leave the ring unchanged and rotate the NOH group

but they are bonded together :(

so they should rotate together.

Sorry for the trouble @JohnRennie. I find stereo chemistry really hard.

But the ends of a bond can rotate relative to each other

With a pi bond the rotation is stiff i.e. you have to put a lot of energy into rotating it, but the bond will rotate.

@JohnRennie how?

I'm not sure what you are asking ...

Do you know that sigma bonds can rotate freely?

yes

OK, so you're happy that single bonds rotate. Presumably you're asking how a double bond can rotate?

No

I am asking that how can one bonded atom rotate.

@JohnRennie In our example we kept the ring fixed and rotated the N.

That's what is causing confusion.

I am unable to understand how we can do that.

Do you mean physically what happens?

I feel that rotation should be simultaneous, that is both the ring and nitrogen should rotate together.

@JohnRennie Yes.

I don't see how this is any different in principle to a sigma bond rotating, and you seem happy about that.

Okay, fine. I saw one visualisation and got it.

@JohnRennie Also what if we rotate it $60^o$?

The N atom.

A pi bond forms when the two p orbitals at either end of the bond line up

This is taken from the web page I linked. The two p orbitals forming the pi bond are in orange.

Suppose I rotate the bond 60° then the two p orbitals won't line up so they can't form a pi bond.

That means to rotate the bond 60° we have to put in energy i.e. we have to put in enough energy to break the pi bond.

@JohnRennie oh yes! it all makes sense now !

But if we carry on rotating then when we reach 180° the p orbitals line up again and the pi bond reforms, so we get the energy back.

yes, thanks for your time!

I guess cis-trans isomerism hadn't been explained to you as a rotation of the double bond, but I've always found that the easiest way to understand it.

yes, it wasn't explained it to me that way.

@JohnRennie but there are cases when there's no double bond..

in cyclic compounds.

How do you get white background in googledraw @JohnRennie?

In a ring it's usually impossible to rotate a bond 180° because the other bonds in the ring make it physically impossible. That is, if you start twisting a bond it distorts the rest of the ring.

@Abcd do you mean the overall white background? If so, I didn't do anything to get the white background - it was white by default.

i always get this background:

Ah, yes, I remember that problem ...

do I need to use another account?

I generally edit an existing drawing that has a white background. I've just tried creating a new drawing and I get the same pattern as you. There must be a setting somewhere. Let me have a look ...

Ah, that was easy :-)

how?

Just right click on the background and choose background from the popup menu and change it to white.

oh yes, done/

I think the pattern is meant as a visual cue to show the background is transparent.

@JohnRennie What about this one? What technique will you use to predict whether or not it'll show GI?

sp3 atoms in full ring..

There's no double bond, so that is just a cyclohexane ring with two OH groups.

yes

The OH groups can be both up, both down or one up and one down.

Up and down mean out of the plane of the paper

ok, simple.

I'm not sure if those are different - I think they probably are. So it's a form of stereoisomerism but it isn't cis-trans isomerism.

@JohnRennie we've been taught that it's geometrical isomerism....

If I attempt a 3D drawing the two forms would be:

@JohnRennie in reality the ring is planar. You made it bent just for convenience right?

@Abcd cyclohexane isn't planar

benzene is planar. Were you thinking of that?

OH!

The carbons in cyclohexane are all sp3 hybridised so they are tetrahedral

@JohnRennie i thought all cyclic compounds are planar :((

If you want I can draw in the hydrogen atoms as well to show this.

if it'd make it clearer...

Wait, I got one/

on Google images.

Dash lines are bonds directed away from us and bold lines are bonds pointing towards us

Each carbon atom has the four bonds arranged tetrahedrally, as you'd expect for sp3

yes

@Abcd are you going to upload an image

@JohnRennie how are they mirror images?

They just don't look like mirror images.

Suppose you rotate the right hand molecule 180º about a vertical axis. Can you do that in your head or shall I try and draw it?

@JohnRennie it becomes upside down?

Perhaps I'd better draw it ...

@JohnRennie I made a model using 4 pens.

And I rotated it and saw.

Cool :-)

Why did author have to turn it and show?

As shown in the diagram the mirror plane is the plane of the paper

Maybe the author thought that was clearer

Oh, okay.