Calling all chemists!

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Hey guys, organic chemistry question here:

organicchemquestion.png


We were talking about this in class today and originally I was thinking all the carbons were coplanar. But, duh, I wasn't aware of how the structure is bent in space. So professor says it's six, but then I was thinking that the bottom, in-plane hydrogen on that methyl group should also be coplanar. I came home and found that my model kit arrived, and it looks to me like that hydrogen actually does lay in the same plane as the six atoms my professor pointed out.

What say you chemists?
 
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My o-chem is incredibly shoddy (I probably only passed because my professor pretty much passed everybody), but I would intuitively think that the least sterically hindered orientation would be for all three hydrogens to not be coplanar with anything (in other words, take what you’re picturing right now and rotate it in the z-plane like 10 degrees). Keep in mind that just because there are lines, bold lines, and dotted lines that that doesn’t mean there are only three planes an object can be in – there are infinite graduations between them.

Of course I’m talking out of my butt and so I could be completely wrong. But logically it seems to me that making that hydrogen coplanar puts unnecessary steric hinderance with that bottom carbon.
 
OK, well, we're still on chapter 1 so I don't know what steric hindrance is yet. But I'll bring it up when I ask my professor about it on the next class.

Any other ideas from any other well-qualified individuals?
 
No, your professor is right. The four groups on the methyl carbon are arranged in a tetrahedral configuration (therefore they are on different planes), because each of the four groups will repel each other equally and as far as possible (VSEPR):

sdfsdl.jpg
 
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I know. But my argument is that you can have at least one hydrogen atom on the methyl group be coplanar at two instances as you rotate it around the bond.

My professor probably is right, but, if he is, I want to know why I'm wrong.
 
Omnis
I know. But my argument is that you can have at least one hydrogen atom on the methyl group be coplanar at two instances as you rotate it around the bond.

My professor probably is right, but, if he is, I want to know why I'm wrong.
Click to expand...

Ah ok, well you can only ever have one hydrogen atom on the methyl group that will be coplanar with the rest of the molecule due to it repelling the other hydrogens on the methyl group. So yes, there is a possibility that one hydrogen will be coplanar - but that possibility is what, 1 in 120? So statistically it's more likely a hydrogen won't be on the same plane. It makes no sense to call that hydrogen co-planar when most of the time it isn't co-planar.
 
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But wouldn't it be held in place by the proximity to the other atoms? I think those two orientations are the farthest you can get the methyl atoms from the cyclohexene ring.
 
I don't think so, since the rotation about the bond between the methyl carbon and the cyclohexene ring isn't influenced by the hydrogens on the methyl. I think the only time a hydrogen on a branch from cyclohexene is guaranteed to to be coplanar is if the first carbon on the branch is doubly bonded to the ring (so you have -CH2 branching off instead).
 
So then the methyl is basically free to rotate? And even though it may be coplanar 1/60th of the time, we can't call it that because it's not fixed?
 
Omnis
So then the methyl is basically free to rotate? And even though it may be coplanar 1/60th of the time, we can't call it that because it's not fixed?
Click to expand...

Yeah, the methyl is free to rotate. I was wrong on the probability though, it is 1/60th like you said, not 1/120th (as the hydrogen will have two chances in a 360 degree rotation to be coplanar). I think your professor wasn't being specific enough, the six atoms he specified are the only ones to be guaranteed co-planar, while the hydrogens on the methyl have a possibility of one being co-planar, but most of the time it isn't.
 
I see. I'll bring this up at the next class then. Thanks for your help.
 
I understand quite literally none of this.
 
F1GTR
Yeah, the methyl is free to rotate. I was wrong on the probability though, it is 1/60th like you said, not 1/120th (as the hydrogen will have two chances in a 360 degree rotation to be coplanar). I think your professor wasn't being specific enough, the six atoms he specified are the only ones to be guaranteed co-planar, while the hydrogens on the methyl have a possibility of one being co-planar, but most of the time it isn't.
Click to expand...

Indeed the methyl group can (and does) rotate, so none of its hydrogen atoms can be considered coplanar.
You got your probability thing by dividing 360° by 3 atoms having 2 coplanar positions each (360/3/2 = 60).
This assumes that the hydrogen can be considered coplanar as soon as the C-C-H angle is under 1°.
But where did you get this value ? What if you consider a 0.001° angle ?
Then the H is coplanar once in 360 000 / 3 / 2 = 60 000 times !
No, the rotation is a continuous movement, you can't discretize it, so this H atom is just never coplanar with the >C=C< bond.
Coplanarity affects the way molecules pile up on each other, so it has a influence at a macroscopic scale on several physical properties like density, boiling point etc.

Btw, there a nice (but small) 3D modelling of the molecure here.
 
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Tedehur
Indeed the methyl group can (and does) rotate, so none of its hydrogen atoms can be considered coplanar.
You got your probability thing by dividing 360° by 3 atoms having 2 coplanar positions each (360/3/2 = 60).
This assumes that the hydrogen can be considered coplanar as soon as the C-C-H angle is under 1°.
But what if you consider a 0.001° angle ?
Then the H is coplanar once in 360 000 / 3 / 2 = 60 000 times !
No, the rotation is a continuous movement, you can't discretize it, so this H atom is just never coplanar with the >C=C< bond.

Btw, there a nice (but small) 3D modelling of the molecure here.
Click to expand...

I did consider smaller increments, yes, I simply used 360° as it would be easier to imagine. I didn't know that the rotation was a continuous movement though (we never did anything more on single bonds other than how they're formed and why they can rotate) so thanks for that little bit of info. 👍

EDIT: Also, why am I talking about Chemistry in my free time when I spent two years waiting to see the back of it? :lol:
 
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The rotation stuff is not obvious because the symbolic representation we all use for molecules is incorrect.
We draw atoms like small balls, all the same size, while an atom of carbon is roughly 12 times bigger than an atom of hydrogen. (6 protons + 6 neutrons for C, one single proton for H)
A better representation would be to imagine the atom of carbon like a planet and the 3 H as satellites turning around it.

Same for the bond. We draw a thin line as if it was fixed in space, while it's actually done by sharing electrons. A cloud of electrons turning around the nuclei.

And yeah, I've studied chemistry for way too long :)
 
Tedehur
Indeed the methyl group can (and does) rotate, so none of its hydrogen atoms can be considered coplanar.
You got your probability thing by dividing 360° by 3 atoms having 2 coplanar positions each (360/3/2 = 60).
This assumes that the hydrogen can be considered coplanar as soon as the C-C-H angle is under 1°.
But where did you get this value ? What if you consider a 0.001° angle ?
Then the H is coplanar once in 360 000 / 3 / 2 = 60 000 times !
No, the rotation is a continuous movement, you can't discretize it, so this H atom is just never coplanar with the >C=C< bond.
Coplanarity affects the way molecules pile up on each other, so it has a influence at a macroscopic scale on several physical properties like density, boiling point etc.

Btw, there a nice (but small) 3D modelling of the molecure here.
Click to expand...

Well I did say may be 1/60th. I didn't think about minutes or seconds for simplicity's sake. ;)

But thanks a ton for your help. You can expect some PMs in the future if I have other questions. Hope you don't mind. :lol:
 
So now I'm looking at a reaction where cyclopentanol turns into cyclopentane. I know that the following reagents can be used to carry this out: TsCl/Pyridine, and H2SO4 [heat] H2/Pt.

Now, shouldn't you also be able to use Pd instead of Pt in the second set of reagents? I think so. I'm more sure about this than I am about this other option below.

Will the TsCl work with benzene instead of pyridine? I'm leaning towards no, but I'm not sure why. To get a free electron pair on the benzene, you'd have a very unstable carbanion and even then I don't see anything that's going to rip the proton off of one of those carbons. And then I don't know if the free electron pair would even be in the same plane as the pair you find on pyridine.

Any thoughts?
 
My chem-o is great, and crap at the same time. I won't be able to tell you off the top of my head, but I'll ring a guy I know. We've known each other for a long time, but I'll see what I can do.

t4
 

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