brew
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Bond rotation
Hi all,
For the molecule say ethylbenzene, the carbons on the ring are sp2 hybridised and due to the pi electrons
and it resonating, the pi orbitol makes the structure unable to twist, form different conformations, and instead it is planar. I grasp that and why,
but I'm unsure of the bond between a carbon of the ring being sp2 and alpha to the ring the carbon would be sp3 hence the bond would be a sigma bond,
but due to there not being a pi bond associated directly to the carbon alpha to the ring, I'm pretty sure there could be bond rotation, but I am
unsure due to one orbital being sp2 hybridised. This would have great benefit to know, so help appreciated.
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12thealchemist
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Your guess is correct. There is free rotation in all the sigma bonds leaving the ring.
A benzene ring is comprised of six sp2 hybridised carbon atoms arranged in a ring. These form the sigma framework. Each carbon atom also
has a pure pi orbital perpendicular to the plane of the ring, formally containing a single electron. These pi orbitals form the pi "cloud" above and
below the benzene ring through delocalisation.
Two of the three sp2 hybrid orbitals of each carbon atom contribute to the sigma framework of the benzene ring, while the third protrudes
away from the ring in the plane of the ring. This can then overlap with another orbital, in this case an sp3 hybridised orbital centred on
the alpha carbon. This overlap forms a cylindrically symmetrical bond, aka a sigma bond. Since there are no p orbitals not already involved in bonding
on either the ring-carbon or the alpha-carbon, no pi bond can form. Hence, the Cring--Calpha bond is sigma in nature, and free
rotation can therefore occur.
I hope this helps; diagrams (which I cannot draw) may be helpful here.
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walruslover69
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There is some amount of pi character in the bond between the benzene ring and the alpha carbon but it is very small. Quickly looking it up it seems
like the barrier to rotation about that bond is ~10kj/mol which sounds about right.
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happyfooddance
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As far as I know what has been said is correct, hindrance about the sigma bond (bond to a phenyl group) certainly depends on the rest of the molecule
as well, for example the pi bonds from a benzoyl C=O exhibit a mesomeric resonance with the pi electrons from the ring, which hinders bond rotation
around the phenyl-carbonyl C-C.
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brew
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Thx for the many responses. Appreciated!
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brew
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So Hsppyfooddance, you are implying that if the alpha carbon was a ketone then it would form a part with a conjugated system, and with resonance the
carbonyl will position itself planar to the aryl group? I think that is what you meant. Earlier on I was wondering if the alpha carbon was a stereo
enter and how free rotation might mess with it forming a specific confirmation with respect to polarised light. It doesn't seem to, as it's only that
one bond rotating in question, with respect to a chiral molecule. It doesn't seem to matter whether it rotates or not, I found it hard to grasp, and
needed to get out the plasticine and matches. Cheers
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walruslover69
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Brew, My organic is a bit hazy but I assume you are referring to a compound like Acetophenone. There is definitely going to be more pi electron
interaction with the aromatic ring in acetophenone than in ethyl benzene. The barrier to rotation for acetophenone is around 22kj/mol compared to the
10kj/mol of ethyl benzene. That being said at room temperature they will freely rotate around the bond.
https://faculty.missouri.edu/~glaserr/8160f07/CNMR_Dynamics....
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brew
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HeyWalruslover69, I was paraphrasing an earlier post that talked about, I am pretty sure, if the alpha to the ring carbon was a ketone, and as rusty
as I am, I assumed with resonance that, I thought would now include the carbonyl, such as the case with acetophenone. This would, I thought prevent
the now sp2-sp2 bond from rotating, and I thought the carbonyl would allign in a planar confirmation with the benzene ring. You have presented the
energy that would be required for movement, which I think is double the amount required for an sp2-sp3 bond. Makes sense, and thanks for going to the
trouble looking it up. Am I right that the carbonyl of scetsphenone would enter the conjugated system with the aryl group. I'm pretty sure a pi bond
can be formed between the ring carbon and the carbonyl carbon, as well as breaking the pi bond of the carbonyl. I'm pretty sure that's correct, but I
might need to get out my text, to remember the details, cheers mate and thx
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walruslover69
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I think you are on the right track. In my opinion its best to try and not think about it in terms of resonance and VESPR but instead in terms of
molecular orbitals and electron density. The pi bond with the carbonyl isn't really being broken, just a small amount of the electron density from
that bond is being delocalized into the molecular orbital. The carbonyl bond is still mostly a double bond, and the ring-carbonyl bond is only
slightly more than a single bond.
It was no trouble at all, I'm actually just starting a research project calculating barriers to rotation and rotation rates using NMR. When you lower
the temperature so that the molecule doesn't have the energy to overcome the barrier to rotation you can see a NMR peak split into multiple peaks
because if they aren't rotating the groups are no longer equivalent.
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brew
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Interesting indeed, I had resonance so drilled into my understanding, but what your implying makes alot of sense. Is there a correlation between bond
rotation/no rotation, and reaction rates. I dare say activation energy is a component. Just thinking out aloud. I'm pleased for you, doing this
research, and best of luck.
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brew
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One thing I am unsure of, and Im sorry to bug you about it, but I hate not grasping stuff, so please understand but you conveyed that when there is
bond rotation, the groups, neighbouring atoms with hydrogens attached, be it shielded, or not so shielded, so with rotation, take a acetylahyde, with
its methyl group protons, and its aldehyde proton, so with rotation, one cannot expect the deshielded aldehyde proton, to exhibit, a greater chemical
shift, with a quartet and the methyl group, less deshielding, hence a doublet, so with bond rotation, one couldn't distinguish this with peaks
according to shielding, less shielding. Am I understanding what you meant. Im stubborn I know, but I like to grasp stuff.
\Also with respect to my abode post, I thought afterwards that the more energetic a system is, the more likely there is to be contact, hence a
reaction, but there is no doubt an optimal temp, and how that relates to bond rotation and reactions, is interesting indeed. I thought that the more
rotation occurring, the less chance for an optimal fit, re reactions, Sorry to waffle on, just bugged me that I wasn't sure what you meant. If you
read this, and answer, thanks in advance. Cheers.
[Edited on 8-12-2018 by brew]
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walruslover69
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You're not bugging me at all,
I don't believe there is any broad relationship between bond rotation and reaction rates or activation energy. I imagine there are situations where
rotation or lack of rotation can effect the reaction mechanism of specific reactions but it would vary for each reaction.
I don't think acetyladhyde is a good example, there seem to be some other factors that lead to the methyl peak not being a doublet.
The best example is DMF. At room temp the methyl groups rotate freely and are therefore equivalent and so they show up as 1 large peak. At low
temperatures there is no rotation around the bond. This means the 2 methyl groups are no longer equivalent and therefore show up as 2 different peaks.
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DraconicAcid
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Just to add- at low temp, the DMF molecule is planar, with one methyl being basically trans to the oxygen, and the other one cis to it. Thus the
different chemical shifts.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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brew
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OOOPs
Talk about not reading other peoples post clearly. I didn't pick up on the word, BASICALLY cis, and trans, as I thought it was totally planar, which
for an sP3 nitrogen, it couldn't exactly be. So the cis, would be deshielded more than the trans orientation, being further away from the carbonyl.
Makes sense. Im still getting my head around the fact, that if bond rotation is occurring, then the methyl group protons, will be seen as one, and
that will cause a single peak, compared to the double peak if the cis, trans protons are different, as you've stated. IM going to completely learn
all this again. I learnt to read spectral data, from reactions I had done, but understanding it, NOT QUITE. Thanks for the input fellas.
[Edited on 20-12-2018 by brew]
[Edited on 20-12-2018 by brew]
[Edited on 20-12-2018 by brew]
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walruslover69
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Here are a couple good resources I used when I did my first lab on the subject
https://www.colby.edu/chemistry/PChem/lab/NMRrotBarrier.pdf
https://www.chem.wisc.edu/areas/reich/nmr/08-tech-03-dnmr.ht...
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brew
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Thanks Walruslover69, appreciated. and AWESOME !!
[Edited on 20-12-2018 by brew]
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