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The volume of a cubane molecule!?!?

I was just wondering how anyone would even go about aprroaching this question. My chemistry teacher told me to think about it because it's the type of thing they ask at interviews.

Essentially cubane is a cube made up of 8 hydrogen atoms with 8 hydrogen atoms coming off each corner of the carbon cube.

So the question is calculate (basically an estimate) the internal volume of the cube, and then would a cyclo-octane molecule fit inside?
Original post by theCreator
I was just wondering how anyone would even go about aprroaching this question. My chemistry teacher told me to think about it because it's the type of thing they ask at interviews.

Essentially cubane is a cube made up of 8 hydrogen atoms with 8 hydrogen atoms coming off each corner of the carbon cube.

So the question is calculate (basically an estimate) the internal volume of the cube, and then would a cyclo-octane molecule fit inside?


Atoms are essentially space filling when bonded together. You need the bond lengths of the cubane molecule, which I imagine willl be similar to the bond lengths in cyclobutane.

Use these and the atomic radii (covalent) to work out the geometric dimensions.

There is no chance of cyclooctane fitting inside!
(edited 10 years ago)
Original post by charco
Atoms are essentially space filling when bonded together. You need the bond lengths of the cubane molecule, which I imagine willl be similar to the bond lengths in cyclobutane.

Use these and the atomic radii (covalent) to work out the geometric dimensions.

There is no chance of cyclooctane fitting inside!



Ah thank you :smile:

So there would be no way of a rough estimate of working out the bond lengths of the C-C bonds without doing it experimentally?
Original post by theCreator
Ah thank you :smile:

So there would be no way of a rough estimate of working out the bond lengths of the C-C bonds without doing it experimentally?


Thats the genius of the written language, you can look it up! Not a simple technique to get bond lengths, you can get very rough estimates from techniques like IR, but that requires a calibration curve formed from similar species.
The most accurate techniques are Neutron and Electron diffraction techniques. As far as I know the only place in the UK where this is possible is at the Diamond light source in Oxfordshire. (So a very, well not necessarily expensive, but rarely used technique as you need to book a time with this lab waaaayyy in advance.)

Your question is merely an exercise in geometry! Should be a more comfortable field for pre-uni level chemists. It does highlight the different bond lengths, throwing out any pre-conceptions that all C-C bonds are equal, but the question, is essentially does an octagon fit in a cube. (If we boil it down to the absolute bare bones)

To take this question to a more advanced level you can then consider the fact that the carbons and H's are not infinitely small dots but occupy a significant space. If we take the radius of the carbon atom to be it's covalent radius, then for cubane we have 4 primitive cubic packed carbons. This has a packing efficiency of ~52%, so even if cyclooctane would fit in a cubane formed from infinitely small carbons, it won't fit in real cubane as the available space is actually much smaller than that of a cube of the same dimensions.

You could advance this even further by considering the actual conformation of cyclooctane. Cyclooctane is not an Octahedron! Check out the major conformers on the wiki page http://en.wikipedia.org/wiki/Cyclooctane
This makes the geometry of the problem more difficult as the cyclooctane is no longer a regular polyhedron but can probably be modeled quite reasonably as a sphere of radius equal to the dimensions of the conformer of cyclooctane.


Original post by charco
Atoms are essentially space filling when bonded together. You need the bond lengths of the cubane molecule, which I imagine willl be similar to the bond lengths in cyclobutane.

Use these and the atomic radii (covalent) to work out the geometric dimensions.

There is no chance of cyclooctane fitting inside!


Yes the bond lengths are very similar to cyclobutane!
Reply 4
Original post by theCreator


So the question is calculate (basically an estimate) the internal volume of the cube, and then would a cyclo-octane molecule fit inside?


No, Cyclo-octane molecule would not fit inside. because of it's boat chair conformation structure or crown conformation structure.
Original post by JMaydom
X


Thanks for your great reply. Could you show me how you would answer it in an interview situation. So for example would you go about the question by saying if we assume that the bond lengths of the c-c bonds are 200nm (for example) and that the carbon atoms are infinitely small dots, this would give us an internal volume of (200^3)nm^3, would this be a viable way to approach it bearing in mind it's an A-level answer?

Also being that you're an Oxford Chemist :biggrin: Could I just ask you why Brady's reagent produces a precipitate when reacted with ketones and aldehydes but not with carboxylic acids or esters. I know it's something to do with electron density but no one can seem to answer me and I can't find a simple clear answer. Thanks alot for your help :smile:
Reply 6
Original post by theCreator
Thanks for your great reply. Could you show me how you would answer it in an interview situation. So for example would you go about the question by saying if we assume that the bond lengths of the c-c bonds are 200nm (for example) and that the carbon atoms are infinitely small dots, this would give us an internal volume of (200^3)nm^3, would this be a viable way to approach it bearing in mind it's an A-level answer?

Also being that you're an Oxford Chemist :biggrin: Could I just ask you why Brady's reagent produces a precipitate when reacted with ketones and aldehydes but not with carboxylic acids or esters. I know it's something to do with electron density but no one can seem to answer me and I can't find a simple clear answer. Thanks alot for your help :smile:


If you've just got the bond length, your approach is perfectly fine. (Normally, when you come across problems like this, you've instead got the average radii and would want to calculate approximate bond lengths / inter-atomic distances for a particular structure. In that case, you would instead use a space-filling model, as charco mentioned -- i.e. say the carbon atoms are spheres with radius 100nm, etc.)

Re Brady's reagent, have you come across resonance structures yet? Perhaps you used them to show why the bond lengths in benzene are all the same, even though there are formally three single and three double bonds.

Well any way, the singly-bonded oxygen in a carboxylic acid can donate one of its lone pairs like so:

R R
| -> | +
O=C-OH (-)O-C=OH


This delocalization of electrons has a stabilizing effect on the starting material, because the electron's negative charge has been spread over a greater volume. After the amine group from Brady's reagent attacks, you get

R
|
(-)O-C-OH
|
(+)NHR'


and the OH oxygen can no longer donate a lone pair (losing O2- would be hugely disfavoured), so the stabilization due to delocalization has been lost. This unfavourable loss of delocalization means that carbonyls with adjacent "mesomeric" electron donating groups (i.e. those groups that can donate a lone pair), such as amides, carboxylic acids and esters, will generally not react with Brady's reagent.
Original post by BJack
X


Sorry for all the questions but why does the oxygen get a negative charge after it's donated a lone pair of electrons? In your diagram the oxygen goes from being double bonded to singly bonded, so does that mean it's lost a bond pair?

But I think I sort of get it, essentially is it because there is resonance in the carboxylic acids due to the interaction of lone pairs, which has a stabilising effect because the charge density has been lowered so it becomes less electrophilic. Thus the carbonyl carbon is not prone to nucleophilic attack??

With regards to cubane, could you just confirm if I was going about it the right way in the example of using the average radii?
If we assume for example that the covalent radii of the carbon atoms is 50nm. That would give us the length of one side which is 100nm, and so the internal volume of the cubane molecule would be (100^3)nm^3.
But since the carbon atoms are spherical this means that part of the carbon atoms (roughly an 1/8 of the sphere) will overlap with the cube and thus reduce the internal colume.
So in order to calculate the actual internal volume we would do (100^3) - (((4/3)*pi*50^3)*(1/8)*8). Correct?

Thanks alot for your help, I really appreciate it :smile:
(edited 10 years ago)
Original post by theCreator
Sorry for all the questions but why does the oxygen get a negative charge after it's donated a lone pair of electrons? In your diagram the oxygen goes from being double bonded to singly bonded, so does that mean it's lost a bond pair?



The resonance model:

One lone pair of electrons from the oxygen in the -OH group moves into a pi orbital in between the oxygen (of the OH group) and the carbon atom. At the same time the pi electrons from the C=O double bond move up onto the carbonyl oxygen, giving it a formal negative charge.
Reply 9
Original post by theCreator
Sorry for all the questions but why does the oxygen get a negative charge after it's donated a lone pair of electrons? In your diagram the oxygen goes from being double bonded to singly bonded, so does that mean it's lost a bond pair?


It's not the carbonyl oxygen that donates a lone pair. The alcoholic oxygen donates a lone pair to form a second bond to the carbon atom. To prevent carbon from becoming hypervalent, a different bond must break; in this instance, the electrons in the O=C pi bond move onto the carbonyl oxygen, so O=C -> (-)O-C. Otherwise, you seem to have it sussed.

With regards to cubane, could you just confirm if I was going about it the right way in the example of using the average radii?


Yes, that's right. :yy:
Reply 10
Original post by theCreator
I was just wondering how anyone would even go about aprroaching this question. My chemistry teacher told me to think about it because it's the type of thing they ask at interviews.

Essentially cubane is a cube made up of 8 hydrogen atoms with 8 hydrogen atoms coming off each corner of the carbon cube.

So the question is calculate (basically an estimate) the internal volume of the cube, and then would a cyclo-octane molecule fit inside?


****ing MASSIVE
Original post by charco
X



Original post by BJack
X


Ahh, such a great feeling now that you've explained that to me, thank you so much for your help. I'll be sure to rep you as soon as my daily rep thing resets. :biggrin:
Original post by JMaydom
Thats the genius of the written language, you can look it up! Not a simple technique to get bond lengths, you can get very rough estimates from techniques like IR, but that requires a calibration curve formed from similar species.
The most accurate techniques are Neutron and Electron diffraction techniques. As far as I know the only place in the UK where this is possible is at the Diamond light source in Oxfordshire. (So a very, well not necessarily expensive, but rarely used technique as you need to book a time with this lab waaaayyy in advance.)

Your question is merely an exercise in geometry! Should be a more comfortable field for pre-uni level chemists. It does highlight the different bond lengths, throwing out any pre-conceptions that all C-C bonds are equal, but the question, is essentially does an octagon fit in a cube. (If we boil it down to the absolute bare bones)

To take this question to a more advanced level you can then consider the fact that the carbons and H's are not infinitely small dots but occupy a significant space. If we take the radius of the carbon atom to be it's covalent radius, then for cubane we have 4 primitive cubic packed carbons. This has a packing efficiency of ~52%, so even if cyclooctane would fit in a cubane formed from infinitely small carbons, it won't fit in real cubane as the available space is actually much smaller than that of a cube of the same dimensions.

You could advance this even further by considering the actual conformation of cyclooctane. Cyclooctane is not an Octahedron! Check out the major conformers on the wiki page http://en.wikipedia.org/wiki/Cyclooctane
This makes the geometry of the problem more difficult as the cyclooctane is no longer a regular polyhedron but can probably be modeled quite reasonably as a sphere of radius equal to the dimensions of the conformer of cyclooctane.




Yes the bond lengths are very similar to cyclobutane!


X-ray crystallography might also be an option, cubane is a solid at SATP.
Original post by chrisawhitmore
X-ray crystallography might also be an option, cubane is a solid at SATP.



Is this because cubane has such an unusual shape? I didn't know X-ray crystallography would work with hydrocarbons.
Original post by theCreator
Is this because cubane has such an unusual shape? I didn't know X-ray crystallography would work with hydrocarbons.


It works with stuff you can crystallise!!!
Original post by theCreator
Is this because cubane has such an unusual shape? I didn't know X-ray crystallography would work with hydrocarbons.


Nah, crystallography can resolve any good crystal.

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