[BE ONE]

I've been told that the staggered conformer of ethane is more stable than the eclipsed due to hyperconjugation between electrons in the sigma bonding orbital of the C-H bond and the sigma antibonding orbital on the C-2. This doesn't make sense to me since it seems that the delocalization of electrons to an antibonding orbital would have the effect of destablizing the molecule.

Please help.
Thanks.

[Valdorod]

I thing the staggered vs eclipsed is simplere when explained in terms of the repulsive interactions between the bonding pairs of electrons.  Bottom line the staggered conformation allows the maximum possible separation of the electron pairs of the six carbon-hydrogen bonds and as such has the lowest energy.  The energy difference between the two is about 2.8 kcal/mol small enough that the torsional barrier can be overcome and the molecule rotates between the two conformations.

In terms of hyperconjugation and the delocalization of charge, perhaps looking at the relative stabilities of carbocations is a better example.  The rule of thumb says that a charged system is stabilized when the charge is dispersed or delocalized.  With carbocations, alkyl groups will shift electron density to the more positive charge.  As the aklyl group attaches itself to a carbocation it obtains part of the positive charge and is able to delocalize the positive charge.

Hyperconjucation usually occurs by the interaction of a bond orbital with a p orbital.  Thus I think in the case you are stating it is the sigma antiboding orbital on the C-2 is not involved, I could be wrong however.  For the case of a methyl group bonding to a carbocation, hyperconjugation occurs between the carbon hydrogen sigma bonds of the methyl group and the vacant p orbital of the carbocation.  This could also be the case for ethane, in which it is the p orbital and not the sigma antiboding in C-2.

Valdorod

[BE ONE]

Thank you for replying. I understand, and concur on what you say about hyperconjugation in carbocations, however, for ethane it seems that there aren't any available, vacant p-orbitals. All of the p orbitals would have been hybridized. That's why "they" explain the delocalization with respect to an anti-bonding orbital. Thanks again for your input, but this is still not making sense to me.

[Valdorod]

Be One,

You are absolutely right, ethane only contains sigma bonds, (duh, I should know better).  I did a little bit of research and found the following articles in nature

Chemistry: A new twist on molecular shape
Nature 411, 539-541 (31 May 2001)
This is just an editorial on the actual article on the same issue

Hyperconjugation not steric repulsion leads to the staggered structure of ethane
Nature 411, 565-568 (31 May 2001) | doi:10.1038/35079036; Received 22 September 2000; Accepted 29 March 2001

After skimming through the article, the conclusion is that while counterintuitive, the quantum mechanics show that quantum superposition (which is weird to begin with) such hyperconjugative CH–CH* interactions lead to a lowering of energy in the staggered conformation.

to quote from the editorial:

"So ethane's energy barriers can be viewed as a form of hyperconjugative 'resonance stabilization' for the electrons of -type single bonds, weaker than, but closely related to, the -electron resonance stabilization in systems with alternating single and double bonds. From this viewpoint, ethane-like molecules adopt staggered conformations not to relieve steric congestion, but to achieve optimal resonance stabilization."

Valdo

[BE ONE]

Valdo,
Thanks for the reply and the research. I'm going to check the article out for myself. It seems that if I accept this explanation, I'd have to throw out what I've already learned about MO theory.   "They" seem to want to have their cake and eat it as well. On the one hand occupation of anti-bonding orbitals is portrayed as highly destabilizing while on the other hand the occupation of antibonding orbitals is said to be a stabilizing effect. Please forgive me for not being able to resolve this seeming paradox.
BE

[phil81]

It's not a paradox. Occupation of an anti-bonding orbital weakens *the bond*. The overall energy of *the molecule* is lowered, though.

In your case, the sigma(C1--H) bond pushes some electron density into the sigma*(C2--H) bond. Thus, C2--H is weakened, but the overall molecular energy is lowered by delocalisation.

[Ahmed Abdullah]

So in staggered form C-H bond energy should be less than
that of eclipsed form. Right?

[Rabn]

This question is an example of the dichotomy that exists in the minds of chemists.  The more chemists I speak to the more I realize that they move seemlessly from static to dynamic models in their minds.  This leads to a lot of unintended confusion.  Keeping a consistently dynamic model of electron density in ones mind is probably the hardest thing a chemist can do.  It is so easy succumb to the pitfalls of the static model that is taught.  Let's be honest...a deprotonated carboxylic acid does not have a minus one charge. Some of that electron density is going to be taken up by the molecule, not to mention the intermolecular interactions massaging the electron density of the entire molecule in all kinds of ways.  It is so hard to visualize the dynamic model in so many cases that it becomes natural to move back and forth between the static and dynamic.  Sheesh....just typing about it makes my mind bend around in crazy ways.