I have just started learning conformational analysis, and a major doubt came into my mind.
In simple alkanes such as ethane, the staggered conformer is much more stable than the eclipsed conformer. This fact may be explained as follows:
- In the staggered conformer, the repulsion between the hydrogens is minimized as they are "anti" with respect to each other.
- We have a case of favorable overlap between sigma bonding MO of the C-H bond, and sigma anti bonding orbital of the C-H bond of the second Carbon.
I have no issues with this part. Then, I started reading about alkenes. In the simple example of propene, two cases may arise, as shown in the figure.
Note that (2) can be obtained from (1) by rotating the C-C bond by 60 degrees. Unlike in the case of alkanes, the eclipsed form is more stable, not the bisected form. This can be easily accounted for: In the eclipsed form, the Hydrogen is oriented in the right way to allow for effective Hyperconjugation. This involves delocalization of electrons from C-H sigma bond to the C=C pi antibonding orbital.
My main question is about when it becomes more complicated. Let me take the example of 1-butene. Here, we have twice the number of eclipsed and bisected conformers. I have labelled them as E1, E2, B1 and B2 (E= eclipsed;B=bisected)
The correct order of stability is E2 > E1 > B1 > B2.
I do not understand this order. When I attempted to find the order of stability, I got: E2 > B1 > B2 > E1. I based this mainly on steric effects.
- E2 is clearly the most stable as we have Hyperconjugation due to the C-H bond which is oriented in just the right way.
- I couldn't get the rest of the order. Why is E1 the second most stable? I understand that it is eclipsed, but shouldn't that lead to greater torsional strain? Is Hyperconjugation the reason for stability, again?
Can Hyperconjugation take place when we have a C-C bond(when there is no alpha Hydrogen with respect to the double bond), or is it restricted to C-H bonds(where we have an alpha Hydrogen)?