An eclipsed conformation is a peak of energy as a C-C bond is turned…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
An eclipsed conformation is a peak of energy as a C-C bond is turned. Most of its extra energy comes from the repulsion between each bond on the front carbon and the bond directly behind it on the back carbon. Chemists estimate the energy of an eclipsed form by adding a standard cost for each such pair. These costs are small numbers that can be tested against measured barriers to rotation.
Use these costs of eclipsing: H with H, ; H with , ; with , .
- Propane, , is eclipsed about the bond between carbon 1 and carbon 2. The front carbon holds three hydrogen atoms. The back carbon holds two hydrogen atoms and one methyl group. Find the cost of the eclipsing.
- In 2,2-dimethylbutane, , the bond between carbon 2 and carbon 3 is eclipsed. The front carbon holds three methyl groups. The back carbon holds two hydrogen atoms and one methyl group. Find the cost of the eclipsing.
Need a reminder? The key ideaShow
A single bond turns, so a molecule has conformations. The relative number of molecules in a form of energy above the lowest is . For butane about the C2-C3 bond: anti , gauche , eclipsed , fully eclipsed . Eclipsing costs: H with H , H with , with .
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Propane. The pairs are H with H, H with H and H with . The cost is .
Step 2. 2,2-Dimethylbutane. The front carbon has three methyl groups. The back groups are , H and H. So the pairs are with , with H and with H. The cost is .
Answer: in propane and in 2,2-dimethylbutane.
Check. The propane value agrees with the measured barrier of about .
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