# What makes an epoxide stable?

Overall ring strain seems to be a big issue when it comes to organic chemistry. That is why cyclopentane may be in an "envelope form" or why cyclobutane may be in a kinked, kite form. Both of these example molecules are not planar. So, why is epoxide (oxacyclopropane) a stable molecule, if at all?

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Chemists talk about conformation, not "form". What do you mean by "stable"? Epoxides seem pretty reactive to me :) –  CHM Apr 30 '12 at 0:42
""a stable molecule, if at all?"" Do You doubt the existence of epoxides? Some molecule being stable without giving conditions means "You can fill that in a bottle and it will suvive for some days at least" in lab slang. –  Georg May 1 '12 at 12:28
You could say it is kinetically stable, but thermodynamically unstable. –  user95 Jul 12 '12 at 5:18

Well, you have to ask: how stable? and compared to what?

Epoxides, including ethylene oxide, are generally considered relatively unstable molecules, with a high chemical activity and involvement in numerous reactions, including polymerization and thermal decomposition. However, ethylene oxide does exist as a molecule, as does cyclopropane, which also contains a three-membered ring.

So, let's compare for example the ring strain in ethylene oxide and cyclopropane. I found some computation data on their relative stability from these lecture notes, which indicate that “ethylene oxide has less ring strain than cyclopropane”. I would attribute this to the fact that the C–O–C angle has less strain because its “relaxed” value would be 104.5°, compared to the “relaxed” C–C–C angle of 109.4°.

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$\newcommand{c}[1]{\mathsf{\small #1}}$ You can't quite compare rings consisting only of carbon atoms with rings with oxygens in them. In the latter, the $\c{C}-\c{O}$ bonds are polar, the electrons are closer to the oxygen atom which means the acuteness at the $\c{C}$ atoms is not as much of a problem as it would be in a hydrocarbon ring.