[Chris1]

Why does cyclohexene oxide (cyclohexane fused with an epoxide) have a half-chair conformation like cyclohexene and not a normal chair like cyclohexane?

I attach an image so that my doubt is understood. A double bond is planar because of the sp2 hybridized carbons. I understand that. Why does the same thing happen in the case of epoxide? And more derived doubts:

1. In a "normal" epoxide (not fused with other cycle), are the four substituents in the same plane also, then?

2. Does the same thing happen with a cyclopropane?

3. Does the same happen with halonium ions?

[mjc123]

One of the C-O bonds would have to be axial and one equatorial. It's not geometrically possible to make the epoxide ring without distorting the chair structure.

[spirochete]

I agree completely with mjc. Also, cyclopropane and halonium ions should have the same issue.

The epoxide (or any three membered ring) has an eclipsed structure automatically. So this should help explain what atoms are co-planar. The carbons are roughly tetrahedral, but distorted to make the strained ring.

This all relates to a thought process for determining the products of the opening of these rings, but I always found them kind of difficult to visualize. The nucleophile has to attack from the side of the half-chair that leads to an actual chair product, which has a trans diaxial arrangement of the new bonds. When drawing the two half-chairs, the nucleophile has a strong preference for attacking the side that sort of "looks" more sterically hindered, although this might be a superficial aspect of the picture.

[RedViper9]

Try drawing in where they hydrogen atoms must be on the two structures. A hand held model will help.