# How is the cavity of a cyclodextrin chiral?

I am studying the host-guest chemistry of cyclodextrins with achiral aromatic molecules, and I found out that a useful way of studying them is by induced circular dichroism spectroscopy. This technique is used by taking an achiral molecule and placing it in a chiral environment to induce a CD signal. Apparently, the cavity of a cyclodextrin is chiral, but for the life of me I cannot see HOW it is chiral. I know the glucose molecules the ring is made of are D-glucose, but I still see mirror planes in the structure, and I cannot figure out how the cavity is chiral! Help??

• – Mithoron Jan 13 '18 at 20:39
• You don't look close enough. From pictures it would seem there's center of symmetry no planes, but rotation is blocked - inside of molecule can't become outside so there's no center and cyclodextrin is chiral. – Mithoron Jan 13 '18 at 20:49

In the image you can see that all the $\ce{CH2OH}$ groups in the center are out of the plane (towards you). These groups are on the upper side of the cavity. If they were below the plane, the cyclodextrin would have other interactions with the host molecule.
• @Mithoron I meant it in general. Like in tartaric acid. In the case of glucose I do not think that it is possible to have a meso form since both ends of the molecule are not the same ($\ce{CHO}$ and $\ce{CH2OH}$). – Raoul Kessels Jan 13 '18 at 21:05