Looking at this structure, I would say this molecule has five chiral carbons. However, its IUPAC name (3R,4R,5S,6R)-3-amino-6-(hydroxymethyl)oxane-2,4,5-triol suggests that there are only four. Which is it?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ It is the hemiacetal carbon atoms. It is chiral but racemization can occur through the open form of the molecules. $\endgroup$– EJCCommented Feb 22, 2016 at 14:43
-
1$\begingroup$ The diagram you drew doesn't indicate the stereochemistry, so I am curious as to how you came up with that IUPAC name. $\endgroup$– orthocresolCommented Feb 22, 2016 at 17:05
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The structure as drawn indeed contains 5 chiral carbon atoms. However, the one carbon that is bound to two oxygen atoms is called the anomeric carbon. It is part of a hemiacetal moiety in the structure you have drawn. That means that at room temperature and in aqueous solution, it will equilibrate reasonably quickly between:
- One enantiomer of hemiacetal
- An open-chain non-cyclic molecule where the anomeric carbon is an aldehyde moiety and is thus not chiral.
- The other enantiomer of hemiacetal.
Thus, for many applications it is probably better to think of this molecule as having only four "fixed" stereocenters.
