# Number of stereoisomers of polysubstituted cubane

The molecule given below is a substituted cubane.

a) How many stereoisomers exist for this molecule?
b) How many pairs of enantiomers are possible?

How do I calculate the number of stereoisomers? Since each carbon is linked to four different groups, can I claim that all are chiral and, therefore, $$2^8$$ stereoisomers are possible?

If that is true, since there is no element of symmetry, can I say that $$2^4$$ enantiomeric pairs exist?

Edit: Looking at the comments clears things up but brings forth another question. If, say, iodine was present at any other position that would not lead to a tetrahedron, what would happen?

• True, but they don't exist. Try to draw some and you'll see. Only 2 isomers are possible. Aug 17 '19 at 10:59
• The molecule u have asked about is very similar to a substituted adamantine...check this out chemistry.stackexchange.com/questions/58727/… Aug 17 '19 at 12:39
• A tetrahedron may be inscribed in a cube, Aug 17 '19 at 13:01

All carbons are $$\mathrm{sp^3}$$-hybridised and three bonds of every carbon are fixed (one corner shares three bonds), so fourth one has only one place in space to maintain an $$\mathrm{sp^3}$$ tetrahedron. This one is optically active, so this and its enantiomer are the only two possible isomers.
Now you get why we don't use the $$2^n$$ formula. And similar thing happens in a bridgehead position.