Number of stereoisomers of polysubstituted cubane

Question

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. — Ivan Neretin, Aug 17 '19 at 10:59
The molecule u have asked about is very similar to a substituted adamantine...check this out https://chemistry.stackexchange.com/questions/58727/chirality-of-substituted-adamantanes — Schwarz Kugelblitz, Aug 17 '19 at 12:39

score 3 · Accepted Answer · edited Sep 02 '19 at 18:33

I hope I am not too late.

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.

Number of stereoisomers of polysubstituted cubane

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