How many stereoisomers are possible for the compound in the picture? I tried to draw a rough picture and indicated the chiral carbons. I'm getting 8 stereoisomers are possible but my book gives the answer as 2. Can someone explain where I'm going wrong?
$\begingroup$ I'm wondering this as well - the only other interpretation I can see is if you had to specify how many types of stereoisomers there are. That would give 2, because you'd have diastereomers and enantiomers but not meso compounds. $\endgroup$– IT TsoiApr 27, 2016 at 7:36
$\begingroup$ Can't there be geometrical isomerism ? $\endgroup$– user14857Apr 27, 2016 at 7:45
$\begingroup$ Geometrical Isomerism doesn't apply in this case since there are no double bonds. $\endgroup$– IT TsoiApr 27, 2016 at 8:09
$\begingroup$ But one could be above the plane and one below...there need not be a double bond $\endgroup$– user14857Apr 27, 2016 at 8:09
$\begingroup$ True, but that would not change the name of the compound regardless of whether the particular group in question takes an axial or Equatorial position. $\endgroup$– IT TsoiApr 27, 2016 at 8:12
The comments on the question are incorrect [edit: the offending comments seem to be gone now], as is the answer given in the book. In general, answers printed at the back of a book should never be taken as the absolute truth. There are typos in all books. Even well-known and well-written textbooks have many errata (usually corrected in subsequent printings).
As you said, there are three chiral centres and therefore 8 stereoisomers.
There are no other elements of symmetry present in this molecule which render any of those 8 stereoisomers equivalent.