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1-(2-methylcyclohexyl)ethan-1-ol

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?

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  • $\begingroup$ 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 Tsoi Apr 27 '16 at 7:36
  • $\begingroup$ Can't there be geometrical isomerism ? $\endgroup$ – user14857 Apr 27 '16 at 7:45
  • $\begingroup$ Geometrical Isomerism doesn't apply in this case since there are no double bonds. $\endgroup$ – IT Tsoi Apr 27 '16 at 8:09
  • $\begingroup$ But one could be above the plane and one below...there need not be a double bond $\endgroup$ – user14857 Apr 27 '16 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 Tsoi Apr 27 '16 at 8:12
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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.

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