The ketone groups of 5-nitrononan-2,8-dione (Figure) can be reduced to the alcohol (OH) function one by one, obtaining a set of isomers of 8-hydroxy-5-nitrononan-2-one, on the one hand, and 5-nitrononan-2,8-diol on the other. Write them all down and say how they are related to each other (enantiomers, diastereoisomers or if they are meso forms).
In compounds 1 there cannot be meso forms because no plane of symmetry can be established that divides the molecule into two equal halves.
1a: (5R, 8S) 1b: (5R, 8R) 1c: (5S, 8S) 1d: (5S, 8R)
1e: (5S, 8R) 1f: (5S, 8S) 1g: (5R, 8R) 1h: (5R, 8S)
Therefore 1a and 1h are identical, 1b and 1g too, as well as 1c and 1f on the one hand and 1d and 1e on the other.
So there are only 4 isomers, as corresponds to the rule of 2 to the power of n, n in this case being 2 (the stereocenters in positions 5 and 8).
1a = 1h is enantiomer of 1d = 1e
1b = 1g is enantiomer of 1c = 1f
All other relationships are diastereoisomerism.
2a = 2d = meso (superimposable object and image)
2b = 2c = meso (superimposable object and image)
2e = 2g (the image of 2e is 2f non-superimposable)
2f = 2h (the image of 2f is 2e non-superimposable)
Although there are three "apparent" stereocenters at positions 2, 5 and 8, only 4 stereoisomers of this structure exist.
2e = 2g is enantiomer of 2f = 2h
All other relationships are diastereoisomerism.
