The open-chain aldohexose $CHO - (CHOH) X_{4} - CH X_{2} OH$ has four stereocentres (C-2 to C-5), none of which is made equivalent by symmetry. How many stereoisomers (counting enantiomers separately) are possible? Give an integer.

Numerical answer type, tolerance 0.5.

Answer: 16

The number of stereoisomers is

2^{n}

with

n = 4

stereocentres. Hence

2^{4} = 16

(8 D-sugars and 8 L-sugars).