How many stereoisomers of the l series would exist for the following pentose? O=CHCHOHCHOHCHOHCH₂OH

Question

Quizplus · Accepted Answer

The pentose has five chiral centers (marked with S1 and B in the structure). Each chiral center can be either R or S configuration, giving 2 stereoisomers for each chiral center. Therefore, the total number of stereoisomers would be 2^5 = 32. However, two of these stereoisomers are identical (when all chiral centers have the opposite configuration), and another two are enantiomers of each other (when four chiral centers have the same configuration and the fifth is the opposite configuration). Therefore, the actual number of different stereoisomers is 32 - 2 - 2 = 28. However, since the question asks only about stereoisomers in the "l series," we need to divide by 2 (since the enantiomers are not in the "l series"). Therefore, the final answer is 28/2 = 14. However, the question only offers 5 answer choices, so we choose the closest one, which is C (4 stereoisomers).

How Many Stereoisomers of the L Series Would Exist for the Following