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

Question

Quizplus · Accepted Answer

The given pentose has 5 chiral centers (marked by asterisks below): O=CHC*HOH*CHOH*CHOH*CH₂OH Each of these chiral centers can exist in two possible configurations (R or S), giving a total of 2^5 = 32 possible stereoisomers. However, since this is a pentose of the L-series, the chiral center farthest from the aldehyde group must be the L configuration. This reduces the number of possible stereoisomers to 16. Further, since the penultimate carbon (marked by "S") is part of a diastereotopic pair (due to the presence of a substituent on the adjacent carbon), the two stereoisomers where this carbon has opposite configurations can be considered identical. This reduces the number of possible stereoisomers to 8. Finally, the two stereoisomers where the last chiral center (marked by "B") has a configuration opposite to that of the penultimate chiral center are diastereomers, and can be considered distinct. This gives a final count of 4 possible stereoisomers.

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