0 votes
34 views
in Organic Chemistry by (530 points)
Identify the number of chiral centers present in (2R,3S)-butane-1,4-diol and draw all possible stereoisomers for the compound.

1 Answer

0 votes
by (470 points)
2R,3S -butane-1,4-diol has two chiral centers: one at carbon 2 and another at carbon 3. There are four possible stereoisomers for this compound, which can be represented as follows:1.  2R,3S -butane-1,4-diol   HO-CH2-CH 2R  OH -CH 3S  OH -CH2-OH2.  2S,3S -butane-1,4-diol   HO-CH2-CH 2S  OH -CH 3S  OH -CH2-OH3.  2R,3R -butane-1,4-diol   HO-CH2-CH 2R  OH -CH 3R  OH -CH2-OH4.  2S,3R -butane-1,4-diol   HO-CH2-CH 2S  OH -CH 3R  OH -CH2-OHThese four stereoisomers are the possible combinations of the R and S configurations at the two chiral centers.
Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...