Question

2) a) Consider 2-bromo-3-chlorobutane. How many stereoisomers are possible for this compound? Draw all stereoisomers, label each stereocenter with the appropriate R/S designation to show handedness, and indicate the relationship (enantiomers or diastereomers) between every possible pairing of compounds. Name each compound according to IUPAC nomenclature, using appropriate R/S designation to show handedness. b) Repeat this exercise for 2,3-Butanediol. What is different about this compound relative to the first example? What word describes the unique stereoisomer? c) Which of the compounds in parts a) and b) above are optically active? Among these, give one example of a pair of compounds that will give a specific rotation of Oif present in equal amounts. Imagine one of them has a specific rotation of +16.9 as a pure sample. What do you expect the specific rotation of the other to be?

Answer 1

a) There are two chiral carbons hence total number of stereo isomers = 2^n = 2^2 = 4 CH3 So the compound will have 4 stereoisomers. H3C Br 2-bromo-3-chlorobutane CI J. SCH3 J...on RCH3 H3CS SCH3 RCH 3 H3C R H2C R H2CS Br Br (A) Br (C) Br (D) (B) (A) 4 Enantiomers (B) Diastereomers (A) OR(B) + (D) OR (C) Enantiomers (D) (C) IUPAC names: A= (2R, 3R)-2-bromo-3-chlorobutane B= (2S,3S)-2-bromo-3-chlorobutane C= (2S,3R)-2-bromo-3-chlorobutane D= (2R,3S)-2-bromo-3-chlorobutane

b) H2C OH There are two chiral carbons hence total number of stereo isomers = 2^n = 2^2 = 4 CH3 So the compound should have 4 stereoisomers. However as the two substituents are same the two meso isomers OH are the same. Hence total stereo isomers are 3. OH OH OH OH R CH3 SCH3 H3CS SCH3 R CH3 H3C ОН OH OH OH (E) identical (F) (G) (G) H2CR HCR Enantiomers (E) + (F) Diastereomers (E) OR (F) + (G) IUPAC names: E= (2R,3R)-butane-2,3-diol F= (2S,3S)-butane-2,3-diol G=(2R,3S)-butane-2,3-diol The unique isomer is termed as meso compound. Because it contains internal plane of symmetry.

c) In part-a) A, B, C and D are optically active compounds. Whereas in part-b) the É and F are optically active compound. The meso isomer G is not optically active. A and B are enantiomers and hence they will have equal and opposite specific rotation. Hence when they are present in equal amount in a solution it gives Oº specific rotation. Assuming (A) has specific rotation [a] = +16.99 Then (B) will have specific rotation = -16.9°