Question

A) how many stereoisomers are possible for this compound? Show how this can be easily determined mathematically

B) draw all stereoisomers for this compound. Use dashes and wedges appropriately at each stereo center. DO NOT SHOW HYDROGEN ATOMS @ EACH STEREOCENTER

C) Identify and label each pair of stereoisomers drawn in part B (above) as either enantiomers or diastereomers

D) identify and label each pair(s) of stereoisomers that would be optically inactive as 50/50 mixture in a polarimeter and explain why this is so.

(24pts) Consider the structure below then answers part (a)-(d) in the space below. (Point total for th estion is max one can earn if all answers or responses below are complete and valid.) Br HOH o Br a) How many stereoisomers are possible for this compound? Show how this can be easily determined mathematically. b) Draw all stereoisomers possible for this compound. Use dashes and wedges appropriately at each stereocenter. Do not show the hydrogen atoms at each stereocenter. c) Identify and label each pair of stereoisomers drawn in part (b) as either enantiomers or diastereomers. (d) Identify and label each pair(s) of stereoisomers that would be optically inactive as a 50/50 mixture in a polarimeter and briefly explain why this is so.

Answer 1

a) The given compound has 2 sterocenters. so It can have mathematically 2² stereoisomers.

The number of stereoisomers = 2ⁿ where n = number of stereocenters.

b)c) I and II are enantiomers.

OH 07 Br 2,3-dibromo-4-hydroxybutanal on korist OH Br I OH OH 07 Karise III Br IV

III and IV are enantiomeric pair.

I,III and I,IV and II,III and II,Iv are diastereomeric pairs.

d)A mixture of I and II and ( III and IV as a separate mixture)  in 50:50 ratio are optically inactive in polarimeter, as the rotation causedd by one is cancelled by other. Such amixture is called RACEMIC mixture /modification.