Question

4. (24pts) Consider the structure below then answers part (a)-(d) in the space below. (Point total for this question is max one can earn if all answers or responses below are complete and valid.) Br OH Y Br o (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

Bro -OH O Bro The above compound contains two different chiral centres and the molecule cannot divided into two equal and similar halves in any possible conformation, @ Thus the total number of stereoisomers. possible for this compound = 21 where, n= Number of Chinal centres7 = 2 Hense, four stereoisomers are possible for this compound.

Bxo Boo OH -OH Bro Во (1) BO OH OH o Br Br © Enantiomer -> 182 -> 384 112 and 3,4 are two pairs of enantiomer,

The diasteromeric pairs are (1,3) Diastexomers → 183 84 283 2 & 4 These are the four diasteromeris pairs.

. (d) The pair 1 and 2 , would be Optically inactive as a 50/50 mixture in a polarimeter. . Also the pair 3 and 4 , would be optically in active as a 50/50 mixture in a polarimeter.

These pairs would be opticallye inactive as a 50/50 mixture in a polarimeter. Because these pains forms racemic mixture when present as 50/50 mixture. Hense in polarimeter thewootation of plane polarised light by one enantion is compensated by the 6- rotation of the other enantiomers,