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Total number of stereoisomers possible for the following compounds is OH OH HO D

Total number of stereoisomers possible for the following compounds is OH OH HO D

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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What is the exact mass of the following molecule? (3 decimal) a. 360.449 b. 362.245 c. 360.194 d. 360 191 e. 361.012 H SI HO OH
Transcribed Image Text:What is the exact mass of the following molecule? (3 decimal) a. 360.449 b. 362.245 c. 360.194 d. 360 191 e. 361.012 H SI HO OH
Total number of stereoisomers possible for the following compounds is HINT: a. 4 How many possible stereoisomers? How do we know how many stereoisomers are possible for a given structure? There is actually a straightforward way to figure this out. All we need to do is count the number of chiral centers and stereogenic alkene groups, the use this following rule: number of stereoisomeric forms = 2" ... where n = the number of chiral centers plus the number of stereogenic alkene groups b. 2 Consider for example a molecule with two chiral centers and one stereogenic alkene. By the rule stated above, we know right away that there must be eight possible stereoisomers. terminal alkene OH stereogenic alkene C. 8 OH d. 3 OH chiral center HO OH 3 2 = 8 T nonstereogenic alkenes (cannot be labelled E or Z) ←
Transcribed Image Text:Total number of stereoisomers possible for the following compounds is HINT: a. 4 How many possible stereoisomers? How do we know how many stereoisomers are possible for a given structure? There is actually a straightforward way to figure this out. All we need to do is count the number of chiral centers and stereogenic alkene groups, the use this following rule: number of stereoisomeric forms = 2" ... where n = the number of chiral centers plus the number of stereogenic alkene groups b. 2 Consider for example a molecule with two chiral centers and one stereogenic alkene. By the rule stated above, we know right away that there must be eight possible stereoisomers. terminal alkene OH stereogenic alkene C. 8 OH d. 3 OH chiral center HO OH 3 2 = 8 T nonstereogenic alkenes (cannot be labelled E or Z) ←
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