Draw the cis and trans isomers for the following: a. 1-bromo-4-chlorocyclohexane b. 1-ethyl-3-methylcyclobutane
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Textbook Solutions for Organic Chemistry
Question
A solution prepared by mixing 10 mL of a 0.10 M solution of the R enantiomer of a compound and 30 mL of a 0.10 M solution of the S enantiomer was found to have an observed specific rotation of +4.8 . What is the specific rotation of each of the enantiomers? ( Hint: mL * M = millimole, abbreviated as mmol) Solution One mmol (10 mL * 0.10 M) of the R enantiomer is mixed with 3 mmol (30 mL * 0.10 M) of the S enantiomer; 1 mmol of the R enantiomer plus 1 mmol of the S enantiomer will form 2 mmol of a racemic mixture, so there will be 2 mmol of S enantiomer left over. Because 2 out of 4 mmol is excess S enantiomer, the solution has a 50% enantiomeric excess. Knowing the enantiomeric excess and the observed specific rotation allows us to calculate the specific rotation. enantiomeric excess = observed specificrotation specific rotation of the pure enantiomer * 100% 50% = +4.8 x * 100% 50 100 = +4.8 x 1 2 = +4.8 x x = 21+4.82 x = 9.6 The S enantiomer has a specific rotation of +9.6 , so the R enantiomer has a specific rotation of -9.6 .
Solution
The first step in solving 4.1 problem number 3 trying to solve the problem we have to refer to the textbook question: A solution prepared by mixing 10 mL of a 0.10 M solution of the R enantiomer of a compound and 30 mL of a 0.10 M solution of the S enantiomer was found to have an observed specific rotation of +4.8 . What is the specific rotation of each of the enantiomers? ( Hint: mL * M = millimole, abbreviated as mmol) Solution One mmol (10 mL * 0.10 M) of the R enantiomer is mixed with 3 mmol (30 mL * 0.10 M) of the S enantiomer; 1 mmol of the R enantiomer plus 1 mmol of the S enantiomer will form 2 mmol of a racemic mixture, so there will be 2 mmol of S enantiomer left over. Because 2 out of 4 mmol is excess S enantiomer, the solution has a 50% enantiomeric excess. Knowing the enantiomeric excess and the observed specific rotation allows us to calculate the specific rotation. enantiomeric excess = observed specificrotation specific rotation of the pure enantiomer * 100% 50% = +4.8 x * 100% 50 100 = +4.8 x 1 2 = +4.8 x x = 21+4.82 x = 9.6 The S enantiomer has a specific rotation of +9.6 , so the R enantiomer has a specific rotation of -9.6 .
From the textbook chapter Cis–Trans Isomers Result From Restricted Rotation you will find a few key concepts needed to solve this.
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