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- 2. A mixture of two enantiomers has as observed rotation of -18°. The specific rotation of the (-)enantiomers is -27°. Find the % of the two enantiomers in the solution.4. The [a] of pure quinine, an antimalarial drug, is -165°. a) Calculate the enantiomeric excess (EE) of a solution with the following [a] values: -83°. H3CO HO Hin Quinine (anti-malaria drug) b) Calculate the percent of each enantiomer present. c) What is [a] for the enantiomer of quinine, and does it rotate light clockwise or counterclockwise? ( d) Identify 2 functional groups in quinine. (.. e) What are 2 properties that are different between the R and S enantiomer.After an attempt to resolve a racemate into its enantiomers, the observed rotation is +22.4 degrees per g/L. If the known specific rotation is +24.2 degrees per g/L,, what is the % enantiomeric excess of the compound?
- Consider a mixture that is 60% (+) enantiomer and 40% (-) enantiomer. In which direction will the mixture rotate plane-polarized light? What is the enantiomeric excess of the mixture?Consider a mixture that is 38% (+) enantiomer and the rest is the (-) enantiomer. a) In which direction would the mixture rotate plane-polarized light? Briefly explain. b) What is the %ee of the mixture? Briefly explain or draw a pie chart like those done in class. c) If the (+) enantiomer has a specific rotation of +28.5°, what is the specific rotation of the mixture? Show whork.1. A solution has 80% (R)-2-bromobutane and 20% (S)-2-bromobutanea. What is the “enantiomeric excess” of (R)-2-bromobutane? b. If pure (R)-2-bromobutane rotates light 100º to the right, how much rotation would occur for a solution with 80% (R)-2-bromobutane and 20% (S)-2-bromobutane c. If a solution has a 50/50 mixture of (R)- and (S)-2-bromobutane, what would be the enantiomeric excess and the optical purity? 2. A mixture of two enantiomers has as observed rotation of -18°. The specific rotation of the (-) enantiomers is -27°. Find the % of the two enantiomers in the solution.
- If the sample of the pure R enantiomer of a molecule has a specific rotation of -40 degrees and your mixture of that molecule has an 82% ee S, what is the observed specific rotation of your sample?A solution is prepared by mixing 5 mL of a 0.20 M solution of the S enantiomer of a compound and 10 mL of a 0.20 M solution of the R enantiomer. The specific rotation of that solution was -7.00. a.) Calculate the millimoles of each enantiomer in the mixture.b.) Calculate the% enantiomer in excess (% e.e).c.) Calculate the specific rotation of each pure enantiomer.b) A synthetic sample of MSG (conc 0.350 g/mL) is found to be a mixture of two enantiomers. What is the percentage of each enantiomer in the sample if the experimental rotation of the sample is found to be -2.15°?
- 7.(a) Draw and label (a) the enantiomer and (b) a pair of diasteromers of the following compound. OH (c) Will a 50:50 mix of the enantiomers of the compound be optically active (rotate plane polarized light). Explain the reason for your answer d) Will a 50:50 mix of the two diasteriomers you drew in (c) be optically active (rotate plane polarized light). Explain the reason for your answerConsider a solution that contains 69.0% R isomer and 31.0% S isomer. If the observed specific rotation of the mixture is -58.0°, what is the specific rotation of the pure R isomer? [a] =1. How can the chirality of compound 1? 2. How the configuration inversion of -OH moiety of compound 2 occur stereoselectively? Thank you....