### Optically Inactive Compounds and Enantiomeric Mixtures**Question 1:** Which two compounds, when mixed in equimolar amounts, would make a solution that is **NOT** optically active?- **Structures:** 1. **I:** A compound with the structure showing COOH on top, CH₃ at the bottom, and Cl and OH on the sides. 2. **II:** A compound with the structure showing COOH on top, CH₃ at the bottom, and OH and Cl on the sides (opposite of I). 3. **III:** A compound with the structure showing COOH on top, CH₃ at the bottom, and the same side configuration as I. 4. **IV:** A compound with the structure showing COOH on top, CH₃ at the bottom, and the same side configuration as II.- **Options:** a. I and II b. I and III c. I and IV d. III and IV **Answer:** The two compounds that, when mixed in equimolar amounts, would result in a solution that is NOT optically active are **a. I and II**.**Explanation:** Compounds I and II are mirror images of each other, thus forming a racemic mixture when combined in equal amounts, which is optically inactive.---**Question 2:** The specific rotation of pure (R)-carvone is -61°. An enantiomeric mixture of (R) and (S) carvone has an observed rotation of +10°. Which enantiomer is in excess?- **Options:** a. (R)-carvone b. (S)-carvone c. Neither, they have equal amounts d. Cannot be determined **Answer:** The enantiomer in excess is **b. (S)-carvone**.**Explanation:** The observed positive rotation indicates that (S)-carvone, which typically has a positive specific rotation (opposite to (R)-carvone), is in excess in the mixture.

The equimolar mixture of two compounds will be optically inactive when they are a pair of enantiomer i.e. they are a non-superimposable mirror of one another. An equimolar molar mixture of enantiomers gives an optically inactive solution because, if one enantiomer has a specific rotation in a clockwise direction, the other will be anti-clockwise and the magnitude will be equal. That is why they cancel the specific rotation, resulting in an optically inactive mixture. In the given question, Compound I and III are a mirror image of each other and non-superimposable. Therefore they are enantiomer. An equimolar mixture of I and III will form an optically inactive solution. So OPTION B is correct.The (R)-carvone has a specific rotation negative (minus) sign, therefore (S)-carvone will have a (+) positive sign. Since the mixture has (+) positive sign, therefore mixture has (S)-carvone in excess. OPTION B is correct.

Science

Chemistry

Which two compounds, when mixed in equimolar amounts, would make a solution that is NOT optically active? COOH ÇOOH ÇOOH ÇOOH H- -O- НО- -H- Но- -H H- -CI H- -CI H- -CI CI- -H H- -OH ČH3 ČH3 ČH3 IV a. I and II b. I and III c. I and IV d. II and IV The specific rotation of pure (R)-carvone is -61°. An enantiomeric mixture of (R) and (S) carvone has an observed rotation of +10°. Which enantiomer is in excess? a. (R)-carvone b. (S)-carvone d. Cannot be determined c. Neither, they have equal amounts

Which two compounds, when mixed in equimolar amounts, would make a solution that is NOT optically active? COOH ÇOOH ÇOOH ÇOOH H- -O- НО- -H- Но- -H H- -CI H- -CI H- -CI CI- -H H- -OH ČH3 ČH3 ČH3 IV a. I and II b. I and III c. I and IV d. II and IV The specific rotation of pure (R)-carvone is -61°. An enantiomeric mixture of (R) and (S) carvone has an observed rotation of +10°. Which enantiomer is in excess? a. (R)-carvone b. (S)-carvone d. Cannot be determined c. Neither, they have equal amounts

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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