Organic Chemistry: A Guided Inquiry
Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN: 9780618974122
Author: Andrei Straumanis
Publisher: Cengage Learning
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Chapter 7, Problem 2CTQ
Interpretation Introduction

Interpretation: The information obtained from each part of the name of molecule in model 1 needs to be explained.

Concept Introduction:

The cis and trans isomerism concept will be applied here. Any molecule will be called cis if it follows the conditions for it that is the same groups attached either above the plane or below the plane.For trans isomer, same groups are attached where one is above the plane and other group is below the plane or vice versa.

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Q2. Answer any TWO of the following parts: (a) Draw the two main conformations that exist for cyclohexane. Explain clearly why one conformer is more stable than the other. Using cis-1-ethyl-3-methylcyclohexane, as an example, explain how ring flipping occurs. Draw both conformers of cis-1-ethyl-3-methylcyclohexane and explain clearly which one predominates. (b) What is polarimetry? The specific rotation of (R)-carvone is - 61°. A chemist prepared a 750 mg mixture of (R)-carvone and its enantiomer in 10 ml of ethanol and placed the solution in a 10 cm polarimeter cell. The observed rotation was - 4.125°. Calculate the specific rotation for the above mixture. What is meant by enantiomeric excess? Then determine the % enantiomeric excess (% ee) in the mixture. (i) (ii) (iii) What % of the mixture is (R)-carvone and (S)-carvone?
Draw all constitutional isomers of all-cis ethylmethylisopropylcyclohexane—that is, in which a methyl group (CH3), an ethyl group (CH2CH3), and an isopropyl group [CH(CH3)2] are all bonded to a cyclohexane ring on the same side of the ring’s plane. Which of those isomers do you think is the most stable? Explain
I don't understand why the first one and second one are cis and trans respectively. Wouldn't the first one be trans-1,2-dimethylcyclobutane because the torsional strain wouldn't allow the carbons to be in the same conformation. Meaning one of the carbons would be up and then the next would be down and so on. Since both methyl groups are equatorial and the first and second carbons are arranged up and down, wouldn't it be trans. Same logic for the second molecule. Carbon 1, which is attached to the methyl is down, carbon 3 which is attached to the methyl should be down also because of torsional strain, and since both methyl are in axial (or equatorial?), it would be cis. Or is it based off of the way the carbons are positioned in the picture?
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