7.4: Solutions to 7.3 Problems

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Adam Carter, Tiffany Culver, & Robert Cichewicz
University of Oklahoma

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Exercises on Bond Line Structures, Drawing, and Isomers

a. Both compounds are 4-ethyl-2-methylhexane. Therefore, they are identical compounds.

b. The first compound is 2,3,5-trimethylheptane, and the second is 3-ethyl-2,5-dimethylhexane. Therefore, these are isomers of one another.

c. The first compound is 2,3,3-trimethylpentane, and the second is also 2,3,3-trimethylpentane. Therefore, these are identical.

a. Rotating the second structure yields the first structure. Therefore these are identical.

b. Non-mirror image and non-superimposable. Therefore, these are diastereomers.

c. Mirror images and non-superimposable on it's mirror image. These are enantiomers.

Functional groups: carboxylic acid, amine, amide, thiol

Amino acids: glutamate/glutamic acid, cysteine, glycine

Stereoisomers: two stereocenters, \(2^2 = 4\) stereoisomers