Chapter 6 - Instructor Supplemental Solutions to Problems...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 6 Principles of Stereochemistry Solutions to In-Text Problems 6.1 (a) This compound is chiral. (b) Methane is achiral. 6.3 (a) Planes of symmetry in methane bisect one set of H—C—H bonds and contain the other H—C—H bonds. (There are four such planes.) (c) One plane of symmetry in ethylene is the plane of the page; the two others are the planes perpendicular to the page. The center of symmetry is the point in the center of the C A C bond. (e) The plane of symmetry in cis- 2-butene is the plane of the page and the plane perpendicular to the page that bisects the C A C bond. (f) The plane of symmetry contains a C—H bond on one carbon as well as the C—C bond, and it bisects an H— C—H bond angle on the other carbon. (There are three such planes.) The center of symmetry is a point at the center of the C—C bond. 6.4 The asymmetric carbon is indicated with an asterisk. (b) 6.5 Remember that there are many different ways to draw a correct line-and-wedge structure. If your structures don’t look like these, and if you’re not sure whether yours is correct, make a model of both and check them for congruency. (When possible, we often adopt a “standard” representation in which the bond to the atom of lowest priority—hydrogen in these examples—is the wedged bond, and it is placed to the left of the asymmetric carbon. This makes it very easy to determine configuration. However, this “standard” representation is not necessary.) (b) 6.6 (b) The asymmetric carbon in the given stereoisomer of malic acid has the S configuration.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 6 2 6.8 (a) Use Eq. 6.1 on text p. 237: a = [ a ] cl = (66.5 deg mL g –1 dm –1 )(1 dm)(5 g) 100 mL = 3.33 degrees (b) The specific rotations of enantiomers must have the same magnitude but the opposite sign. Therefore, the enantiomer of sucrose must have specific rotation = –66.5 deg mL –1 g –1 dm –1 . 6.11 The racemate has no effect on the observed rotation other than to dilute the sample. Hence, after addition of the racemate, the concentration of the excess ( R )-2-butanol is 0.75 M . This corresponds to (0.75 mol L –1 )(74.12 g mol 1 )(0.001 L mL –1 ) = 0.055 g mL –1 . Use this as the value of c in Eq. 6.1 with [ ± ] = –13.9 deg mL g –1 dm –1 : a = (–13.9 deg mL g –1 dm –1 )(0.0556 g mL –1 )(1 dm) = –0.773 deg 6.13 Proceed in the manner suggested by the solution to Problem 6.12. The absolute configuration of the alkene in Eq. 6.2 is known. Carry out the following catalytic hydrogenation:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

Chapter 6 - Instructor Supplemental Solutions to Problems...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online