1. When (R)-1-chloro-2-methylbutane undergoes free radical chlorination, five dichloro constitutional isomers are formed. What are these structures? Draw them. Be explicit as to diastereomers, enantiomers, racemates, etc.
The five constitutional isomers (different atom connectivity) are: 2, (3, 4), 5, 6 and 7. Still optically active: 2 (R), 3 (2S, 3S), 4 (2S, 3R) and 5 (R). Notice in 3 and 4, which are diastereomers, the (R, S)-configuration has changed at C₂ from the monochloro compound although the asymmetric center has not been altered. This is a change in group priority. #6 is racemic because abstraction of the tertiary hydrogen in the first propagation step creates a planar radical. In the second propagation step, chlorination occurs with equal facility on either face of the planar radical. #7 is achiral; two identical groups, -CH₂Cl, in the molecule.

2. A 4:1 mixture of enantiomers has [α]_D = +120^o. What is the rotation of the levorotatory enantiomer? The dextrorotatory enantiomer?   Show work. Because the net rotation is positive, the major enantiomer must be dextrorotatory; the minor one, levorotatory. A 4:1 mixture is 80% dextro-, 20% levo. Thus, +120 = 0.80x(+rot.) + 0.20x(-rot.); +120 = 0.60x(+rot.); +rot. = 120/0.60 = +200^o for the dextrorotatory enantiomer; -200^o for the levorotatory enantiomer.

3. a) 1,2-Dibromoethane is optically inactive yet it has a dipole moment. Explain and illustrate. [Hint:  Draw the staggered conformations and assess optical activity and dipole moment for each.]
Optical activity is an algebraically additive property; dipole moments are cumulative in nature. The three staggered conformations of 1,2-dibromobutane are shown on the right in both Newman projections and sawhorse views. The anti conformation has no net dipole. The bond dipoles cancel. [Only the green C-Br bond dipoles are shown.] The two gauche conformations have net dipoles. The vector sum of the bond dipoles gives the molecular dipole. As to optical activity, the anti conformation is achiral [center of symmetry]. The gauche conformations are chiral and form a racemate. No net optical activity.

b) meso-2,3-Dibromobutane exists in three staggered conformations, none of which has a plane of symmetry. Yet the compound is optically-inactive. Indeed, the only conformation that has a plane of symmetry is quite unstable. Explain and illustrate.

Gauche 1 and 2 form a racemic pair (external comparison). The anti conformation has a center of symmetry at the center of the C₂-C₃ bond (internal comparison). The eclipsed conformation of the meso compound in which all groups are paired is unstable. The eclipsed conformation is used as a test of whether or not there is a plane of symmetry in the the molecule.