Chem2O6 - 1997/98

Problem Set #5 ANSWERS October 24, 1997

1. As you discover in part (c), there are two conformers in which both the A and B rings in this molecule are chairs. They are shown below, along with the answers to (a) and (b) given for each.

(a) With respect to the A-ring,
(i) the NH2 is axial
(ii) H-5 is axial
(iii) H-10 is equatorial
(iv) bond 9-10 is axial
(v) bond 5-6 is equatorial

(b) With respect to the B-ring,
(i) the COOCH3 is axial
(ii) H-5 is equatorial
(iii) H-10 is axial

 (a) With respect to the A-ring,
(i) the NH2 is equatorial
(ii) H-5 is equatorial
(iii) H-10 is axial
(iv) bond 9-10 is equatorial
(v) bond 5-6 is axial

(b) With respect to the B-ring,
(i) the COOCH3 is equatorial
(ii) H-5 is axial
(iii) H-10 is equatorial

Since the structure on the right has the NH2 and COOCH3 groups equatorial rather than axial, it is the more stable conformer.

2.

3.

4. (-)-3-chloropentanoic acid, (R)-3-chloropentanoic acid, has [Greek alpha] D25 -27.8o in water.

(a) What would be the observed optical rotation of a sample of 2.5 g of the compound in 100 mL of water, measured in a 40-cm tube?

Answer:

Since [Greek alpha]D25 = Greek alpha / lc,

the observed optical rotation Greek alpha = [Greek alpha]D25 X (lc) = (-27.8)(0.025)(4) = -2.78o

(b) A sample of 3-chloropentanoic acid having [Greek alpha] D25 +12.7o was recovered by resolution of a racemic mixture of the acid. What is the enantiomeric excess of the sample recovered from the racemic mixture?

Answer:

ee = (Greek alpha / [Greek alpha]D25) X 100%
  = 45.6% of (+)-3-chloropentanoic acid

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24oct97; wjl