Problem Set 2 Answers

2.6.  2.6 answer
Each is symmetric; the maximum energy has a12 strain between the acetyl methyl and a substituent on C3.


2.8  Answer to 2.8

Orient in diamond lattice and count 1,6 H-H interactions in each form.

2.9

2.9a

MM calculated difference:  0.5 kcal/mol
2.9b

MM calculated difference:  3.5 kcal/mol
2.9c

MM calculated difference:  1.5 kcal/mol
2.9d

MM calculated difference:  3.5 kcal/mol


2.25
a, b.  The methyl groups are enantiotopic.

c.  The two methyls are diastereotopic.

d.  The two methyls are diastereotpic.

e.  The hydrogens in each CH2 group are diastereotopic.

f.  The ethoxy groups, and the hydrogens in the CH2, are each enantiotopic.

B.1

Comments:

a.  A simple starting point is b-D-glucopyranose, which has all substituents equatorial.  Mannose is the 2-epimer of glucose; placing one group axial is better than 4.

b.  Likewise, in this compound the 4-position is opposite that in glucose, and there is no 2- or 6-substitution.  One axial methoxy is better than 3 others, especially where the other three are all 1,3-related.

c.  Start with the mirror image of the b-pyranose ring to capture the L configuration.  The axial methyl group is a problem, but the ring flip isomer is worse.

B.2.

B.3 a.

b.

 

B.4.  The cis comes from the "crown" form, whereas the trans comes from a distorted boat-chair.  (A syn elimination is also possible that does not require as much distortion in the ring.)  Note that there are syn eliminations also; one leading to each.