1). (24 points) Compound 1 reacts with H2 and Pt catalyst to give two products. (Note that the two hydrogens add from the same face.) Draw the structures of the two products, name them, and indicate how one could readily distinguish them via their NMR spectra.
2) (10 points) How many stereoisomers are there of 2,3,4-pentanediol? Draw them all.
6
3) (24 points) Consider 2,6-dimethyl-2’-bromo-6’-chlorobiphenyl shown below. Because of the steric interactions of the substituents at the ortho positions, the phenyl rings are approximately perpendicular to each other.
a) Is this compound chiral or achiral? Why?
achiral, it possesses a mirror plane of symmetry
b) Are the two methyl groups homotopic, enantiotopic or diaseriotopic? If they are enantiotopic or diasteriotopic, label them in the drawing above as pro-R or pro-S.
enantiotopic, since related by a mirror plane
c) This biphenyl is oxidized under phase-transfer conditions. The oxidation is done by mixing a benzene solution of the biphenyl with an aqueous solution of KMnO4 and then adding 10 mole % of (R)-methylnaphthylphenyloctadecylammonium bromide:
In this case, the inorganic reactant (KMnO4) is carried into the organic phase by the ammonium ion. The product of the oxidation is a monocarboxylic acid (one of the methyl groups is oxidized to COOH). Will this product show optical activity?
yes, achiral oxidizing conditions will likely produce an excess of one enantiomer over the other, since the transition states will be diasterotopic
d) Draw the structure of one of the possible moncarboxylic acid products resulting from the above procedure and indicate whether it is R or S.
4) (12 points) a) When compound 2 is heated, in principal, two products 3 and 4 can be obtained. What is the relationship of the transition states leading to these two products?
diasteriotopic transition states
b) Heating compound 5 can also give the same two products, 3 and 4. However, 2 only gives 3 and 5 yields only 4. What term(s) describes the outcomes of these reactions?
5) (30 points) Compound 6 is a neutral hydrocarbon, yet it displays a dipole moment. Using the results of the Huckel calculation given below, explain why 6 has a non-zero dipole moment. Predict the orientation of the dipole moment. Can you rationalize this result using resonance structures?
|
MO |
energy |
C1 |
C2 |
C3 |
C4 |
C5 |
C6 |
C7 |
C8 |
|
1 |
a + 2.36b |
0.38 |
0.38 |
0.51 |
0.46 |
0.28 |
0.21 |
0.21 |
0.28 |
|
2 |
a + 1.82b |
0.38 |
0.38 |
0.31 |
|
-0.34 |
-0.41 |
|
-0.34 |
|
3 |
a + 0.68b |
0.26 |
0.26 |
-0.08 |
-0.59 |
-0.16 |
|
0.48 |
-0.16 |
|
4 |
a + 0.62b |
0.00 |
|
0.00 |
0.00 |
-0.60 |
|
0.37 |
|
|
5 |
a - 0.87b |
0.34 |
0.34 |
-0.63 |
-0.12 |
0.37 |
-0.20 |
-0.20 |
0.37 |
|
6 |
a - 1.00b |
0.71 |
-0.71 |
0.00 |
0.00 |
0.00 |
0.00 |
0.00 |
0.00 |
|
7 |
a - 1.62b |
0.00 |
0.00 |
0.00 |
0.00 |
0.37 |
-0.60 |
0.60 |
-0.37 |
|
8 |
a - 1.98b |
0.161 |
|
-0.48 |
|
-0.38 |
0.13 |
0.13 |
-0.38 |
pop(k) = S Nik cik2
sum over the first 4 MOs only - these are the occupied ones
|
atom |
pop |
charge |
|
1 |
2*(.38)2+2*(.38)2+2*(.26)2+2*(0.0)2 = 0.71 |
1-0.71 = +0.29 |
|
2 |
2*(.38)2+2*(.38)2+2*(.26)2+2*(0.0)2 = 0.71 |
1-0.71 = +0.29 |
|
3 |
2*(.51)2+2*(.31)2+2*(-.08)2+2*(0.0)2 = 0.72 |
1-0.72 = +0.28 |
|
4 |
2*(.46)2+2*(-.20)2+2*(-.59)2+2*(0.0)2 = 1.20 |
1-1.20 = -0.20 |
|
5 |
2*(.28)2+2*(-.34)2+2*(-.16)2+2*(-.60)2 = 1.16 |
1-1.16 = -0.16 |
|
6 |
2*(.21)2+2*(-.41)2+2*(.48)2+2*(-.37)2 = 1.19 |
1-1.19 = -0.19 |
|
7 |
2*(.21)2+2*(-.41)2+2*(.48)2+2*(.37)2 = 1.19 |
1-1.19 = -0.19 |
|
8 |
2*(.28)2+2*(-.34)2+2*(-.16)2+2*(.60)2 = 1.16 |
1-1.16 = -0.16 |
so the charge distribution is like so:
Rationalize by the following resonance structures:
