16 = A, 12 = B, 8 = C, 4 = D. (Ignores +/-.) 

To estimate your current course grade, average your two test letter grades so far. If relevant, round toward the most recent test. I will apply the same basic strategy throughout the course. (I am happy to discuss individual situations with you.)

A brief, partial answer key follows. Alternative answers are possible in some cases. I have tried to minimize drawing structures here. (You can ask me about any of these, in class or privately.)

General comment: I’m somewhat concerned about “sloppy” structures. If you are going to attach a hydroxymethyl group, -CH2OH, to something, show it with the attachment to the C ‑‑ not to an H or O. I realize how easy it is to just aim a bond in the general direction of the group, but it does not look good. If you show H atoms in a structure (rather than just the C-skeleton), show them correctly. C forms 4 bonds -- almost always. And sometimes you need the H at certain places to see what is going on. If I noted things such as these more than occasionally on your test, please try to be more careful.

1. 1,4-dipropylbenzene + Cl2 (with Fe or FeCl3)  2-chloro-1,4-dipropylbenzene + HCl. In the following, the underlined atoms switch places.

2. a. 2-heptanone (oxidation of secondary alcohol)
b. No reaction. Cannot oxidize tertiary alcohol; there is no H available on the alcohol C.
c. 1-hexene (elimination, dehydration)

3. Enantiomer is the mirror image; reverse left-right at all stereocenters.

(The sugar given in the question is L-arabinose. The enantiomer, at right, is D‑arabinose.)
	      HCO
          
 HOCH
          
   HCOH
          
   HCOH
          
          CH2OH	

4. a. secondary
b. & c. Multiple possibilities, but all are ethers, such as ethyl propyl ether.
d. The alcohol has the higher BP, because it can form intermolecular H-bonds with other molecules of itself. Both molecules have H-bond acceptors. However, the alcohol has an H‑bond donor, whereas the ether does not. Therefore, alcohol molecules can H-bond with each other, whereas ether molecules cannot. The H-bond donor (the alcohol) results in a major increase in boiling point. (This effect is larger than any possible shape effect, no matter which ether isomer you drew.)
e. No. The original compound is saturated: no double bonds and no rings. Thus any isomer must also be saturated.

5.

6. a. From left to right: phenol, ether.
b. The phenol group is (weakly) acidic. There are no basic groups. (We have not yet encountered any basic groups in organic chemicals. Amines, Ch 14, are basic.)
c. Just to the right of the ether oxygen.

d. In the following structure I have shown the stereochemistry that is relevant to the paper listed below (and to question B2). But the stereochemistry is not relevant at this point.

e. See structure above. The two right-most stereocenters are relevant to this part. Note that hydrogenation of the terminal double bond does not give a stereocenter.
f. We now have three stereocenters, so 23 = 8 stereoisomers are possible.
g. The only “double bonds” on the rings are part of an aromatic system, with delocalized  bonds, thus are relatively unreactive.

S Bell et al, Asymmetric hydrogenation of unfunctionalized, purely alkyl-substituted olefins. Science 311:642, 2/3/06. (+ News, p 619.) The goal of the paper was to develop new methods for stereo-specific hydrogenation of alkenes; synthesis of an important biological molecule was part of their demonstration of success. The structure given in the question leads to -tocopherol, one of the forms of vitamin E (the forms vary in the methylation pattern on the aromatic ring). The product I showed above in part d is the actual stereoisomer they made -- the one that is -tocopherol.

7. C #1 is the top right C; that gives the smallest number for the first substituent.

Naming the compound, ignoring stereochemistry: 1,5-dibromo-7-methyloctane.

For ranking the four groups at the stereocenter: Br is #1 and H is #4, with the two C-groups in between; this is simply by atomic number. To rank the two C-groups, we need to look at the 2nd C. For the upper group, it has CHH; for the lower group, it has CCH. Thus the lower group is #2. (Remember, you are looking for first point of difference between the groups; p 181 rule 2, which repeats what you learned in Ch 4, p 108. Don’t be distracted by that 1-Br, which is irrelevant to the issue here.) 

Having ranked the four groups Hold with #4, the H, away from you. The other 3 groups are in priority order counter-clockwise. Thus the configuration is S, and the complete name is (S)-1,5-dibromo-7-methyloctane.

B1. a. The only non-aromatic C is the one at the top middle.
b. 4. The two rings are equivalent, but the four H on each ring are not (because of that non-aromatic C in the middle, which results in a top-bottom asymmetry).
c. Hydroxylation of the aromatic ring, to make a phenol. (p 158; also ok to note the formation of the epoxide.)

H Habe et al, Characterization of the upper pathway genes for fluorene metabolism in Terrabacter sp. Strain DBF63. J Bacteriol 186:5938-5944, 9/04. The paper discusses two pathways used by bacteria to degrade this compound; neither starts with phenol formation, which is typical of mammalian liver detoxification.

B2. a. No effect. An O forms two bonds, and just fits between two other atoms. If you aren’t convinced by the logic, try it with simple compounds such as methanol.
b. 16 H. There are 2 rings and 6 double bonds; each “costs” 2 H.

c. There is an O-group and 3 C-groups. The O-group is high priority (#1), by atomic number.

Comparing the 3 C-groups, one is simply methyl, and the others are more complex. More rigorously, the methyl C has HHH on it, whereas the others both have CHH. Thus the methyl C is low priority (#4).

Now we have left the two large C-groups. The first 2 C of each are the same: -CH2-CH2-. Thus we must look at the 3rd C. For the long tail, the 3rd C has CCH. For the upper group, the 3rd C is in the aromatic ring, and has CCC. Thus the top group is 2nd priority, and the side chain is 3rd.

d. See 6.d. for my choice of stereoisomer. This compound is R at the ring stereocenter (and at all three stereocenters).

cq\bq2a	Spring 2006
4/2/06