16 = A, 12 = B, 8 = C. (ignoring +/-) (same for Tests 1-3)
Mean = 16.2, median = 17.5. Top scores: 19, 2 at 18.5. Not bad!

To estimate your current course grade, average your three test letter grades so far. If relevant, round toward Test #3. I will apply the same basic strategy after the final exam: average test letter grades, round towards the final if relevant. (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.)

1. a. Ethanol. The alcohol, with -OH, has both an H-bond donor and H-bond acceptor. Aldehydes lack an H-bond donor. Thus alcohol molecules have stronger intermolecular attraction, with other molecules of self. (Water is irrelevant. And it is sloppy to simply say that aldehydes can’t H-bond with other molecules. You want to be clear that, since this is a question about boiling, you are talking about other molecules of self.)

b. Propanoic acid. Carboxylic acids have even more possibilities for H-bonding, with one more O. In fact, they can form H-bonded cyclic dimers (p 336, bottom).

2. The upper (or “right”) ring is an amino sugar. Sugars are polyhydroxy carbonyl compounds; the carbonyl group appears as a hemiacetal or acetal in the cyclic forms. The upper ring has the characteristic ring O, which is part of the acetal group (glycosidic bond). Otherwise, each of its C has an -OH, except for the one -OH that has been replaced by -NH2.

The lower ring is a cyclohexane ring, and is not convertible to a sugar by any simple manipulation. (It is inositol, a particular stereoisomer of 1,2,3,4,5,6-hexahydroxycyclohexane. You met inositol later as a component of phospholipids; it -- in the form of various phosphate esters -- is also a signaling molecule.)

The -CH2OH (hydroxymethyl) group is not relevant. Some sugars have them, some do not (e.g., Exercise 11.15c). If there is one, it is simply because the chain is longer than the part involved in the ring; no fundamental relevance.

G L Newton et al, The glycosyltransferase gene encoding the enzyme catalyzing the first step of mycothiol biosynthesis (mshA). J Bacteriol 185:3476-3479, 6/03.

3. C60 is entirely non-polar; it would be expected to be quite insoluble in water. Sugars, of course, are highly polar (lots of -OH) and correspondingly are very soluble in water. Thus the sugared-C60 should be significantly more soluble in water than the parent C60.

Noted briefly in Science 289:1109, 8/18/00.

4. The alcohol O of one molecule attacks the carboxyl C of the other, to form the ester. Those key atoms are underlined in both reactants and product below; the doubly-underlined atoms end up as water. (p 351, Fischer esterification, a condensation reaction)

  OH  OH      OH  OH        OH  O   CH3  OH 
                                 
C-C-C-C=O + C-C-C-C=O   C-C-C-C-O-C-C-C=O + H2O

The question says it is about esters, and asks for a condensation reaction. (Making an anhydride is a bad reaction.)

5. a. You get cotinine, shown at the right:

b. This is an amide (or, since it is cyclic, a lactam).	
	
A E Witter et al, Quantitative determination of nicotine and cotinine in urine and sputum using a combined SPME-GC/MS Method. J Chem Educ 79:1257, 10/02.

6. a. cyclohexanol				b. No reaction; ketones resist “mild” oxidation.
c. Ethanoate (acetate) ion, CH3COO-.	d. isobutyl pentanoate

7. a. Alkene (C=C) and ester (or lactone, since it is cyclic).
b. The ester will hydrolyze to acid + alcohol.

K M George et al, Mycolactone: A polyketide toxin from Mycobacterium ulcerans required for virulence. Science 283:854, 2/5/99.

8. a. The formula shows that the compound is short 2 H of maximum. Thus it has a ring or a double bond. The double bond could be C=C or C=O.

Lack of reaction with bromine eliminates C=C. Thus it has either a ring or a C=O. (Note that nothing that follows resolves this.)

Oxidation indicates that the O is a 1 or 2 alcohol or an aldehyde. (Eliminates 3 alcohol, ether, ketone.)

Oxidation of 1 alcohol or aldehyde would yield an acidic product. (Eliminates 2 alcohol.)

b. From all the evidence above, we are left with two general possibilities. It is either an aldehyde with no rings, or a 1 alcohol with one ring. No information given here distinguishes them. 

Many possible answers within those two classes. A simple example of the first is pentanal, and a simple example of the second is cyclobutylmethanol.

Most of you reached a reasonable conclusion, coming up with a compound that fit the data. So you got credit for part b. (You favored the aldehyde, by about 4:1.) Please read the comments I wrote you for part a. I gave credit for some things that were significantly incomplete, and you should understand the more complete arguments. Overall, you did quite well with this question.

9. a & b. “Flip” the 5 position of D-mannose, to get L-gulose (the enantiomer of D-gulose).
c. Oxidize the “bottom” (#6) position of the D-mannose from -CH2OH to -COOH.
d. You need to know how they are connected -- which positions and which anomer.

B1. Think backwards. The last step must be isopropyl alcohol + acetic acid  the ester. To make those:

Propene  2-propanol (acid catalyzed hydration, with M-rule giving the desired alcohol).

Ethene  ethanol (acid catalyzed hydration)  acetic acid (oxidation of a 1 alcohol).

B2. H-Kr-CC-H

This compound was reported just recently -- while our course was in progress. Aside from the curiosity factor -- a surprisingly stable Kr compound -- it may be useful in opening up new reaction pathways with alkynes. L Khriachtchev et al, A gate to organokrypton chemistry: HKrCCH. JACS 125:6876, 6/11/03.

B3. a. At the right are the structures of the two compounds given in the question and their ester. The ester is pyrethrin.

(Note that there are some stereochem issues, which I have not tried to resolve.)		

b. Do a common reaction with any one functional group. Whether you do it before or after the esterification would depend on whether the feature exists in both subunits (and whether the change you make would survive the esterification reaction).

Example: Reduce the C=C in the propenyl side chain of the acid. This must be done before ester formation, because there are other groups in the alcohol that would also react. (Actually, not as easy as it sounds, because the very strained cyclopropane ring may also react. Nevertheless, this illustrates how organic chemists would approach such a problem.)

B4. There are three stereocenters, all in the big ring. They are where R1, R2 and the -CH3 just above R2 are attached.

Some structures above were drawn with ISIS/Draw. See my web site for more information and a “tutorial” on this useful drawing program.

cq\bq3a	(Summer 2003) 
8/17/03