16 = A, 12 = B, 8 = C, 4 = D. (Ignores +/-.) Same as Test 2.

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. -OH groups (alcohol groups). The -ose ending of the compound given indicates it is a sugar. Sugars are polyhydroxy carbonyl compounds. -OH groups are very common; there is only one carbonyl group per monomer sugar.

S Chaillou et al, Cloning, sequence analysis, and characterization of the genes involved in isoprimeverose metabolism in Lactobacillus pentosus. J Bacteriol 180:2312, 5/98. Isoprimeverose is a disaccharide of xylose and glucose.

2. a & b. Products are methyl ethanoate (acetate) + ethanoic (acetic) acid.

c. For example, 3-oxobutanoic acid. One strategy you might use: The given compound is C4 and O3, with two double bonds. We are asked for a carboxylic acid. So let’s start with butanoic acid. That takes care of the C4 and two of the O and one of the double bonds. We still need one O and one double bond (or ring). An oxo takes care of both of those. Check the H count to be sure this comes out right.

3. a. 2-butanol. (Reduce ketone to alcohol. The only C4 ketone is 2-butanone = butanone.)

b. propanoic (propionic) acid. It has approximately the same mass as 1-butanol (same number of C + O). Although alcohols have high boiling points (BP), carboxylic acids have even higher BP, because of the ability to form a cyclic dimer with two hydrogen bonds between two carboxyl groups.

4. a. (Folks Show the aldehyde H. It is a defining part of the functional group.)

    2
C-C=C-C=O
     
  H3C  H

b. Configuration at double bond (E,Z) is not specified.

E M Blass, Reproductive biology: Mammary messages. Nature 424:25, 7/3/03. The pheromone appears to be species-specific. Why did I write 2‑methylbut-2-enal instead of 2‑methyl-2-butenal? Because they did; anyway, this is the IUPAC preference. As to the isomer I don’t think they say. But further sleuthing suggests it is the E-isomer, also known -- though with some ambiguity -- as trans. (This compound is very similar to that shown in question 7 on the practice quiz for alkenes. That question illustrates the ambiguity of “trans” here.)

5. It is D-gulose, shown at the right.

d-gulose is incorrect. In fact, D-gulose is l‑gulose, meaning (-)-gulose; see the chart I handed out. But the d/l usage is “archaic”.
	    CHO
        
  H-C-OH
        
  H-C-OH  (#3)
        
 HO-C-H
        
  H-C-OH
        
    CH2OH	

6. It is an ester, formed between the -OH of one molecule and the carboxyl group of the next. The LHS of the following picture shows two molecules of lactic acid lined up to form an ester. The RHS shows the ester dimer, plus the molecule of water from the underlined atoms. Note that both ends of the dimer can extend further; thus this is the basis of a polymer.

E Matsuda et al, Gene cloning and molecular characterization of an extracellular poly(L-lactic acid) depolymerase from Amycolatopsis sp. Strain K104-1. J Bacteriol 187:7333-7340, 11/05.

7. a. One is at the top right, and one is at the bottom of the big ring. The proper bond to mark is the one between the alcohol O and the carbonyl group. (p 353)

b. Contains neither. Acid groups are created by hydrolysis of the esters.

c. The strained 3-rings (the epoxides).

d. Not so good. The chemical shown is rhizoxin, which causes rice seedling blight.

L P Partida-Martinez & C Hertweck, Pathogenic fungus harbours endosymbiotic bacteria for toxin production. Nature 437:884, 10/6/05. (+ News story, p 823.) The point here is that the toxin, long thought to be made by a fungus (Rhizopus), is actually made by a bacterium (Burkholderia) endosymbiotic with the fungus (i.e., living within the fungal cells). The toxin interferes with mitosis, by binding the spindle fibers; it has been considered as a possible anti-cancer drug. 

8. hemiacetal + alcohol  acetal.  anomer has the alcohol R attaching “up”.

9. a. First, figure out what -D-Gal is in the cyclic form. -D-Glc is given; -D-Gal differs at the 1 and 4 positions. Now, attach this, by its 6-position (its top C), to the given sialic acid (SA), at the 2-position, . The SA is given in its  form.

The key things I looked for: acetal linkage between 2 position of the SA and 6 position of the galactose.  linkage -- and the given SA hemiacetal is shown . -Gal: distinctive features are C#1 and C#4, both with -OH up.

b. ketononose. “Keto” is based on where hemiacetal (hemiketal) formation occurred. “Non”? C9. The given structure shows C6, and R has C3 -- exactly the kinds of C found in an ordinary sugar (polyhydroxy).

K Shinya et al, Avian flu: Influenza virus receptors in the human airway. Nature 440:435, 3/23/06. SAGal structures are found on cell surfaces -- with various linkages. Human flu virus preferentially attaches to such structures when linked -2,6 (as in this question), whereas bird flu virus preferentially attaches to such structures linked -2,3. The paper listed here shows that there are some receptors for bird flu virus in the human respiratory system, but rather deep down. This may help explain why transmission to humans is low, why transmission from humans is low, but perhaps also why the virus causes a more serious illness if we do get it.

10. 2-propylpentanoic acid

Depletion of latent HIV-1 infection in vivo: a proof-of-concept study. Lancet 366:549-55, 8/13/05. ( + Comment, Routy, p 523.) The study uses valproic acid to activate latent virus, making it susceptible to normal treatment. Valproic acid, a drug already approved for other purposes, affects chromatin structure; there is some logic to choosing it to try to activate latent HIV, though in the end data is needed to show it really works. The data here, while encouraging, should be considered very preliminary. Also see Science 309:999, 8/12/05, for a brief news story about this paper.

11. a. acetal (a C with two -OR attached). (If you are inclined to say ether here, then part b makes no sense, so that should be a clue this is not an ether.)
b. 1,2-ethanediol (ethylene glycol). (Acetals are unstable in acid; the resulting hemiacetal is unstable, since it is not cyclic.)

S H Chanteau et al, Arts and sciences unite in Nanoput: Communicating synthesis and the nanoscale to the layperson. J Chem Educ 80:395, 4/03. Also see: http://www.NanoArtWorks.com

B1.
a. 
(The two ‑OH are not bonded to each other.)		b. 		

c. It is an enol. It would be more stable as the aldehyde tautomer. (p 288)

d. Three of them, all on the 5-ring: the top two C and the lower right C.

Nepetalactone is the active ingredient in catnip.

B2. Treat the acetal with acid, and you are left with an aldehyde group there. You should be able to name this aldehyde as a derivative of benzaldehyde; it will just take considerable patience -- and parentheses. Give it a try and show me what you get. 

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	(Spring 2006) 
5/6/06