bq3aj03 - X402 Test 3, Summer 2003. Answer key. 16 = A, 12...

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X402 Test 3, Summer 2003. Answer key. 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 inter molecular 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 Otherwise, each of its C has an -OH, except for the one -OH that has been replaced by -NH 2 . The lower ring is a cyclohexane ring, and is not convertible to a sugar by any simple
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bq3aj03 - X402 Test 3, Summer 2003. Answer key. 16 = A, 12...

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