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# bq3as06 - X402 Test 3 Spring 2006 Answer key 16 = A 12 = B...

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X402 Test 3, Spring 2006. Answer key. 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 C 4 and O 3 , with two double bonds. We are asked for a carboxylic acid. So let’s start with butanoic acid. That takes care of the C 4 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 C 4 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.

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bq3as06 - X402 Test 3 Spring 2006 Answer key 16 = A 12 = B...

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