Bq2aj03 - X402 Test 2 Summer 2003 Answer key 16 = A 12 = B 8 = C same as Test 1(Ignores Mean = 16.9 median = 17.5 Top scores 2 at 19.5 To estimate

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X402 Test 2, Summer 2003. Answer key. 16 = A, 12 = B, 8 = C; same as Test 1. (Ignores +/-) Mean = 16.9, median = 17.5. Top scores: 2 at 19.5. To estimate your current course grade, average your two test letter grades so far. If relevant, round toward Test #2. I will apply the same basic strategy throughout the course. (I am happy to discuss individual situations with you.) Test 3 has a reputation of being considerably harder, presumably because of the amount of material behind us -- and the complexity of certain things such as acetal formation. You are forewarned. 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. 2-decanol, due to the polarity (specifically, ability to hydrogen bond) of the alcohol (-OH). b. 2-propanol. Each molecule has the same polar group. The smaller propanol has a smaller nonpolar part; the nonpolar part tends to reduce the solubility. c. 2-decanol. It is more polar than the decane, and heavier than either of the others. Thus it can stick together by both London forces and hydrogen bonding, more than either of the others. Actual data for BP (in ° C): 2-decanol, 211; decane, 174; 2-propanol, 82. For solubility, both of the C 12 compounds are listed as insoluble, but we would expect a (very) slight solubility of the alcohol. 2-propanol is miscible with water. 2. C-C-C-C-CH 2 OH º ³ O C-C Some of you had the wrong starting compound here. If your starting compound did not include both a 1 ° and 2 ° -OH, then you were missing the point of the
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This note was uploaded on 10/24/2011 for the course ANP 101 taught by Professor Jaques during the Spring '11 term at Queens University.

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Bq2aj03 - X402 Test 2 Summer 2003 Answer key 16 = A 12 = B 8 = C same as Test 1(Ignores Mean = 16.9 median = 17.5 Top scores 2 at 19.5 To estimate

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