Chem316F07MT2a - Chem 316 Fall 2007 Page 1 of 6 Kantorowski...

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Chem 316 Fall 2007 Page 1 of 6 Kantorowski Exam II Key 25 October 2007 Name: ___________________________________________ This exam contains 5 pages of questions – confirm this once you begin . You will have 50 minutes An abbreviated Periodic Table can be found on the last page. No calculators or models are permitted. Read all questions carefully – answer the question that is asked! Illegible or indecipherable answers may not receive potential partial credit. Good luck! 1. (6 pts) Calculate the index of hydrogen deficiency (IHD) for each of the following molecular formulas. a) C 10 H 16 O 2 __ 3 __ b) C 8 H 15 ClO 3 __ 1 __ d) Assume that you have determined that the compound in (a) possesses an ester, but no alkenes nor alkynes. What other information could you conclude about its structure? The ester FG (= 1 IHD) employs both oxygens. Each of the remaining IHD units must be in the form of a ring. Therefore, there are two rings present. 2. (8 pts) Indicate whether each of the following is cis or trans. OCH 3 Cl H H H Br SH H H H H H Cl Cl trans trans trans trans H H H H 3. (5 pts) The eclipsed conformation of ethane is 3.0 kcal/mol less stable than its staggered form. The eclipsed conformation of CH 3 CH 2 Br is 3.8 kcal/mol less than its staggered form. Based on these observations, calculate the H,Br eclipsing energy. H H H H H H ethane (eclipsed) each H-H = 1.0 kcal/mol Total destabilization = 3.0 kcal/mol 1.0 1.0 1.0 H H H H H Br 1.0 1.0 x chloroethane (eclipsed) each H-H = 1.0 kcal/mol = 2.0 kcal/mol Total destabilization = 2.0 + x = 3.8 kcal/mol Therefore, the H,Br interaction is worth 1.8 kcal/mol (Note: a value of 0.8 would not make sense - the H,Br interaction should be larger than an H-H interaction!)
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