Exam 1 FS10 - CEM 351, Fall 2010 Midterm Exam 1 Friday,...

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CEM 351, Fall 2010 Midterm Exam 1 Friday, October 1, 2010 1:50 – 2:40 p.m. Room 138, Chemistry Name (print) Wright N. Swers Signature Student # Section Number (2 pts extra credit) Grade this question? YES NO! . . 1.(20 pts.) . . 2.(20 pts.) . . 3.(20 pts.) . . 4.(20 pts.) . . 5.(20 pts.) . . 6.(20 pts.) TOTAL (100 pts.) Score Note : Answer any 5 questions for a total score of 100 pts. Be sure to look over all the questions first before beginning the exam, and indicate which five questions are to be graded by checking the corresponding box. Recitation Day and Time Instructors: Room 1. T 9:10-10:00 a.m. Jason Lam 85 2. M 10:20-11:10 a.m. Carmin Burrell 281 3. T 10:20-11:10 a.m. Carmin Burrell 85 4. Th 10:20-11:10 a.m. Arvind Jaganathan 287 5. T 10:20-11:10 a.m. Jason Lam 283 6. T 11:30-12:20 p.m. Jason Lam 283 7. Th 11:30 - 12:20 Arvind Jaganathan 85 8. M 4:10-5:00 p.m. Carmin Burrell 110 9. Th 4:10-5:00 p.m. Arvind Jaganathan 183
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2 1. (20 pts) (a) (12 pts) Draw structures to represent the following IUPAC names (i) 2-isopropoxypropane O (ii) cis -4-methylcyclohexanol OH (iii) 2,2,3,3-tetramethylbutane or (b) (2 pts) Which of the compounds in (a) has no dipole? Please circle your answer (i) (ii) (c) (4 pts) Are any of the groups on the cyclohexane in (ii) axial? Please explain with a drawing and a few words; here is a model to work from but you must make your own drawing in the box provided. Explanation: (d) (2 pts) How many resonances would you expect in the 1 H NMR of (i)? of (iii)? Explanation: (i) Two resonances: (a) a doublet [at ca. 1.1] and (b) a septet [at ca. 3.5 ppm], with the a:b integral ratio of 6:1 (= 12:2). The 6 equivalent H ʼ s in the isopropyl group ʼ s CH 3 groups split the one H on the CH between them into a septet, while they are split into doublets. (iii) One singlet [at about 0.8 ppm]. All six of those CH 3 groups are equivalent. Note: though rough chemical shifts are given above, they weren ʼ t required for the answer. The cis related groups are on the same face of the ring [imagine it flat, as in the answer to (ii)]. Here the –CH 3 and –OH groups point “up”; the H ʼ s on carbons 1 and 4 then show the contrasting “down” (other face) position. OH H 3 C H H 1 4 2 3
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3 2. (20 pts) (a) (3 pts each) Form acceptable Lewis structures for the following uncharged compounds by adding the necessary nonbonding electrons and multiple bonds: : N C O N : H H Urea H C N : Hydrogen cyanide H H (b) (3 pts each) Assign formal charges as necessary to the atoms in the following Lewis structures.
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This note was uploaded on 02/17/2012 for the course MTH 133 taught by Professor Staff during the Fall '08 term at Michigan State University.

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Exam 1 FS10 - CEM 351, Fall 2010 Midterm Exam 1 Friday,...

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