14C_s10_lec02_e02_key

14C_s10_lec02_e02_key - Chemistry 14C Lecture 2 Spring 2010...

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Chemistry 14C Lecture 2 Spring 2010 Exam 2 Solutions Page 1 Statistics : High score, average and low score will be posted on the course web site after exam grading is complete. The exam is ready to be picked up when these numbers are posted. A note about exam keys : The answers presented here may be significantly longer than expected from a student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as well. To see the projected course grade cutoffs, consult the grading scale on the Chem 14C course web page. 1. (a) (10 x 12) + (13 x 1) + (3 x 14) + (5 x 16) + (1 x 32) = 287 (b) (10 x 1.107%) + (13 x .015%) + (3 x 0.366%) + (5 x 0.037%) + (1 x 0.76%) = 13.3% (c) One sulfur (1 x 4.22%) + five oxygens (5 x 0.204%) + tiny contributions for 14 C, etc. = 5.24% = ~6%. 2. Molecule A , molecule C , molecule D , molecule E , and Nifurtimox. An ion with m/z = 96 could be a molecular ion or fragment. Only molecule B cannot produce an ion of this m/z . 3. None of these. Every molecule obeys the nitrogen rule . 4. Three DBE. One for the ring, and one for each of the S=O bonds . It is not necessary to use the molecular formula and DBE formula to determine the DBE count. If the molecular structure is available, we need only count the number of rings and pi bonds . 5. Alcohol : Absent. The peak in zone 1 is too narrow to be an alcohol . It is probably a C–H stretch. Alkene : Cannot determine. The peak at ~1600 cm -1 could be an alkene . Alkyne : Cannot determine. The peak at ~2200 cm -1 could be alkyne or nitrile . Ether : Cannot determine. There is no peak in an IR spectrum that is characteristic for an ether. The presence of an ether is determined by eliminating other possibilities . Ketone : Absent. There is no strong C=O peak in zone 4 . Nitrile : Cannot determine. The peak at ~2200 cm -1 could be alkyne or nitrile . 6. None of these. None of the choices can account for the triple bond peak in zone 3, or the benzene ring peaks in zone 5 . 7. N–N less intense than S=O. The N–N bond is much less polar than the S=O bond . 8. (a) The 1 H-NMR spectrum has eleven signals. The hydrogens of each CH 2 group are not equivalent to each other, as discussed in question 10 .
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This note was uploaded on 09/06/2011 for the course CHEM 140A taught by Professor Whiteshell during the Fall '04 term at UCSD.

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14C_s10_lec02_e02_key - Chemistry 14C Lecture 2 Spring 2010...

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