finalexam-2003-answers - Name _ Final Exam: CHEM/BCMB...

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Name _________________________ Final Exam: CHEM/BCMB 4190/6190/8189 1 Final Exam: CHEM/BCMB 4190/6190/8189 (290 pts) Tuesday, 16 December, 2003 1). ( 15 pts ) For ‘a’ and ‘b’ below, indicate relative intensities of signals in spectra, and indicate coupling constants (in terms of J CH ) where appropriate. Mark the Larmor frequencies on your spectra with ν H and ν C . Identify individual transitions (i.e. M C H α , etc.) a. Sketch the 1D, 1 H NMR spectrum of chloroform, CHCl 3 (no decoupling). b. Sketch the 1D, 13 C NMR spectrum of chloroform (no decoupling). c. If you were going to perform an SPI experiment to enhance the intensities in the 13 C NMR spectrum, indicate (circle) one transition in one of the signals above that you might excite to accomplish this. In the SPI experiment, either the M H C α or M H C β transition in the 1 H NMR signal (but not both!) could be excited to enhance the intensities of the components of the 13 C signal. I’ve arbitrarily circled one of them. J CH ν H 1 H- 12 C- M H C β M H C α M C H α M C H β J CH ν C
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Name _________________________ Final Exam: CHEM/BCMB 4190/6190/8189 2 2). ( 20 pts ) In the DEPT experiment, the phase angle (pulse width) of the third 1 H pulse (applied along the ‘y’ axis) can be set to any value, in order to achieve the desired result. Shown in the figure below are the intensities of signals from CH, CH 2 , and CH 3 groups as a function of the phase angle Θ y: a. The DEPT(135) spectrum for isopentyl acetate is shown below. Draw the DEPT(45)- DEPT(135) spectrum for isopentyl acetate. Make sure to identify (label) the signals in the same manner that they are labeled in the DEPT(135) spectrum. Write a short explanation of why the DEPT(45)-DEPT(135) spectrum looks as it does. With the phase angle y set at 45 ° , the intensities of –CH and –CH 3 signals (DEPT(45)) are almost identical to their intensities when y set at 135 (DEPT(135)). Thus in the DEPT(45)- DEPT(135) spectrum, signals from these groups will be absent. The intensities of signals from –CH 2 groups in the DEPT(45) spectrum are very large and positive, whereas signals from –CH 2 groups in DEPT(135) spectra are very large and negative. Subtracting the latter from the former results in very large positive signals (DEPT(45)-DEPT(135)). E D O || C 1 H 3 -C 2 -O-C 3 H 2 -C 4 H 2 -C 5 H-C 6 H 3 | C 7 H 3 isopentyl acetate A C B D E 80 60 40 20 0 13 C, δ (ppm) DEPT(135) 80 60 40 20 0 13 C, δ (ppm) DEPT(45)-DEPT(135)
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Name _________________________ Final Exam: CHEM/BCMB 4190/6190/8189 3 b. Indicate in the table below whether each signal (A-E) arises from a –CH 3 , -CH 2 , -CH, or quaternary carbon by placing an ‘X’ in each row in the appropriate column. signal -CH 3 -CH 2 -CH quaternary A (21 ppm) X B (23 ppm) X C (25 ppm) X D (35 ppm) X E (65 ppm) X c. In the table below, assign each signal in the DEPT(135) spectrum of isopentyl acetate by placing the appropriate integer (1-7) in the ‘assignment’ column, corresponding to the correct carbon atom in the structure of isopentyl acetate shown above. Give a concise but thorough rationale for the assignment.
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This note was uploaded on 11/13/2011 for the course CHEM 4190 taught by Professor Staff during the Fall '08 term at University of Georgia Athens.

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finalexam-2003-answers - Name _ Final Exam: CHEM/BCMB...

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