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Unformatted text preview: Name _________________________ Exam 3, 2006: CHEM/BCMB 4190/6190/8189 1 Exam 3: CHEM/BCMB 4190/6190/8189 (156 points) Tuesday, 7 November, 2006 1 ). 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 y: a. The DEPT(135) spectrum for 4-ethoxyphenylacetic acid is shown below. The chemical shifts for the peaks are indicated. Draw the DEPT(45)-DEPT(135) spectrum for 4-ethoxyphenylacetic acid. Explain why the DEPT(45)-DEPT(135) spectrum looks as it does. ( 8 points ) 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)). 4-ethoxyphenylacetic acid 200 160 120 80 40 0 13 C, (ppm) DEPT(135) 200 160 120 80 40 0 13 C, (ppm) DEPT(45) - DEPT(135) 15 ppm 41 ppm 63 ppm 115 ppm 130 ppm Name _________________________ Exam 3, 2006: CHEM/BCMB 4190/6190/8189 2 b. There are 8 peaks in the 13 C spectrum of 4-ethoxyphenylacetic acid. The chemical shifts are shown in the table below. Indicate in the table whether the signal arises from a CH 3 , -CH 2 , - CH, or quaternary carbon. In the assignment column, put the letter(s) corresponding to the correct nucleus or nuclei in the molecule that gives rise to that particular signal. In the rationale column provide a complete justification for your assignment (you can use the back of this page if necessary). ( 24 points ) chemical shift -CH 3 , -CH 2 , - CH, quaternary assignment rationale 15 ppm -CH 3 J The methyl group in the molecule should give a peak in DEPT(135) that is up and with an intensity larger than the -CH groups. Also, the methyl group in the molecule would be expected to be shielded (upfield). Thus, this is carbon J (-CH 3 ). 41 ppm -CH 2 B In DEPT(135), -CH 2 groups are down, so the methyl groups B and I are at 41 and 63 ppm. We expect I to be more deshielded because it is bonded to the electronegative oxygen. 63 ppm -CH 2 I Same as above. 115 ppm -CH F, G In DEPT(135), -CH groups are up and less intense than -CH groups. Thus, the peaks at 115 and 130 are the -CH groups F/G and D/E. Because there are only 8 peaks in the 13 C spectrum, F and G are equivalent as are D and E. F and G are more shielded due to delocalization of electrons at these positions (resonance structures with ethoxy group), thus F/G is at 115 (D/E is 130)....
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