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exam4-2009 - Name Exam 4 CHEM/BCMB 4190/6190/8189(100...

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Name _________________________ Exam 4, 2009: CHEM/BCMB 4190/6190/8189 1 Exam 4: CHEM/BCMB 4190/6190/8189 (100 points) Tuesday, 15 December, 2009 1 ). In the DEPT experiment, the pulse angle ( Θ y , 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 diagram below (right) are the intensities of signals from –CH, -CH 2 , -CH 3 groups as a function of the phase angle Θ y . Below are shown a (top) and a DEPT spectrum (bottom) for the compound propyl benzoate (shown in the inset). a. What, most likely, is the phase angle, Θ y , in the DEPT spectrum shown. You will have to provide a complete explanation for credit. ( 4 points ) The -CH 2 groups are of opposite phase (C2 and C3, phased down) with respect to both the -CH 3 (C1) and -CH (C6, C7, and C8) groups. Thus, the angle must be somewhere between 90 and 180. Given that the amplitude of the –CH3 signal (C1) and the signal from the -CH of C8 are approximately equal in size, this would suggest that the phase angle is perhaps 120 ° (where the -CH and -CH 3 signals have approximately equal amplitudes, according to the figure above, right). The -CH signals from C7 and C6 would be expected to be about twice as large because there are two of each in the benzene ring. b. What does the signal at ~77 ppm in the 13 C spectrum arise from? Why does it not show up in the DEPT spectrum? ( 4 points ) This is from solvent, probably CDCl 3 . It doesn’t show up in the DEPT because in DEPT there is polarization transfer from 1 H to 13 C, and there are no hydrogens in CDCl 3 . CH CH 2 CH 3
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Name _________________________ Exam 4, 2009: CHEM/BCMB 4190/6190/8189 2 c. Positions 4 and 5 are quaternary carbons. Why do signals from these not show up in the DEPT spectrum? ( 4 points ) They doesn’t show up in the DEPT because in DEPT there is polarization transfer from 1 H to 13 C, and there are no hydrogens attached to these carbon nuclei. d. Positions 4 and 5 are quaternary carbons. Why are the signals from these in the 13 C spectrum much less intense than the signals from the other carbons? ( 4 points ) Quaternary carbon ( 13 C) nuclei relax slowly. If the relaxation delay for the experiment is not long enough, there will be incomplete relaxation (compared to other 13 C nuclei in the molecule), and, therefore, less coherent transverse magnetization created following the next excitation pulse, resulting in a signal of lower intensity. So, most likely, the relaxation delay in the experiment was not long enough to accommodate the long relaxation times for these nuclei.
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Name _________________________ Exam 4, 2009: CHEM/BCMB 4190/6190/8189 3 2 ). The 2D 1 H, 13 C-HSQC spectrum of propyl benzoate is shown (below). An inset expands a region of the spectrum from about 125-135 ppm ( 13 C). The observed signals in the 1 H spectrum are 8.05, 7.55, 7.40, 4.20, 1.75, and 1.0 ppm. Put the correct 1 H chemical shift in the appropriate cell in the table. Provide any explanation you think necessary. ( 8 points ) signal 13 C (ppm) 1 H (ppm) 1 10.5 1.0 2 22.0 1.75 3 66.5 4.20 4 167.0 no H 5 130.5 no H 6 129.5 8.05 7 128.3 7.40 8 133.0 7.55
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Name _________________________ Exam 4, 2009: CHEM/BCMB 4190/6190/8189 4 3 ). The 2D 1 H- 1 H COSY spectrum of propyl benzoate is shown below.
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