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# midterm-2003-answers - Name Midterm Exam CHEM/BCMB...

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Name _________________________ Midterm Exam: CHEM/BCMB 4190/6190/8189 1 Midterm Exam: CHEM/BCMB 4190/6190/8189 (155 pts) Tuesday, 7 October, 2003 1). In the example (right), the effect of a 90 ° ( π /2) pulse applied along the “x” axis (90 ° x) is shown. For a-f below, show the effect of the indicated pulses on the indicated magnetization vectors by drawing the resulting vectors on the coordinate axes. (12 pts) a. d. b. e. c. f. 2). A diagram of the Zeeman levels/states for 13 C is shown to the right. Draw the Zeeman diagram for 10 B ( I = 3). (4 pts) 90 ° x M 0 y z x y z x 0 E m +1/2 -1/2 E = +1/2 γ hB 0 E = -1/2 γ hB 0 E = 0 γ hB 0 = 0 E = -1 γ hB 0 E = -2 γ hB 0 E = -3 γ hB 0 E = +1 γ hB 0 E = +2 γ hB 0 E = +3 γ hB 0 E m +1 -1 +3 0 -2 +2 -3 90 ° -x y z x y z x 90 ° -y y z x y z x 180 ° x y z x y z x 270 ° y y z x y z x 360 ° -y y z x y z x 90 ° -y y z x y z x

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Name _________________________ Midterm Exam: CHEM/BCMB 4190/6190/8189 2 3). If we directly compare 10 B spins and 13 C spins; a. What is the ratio of the bulk/macroscopic magnetization ( M 0 10B / M 0 13C ) produced for equal numbers of nuclei? (6 pts) b. How does the sensitivity of the NMR signal compare for 10 B and 13 C spins? (6 pts) a. M 0 10B M 0 13C = N γ 10B 2 h 2 B 0 I 10B (I 10B +1) 3k B T N γ 13C 2 h 2 B 0 I 13C (I 13C +1) 3k B T = γ 10B 2 × 3 3 + 1 ( ) γ 13C 2 × 1 2 1 2 + 1 = (2.8747 × 10 7 rad/Ts) 2 × 12 (6.7283 × 10 7 rad/Ts) 2 × 3 4 = 2.9207 So, the bulk magnetization is about 2.9 times as large for 10 B as for 13 C. b. Sensitivity is proportional to the electromotive force ( ε ) induced in the receiver coil by the bulk magnetic moment. The magnitude of ε is proportional to the rate of change in the magnetic moment ( ε dM/dt = γ M 0 B ). Remember, M 0 = N γ 2 h 2 B 0 I(I +1) 3k B T (“h” here is “h-bar”, Planck’s constant divided by 2 π ), so ε γ M 0 B = N γ 3 h 2 B 0 2 I(I +1) 3k B T . So, whereas the magnitude of the bulk magnetization was dependent on γ 2 , the sensitivity is dependent on γ 3 . Thus, ε 10B ε 13C = N γ 10B 3 h 2 B 0 2 I 10B (I 10B +1) 3k B T N γ 13C 3 h 2 B 0 2 I 13C (I 13C +1) 3k B T = γ 10B 3 I 10B (I 10B + 1) γ 13C 3 I 13C (I 13C + 1) = (2.8747 × 10 7 rad/Ts) 3 × 12 (6.7283 × 10 7 rad/Ts) 3 × 3 4 = 1.248 So, 10 B nuclei are about 1.25 times more sensitive than 13 C nuclei.
Name _________________________ Midterm Exam: CHEM/BCMB 4190/6190/8189 3 4). The frequency difference ( ∆ν ) between two 1 H signals in a 1 H spectrum is 4320 Hz when the spectrum is acquired with a magnetic field strength of 9.4 T (corresponding to a 1 H resonance/observe frequency of 400 MHz). a. What would be the frequency difference between the two 1 H signals if the spectrum was acquired with a 18.79 T magnet? (5 pts) b. What is the chemical shift difference ( ∆δ ) between the two signals? (5 pts) a. We know that 18.79 T 800 MHz, δ 400 = ν 400 400 MHz × 10 6 and δ 800 = ν 800 800 MHz × 10 6 . We also know that, by definition, ∆δ 400 = ∆δ 800 (chemical shift is independent of B 0 ). So, the ratio of the frequencies is equal to the ratio of the field strengths: ν 400 ν 800 = 400 MHz 800 MHz , therefore , ν 800 = 4320 Hz × 800 MHz 400 MHz = 8640 Hz b. δ = ν observe frequency × 10 6 = 4320 Hz 400 MHz × 10 6 = 4320 Hz 400 Hz = 10.8 "ppm" 5). a. For the hydrocarbon below, the T 1 values for the 13 C nuclei (in seconds, shown below the respective carbon atoms) increase towards the ends of the hydrocarbon chain.

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