ν 1H γ 1H ν 13C = γ 1H 2π B 0 and ν 13C = γ 13C 2π B 0,therefore = π γ 13C =150.9 MHz = MHz 500 MHz ν 1H, 11.

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1 Problem Set #1, CEM/BCMB 4190/6190/8189 1). Which of the following statements are rue, False, or Possibly rue, for the hypothetical element X? he ground state spin is I0 for 5 4 b. he ground state spin is I0 for c. he ground state spin is I1/ for 45 1 d. he ground state spin is I1 for 48 e. he ground state spin is I1/ for 9 14 rue: both the mass number and the atomic number are even. herefore, I0. b. False: the mass number is even and the atomic number is odd. herefore, I is integral and non-zero. c. Possibly rue: the mass number is odd, therefore the spin is half-integral (might be 1/, might also be 3/ or 5/, etc.). d. False: both the mass number and the atomic number are even. herefore, I0. e. Possibly rue: same rationale as c. ). he resonance/precession frequency of a 1 nucleus (ν 1 ) in a (esla) magnet is Mz. What is the resonance frequency of a 13 C nucleus (ν 13C ) in the same magnet? (γ x 10 7 rad/s, γ 13C x 10 7 rad/s). b. What is the resonance frequency of the 1 nucleus in a magnet? ν 1 γ 1 π B 0 and ν 13C γ 13C π B 0,therefore b. ν Mz ν 13C ν 13C ν 13C Mz γ 1 π γ 13C π ν 1, Mz ν 1, ν 1, Problem Set #1, CEM/BCMB 4190/6190/ B 0 γ B 0 γ 13C γ 1 π γ π ν 1, Mz 500 Mz

2 3). ow does the magnitude of the bulk/macroscopic magnetization compare for: 1 and 13 C spins? b. 1 and 13 C spins at natural abundance (assume the natural abundance of 1 is 100% and 13 C is 1.11%)? c. 1 and 17 spins? (γ x 10 7 rad/s, γ 13C x 10 7 rad/s, γ x 10 7 rad/s) M 0 Nγ I(I ( h here is h-bar, Planck s constant divided by π) M 01 M 0 13C Nγ 1 I 1 (I 1 Nγ 13C I 13C (I 13C γ 1 γ ( rad/s) So, the bulk 13C ( rad/s) magnetization is about 16 times as large for 1 compared to 13 C, or 13 C magnetization is about 6.3% that of 1. b. M 01 M 0 13C Nγ 1 I 1 (I 1 Nγ 13C I 13C (I 13C (100) γ 1 (1.11) γ (100) ( rad/s) 13C (1.11) ( rad/s) 144 So, the bulk magnetization is about 358 times as large for 1 compared to 13 C at natural abundance, or 13 C magnetization is about 0.07% that of 1. c. M 01 M 0 17 Nγ 1 I 1 (I 1 Nγ 17 I 17 (I 17 γ γ ( rad/s) 0.75 ( rad/s) So, the bulk magnetization is about 4.7 times as large for 1 as for 17, or 17 magnetization is about 1.4% that of 1. 4). ow does the sensitivity of the NMR signal compare for 1 and 13 C spins? b. For equivalent numbers of 1 and 13 C nuclei, how many more scans are necessary to produce 13 C spectra with signal-to-noise equivalent to 1 spectra? Problem Set #1, CEM/BCMB 4190/6190/8189

3 Sensitivity is proportional to the electromotive force (ε) induced in the receiver coil by the bulk magnetic moment. he magnitude of ε is proportional to the rate of change in the magnetic moment: ε dm/dt γm 0 B Remember, M 0 Nγ I(I ( h here is h-bar, Planck s constant divided by π), so ε γm 0 B Nγ 3 h B 0 I(I. So, whereas the magnitude of the bulk magnetization was dependent on γ, the sensitivity is dependent on γ 3. hus, ε 1 ε 13C Nγ 3 1 h B 0 I(I Nγ 3 13C h B 0 I(I γ 3 1 γ ( rad/s) So, 1 13C ( rad/s) 3 nuclei are about 63 times more sensitive than 13 C nuclei, or the sensitivity of a 13 C nucleus is about 1.59% that of a 1 nucleus. b. In NMR, signal-to-noise (S/N) improves with the square root of the number of scans (N 1/ ). So, since the 1 sensitivity S/N from a above is times better that 13 C, it will take (6.856) C scans for every 1 scan to get equivalent S/N for equivalent numbers of 1 and 13 C nuclei. 5). A diagram of the Zeeman levels/states for 1 is shown to the right. ow do you determine the number of levels/states for a given nucleus? E b. ow many levels/states are there for 17? m c. What are the magnetic quantum numbers for the 17 states? -1/ d. ow do you determine the energy of a given state? E +1/γhB 0 e. Draw the Zeeman diagram for / E -1/γhB 0 he number of Zeeman levels is equal to I+1, where I is the spin quantum number. b. For 17 there are (5/)+1 6 levels c. m (-I, -I+1,, I-1, I), so, for I 5/, these are 5/, -3/, -1/, 1/, 3/, and 5/ d. E µ z B 0 -mγhb 0 m E e. -5/ -3/ -1/ +1/ +3/ +5/ Problem Set #1, CEM/BCMB 4190/6190/ E +5/γhB 0 E +3/γhB 0 E +1/γhB 0 E -1/γhB 0 E -3/γhB 0 E -5/γhB 0

4 6). he frequency difference ( ν) between two 1 signals in a 1 spectrum is 3000 z when the spectrum is acquired with a magnetic field strength of (corresponding to a 1 resonance/observe frequency of 500 Mz). What is the chemical shift difference ( δ) between the two signals? b. What would be the frequency difference between the two 1 signals if the spectrum was acquired with a magnet? c. If the spectrum was acquired with a magnetic field strength of (600 Mz 1 resonance/observe frequency), what would be the chemical shift difference? δ b. ν ν 3000 z 3000 z observe frequency Mz "ppm" ( parts per million ) 500 z δ observe frequency Mz 3600 z c. he chemical shift difference would still be 6.0 ppm. he chemical shift is independent of magnetic field strength. ν 3600 z 3600 z δ observe frequency Mz "ppm" 600 z 7). he measured magnitude of J, for a particular AX spin system using a 500 Mz (11.74) magnet is 4.5 z. What is the magnitude of this coupling using a 600 Mz (14.09) magnet? he measured magnitude for the coupling is the same (4.5 z). he magnitudes of scalar couplings are independent of the B 0 field strength. 8). You have found that τ ρ (pulse width/length) for a 90 (π/) pulse for your sample is 10 µs. Sketch a bar graph of signal amplitude versus τ ρ for your sample for τ ρ values of 0, 5, 10, 15, 0, 5, 30, 35, and 40 µs. b. Explain, in general, the relationship between the transverse component of the bulk magnetization, My, τ ρ, the pulse angle α, and the NMR signal amplitude. signal amplitude τ ρ (µs) Problem Set #1, CEM/BCMB 4190/6190/8189 4

5 b. he amplitude of the signal induced in the receiver coil of an NMR instrument is proportional to My, the transverse component of the bulk magnetization. In turn, both My and α are dependent on τ ρ. For a given pulse power, α increases from 0 to 90 with incremental increases in τ ρ and My increases as M sin α. 9. he 1 and 13 C NMR spectra for phenol are shown below. he chemical shifts for the 1 signals are 5.4, 6.8, 6.9 and 7.5 ppm. he chemical shifts for the 13 C signals are 115, 11, 130, and 155 ppm. he 1 and 13 C chemical shifts for benzene are 7.7 and 18 ppm respectively. Match the chemical shifts with the correct proton and carbon atoms: C C C C C C 6 b. Justify your answers to a. Draw resonance structures, state which nuclei are shielded or deshielded relative to one another and to benzene and why C C C C C C b. In the 1 spectrum, the signals around 7 ppm are due to the aromatic protons, as in benzene. he hydroxyl signal is the signal at 5.4 ppm, and is deshielded relative to aliphatic protons due to the attached (electronegative) oxygen atom. Likewise, in the 13 C spectrum, the signals at are due to 13 C atoms -6, as in benzene, whereas the signal at 155 is due to 13 C 1, which is deshielded relative to benzene due to attachment to the oxygen atom. Resonance structures can be drawn (below) where the lone pair of electrons on the oxygen is donated to the C- bond, Problem Set #1, CEM/BCMB 4190/6190/8189 5

6 thus localizing a negative charge at either the ortho ( or 6) or para (4) positions. Because the charge density, and thus the shielding, at the ortho and para positions will be greater than at the meta position, the 1 and 13 C chemical shifts at the meta positions will be expected to be larger. hus, δ 1 3 δ , and δ 13 C 3 δ 13 C Finally, one would expect that the charge density at the ortho position would be larger than at the para position because of its proximity to the oxygen and the C double bond, and thus would be more shielded. So, the 1 and 13 C chemical shifts at the ortho position (δ 1 δ , δ 13 C δ 13 C 6 115) are smaller than at the para position (δ , δ 13 C 4 11). 10). If the longitudinal relaxation time for a given nucleus is 10 seconds, how long after a 90 pulse will we have to wait until the magnitude of the z component of the bulk magnetization (M z ) is equal to 95% of the magnitude of the bulk magnetization before the pulse (M 0 )? We assume, as did Bloch, that this relaxation process is first order, and M z M 0 (1 e t / 1 ). M z / M (1 e t / 1 ) (1 e t /10s ) e t /10s 0.05 t /10s.9957 t 30s Problem Set #1, CEM/BCMB 4190/6190/8189 6