Introduction to NMR Spectroscopy and Imaging Assignment for Chapter 02: Chemical shift and J Coupling

Transcription

1 Introduction to NMR Spectroscopy and Imaging Assignment for Chapter 02: Chemical shift and J Coupling 0. Choose the correct one(s) from the following statements or explain briefly your supporting reason if you decide that one is incorrect: a. Chemical shift is zero if the electronic distribution around the nucleus is spherical. N b. Anything that causes a change of the local magnetic field at a nucleus causes shift or splitting or broadening of the resonance frequency. Y c. The 1 H chemical shift of a free hydrogen atom is larger than that of water because the proton in free hydrogen atom is more shielded by its electron. N d. Anti-shielding means the chemical shift is negative. N e. The fact that 1 H chemical shift of water vapor is smaller than that of liquid water means that van der Waals force plays an important role in chemical shift. Y f. In liquid state NMR, only the isotropic chemical shift, or, the average of the three principal values of the chemical shift tensor, is observed. Y g. The 1 H, 13 C chemical shift of organometallic compounds involving transition metals can be as large as 100 ppm and 500 ppm, respectively. Y h. Chemical shift is anisotropic because the electrons in the molecular orbitals around a nucleus are in different states in different orientations. Y i. Two spins always have a magnetic coupling as long as they are close to each other. This coupling is called direct dipolar coupling. Y j. J coupling can be understood as an additional dipolar interaction between two spins mediated by the electrons close to them. Y k. J-coupling is caused by the shielding of the electrons around a nucleus against the magnetic field generated by the coupling partner (another spin). Y l. Weak coupling means the J coupling between two spins is much smaller than the chemical shift difference of the two spins. Y m. For a pair of heteronuclear spins, weak coupling condition is always satisfied. Y n. Weak coupling condition can be violated only in homonuclear spin systems. Y o. Spin pairs with cis-configuration usually show smaller 3 J couplings than trans-configuration. Y

2 p. Karplus formula applies to a pair of spins no matter whether they are directly bonded or several bonds away from each other. N q. Hyperconjugation of the second kind is responsible for Karplus equation. Y r. Decoupling makes the local field generated by other spins change direction constantly, giving an average of zero. Y 1. What is the difference between shielding and shift? Shielding describes the phenomenon that the local magnetic field is different from the applied magnetic field as a result of the shieling of the surrounding electrons. Shift is the resonance frequency of a nucleus in a chemical environment differs from the Zeeman frequency of the nucleus. Shielding is the cause and shift is the effect. They differ by a sign: the larger the shielding, the smaller the chemical shift: B() r = B B = B (1 σ ()) r B B B 0 s 0 σ = = δ = σ = ω ω ω ω0 ω ω Why is TMS chosen for 1 H chemical shift standard? TMS is chosen as the 1 H chemical shift standard because its chemical shift is smaller than most of hydrogen-containing compounds and is chemically stable, inexpensive and relatively safe. 3. Which of the two chemically different types of protons in CH2Cl and CHCl 2 resonate at higher frequency? CHCl 2 or CH2Cl? Explain your choice. The proton in CHCl 2 has higher resonance frequency because it is less shielded by electrons as the result of the two chlorines which attract the electrons to them. 4. A spectrometer with proton operating frequency of 100 MHz was used to measure the frequency separation of the resonances of chloroform CHCl3, and TMS, which was found to be 730 Hz, the CHCl3 being to high frequency. What is the chemical shift of chloroform on the δ scale (i.e. in ppm)? What would the frequency separation and chemical shift be if the sample were measured in a spectrometer operating at 300 MHz?

3 ω ω ref 730 The chemical shift δ = = 6 = 7.3 ppm. ω ref In a proton spectrum the peak from TMS is found to be at MHz. What is the shift, in ppm, of a peak which has a frequency of MHz? Recalculate the shift using the spectrometer frequency, ν spec quoted by the manufacturer as MHz rather than ν TMS in the denominator of this Eq: δppm = 10 6 (ν ν TMS )/νspec. Does this make a significant difference to the value of the shift? Two peaks in a proton spectrum are found at 1.54 and 5.34 ppm. The spectrometer frequency is quoted as MHz. What is the separation of these two lines in Hz and in rad s 1? δppm = 10 6 (ν ν TMS )/ν TMS = 10 6 ( )/ = ppm. δppm = 10 6 (ν ν TMS )/ν spec = 10 6 ( )/ = ppm. Therefore, there is no significant difference between the above two. Δ δppm= = 3.8 ppm = 3.8x10-6 x400.13x10 6 Hz = Hz = x2π rad s -1 = rad s Calculate the Larmor frequency (in Hz and in rad s 1 ) of a carbon-13 resonance with chemical shift 48 ppm when recorded in a spectrometer with a magnetic field strength of 9.4 T. The gyromagnetic ratio of carbon-13 is rad s 1 T 1. Write the 1 H chemical shifts of HF, HCl, HBr and HI in ascending order and give your reasons. ω = γ B = rad s T 9.4T = rad s = M Hz. The proton in HF is least shielded and the proton in HI is most shielded: δ > δ > δ > δ HF HCl HBr HI 7. How does hydrogen bonding affect chemical shift? Explain your argument using a few compounds.

4 Hydrogen bonds attract the electrons away from the proton hence the proton is less shielded and proton chemical shift is increased as a result of hydrogen bonding. 8. List the most important factors that would affect the measured chemical shift values of a compound. The most important factors: (i) The lone pairs and bonding pairs; The more lone pairs and more binding pairs, the larger the shielding and smaller the chemical shift; (ii) solvent (concentration, composition, ph). The lesser important factors: (i) temperature; (ii) isotope effect; (iii) pressure. 9. How does temperature generally affect chemical shift? Write out your arguments. The higher the temperature, the faster the molecular vibration less shielding larger chemical shift. 10. Try to explain the trend shown in the following table: Solvent H 2 O D 2 O DMSO acetone CD Cl 3 C 6 D 6 Shift *(H 2 O) * Relative to TMS. As the solvent becomes less polarized, the smaller the chemical shift of water protons (more shielded as a result of weakened hydrogen bonds).

5 11. Explain why the 1 H chemical shift of HDO is ppm upfield of H2O. Which 1 H chemical shift is larger, H 17 OH or H 16 OH? Vibration reduces shielding and increases chemical shift. HDO has lower vibrational frequency hence larger shielding (smaller chemical shift or upfield). For the same reason, the protons in H 17 OH has smaller chemical shift than that in H 16 OH. 12. Which 14 N chemical shift is larger 14 NH or 14 ND? Which 15 N chemical shift is larger 15 NH or 15 ND? Which 1 H chemical shift is larger 14 NH or 15 NH? Similar to Problem N chemical shift is larger in 14 NH than that in 14 ND; 15 N chemical shift is larger in 15 NH than that in 15 ND; The 1 H chemical shift is larger 14 NH than that in 15 NH. 13. Why are the chemical shifts of many organometallic compounds involving transition metals much larger than that of most organic compounds? The paramagnetic field produced by the (unpaired d electrons in) transition metals may cause unusually larger chemical shifts. 14. How many proton peaks can be observed on each of the following samples: (a) H 2 O, (b) HDO, (c) CH 3 COOH, and (d) CH 2 DCOOH? (a) 1; (b) 3 (H peak is split by D into a triplet); (c) 1(CH 3 )+1 (OH)=2; (d) 3 (CH 2 D)+1 (OH)= Consider a system of two weakly coupled spins. Let the Larmor frequency of the first spin be 100 Hz and that of the second spin be 200 Hz, and let the coupling between the two spins be 5 Hz. Compute the frequencies of the lines in the normal (single quantum) spectrum. Make a sketch of the spectrum, roughly to scale, and label each line with the energy levels involved (i.e. 1 2 etc.). Also indicate for each line which spin flips and the spin state of the passive spin.

6 The resonance frequencies of the two spins are ω = ± /2 i ω0, i J ω = 100 ± ( 5) / 2 = 102.5, 97.5 Hz 1 ω = 200 ± ( 5) / 2 = 202.5, Hz Interpret the spectra shown in the following attached diagrams (i.e., find the correct molecular structure according to the given molecular formula and its proton NMR spectrum). State your reasoning. (Spectra and answers can be found from Akitt s book).

7 C 3 H 4 SO 2 : H H CH 3 SO 2 C CH HOCH 2 SOC CH H S (A) (B) 2 C (C) The spectrum shows three types of protons with intensities 2:1:1 and the CH 2 signal is a doublet with a tiny splitting and one CH is doublet and the other two triplet. This means one CH is coupled with another CH and weakly coupled with CH 2. Therefore, only C is the correct structure. C 2 F 3 H 3 O: CF 2 HCHFOH CF 3 CH 2 OH CH 2 FCF 2 OH (A) (B) (C) B is the correct strcture. A would show three types of protons and the CH 2 in C would show two triplets.

8 C 3 F 4 H 4 O: CF 2 HCH 2 CF 2 OH CF 3 CHFCH 2 OH CF 2 HCF 2 CH 2 OH CH 3 CF 2 CF 2 OH (A) (B) (C) (D) Notice that proton can be coupled with both proton and fluorine. From the spectrum, we see three types of protons, one being singlet, one triplet and two equivalent triplets. C is the correct structure. A and B would show much more complicated splittings because CH and CH 2 are coupled and they are both coupled with fluorine.

9 C 3 H 4 O: CH 2 =CHCHO HC CCH 2 OH (A) (B) A is the correct structure because B would show neither doublet (CH 2 ) nor triplet (CH) C 14 H 14 O: A has no methyl group while the methyl group in D would show a triplet. The methyl group in B is bonded with an oxygen and would show a larger chemical shift. Therefore, only C is the correct structure.

10 C 10 H 6 O 2 : (A) (B)

11 A is the correct structure because it has two equivalent protons which show singlet and other four-spin system showing 16 peaks. B cannot account for the singlet (2 protons) at ~7.0 ppm. C 5 H 8 O 2 : (CH 3 ) 2 C=CHCOOH (A) (B) (C) A is the correct structure because B would show a CH 3 doublet while C has only one methyl group. C 5 H 9 NO 2 : ONC(CH 3 ) 2 COCH 3 CH 3 ON=C(CH 3 )COCH 3 (CH 3 ) 2 C=C(CH 3 )NO 2 (A) (B) (C) Obviously only B can account for the singlet for all three methyl groups.

12 C 8 H 19 NP + I - :

13 (A) (B) (C) A is the correct structure because only it and C have two CH 3 quadruplets, but C cannot account for the three protons bonded with P which causes a doublet for the 3 protons. C 4 H 9 NO 2 : (A) (B) (C) (B) A is the correct structure. C has only one CH 3 ; in B, CH 2 would show a singlet. C 3 H 6 S: CH 3 CH=CHSH CH 2 =CHSCH 3 CH 2 =CCH 3 SH (A) (B) (C) B is the correct structure because A would show split CH 3 peaks whereas C would mean a single CH 2 peak.

14 C 6 H 5 NO 3 : (A) (B) (C) The above spectra show four CH on the aromatic ring (OH is not seen). B is the correct structure because A has an uncoupled CH whereas C would give symmetrical spectra showing two groups of identical CHs. 17. For a three spin system, draw up a table similar to that on page 2 10 showing the frequencies of the four lines of the multiplet from spin 2. Then, taking ν0,2 = 200 Hz, J23 = 4 Hz and the rest of the parameters as in Fig (from Prof Keeler s online book), compute the frequencies of the lines which comprise the spin 2 multiplet. Make a sketch of the multiplet (roughly to scale) and label the lines in the same way as is done in Fig How would these labels change if J23 = 4 Hz? On an energy level diagram, indicate the four transitions which comprise the spin 2 multiplet, and which four comprise the spin 3 multiplet. The center of gravity of the subspectrum of spin 2 is at -200 Hz. The J couplings are J 12 = (or, ) = 20 Hz (found from above diagram), J 23 = 4 Hz. One

15 then readily draw the sketch subspectrum of spin 2: 18. Use Karplus equation to calculate the 3 J coupling of the two protons belonging to two adjacent C-H bonds with a dihedral angle of 75 o. We may use the Karplus equation for the vicinal protons 3 J HH (φ) = 8.5 cos 2 φ for 0 ϕ 90 3 J HH (φ) = 9.5 cos 2 φ for 90 ϕ 180 Plugging φ= 75 o into the first equation, one immediately gets 3 J HH = 0.57 Hz. ( you may use other parameters for the Karplus equation but be warned of getting unreasonable values for cosφ such as larger than 1) 19. The 3 J coupling constant between a pair of vicinal protons is measured to be 7.8 Hz. Find the dihedral angle φ. We may use the Karplus equation for the vicinal protons 3 J HH (φ) = 8.5 cos 2 φ for 0 ϕ 90 3 J HH (φ) = 9.5 cos 2 φ for 90 ϕ 180 Plugging in 3 J HH = 7.8 Hz. one immediately gets cosφ = 7.8/8.5 or 7.8/9.5 and one gets φ= o or o ( you may use other parameters for the Karplus equation) 20. Suppose the line-width of a carbon-13 resonance is 2 khz which is caused by the dipolar coupling between carbon-13 and its neighboring hydrogen nuclei. How strong

16 the decoupling field should be applied to the proton channel to obtain satisfactory decoupling? The proton decoupling power should be equal to, or largerthan, the dipolar coupling, 2 khz, in this case. 21. What is hyperconjugation of the first kind? What is hyperconjugation of the second kind? How are they related to Karplus equation? Hyperconjugation of the first kind does not consider electron correlations, i.e., the probability amplitude that to electrons be delocalized simultaneously from two occupied MOs to two virtual MOs is set to zero. It was found that the hyperconjugation of the first kind to be of minor importance to Karplus behavior. Hyperconjugation of a second kind can be understood as related to the probability amplitude that two electrons be delocalized simultaneously from two occupied MOs to two virtual MOs. Of special importance for NMR J-couplings are that excitations concerning two pairs of MOs that are localized in different regions of the molecule like the two simultaneous excitations σ 1 σ 1 *,σ 2 σ 2 *. These excitations arise from the hyperfine interaction at the site of each coupled nuclei when the Fermi contact mechanism is taken into account.