1. Instrumental Analysis NMR (II) 1 Tutorial 7

2. Assignment 2 The assignment should be submitted on individual basis (no group assignment). Only one assignment grade will count to your total score (best one out of two assignments). The assignment should be delivered after 1 week from your tutorial session to your instructor in his office (Dr. Rasha Hanafi: B5-101 and Ms. Menna Hammam B1-203) 2

3. According to the table at the end of the document and using the NMR data shown below; (1) calculate the IHD and (2) predict the structure of the compound assigned to you: 3

4. N.B. the assignment should be submitted on individual basis (no group assignment). Only one assignment grade will count to your total score (best one out of two assignments). The assignment should be delivered after 1 week from your tutorial session 4 GroupFirst Compound no.and second Compound no. 1ht 2iu 3jw 4kx 5ly 6nz 7oaa 8qbb 9rcc 10sr

5. 5 INFORMATION WE CAN GET FROM NMR SPECTRUM 1.Number of 1 H-NMR signals = Number of kinds of protons 2.Chemical shift gives the electronic environment of protons (Shielding and Deshielding) 3.NMR spectrum can reveal the number of protons assigned for each signal (integral lines : no. of protons in each signal) 4.NMR spectrum indicates the carbon skeleton Spin-spin splitting gives the number of equivalent protons on adjacent carbon atoms

6. 6 The signal for proton H b is split into triplet by the two hydrogens (H a ) on the adjacent carbon The H a is split into a quintet by the four equivalent H b protons on its adjacent carbons because the two set of protons are equivalent Splitting of the signals by two adjacent groups of equivalent protons

7. 7 split into doublet by the H e protons split into septet by the H b protons split into triplet by the H c protons Shows five signals The signal for H c protons will split by both the H a and H d protons. Because H a and H d are not equivalent, the n+1 rule has to be applied separately to each set. Thus the H c proton will split into (n a +1)(n d +1)=(4)(3) = 12 It is called multiplet in this case Splitting of the signals by two adjacent groups of nonequivalent protons

8. 8 Protons in Substituted Aromatic Benzene Ring a c b d c d e Suppose to have five signals split into triplet by the H b protons split into quartet by the H a protons split into doublet by the H d protons split into triplet by the two H d protons H c and He protons are not equivalent protons. H d protons will split into doublet by H e proton and then each of the doublet peak split into another doublet by H c protons forming a doublet of doublet (split into four signals of equal intensities Since the electronic effect (electron-donating ability) of an ethyl substituent is not sufficiently large to cause a great effect on the environment of H b, H c, and H d, the signals of these protons are overlapped to form a multiplet signal We can generally say that the protons in mono- substituted benzene ring show a multiplet signal doublet of doublet

9. 9 In contrast to benzene protons of ethyl benzene, the benzene protons in nitrobenzene (H a, H b, and H c ) show three distinct signals in the region of  =7-8. That is because of the powerful electron withdrawing effect of the nitro group. c a c a b strong electron withdrawing group

10. 10 1- How many signals would you expect to see in the 1 H-NMR spectrum of each of the following compounds? 2 1 3 3 Least shielded most shielded Least shielded most shielded Least shielded most shielded 2-Which set of protons in EACH of following compounds is the least shielded? Which set of protons in each compound is the most shielded?

11. 11 3- The following 1 H-NMR spectrum corresponds to one of the compounds shown below. Which compound is responsible for this spectrum? Hint: This question can be solved by considering the integration of each peak

12. 12 4- How would the integration distinguish the 1 H-NMR spectra of the following compounds? (9:2) (6:2 or 3:1) (6:3 or 2:1) The three compounds show two signals in their NMR spectra that integrate to: 5- [18]-annulene shows two signals in its 1 H-NMR spectrums: one at 9.25 ppm and the other at far upfield (beyond TMS) at  2.88 ppm. What hydrogens are responsible for each signal? Justify your answer. at 9.25 ppm Outside the ring at 9.25 ppm Outside the ring Because of the magnetic anisotropy of the circulating  -electrons that add to the applied field for the protons located outside the ring and let them absorb at higher . On the other hand the magnetic field created by the  -cloud oppose the external field for the protons located inside the ring and let them appear highly upfield even if compared with TMS. at -2.88 ppm Inside the ring at -2.88 ppm Inside the ring

13. 13 6- The protons in acetonitrile (CH 3 CN) have resonance at  1.97 while methyl chloride (CH 3 Cl) has a resonance at  3.05, even though the dipole moment of acetonitrile is 3.92 while that of methyl chloride is only 1.85. The larger dipole moment for the cyano-group suggests that the electronegativity of this group is larger than that of chloride atom. Explain why the methyl protons on acetonitrile are more shielded than those in methyl chloride, in contrast to the expected results based upon electronegativity. Acetonitrile shows similar magnetic anisotropic behavior to acetylene. The methyl hydrogens are in the shielding region of the magnetic field generated by the induced circulation of the  -electrons of the cyano group. The geometry of such a field makes the methyl hydrogens shielded and thus, resonate at lower frequency-upfield than expected. i.e. at lower  of 1.97 ppm. CH 3 N

14. 14 SOME COMMON SPLITTING PATTERNS

15. 15 a- C 4 H 8 O:  1.0 ppm (triplet, 3H),  2.1 ppm (singlet, 3H),  2.5 ppm (quartet, 2H). b- C 3 H 8 O:  1.2 ppm (doublet, 6H),  2.0 ppm (singlet, 1H),  4.0 ppm (septet, 1H).The signal at  2.0 ppm disappears upon shaking with D 2 O. 7- Deduce the structure that is consistent with the following NMR data: Notice that the hydroxyl protons (amine protons) appear as singlet in the NMR spectrum (not split by neighboring H)

16. 16 8- Identify the compound from its molecular formula and its 1 H-NMR spectrum. Propyl benzene 3H 2H

17. 17 9- What arrangement of protons would give two triplets of equal area (integrals)? Answer:X-CH 2 -CH 2 -Y 10-Which of the compounds shown below would be most consistent with the 1 H-NMR spectrum shown? Justify your answer. Answer: The chosen compound is the only one that can show two signals; one is quintet and the other is triplet 

18. 18 11-Consider the two isomeric compounds to the right. a. In compound A, which proton group would provide the farthest downfield  value? __. What multiplicity would it show ___ ___ ___ b. In compound B, which proton group would correlate with the largest numerical value for  ? __. What would be its multiplicity? ___ ___ c. Based on your answers to parts a and b, briefly describe how you would use 1 H-NMR to distinguish between A and B. In other words, if you were given a sample and told that it was either A or B, what would you look for in its 1 H-NMR to tell you which one you had? If farthest downfield proton group is a quartet that integrates for two protons, you have compound A. Or, if it's a singlet integrating for three protons, you've got compound B. b 4 (quartet) c singlet

19. 19 C4H8O2C4H8O2 12- Show the structure of the compound represented by the following NMR spectrum. 3H 2H C4H8O2C4H8O2 Why? Answer

20. 20 13- List in tabular form the expected signals for 1 H-NMR spectrum of the compound shown below. Make sure that your table of data includes (for each type of protons): chemical shift (approximate value), integration, and splitting pattern (multiplicity). Then sketch the 1 H NMR of the compound. Be as clear as possible, and be sure to label each type of proton in the structure, in your table, and in your spectrum. Answer: a c b Protons a:  0-2 (t, 3H) Protons b:  1.5-3 (q, 2H) Protons c:  3-5 (s, 3H) 1 H-NMR Spectrum 

21. 21 14- Predict the structure of the compound that show the NMR data showing below: i.C 9 H 10 O  1.2 (triplet, 3H),  3.0 (quartet, 2H),  7.4-8.0 (multiplet, 5H) ii.C 10 H 14  1.3 (singlet, 9H),  7.2 (multiplet, 5H) iii.C 10 H 12 O 2  2.0 (singlet, 3H),  2.9 (triplet, 2H),  4.2 (triplet, 2H),  7.3 (multiplet, 5H) iv.C 8 H 7 N  3.7 (singlet, 2H),  7.2 (multiplet, 5H) v.C 4 H 6 Cl 2 O 2  1.4 (triplet, 3H),  4.3 (quartet, 2H),  5.9 (singlet, 1H) vi.C 7 H 14 O  0.9 (triplet, 6H),  1.6 (sextet, 4H),  2.4 (triplet, 4H) vii.C 3 H 6 Br 2  2.4 (quintet, 2H),  3.5 (triplet, 4H) viii.C 4 H 8 O 2  1.4 (doublet, 3H),  2.2 (singlet, 3H),  3.7 (singlet, 1H),  4.3 (quartet, 1H) The singlet signal at  3.7 is a broad one that disappears upon shaking with D 2 O ix. C 10 H 14  1.2 (doublet, 6H),  2.3 (singlet, 3H),  2.9 (septet, 1H),  7.0 (multiplet, 4H),

22. 22 i- ii- iii- iv- v- vi- vii- viii- Ethyl phenyl ketone ix- Answer of Q 14: