10 problems in nmr

Problems in NMR

1. Problems in NMR U.A. Deokate Lecturer Govt. College of Pharmacy, Amravati Copyright © by U. A. Deokate, all rights reserved.

2. Ring rule The Index of Hydrogen Deficiency (IHD), is a count of how many molecules of H2 need to be added to a structure in order to obtain the corresponding saturated, acyclic species It gives idea regarding deficiency of H i.e. presence of double bond or ring in a molecule. Known as HID (Hydrogen ion deficiency) Hence it takes a count of how many rings and multiple bonds are present in the structure. 11/8/2016 2 Deokate U A

3. Index of Hydrogen ion deficiency If you have a molecular formula, C c H h N n O o X x , then the following equation can be derived: IHD = [2c+2-h-x+n] 2 HID= 2n+2- No of hydrogen including halogens 2 11/8/2016 3 Deokate U A

4. Where does this equation come from ? The maximum number of hydrogen atoms for "c" carbon atoms is 2c+2 (think of the formulae of saturated hydrocarbons such as ethane, propane etc.). From this number, subtract the "h" hydrogen that you have. Since, like hydrogen, a halogen only forms one bond, then they can be treated as if they are hydrogen, so subtract them as well. Oxygen forms two bonds, therefore it has no impact (compare H count for methane, CH4, and methanol, CH3OH). Nitrogen forms three bonds. This means for "n" nitrogens, "n" extra hydrogen atoms are needed (compare the H count for methane, CH4, and methyl amine, CH3NH2), therefore, add "n". The factor of ½ accounts for us counting H atoms , but adding hydrogen, H2 , molecules . 11/8/2016 4 Deokate U A

5. Determining the IHD for molecules can be useful for the following reasons: Seeing what types of structural units maybe possible Quickly checking structures to see if they fit the molecular formula rather than simply counting H What is the IHD for each of the following molecular formulae ? C 6 H 10 C 6 H 6 C 4 H 8 O C 4 H 9 N C 2 H 2 Cl 2 2 4 1 1 1 11/8/2016 5 Deokate U A

6. EXAMPLE 2 11/8/2016 6 Deokate U A

7. NMR problem 01 C 3 H 5 Cl 5 1. triplet δ 4.52 = 1H 2. doublet δ 6.02 = 2H HID = 0 Cl Cl Cl --CH2—CH- -- C – Cl Cl 11/8/2016 7 Deokate U A

8. NMR problem 02 C 3 H 5 Cl 3 1. Singlet δ 2.20 = 3H 2. Singlet δ 4.02 = 2H HID = 0 Cl CH3—C---CH2– Cl Cl 11/8/2016 8 Deokate U A

9. NMR problem 02 C 4 H 9 Br 1. Doublet δ 1.04 = 6H 2. Multiplet δ 1.95 = 1H 3. Doublet δ 3.33 = 2H HID = 0 CH3 CH–CH2– Br CH3 11/8/2016 9 Deokate U A

10. NMR problem 03 C 10 H 14 1. Singlet δ 1.30 = 9H 2. Singlet δ 7.20 = 5H HID =4 CH3 C CH3 CH3 11/8/2016 10 Deokate U A

11. NMR problem 04 C 10 H 14 1. Doublet δ 0.88 = 6H 2. Multiplet δ 1.86 = 1H 3. Doublet δ 2.45 = 2H 4. Singlet δ 7.12 = 5H HID =4 CH3 CH2--CH CH3 11/8/2016 11 Deokate U A

12. NMR problem 05 C 9 H 10 1. Quintet δ 2.04 = 2H 2. triplet δ 2.91 = 4H 3. Singlet δ 7.17 = 4H HID = 05 11/8/2016 12 Deokate U A

13. NMR problem 06 C 10 H 13 Cl 1. Singlet δ 1.57 = 6H 2. Singlet δ 3.07 = 2H 3. Singlet δ 7.27 = 5H CH 3 CH 2 —C—CH 3 Cl 11/8/2016 13 Deokate U A

14. NMR problem 08 C 10 H 12 1. Multiplet δ 0.65 = 2H 2. Multiplet δ 0.81 = 2H 3. Singlet δ 1.37 = 3H 4. Singlet δ 7.17 = 5H 11/8/2016 14 Deokate U A

15. NMR problem 09 C 3 H 5 ClO 2 Doublet δ 1.73 = 3H Quintet δ 4.47 = 1H Singlet δ 11.22 = 1H 11/8/2016 15 Deokate U A

16. NMR problem 10 C 3 H 5 ClO 2 Singlet δ 3.81 = 3H Singlet δ 4.08 = 2H 11/8/2016 16 Deokate U A

17. NMR problem 11 C 4 H 7 BrO 2 Triplet δ 1.30 = 3H Singlet δ 3.77 = 2H Quartet δ 4.23 = 2H 11/8/2016 17 Deokate U A

18. NMR problem 12 C 4 H 7 BrO 2 Triplet δ 1.08 = 3H Quintet δ 2.07 = 2H Triplet δ 4.23 = 1H Singlet δ 10.97 = 1H 11/8/2016 18 Deokate U A

19. NMR problem 13 C 4 H 8 O 3 Triplet δ 1.27 = 3H Quartet δ 3.66 = 2H Singlet δ 4.13 = 2H Singlet δ 10.95 = 1H 11/8/2016 19 Deokate U A

20. NMR problem 14 C 9 H 11 Br Quintet δ 2.15 = 2H Triplet δ 2.75 = 2H Triplet δ 3.38 = 2H Singlet δ 7.22 = 5H 11/8/2016 20 Deokate U A

21. NMR problem 15 C 3 H 5 ClF 2 Triplet δ 1.75 = 3H Triplet δ 3.63 = 2H 11/8/2016 21 Deokate U A

22. 13 C - NMR Problems- 11/8/2016 Deokate U A 22

23. CMR problem 1 C 3 H 5 ClF 2 Triplet δ 1.75 = 3H Triplet δ 3.63 = 2H 11/8/2016 23 Deokate U A