IB Chemistry on HNMR Spectroscopy and Spin spin coupling

IB Chemistry on HNMR Spectroscopy and Spin spin coupling

1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok Tutorial on Nuclear magnetic resonance spectroscopy (NMR) and Spin spin Coupling

2. Spectroscopy measures interaction of molecules with electromagnetic radiation Particles (molecule, ion, atom) can interact/absorb a quantum of light Spectroscopy Electromagnetic Radiation Nuclear spin High Energy Radiation Gamma/X ray Transition of inner electrons UV or visible Transition of outer most valence electrons Infrared Molecular vibration Microwave Molecular rotation Radiowaves Low Energy Radiation Infrared Spectroscopy Nuclear Magnetic Resonance Spectroscopy Ultra Violet Spectroscopy Atomic Absorption Spectroscopy Velocity of light (c ) = frequency (f) x wavelength ( λ ) - c = f λ All electromagnetic waves travel at speed of light (3.00 x 10 8 ms -1 ) Radiation with high ↑ frequency – short ↓ wavelength Electromagnetic radiation/photon carry a quantum of energy given by E = hf h = plank constant = 6.626 x 10 -34 Js f = frequency λ = wavelength Click here notes spectroscopy

3. Electromagnetic Radiation and Spectroscopy Radiowaves Nuclear spin Nuclear Magnetic Resonance Spectroscopy Organic structure determination MRI and body scanning Infrared Molecular vibration Infrared Spectroscopy UV or visible Transition of outer valence electron Organic structure determination Functional gp determination Measure bond strength Measure degree unsaturation in fat Measure level of alcohol in breath Electromagnetic Radiation UV Spectroscopy Atomic A Spectroscopy Quantification of metal ions Detection of metal in various samples Electromagnetic Radiation Interact with Matter (Atoms, Molecules) = Spectroscopy

4. Nuclear Magnetic Resonance Spectroscopy (NMR) Involve nucleus (proton + neutron) NOT electron Proton + neutron = Nucleons Nucleons like electron have spin and magnetic moment (acts like tiny magnet) Nuclei with even number of nucleon ( 12 C and 16 O) Even number of proton and neutron – NO net spin Nucleon spin cancel out each other –Nucleus have NO overall magnetic moment – NOT absorb radiowave Nuclei with odd number of nucleon ( 1 H, 13 C, 19 F, 31 P) Nucleon have net spin – Nucleus have NET magnetic moment – Absorb radiowave Nuclei with net spin – magnetic moment will interact with radiowaves Nuclei have a “spin” associated with them (i.e., they act as if they were spinning about an axis) due to the spin associated with their proton and neutron. Nuclei are positively charged, their spin induces a magnetic field NMR spectroscopy does not work for nuclei with even number of protons and neutrons - nuclei have no net spin . Nuclear Magnetic Resonance Spectroscopy (NMR) Spin cancel each other Net spin

5. Main features of HNMR Spectra 1. Number of diff absorption peak – Number of diff proton/chemical environment 2. Area under peak - Number of hydrogen in a particular proton/chemical environment (Integration trace) - Ratio of number of hydrogen in each environment 3. Chemical shift - Chemical environment where proton is in - Spinning electron create own magnetic field, creating a shielding effect - Proton which are shielded appear upfield . (Lower frequency for resonance to occur) - Proton which are deshielded appear downfield. (Higher frequency for resonance to occur) - Measured in ppm (δ) 4. Splitting pattern - Due to spin-spin coupling - N umber of peak split is equal to number of hydrogen on neighbouring carbon+1 (n+1) peak Chemical Shift NMR spectrum CH 3 CH 2 Br Number of peaks Area under peaks Chemical shift Splitting pattern Nuclear Magnetic Resonance Spectroscopy (NMR) Click here khan NMR videos.

6. NMR spectrum CH 3 CH 2 Br Number of peaks Splitting pattern Click here khan NMR videos. Number equi H Multiplicity Ratio Singlet 1 1 Doublet 1: 1 2 Triplet 1 : 2 : 1 3 Quartet 1 : 3 : 3 : 1 4 Quintet 1 : 4 : 6 : 4 : 1 5 Hextet 1 : 5 : 10 : 10 : 5 : 1 6 Septet 1 : 6 : 15 :20 : 15 : 6 : 1 Splitting Pattern Singlet – Neighbouring Carbon with No H Doublet – Neighbouring Carbon with 1H Triplet – Neighbouring Carbon with 2H Quartet – Neighbouring Carbon with 3H Equiv H in same chemical environment have no splitting effect on each other Equiv H do not split each other All Equiv H in same chemical environment will produce a same peak Spin spin coupling – occur when proton have diff chemical shift Splitting not observed for proton that are chemically equivalent/same chemical shift High Resolution (NMR) Main features of HNMR Spectra 1. Number of diff absorption peak – Number of diff proton/chemical environment 2. Area under peak - Number of hydrogen in a particular proton/chemical environment (Integration trace) - Ratio of number of hydrogen in each environment 3. Chemical shift - Chemical environment where proton is in - Spinning electron create own magnetic field, creating a shielding effect - Proton which are shielded appear upfield . (Lower frequency for resonance to occur) - Proton which are deshielded appear downfield. (Higher frequency for resonance to occur) - Measured in ppm (δ) 4. Splitting pattern - Due to spin-spin coupling - N umber of peak split is equal to number of hydrogen on neighbouring carbon+1 (n+1) peak

7. Br ׀ H – C – Br ׀ H – C – H ׀ H (n + 1 rule) Equiv H in same environment do not split each other. If H has n equiv proton on neighboring carbon, signal for H will split to n + 1 peak. H nuclei split neighbouring H in CH 3 to 2 peak, (doublet). H nuclei split CH 3 methyl gp to doublet H can align with EMF or against EMF. CH 3 will experience 2 diff EMF One lower, one higher EMF Split to doublet EMF align against MF produce by H Overall MF experience CH 3 lower H from CH 3 will absorb at lower freq ( upfield ) EMF EMF align with MF produce by H Overall MF experience by CH 3 higher H from CH 3 will absorb at higher freq (downfield) EMF CH 3 spilt to doublet by 1 adj H CH 3 experience 2 slightly diff MF due to neighbouring H MF MF Split with relative intensity of 1 : 1 Downfield Upfield High Resolution (NMR) Br ׀ H – C – Br ׀ H – C – H ׀ H Br ׀ H – C – Br ׀ H – C – H ׀ H Br ׀ H – C – Br ׀ H – C – H ׀ H doublet Splitting peaks occur as effective MF experience by H nuclei is modified by MF produced by neighbouring proton/H

8. (n + 1 rule) If H has n equiv proton on neighboring carbon, signal for H, split to n + 1 peak. 2H nuclei split neighbouring H in CH 3 to 3 peak, (triplet). 2H nuclei split CH 3 methyl to triplet H can align with EMF or against EMF. CH 3 will experience 3 diff EMF One lower, one higher , one no net change Split to triplet (ratio 1 : 2 : 1 ) EMF align against MF by H Both align against EMF (Net lower EMF) Overall MF experience by CH 3 lower H from CH 3 , absorb lower freq EMF EMF align with MF produce by H Both H align with EMF (Net greater EMF) Overall MF experience by CH 3 higher H from CH 3 , absorb at higher freq EMF EMF MF MF MF MF EMF align with/against MF produce by H 1 align with and 1 against EMF MF by H cancel each other Overall MF experience by CH 3 the same Split with relative intensity of 1 : 2 : 1 CH 3 spilt to triplet by 2 adj H CH 3 experience 3 diff MF due to 2 adjacent H Downfield Upfield H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H Br ׀ H – C – H ׀ H – C – H ׀ H triplet High Resolution (NMR) Splitting peaks occur as effective MF experience by H nuclei is modified by MF produced by neighbouring proton/H

9. 3H nuclei split CH 2 methylene to quartet H can align with EMF or against EMF. CH 2 will experience 4 diff EMF Split to quartet (ratio 1 : 3 : 3 : 1 ) EMF align against MF by H 3 H align against EMF ( L ower EMF) Overall MF experience CH 2 lower H from CH 2 , absorb at lower freq EMF EMF align with MF by H 3 H align with EMF (Net greater EMF ) Overall MF experience by CH 2 higher H from CH 2 , absorb at higher freq EMF EMF MF MF EMF align with/against MF by H 2 align with and 1 against EMF (higher) 2 align against and 1 with EMF (lower) 2 diff MF experience by CH 2 in 3 : 3 ratio Split with relative intensity of 1 : 3 : 3 : 1 CH 2 spilt to quartet by 3 adjacent H CH 2 experience 4 diff MF due to 3 adjacent H (n + 1 rule) If H has n equiv protons on neighboring carbons, signal for H, split to n + 1 peak. 3H nuclei split neighbouring H in CH 3 into 4 peak, called quartet. H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H H – C – H ׀ H – C – H ׀ H quartet High Resolution (NMR) Splitting peaks occur as effective MF experience by H nuclei is modified by MF produced by neighbouring proton/H

10. Singlet peak H nuclei attach to electronegative atom, O - NO splitting – Singlet H nuclei attach to neighbouring C without any H - NO splitting – Singlet Equiv H nuclei do not split each other but will split neighbouring H CH 3 spilt to triplet by 2 adj H CH 3 experience 3 diff MF due to 2 adj H CH 2 spilt to quartet by 3 adj H CH 2 experience 4 diff MF due to 3 adj H No signal splitting from coupling bet hydroxyl proton and methylene proton of CH 2 – despite 2 adjacent H H attached to O, undergo rapid chemical exchange, transfer rapidly from each other /loss of H Spin coupling due to H (OH) on methylene proton CH 2 is negligible/not seen. NO triplet split on OH due to 2 adjacent H from CH 2 – Only singlet H H ׀ ׀ H O- C- C- H ׀ ׀ H H CH 3 chemical shift ≈ 1 integration = 3 H split into 3 CH 2 chemical shift ≈ 3.8 integration = 2 H split to 4 OH chemical shift ≈ 4.8 integration = 1 H No split (Singlet) 3 2 1 Triplet split Quartet split Singlet split triplet quartet High Resolution (NMR)

11. - Equiv H in same chemical environment have no splitting effect on each other - All Equiv H produce same signal O ‖ CH 3 -C-O-CH 2 -CH 3 HO-CH 2 -CH 3 O ‖ HO-C-CH 2 -CH 3 O ‖ CH 3 -C-CH 2 -CH 2 -CH 3 Equivalent Hydrogen in same chemical Environment (chemical Shift) 4 diff chemical environment 4 peak ratio 3:2:3:2 3 diff chemical environment 3 peak ratio 3:2:1 3 diff chemical environment 3 peak ratio 3:3:2 3 diff chemical environment 3 peak ratio 3:2:1 12 3 3 2 3 2 3 2 1 3 2 3 2 equiv H 3 equiv H 3 equiv H 2 equiv H 2 1 3 equiv H 2 equiv H 1 equiv H 3 equiv H 2 equiv H 3 equiv H 3 equiv H 2 equiv H 1 equiv H

12. Equivalent Hydrogen in molecule with plane of symmetry Equiv H - Hydrogen attach to carbon in particular chemical environment Equiv H in same environment have no splitting effect on each other H on neighbouring carbon can be equiv if they are in same environment All Equiv H in same environment will produce a same signal. CH 3 | CH 3 – C -CH 3 | CH 3 1 chemical environment 1 peak O ║ CH 3 -CH 2 -C-CH 2 - CH 3 2 diff chemical environment 2 peak ratio 3:2 CI ׀ CH 3 -C-CH 3 ׀ H CH 3 ׀ HO-CH 2 - C- H ׀ CH 3 2 4 2 3 2 12 6 1 2 1 6 1 12 equiv H 2 diff chemical environment 2 peak ratio 6:1 4 diff chemical environment 4 peak ratio 6:1:1:2 1 equiv H 6 equiv H 6 equiv H 1 equiv H 2 equiv H 3 equiv H 2 equiv H

13. O CH 3 ‖ ׀ H-C- C-CH 3 ׀ CH 3 CH 3 | H-C-OH | CH 3 O CH 3 ‖ ׀ CH 3 -C-O-C-H ׀ CH 3 H CH 3 ׀ ׀ CI- C- C- CH 3 ׀ ׀ H CH 3 9.7 9 1 6 1 1 9 2 6 3 1 Equivalent Hydrogen in molecule with plane of symmetry Equiv H - Hydrogen attach to carbon in particular chemical environment Equiv H in same environment have no splitting effect on each other H on neighbouring carbon can be equiv if they are in same environment All Equiv H in same environment will produce a same signal. 3 diff chemical environment 3 peak ratio 6:1:1 2 diff chemical environment 2 peak ratio 9:1 1 equiv H 6 equiv H 1 equiv H 9 equiv H 1 equiv H 3 diff chemical environment 3 peak ratio 6:3:1 2 diff chemical environment 2 peak ratio 9:2 6 equiv H 3 equiv H 1 equiv H 9 equiv H 2 equiv H

14. CI CI | | C = C | | H H CI CI CI ׀ ׀ ׀ H- C- C - C- H ׀ ׀ ׀ CI H CI H H ׀ ׀ CI- C- C- CI ׀ ׀ H H H H ׀ ׀ H - C- C - H ׀ ׀ H H 4.5 6.1 2 2 1 6 4 Equivalent Hydrogen in molecule with plane of symmetry Equiv H - Hydrogen attach to carbon in particular chemical environment Equiv H in same environment have no splitting effect on each other H on neighbouring carbon can be equiv if they are in same environment All Equiv H in same environment will produce a same signal. 2 diff chemical environment 2 peak ratio 1:2 1 chemical environment 1 peak 2 equiv H 1 equiv H 2 equiv H 1 chemical environment 1 peak 4 equiv H 1 chemical environment 1 peak 6 equiv H

15. O ‖ CH 3 -C-O-CH 2 -CH 3 HO-CH 2 -CH 3 O ‖ HO-C-CH 2 -CH 3 O ‖ CH 3 -C-CH 2 -CH 2 -CH 3 12 Equiv H in same environment have no splitting effect on each other Equiv H do not split each other All e quiv H in same environment will produce a same peak . Triplet 2 adj H Septet 5 adj H Singlet No adj H Triplet 2 adj H Triplet 2 adj H Quartet 3 adj H Singlet OH – No split Triplet 2 adj H Singlet No adj H Quartet 3 adj H Triplet 2 adj H Quartet 3 adj H Singlet No adj H Splitting Pattern by neighbouring H 3 2 1 3 2 3 2 3 2 1 3 3 2 4 chemical environment 4 peak ratio 3:2:3:2 3 chemical environment 3 peak ratio 3:2:1 3 chemical environment 3 peak ratio 3:3:2 3 chemical environment 3 peak ratio 3:2:1

16. CH 3 ׀ HO-CH 2 - C- H ׀ CH 3 CH 3 | CH 3 – C -CH 3 | CH 3 O ║ CH 3 -CH 2 -C-CH 2 - CH 3 CI ׀ CH 3 -C-CH 3 ׀ H 2 4 Singlet No adj H Triplet 2 adj H Quartet 3 adj H Doublet 1 adj H Heptet 6 adj H Doublet 1 adj H Doublet 1 adj H Singlet OH- No split Nonet 8 adj H 3 2 12 6 1 1 2 6 1 Splitting Pattern by neighbouring H Equiv H in same environment have no splitting effect on each other Equiv H do not split each other All e quiv H in same environment will produce a same peak . 2 chemical environment 2 peak ratio 6:1 1 chemical environment 1 peak 2 chemical environment 2 peak ratio 3:2 4 chemical environment 4 peak ratio 6:1:1:2

17. O CH 3 ‖ ׀ H-C- C-CH 3 ׀ CH 3 CH 3 | H-C-OH | CH 3 O CH 3 ‖ ׀ CH 3 -C-O-C-H ׀ CH 3 H CH 3 ׀ ׀ CI- C- C- CH 3 ׀ ׀ H CH 3 9.7 Heptet 6 adj H Singlet OH- No split Doublet 1 adj H Singlet No adj H Doublet 1 adj H Heptet 6 adj H Singlet No adj H Singlet No adj H Singlet No adj H Singlet No adj H 9 1 6 1 1 9 2 6 1 3 Splitting Pattern by neighbouring H Equiv H in same environment have no splitting effect on each other Equiv H do not split each other All e quiv H in same environment will produce a same peak . 3 chemical environment 3 peak ratio 6:1:1 2 chemical environment 2 peak ratio 9:1 3 chemical environment 3 peak ratio 6:3:1 2 chemical environment 2 peak ratio 9:2

18. Singlet Splitting Pattern Equiv H in same environment have no splitting effect on each other All e quiv H in the same environment will produce a same peak . Singlet can be due to presence of OH or no adjacent H CH 3 | CH 3 – C -CH 3 | CH 3 Singlet No adj H O CH 3 ‖ ׀ H-C- C-CH 3 ׀ CH 3 Singlet No adj H 9.7 Singlet No adj H H CH 3 ׀ ׀ CI- C – C- CH 3 ׀ ׀ H CH 3 Singlet No adj H Singlet No adj H H H ׀ ׀ CI- C – C- CI ׀ ׀ H H Singlet No adj H 9 2 4 12 9 1 Singlet due to Equiv H in same environ No adj H Singlet due to Equiv H in same environ No adj H Singlet due to Equiv H in same environ Equiv H do not split each other Singlet due to Equiv H in same environ No adj H

19. Singlet All equiv H Singlet No adj H Singlet No adj H H H ׀ ׀ H - C- C- H ׀ ׀ H H CH 3 ׀ CH 3 –O-C-CH 3 ׀ CH 3 O ‖ HO-C-CH 3 12 Singlet No adj H Singlet due to OH in COOH No adj H 2 Singlet No adj H O ‖ HO-C-H Singlet due to OH in COOH H in CHO Singlet No adj H 10.6 8.3 Singlet No adj H 3 1 1 1 6 9 3 Singlet Splitting Pattern Equiv H in same environment have no splitting effect on each other All e quiv H in same environment will produce a same peak . Singlet can be due to presence of OH or no adjacent H Singlet due to Equiv H in same environ Equiv H do not split each other Singlet due to Equiv H in same environ No adj H

20. 2 diff proton environment, Ratio H – 3 : 5 Peak A – No split ( No H on adj C) Peak B – split to 3 (2H on adj C) Peak C – split to 3 (2H on adj C) Peak D – split to 2 (1H on adj C) A B 3 5 2 1 2 C D 7.3 8 All H in benzene consider as 1 proton environment 7.3 8 2 E 1 D 2 5 C 2 3 2 A B 3 diff proton environment, Ratio H – 3: 2 : 5 Peak A – split to 3 (2H on adj C) Peak B – split to 4 (3H on adj C) Peak C – split to 3 (2H on adj C) Peak D – split to 3 (2H on adj C) Peak E – split to 2 (1H on adj C) Molecule with benzene ring Molecule with benzene ring All H in benzene consider as 1 proton environment High Resolution (NMR)

21. A C 3 5 2 1 2 D E 7.3 8 7.3 8 2 F 1 E 2 5 D 3 1 2 A B 4 diff environment, Ratio H – 1 : 2 : 2 : 5 Peak A – No split for OH Peak B – split to 3 (2H on adj C) Peak C – split to 3 (2H on adj C) Peak D – split to 3 (2H on adj C) Peak E – split to 3 (2H on adj C) Peak F – split to 2 (1H on adj C) 2 B 3 diff proton environment, Ratio H – 3 : 2 : 5 Peak A – split to 3 (2H on adj C) Peak B – split to 4 (3H on adj C) Peak C – split to 3 (2H on adj C) Peak D – split to 3 (2H on adj C) Peak E – split to 2 (1H on adj C) 3 4 C 2 High Resolution (NMR) Molecule with benzene ring All H in benzene consider as 1 proton environment All H in benzene consider as 1 proton environment Molecule with benzene ring

22. A C 6 5 2 1 2 D E 7.3 8 1 B 3 diff environment, Ratio H – 6 : 1 : 5 Peak A – split to 2 (1H on adj C) Peak B – split to 7 (6H on adj C) Peak C – split to 3 (2H on adj C) Peak D – split to 3 (2H on adj C) Peak E – split to 2 (1H on adj C) 5 High Resolution (NMR) Molecule with benzene ring All H in benzene consider as 1 proton environment Unknown X have MF of C 3 H 6 O 2 with HNMR shown Chemical shift/ppm Number H atoms Splitting pattern 1.3 3 3 4.3 2 4 8 1 1 i . Deduce IHD 3 2 8 4.3 1.3 Triplet 2 adj H Quartet 3 adj H Singlet No adj H 1 3 diff environment 3 peak/chemical shift O ‖ HO-C-CH 2 -CH 3 ii . Deduce molecular structure Molecule Index H 2 Deficiency C 3 H 6 O 2 1 IHD = 1 (1 double bond or ring) H in COO H – singlet – 8 ppm (next to COO ) - No adj H bond neighbour C H in C H 3 - triplet – 1.3 ppm (next to C H 2 ) - 2 adj H bond neighbour C H in C H 2 - quartet – 4.3 ppm (next to C=O) - 3 adj H bond neighbour C O ‖ HO-C-CH 2 -CH 3

23. Unknown X have mass composition of 85.6% C, 14.4% H Mass spectra, IR and NMR shown below. % composition mass C 85.6 H 14.4 m/e IB Question 10 20 30 40 50 60 70 80 90 i . Deduce EF and MF of compound abundance 28 42 56 84 2 1 0 Singlet All equiv H 12 Empirical formula = CH 2 Molecular ion, M + = RMM = 84 n (EF) = MF n (CH 2 )= 84 n (12 + 2.01) = 84 n = 6 MF = C 6 H 12 C – H stretch (2840 – 3000) C – H bend (1200) Mass spec IR spec HNMR spec Molecule Index H 2 Deficiency C 6 H 12 1 M + peak M + = C 6 H 12 + = 84 ii. Deduce IHD of compound IHD = 1 (1 double bond or ring) iii. Deduce the molecular structure IR spec → C-H stretch and C – H bend No C=C absorption at 1610 No C =O/C-O/OH functional gp HNMR spec → 1 singlet peak All equiv H at same chemical environment Molecular Formula H in C H 2 - singlet – 1 ppm All 12 H in same chemical environment (symmetry)

24. O ‖ CH 3 C-OH % composition mass C 40 H 6.7 O 53.3 m/e IB Question i . Deduce EF and MF of compound abundance 28 45 60 2 Singlet No adj H 3 Empirical formula = CH 2 O Molecular ion, M + = RMM = 60 n (EF) = MF n(CH 2 O)= 60 n (12 + 2.01 + 16) = 60 n = 2 MF = C 2 H 4 O 2 C – H stretch (2840 – 3000) C – H bend (1200) Molecule Index H 2 Deficiency C 2 H 4 O 2 1 M + peak M + = C 2 H 4 O 2 + = 60 ii. Deduce IHD of compound IHD = 1 (1 double bond or ring) iii. Deduce the molecular structure IR spec → C-H /O-H stretch (broad absorption) C=O absorption at 1680 C-O absorption at 1200 C=O/C-O/OH functional gp HNMR spec → 2 singlet peak 2 peak/diff environment, ratio 3 : 1 Molecular Formula Unknown X mass composition of 40% C, 6.7% H, 53.3% O Mass spectra, IR and NMR shown below. 10 20 30 40 50 60 70 80 90 15 17 O – H stretch (3230 -3550) C – O stretch (1000-1300) C = O stretch ( 1680 -1740) Singlet No adj H 12 1 C – H stretch (2840 – 3000) O ‖ CH 3 C-OH H in COO H – singlet – 12ppm (next to COO) H in C H 3 - singlet – 2 ppm (next to C=O)

25. m/e IB Question i . Deduce structural formula X 29 45 74 1 3 Molecular ion, M + = RMM = 74 MF = C 3 H 6 O 2 C – H stretch (2840 – 3000) C – H bend (1200) Molecule Index H 2 Deficiency C 3 H 6 O 2 1 M + peak M + = C 3 H 6 O 2 + = 74 ii. Deduce IHD of compound IHD = 1 (1 double bond or ring) iii. Deduce the molecular structure IR spec → C-H /O-H stretch (broad absorption) C=O absorption at 1680 C-O absorption at 1200 C=O/C-O/OH functional gp HNMR spec → triplet/quartet – CH 3 CH 2 present → singlet – at 11ppm - COOH 3 peak/diff environment, ratio 3:2:1 Molecular Formula 10 20 30 40 50 60 70 80 90 15 17 O – H stretch (3230 -3550) C – O stretch (1000-1300) C = O stretch ( 1680 -1740) Singlet No adj H 11 1 C – H stretch (2840 – 3000) Unknown X have MF C 3 H 6 O 2 Mass spectra, IR and NMR shown below. O ‖ CH 3 CH 2 -C-OH 2 2 Triplet 2 adj H Quartet 3 adj H C 2 H 5 + = 29 COOH + = 45 CH 3 + = 15 OH + = 17 O ‖ HO-C-CH 2 -CH 3 H in COO H – singlet – 11ppm (next to COO) H in C H 2 - quartet – 2 ppm (next to CH 3 ) H in C H 3 - triplet – 1 ppm (next to CH 2 ) O ‖ CH 3 CH 2 -C-OH CH 3 + = 15 C 2 H 5 + = 29 COOH + = 45

26. H O H H ׀ ‖ ׀ ׀ H - C – C –O – C – C– H ׀ ׀ ׀ H H H H O H H ׀ ‖ ׀ ׀ H - C – C –O – C – C– H ׀ ׀ ׀ H H H % composition mass C 40 H 6.7 O 53.3 m/e IB Question i . Deduce EF and MF of compound 29 45 88 1 3 Empirical formula = C 2 H 4 O Molecular ion peak, M + = RMM = 88 n (EF) = MF n(C 2 H 4 O)= 88 n (24 + 1 .01 x 4 + 16) = 88 n = 2 MF = C 4 H 8 O 2 C – H stretch (2840 – 3000) Molecule Index H 2 Deficiency C 4 H 8 O 2 1 M + peak M + = C 4 H 8 O 2 + = 88 ii. Deduce IHD of compound IHD = 1 (1 double bond or ring) iii. Deduce the molecular structure IR spec → No O-H stretch (No broad absorption) C=O absorption at 1680 C-O absorption at 1200 C=O/C-O, functional gp HNMR spec → triplet/quartet – CH 3 CH 2 present → singlet – at 2 ppm – CH 3 next to C=O 3 peak/diff environment, ratio 3:3:2 Molecular Formula 10 20 30 40 50 60 70 80 90 15 C – O stretch (1000-1300) C = O stretch ( 1680 -1740) 4 2 3 2 Triplet 2 adj H Quartet 3 adj H C 2 H 5 + = 29 C 2 H 5 O + = 45 CH 3 + = 15 CH 3 CO + = 43 H in C H 3 - triplet – 1 ppm (next to CH 2 ) H in C H 3 – singlet – 2 ppm (next to C=O) H in C H 2 - quartet – 4 ppm (next to O ) Unknown X have EF C 2 H 4 O Mass spectra, IR and NMR shown below. 43 Singlet No adj H E ster group

27. O H H ‖ ׀ ׀ H–C –O – C – C– H ׀ ׀ H H % composition mass C 48.63 H 8.18 O 43.19 m/e IB Question i . Deduce EF and MF of compound 29 45 74 1 3 Empirical formula = C 3 H 6 O 2 Molecular ion peak, M + = RMM = 74 n (EF) = MF n(C 3 H 6 O 2 )= 74 n ( 12 x 3 + 1 .01 x 6 + 16 x 2) = 74 n = 1 MF = C 3 H 6 O 2 C – H stretch (2840 – 3000) Molecule Index H 2 Deficiency C 3 H 6 O 2 1 M + peak M + = C 3 H 6 O 2 + = 74 ii. Deduce IHD of compound IHD = 1 (1 double bond or ring) iii. Deduce the molecular structure IR spec → No O-H stretch (No broad absorption) C=O absorption at 1680 C-O absorption at 1200 C=O/C-O, functional gp HNMR spec → triplet/quartet – CH 3 CH 2 present → singlet – at 8 ppm – H next to COO 3 peak/diff environment, ratio 3: 2: 1 Molecular Formula 10 20 30 40 50 60 70 80 90 15 C – O stretch (1000-1300) C = O stretch ( 1680 -1740) 8 2 1 4 Triplet 2 adj H Quartet 3 adj H C 2 H 5 + = 29 C 2 H 5 O + = 45 CH 3 + = 15 HCOO + = 45 H in C H O – singlet – 8 ppm (next to COO) H in C H 2 - quartet – 4 ppm (next to O) H in C H 3 - triplet – 1 ppm (next to CH 2 ) Singlet No adj H E ster group Unknown X mass composition of 48.63% C, 8.18 % H, 43.19 % O Mass spectra, IR and NMR shown below. O H H ‖ ׀ ׀ H–C –O – C – C – H ׀ ׀ H H

28. % composition mass C 15.4 H 3.24 I 81.36 m/e IB Question i . Deduce EF and MF of compound abundance 29 127 156 3 1 3 Empirical formula = C 2 H 5 I Molecular ion peak, M + = RMM = 156 n (EF) = MF n(C 2 H 5 I)= 156 n (12 x 2 + 1.01 x 5 + 127) = 156 n = 1 MF = C 2 H 5 I C – H stretch (2840 – 3000) C – H bend (1200) Molecule Index H 2 Deficiency C 2 H 5 I M + peak M + = C 2 H 5 I + = 156 ii. Deduce IHD of compound IHD = 0 (Saturated) iii. Deduce the molecular structure IR spec → C-H stretch and C – H bend No C=C absorption at 1610 No C =O/C-O/OH functional gp HNMR spec → triplet/quartet - CH 3 CH 2 present 2 peak/ diff environment, ratio 3:2 Molecular Formula 20 30 40 120 150 Unknown X have mass composition 15.4% C, 3.24% H, 81.36% I Mass spectra, IR and NMR shown below. C – CI stretch (700-800) H H ׀ ׀ H - C- C - I ׀ ׀ H H C 2 H 5 + = 29 I + = 127 Triplet 2 adj H Quartet 3 adj H 2 H in C H 3 - triplet – 1 ppm (next to CH 2 ) H in C H 2 - quartet – 3 ppm (next to I ) H H ׀ ׀ I - C- C - H ׀ ׀ H H

29. Tetramethyl Silane (TMS) as STD Strong peak upfield (shielded) Silicon has lower EN value < carbon Electron shift to carbon H in CH 3 more shielded Experience lower EMF, absorb ↓ freq UPFIELD ≈ 0 Click here for more complicated proton chemical shift 3 diff proton environment Ratio of 3:2:1 CH 3 chemical shift ≈ 1 integration = 3 H split to 3 CH 2 chemical shift ≈ 3.8 integration = 2 H split to 4 OH chemical shift ≈ 4.8 integration = 1 H No split (Singlet) 3 2 1 Upfield 12 Advantages using TMS Volatile and can be removed from sample All 12 hydrogen in same proton environment Single strong peak, upfield , wont interfere with other peak All chemical shift, in ppm (δ) are relative to this STD, ( zero) Nuclear Magnetic Resonance Spectroscopy (HNMR) HO-CH 2 -CH 3 CH 3 ׀ H 3 C – Si – CH 3 ׀ CH 3 Click here Spectra database (Ohio State) Click here Spectra database (NIST) TMS Downfield

30. 1 H NMR Spectrum O ‖ HO-C-CH 2 -CH 3 3 diff environment, Ratio H - 3:2:3 Peak A – split to 3 (2H on neighbour C) Peak B - No split Peak C – split to 4 (3H on neighbour C) 3 diff environment, Ratio H - 3:2:1 Peak A – split to 3 ( 2H on neighbour C) Peak B – split to 4 (3H on neighbour C) Peak C – No split A B C B A C 12 3 2 3 3 2 1 O ‖ CH 3 -C-O-CH 2 -CH 3

31. O ‖ CH 3 -C-CH 2 -CH 2 -CH 3 3 diff environment, Ratio H - 3:2:1 Peak A – split to 3 (2H on neighbour C) Peak B – split to 4 (3H on neighbour C) Peak C – No split 4 diff environment , Ratio H - 3:2:2:3 Peak A – split to 3 (2H on neighbour C) Peak B – split to 6 (5H on neighbour C) Peak C – No split Peak D – split to 3 (2H on neighbour C) A B C 3 B A C D 2 1 3 2 2 3 HO-CH 2 -CH 3 1 H NMR Spectrum

32. O ‖ H-C-CH 3 4 diff environment , Ratio H – 3:2:2:3 Peak A – split to 3 (2H on neighbour C) Peak B – split to 6 ( 5H on neighbour C) Peak C – No split Peak D – split to 3 (2H on neighbour C) A B C D 2 diff environment, Ratio H - 3:1 Peak A – split to 2 (1H on neighbour C) Peak B – split to 4 (3H on neighbour C) 9.8 A B 3 2 2 3 3 1 O ‖ CH 3 -C-O-CH 2 -CH 2 -CH 3 1 H NMR Spectrum

33. 3 diff environment, Ratio H - 6:1:1 Peak A – split to 2 ( 1H on neighbour C) Peak B – No split Peak C – split to 7 (6H on neighbour C) O CH 3 ‖ ׀ CH 3 -C-O-C-H ׀ CH 3 A B C A B C 3 diff environment, Ratio H - 6:3:1 Peak A – split to 2 ( 1H on neighbour C) Peak B – No split Peak C – split to 7 (6H on neighbour C) Molecule with plane of symmetry 6 1 1 6 3 1 CH 3 ׀ H-C-OH ׀ CH 3 Molecule with plane of symmetry 1 H NMR Spectrum

34. 2 diff environment, Ratio H – 6:4 Peak A – split to 3 (2H on neighbour C) Peak B – split to 4 (3H on neighbour C) A B A B 6 4 9 1 2 diff environment, Ratio H – 9:1 Peak A – No split Peak B – No split Molecule with plane of symmetry O ‖ CH 3 -CH 2 -C-CH 2 -CH 3 Molecule with plane of symmetry O CH 3 ‖ ׀ H-C-C-CH 3 ׀ CH 3 1 H NMR Spectrum

35. 4 diff environment, Ratio H- 6:1:1:2 Peak A – split to 2 ( 1H on neighbour C) Peak B – split to 7 (6H on neighbour C) Peak C – No split Peak D – split to 2 ( 1H on neighbour C) A B D C 2 diff environment, Ratio H – 6:1 Peak A – split to 2 (1H on neighbour C) Peak B – split to 7 (6H on neighbour C) A B 6 1 1 2 6 1 Molecule with plane of symmetry CH 3 ׀ HO-CH 2 -CH ׀ CH 3 Molecule with plane of symmetry CH 3 -CH-CH 3 ׀ CI 1 H NMR Spectrum

36. Acknowledgements Thanks to source of pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/ Prepared by Lawrence Kok Check out more video tutorials from my site and hope you enjoy this tutorial http://lawrencekok.blogspot.com