IB Chemistry on HNMR Spectroscopy and Spin spin coupling

Biology 2005N Paper 3 Option H

http://lawrencekok.blogspot.com

Prepared by Lawrence KokTutorial on Nuclear magnetic resonance spectroscopy (NMR) and Spin spin Coupling

Spectroscopy measures interaction of molecules with electromagnetic radiation Particles (molecule, ion, atom) can interact/absorb a quantum of light

SpectroscopyElectromagnetic Radiation Nuclear spinHigh Energy RadiationGamma/X ray Transition of inner electronsUV or visible Transition of outer most valence electronsInfrared Molecular vibrationMicrowave Molecular rotationRadiowavesLow Energy RadiationInfrared SpectroscopyNuclear Magnetic Resonance Spectroscopy Ultra Violet SpectroscopyAtomic Absorption Spectroscopy

Velocity of light (c ) = frequency (f) x wavelength () - c = f All electromagnetic waves travel at speed of light (3.00 x 108ms-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 = wavelength

Click here notes spectroscopy

Electromagnetic Radiation and SpectroscopyRadiowaves Nuclear spinNuclear Magnetic Resonance Spectroscopy Organic structure determination MRI and body scanning Infrared Molecular vibrationInfrared SpectroscopyUV 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 breathElectromagnetic Radiation

UV SpectroscopyAtomic A Spectroscopy Quantification of metal ions Detection of metal in various samplesElectromagnetic Radiation Interact with Matter (Atoms, Molecules) = Spectroscopy

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 (12C and 16O) 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 (1H, 13C, 19F, 31P)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 otherNet spin

Main features of HNMR Spectra1. Number of diff absorption peak Number of diff proton/chemical environment2. Area under peak - Number of hydrogen in a particular proton/chemical environment (Integration trace) - Ratio of number of hydrogen in each environment3. 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 ()4. Splitting pattern - Due to spin-spin coupling - Number of peak split is equal to number of hydrogen on neighbouring carbon+1 (n+1) peak

Chemical ShiftNMR spectrum CH3CH2BrNumber of peaksArea under peaksChemical shiftSplitting pattern

Nuclear Magnetic Resonance Spectroscopy (NMR)

Click here khan NMR videos.

NMR spectrum CH3CH2BrNumber of peaksSplitting pattern

Click here khan NMR videos.Number equi HMultiplicityRatio0Singlet11Doublet1: 12Triplet1 : 2 : 13Quartet1 : 3 : 3 : 14Quintet1 : 4 : 6 : 4 : 15Hextet1 : 5 : 10 : 10 : 5 : 16Septet1 : 6 : 15 :20 : 15 : 6 : 1

Splitting PatternSinglet Neighbouring Carbon with No HDoublet Neighbouring Carbon with 1H Triplet Neighbouring Carbon with 2HQuartet 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 shiftHigh Resolution (NMR)Main features of HNMR Spectra1. Number of diff absorption peak Number of diff proton/chemical environment2. Area under peak - Number of hydrogen in a particular proton/chemical environment (Integration trace) - Ratio of number of hydrogen in each environment3. 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 ()4. Splitting pattern - Due to spin-spin coupling - Number of peak split is equal to number of hydrogen on neighbouring carbon+1 (n+1) peak

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 CH3 to 2 peak, (doublet). H nuclei split CH3 methyl gp to doublet H can align with EMF or against EMF. CH3 will experience 2 diff EMF One lower, one higher EMF Split to doublet

EMF align against MF produce by H Overall MF experience CH3 lower H from CH3 will absorb at lower freq (upfield)EMFEMF align with MF produce by H Overall MF experience by CH3 higher H from CH3 will absorb at higher freq (downfield)

EMF

CH3 spilt to doublet by 1 adj H CH3 experience 2 slightly diff MF due to neighbouring HMFMFSplit with relative intensity of 1 : 1DownfieldUpfieldHigh 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

doubletSplitting peaks occur as effective MF experience by H nuclei is modified by MF produced by neighbouring proton/H

(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 CH3 to 3 peak, (triplet). 2H nuclei split CH3 methyl to triplet H can align with EMF or against EMF. CH3 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 CH3 lower H from CH3, absorb lower freq EMFEMF align with MF produce by H Both H align with EMF (Net greater EMF) Overall MF experience by CH3 higher H from CH3, absorb at higher freq

EMF

EMFMFMFMFMFEMF align with/against MF produce by H 1 align with and 1 against EMF MF by H cancel each other Overall MF experience by CH3 the sameSplit with relative intensity of 1 : 2 : 1 CH3 spilt to triplet by 2 adj H CH3 experience 3 diff MF due to 2 adjacent HDownfieldUpfield H C H H C H H H C H H C H HH 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

tripletHigh Resolution (NMR)Splitting peaks occur as effective MF experience by H nuclei is modified by MF produced by neighbouring proton/H

3H nuclei split CH2 methylene to quartet H can align with EMF or against EMF. CH2 will experience 4 diff EMF Split to quartet (ratio 1 : 3 : 3 : 1 )

EMF align against MF by H 3 H align against EMF (Lower EMF) Overall MF experience CH2 lower H from CH2, absorb at lower freq EMFEMF align with MF by H 3 H align with EMF (Net greater EMF) Overall MF experience by CH2 higher H from CH2, absorb at higher freq

EMF

EMFMFMFEMF 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 CH2 in 3 : 3 ratio

Split with relative intensity of 1 : 3 : 3 : 1 CH2 spilt to quartet by 3 adjacent H CH2 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 CH3 into 4 peak, called quartet. H C H H C H HH C H H C H HH C H H C H HH C H H C H HH C H H C H HH C H H C H HH C H H C H HH C H H C H HH C H H C H HH C H H C H H

quartetHigh Resolution (NMR)Splitting peaks occur as effective MF experience by H nuclei is modified by MF produced by neighbouring proton/H

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 CH3 spilt to triplet by 2 adj H CH3 experience 3 diff MF due to 2 adj H CH2 spilt to quartet by 3 adj H CH2 experience 4 diff MF due to 3 adj H No signal splitting from coupling bet hydroxyl proton and methylene proton of CH2 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 CH2 is negligible/not seen. NO triplet split on OH due to 2 adjacent H from CH2 Only singlet H H HO- C- C- H H HCH3 chemical shift 1 integration = 3 H split into 3 CH2 chemical shift 3.8 integration = 2 H split to 4 OH chemical shift 4.8 integration = 1 H No split (Singlet)

321Triplet splitQuartet splitSinglet split

tripletquartetHigh Resolution (NMR)

- Equiv H in same chemical environment have no splitting effect on each other- All Equiv H produce same signal O CH3-C-O-CH2-CH3HO-CH2-CH3 O HO-C-CH2-CH3 O CH3-C-CH2-CH2-CH3Equivalent Hydrogen in same chemical Environment (chemical Shift)4 diff chemical environment 4 peak ratio 3:2:3:23 diff chemical environment 3 peak ratio 3:2:13 diff chemical environment 3 peak ratio 3:3:23 diff chemical environment 3 peak ratio 3:2:1

12332323213232 equiv H3 equiv H3 equiv H2 equiv H213 equiv H2 equiv H1 equiv H3 equiv H2 equiv H3 equiv H3 equiv H2 equiv H1 equiv H

Equivalent Hydrogen in molecule with plane of symmetryEquiv 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. CH3 |CH3 C -CH3 | CH31 chemical environment 1 peak O CH3-CH2-C-CH2- CH32 diff chemical environment 2 peak ratio 3:2 CI CH3-C-CH3 H CH3 HO-CH2- C- H CH3

2

42321261216112 equiv H2 diff chemical environment 2 peak ratio 6:14 diff chemical environment 4 peak ratio 6:1:1:21 equiv H6 equiv H6 equiv H1 equiv H2 equiv H 3 equiv H2 equiv H

O CH3 H-C- C-CH3 CH3 CH3 |H-C-OH | CH3 O CH3 CH3-C-O-C-H CH3 H CH3 CI- C- C- CH3 H CH3

9.7

9161192631Equivalent Hydrogen in molecule with plane of symmetryEquiv 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 6:1:12 diff chemical environment 2 peak ratio 9:11 equiv H6 equiv H1 equiv H9 equiv H1 equiv H3 diff chemical environment 3 peak ratio 6:3:12 diff chemical environment 2 peak ratio 9:26 equiv H3 equiv H1 equiv H9 equiv H2 equiv H

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

4.56.122164Equivalent Hydrogen in molecule with plane of symmetryEquiv 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 1:21 chemical environment 1 peak2 equiv H1 equiv H2 equiv H1 chemical environment 1 peak 4 equiv H1 chemical environment 1 peak6 equiv H

O CH3-C-O-CH2-CH3HO-CH2-CH3 O HO-C-CH2-CH3 O CH3-C-CH2-CH2-CH3

12 Equiv H in same environment have no splitting effect on each other Equiv H do not split each other All equiv H in same environment will produce a same peak . Triplet 2 adj H Septet 5 adj H Singlet No adj HTriplet 2 adj H Triplet 2 adj H Quartet 3 adj H Singlet OH No split Triplet 2 adj H Singlet No adj HQuartet3 adj H Triplet 2 adj H Quartet 3 adj H Singlet No adj HSplitting Pattern by neighbouring H 32132323213324 chemical environment 4 peak ratio 3:2:3:23 chemical environment 3 peak ratio 3:2:1 3 chemical environment 3 peak ratio 3:3:23 chemical environment 3 peak ratio 3:2:1

CH3 HO-CH2- C- H CH3 CH3 |CH3 C -CH3 | CH3 O CH3-CH2-C-CH2- CH3 CI CH3-C-CH3 H

2

4SingletNo 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 H3212611261Splitting Pattern by neighbouring H Equiv H in same environment have no splitting effect on each other Equiv H do not split each other All equiv H in same environment will produce a same peak .2 chemical environment 2 peak ratio 6:11 chemical environment 1 peak2 chemical environment 2 peak ratio 3:2 4 chemical environment 4 peak ratio 6:1:1:2

O CH3 H-C- C-CH3 CH3 CH3 |H-C-OH | CH3 O CH3 CH3-C-O-C-H CH3 H CH3 CI- C- C- CH3 H CH3

9.7

Heptet 6 adj H SingletOH- No splitDoublet1 adj HSingletNo adj HDoublet 1 adj H Heptet 6 adj H Singlet No adj H Singlet No adj H Singlet No adj H Singlet No adj H9161192613Splitting Pattern by neighbouring H Equiv H in same environment have no splitting effect on each other Equiv H do not split each other All equiv H in same environment will produce a same peak .3 chemical environment 3 peak ratio 6:1:12 chemical environment 2 peak ratio 9:13 chemical environment 3 peak ratio 6:3:12 chemical environment 2 peak ratio 9:2

Singlet Splitting Pattern Equiv H in same environment have no splitting effect on each other All equiv H in the same environment will produce a same peak . Singlet can be due to presence of OH or no adjacent H CH3 |CH3 C -CH3 | CH3

SingletNo adj H O CH3 H-C- C-CH3 CH3

Singlet No adj H

9.7 Singlet No adj H H CH3 CI- C C- CH3 H CH3

Singlet No adj H Singlet No adj H H H CI- C C- CI H H

Singlet No adj H9241291Singlet due to Equiv H in same environ No adj HSinglet due to Equiv H in same environ No adj HSinglet due to Equiv H in same environ Equiv H do not split each otherSinglet due to Equiv H in same environ No adj H

Singlet All equiv H

Singlet No adj H Singlet No adj H H H H - C- C- H H H CH3 CH3O-C-CH3 CH3 O HO-C-CH3

12 Singlet No adj H

Singlet due to OH in COOH No adj H2 Singlet No adj H O HO-C-HSinglet due to OH in COOH H in CHO

Singlet No adj H10.68.3 Singlet No adj H3111693Singlet Splitting Pattern Equiv H in same environment have no splitting effect on each other All equiv H in same environment will produce a same peak . Singlet can be due to presence of OH or no adjacent HSinglet due to Equiv H in same environ Equiv H do not split each otherSinglet due to Equiv H in same environ No adj H

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)AB3

5212

CD7.38All H in benzene consider as 1 proton environment

7.382E1D25

C232AB3 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 ringMolecule with benzene ringAll H in benzene consider as 1 proton environmentHigh Resolution (NMR)

AC3

5212

DE7.38

7.382F1E25

D312AB4 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)2B3 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)34C2High Resolution (NMR)Molecule with benzene ringAll H in benzene consider as 1 proton environmentAll H in benzene consider as 1 proton environmentMolecule with benzene ring

21

AC6

5212

DE7.381B3 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)5High Resolution (NMR)Molecule with benzene ringAll H in benzene consider as 1 proton environmentUnknown X have MF of C3H6O2 with HNMR shownChemical shift/ppmNumber H atomsSplitting pattern1.3334.324811

i. Deduce IHD328 4.3 1.3 Triplet 2 adj H Quartet 3 adj H Singlet No adj H13 diff environment 3 peak/chemical shift O HO-C-CH2-CH3

ii. Deduce molecular structureMoleculeIndex H2 DeficiencyC3H6O21

IHD = 1 (1 double bond or ring)H in COOH singlet 8 ppm (next to COO) - No adj H bond neighbour CH in CH3 - triplet 1.3 ppm (next to CH2) - 2 adj H bond neighbour CH in CH2 - quartet 4.3 ppm (next to C=O) - 3 adj H bond neighbour C O HO-C-CH2-CH3

22

Unknown X have mass composition of 85.6% C, 14.4% HMass spectra, IR and NMR shown below.% composition massC85.6H14.4

m/eIB Question10 20 30 40 50 60 70 80 90 i. Deduce EF and MF of compoundabundance

28425684

2 1 0 Singlet All equiv H12Empirical formula = CH2Molecular ion, M+ = RMM = 84n(EF) = MFn(CH2)= 84n (12 + 2.01) = 84n = 6MF = C6H12 C H stretch(2840 3000) C H bend(1200)

Mass specIR specHNMR spec

MoleculeIndex H2 DeficiencyC6H121

M+ peakM+ = C6H12+ = 84ii. Deduce IHD of compoundIHD = 1 (1 double bond or ring)iii. Deduce the molecular structureIR spec C-H stretch and C H bendNo C=C absorption at 1610No C =O/C-O/OH functional gp

HNMR spec 1 singlet peak All equiv H at same chemical environmentMolecular Formula

H in CH2- singlet 1 ppmAll 12 H in same chemical environment(symmetry)

O CH3C-OH % composition massC40H6.7O53.3

m/eIB Questioni. Deduce EF and MF of compoundabundance2845602 Singlet No adj H3Empirical formula = CH2OMolecular ion, M+ = RMM = 60n(EF) = MFn(CH2O)= 60n (12 + 2.01 + 16) = 60n = 2MF = C2H4O2 C H stretch(2840 3000) C H bend(1200)

MoleculeIndex H2 DeficiencyC2H4O21

M+ peakM+ = C2H4O2+ = 60ii. Deduce IHD of compoundIHD = 1 (1 double bond or ring)iii. Deduce the molecular structureIR spec C-H /O-H stretch (broad absorption)C=O absorption at 1680C-O absorption at 1200C=O/C-O/OH functional gp

HNMR spec 2 singlet peak 2 peak/diff environment, ratio 3 : 1Molecular 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 1517

O H stretch(3230 -3550) C O stretch(1000-1300) C = O stretch(1680 -1740) Singlet No adj H121 C H stretch(2840 3000) O CH3C-OH H in COOH singlet 12ppm (next to COO)H in CH3 - singlet 2 ppm (next to C=O)

m/eIB Questioni. Deduce structural formula X2945741 3Molecular ion, M+ = RMM = 74MF = C3H6O2 C H stretch(2840 3000) C H bend(1200)


M+ peakM+ = C3H6O2+ = 74ii. Deduce IHD of compoundIHD = 1 (1 double bond or ring)iii. Deduce the molecular structureIR spec C-H /O-H stretch (broad absorption)C=O absorption at 1680C-O absorption at 1200C=O/C-O/OH functional gp

HNMR spec triplet/quartet CH3CH2 present singlet at 11ppm - COOH3 peak/diff environment, ratio 3:2:1Molecular Formula

10 20 30 40 50 60 70 80 90 1517

O H stretch(3230 -3550) C O stretch(1000-1300) C = O stretch(1680 -1740) Singlet No adj H111 C H stretch(2840 3000) Unknown X have MF C3H6O2Mass spectra, IR and NMR shown below. O CH3CH2-C-OH 22 Triplet 2 adj H Quartet 3 adj HC2H5+ = 29COOH+ = 45CH3+ = 15OH+ = 17 O HO-C-CH2-CH3H in COOH singlet 11ppm (next to COO)H in CH2 - quartet 2 ppm (next to CH3)H in CH3 - triplet 1 ppm (next to CH2) O CH3CH2-C-OH

CH3+ = 15C2H5+ = 29COOH+ = 45

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 massC40H6.7O53.3

m/eIB Questioni. Deduce EF and MF of compound2945881 3Empirical formula = C2H4OMolecular ion peak, M+ = RMM = 88n(EF) = MFn(C2H4O)= 88n (24 + 1.01 x 4 + 16) = 88n = 2MF = C4H8O2 C H stretch(2840 3000)


M+ peakM+ = C4H8O2+ = 88ii. Deduce IHD of compoundIHD = 1 (1 double bond or ring)iii. Deduce the molecular structureIR spec No O-H stretch (No broad absorption)C=O absorption at 1680C-O absorption at 1200C=O/C-O, functional gp

HNMR spec triplet/quartet CH3CH2 present singlet at 2 ppm CH3 next to C=O3 peak/diff environment, ratio 3:3:2Molecular Formula

10 20 30 40 50 60 70 80 90 15 C O stretch(1000-1300) C = O stretch(1680 -1740) 4232 Triplet 2 adj H Quartet 3 adj HC2H5+ = 29C2H5O+ = 45CH3+ = 15CH3CO+ = 43H in CH3 - triplet 1 ppm (next to CH2)H in CH3 singlet 2 ppm (next to C=O)H in CH2 - quartet 4 ppm (next to O)Unknown X have EF C2H4OMass spectra, IR and NMR shown below.43 Singlet No adj H

Ester group

O H H HC O C C H H H

% composition massC48.63H8.18O43.19

m/eIB Questioni. Deduce EF and MF of compound2945741 3Empirical formula = C3H6O2Molecular ion peak, M+ = RMM = 74n(EF) = MFn(C3H6O2)= 74n (12 x 3 + 1.01 x 6 + 16 x 2) = 74n = 1MF = C3H6O2 C H stretch(2840 3000)


M+ peakM+ = C3H6O2+ = 74ii. Deduce IHD of compoundIHD = 1 (1 double bond or ring)iii. Deduce the molecular structureIR spec No O-H stretch (No broad absorption)C=O absorption at 1680C-O absorption at 1200C=O/C-O, functional gp

HNMR spec triplet/quartet CH3CH2 present singlet at 8 ppm H next to COO3 peak/diff environment, ratio 3: 2: 1Molecular Formula

10 20 30 40 50 60 70 80 90 15 C O stretch(1000-1300) C = O stretch(1680 -1740) 8214 Triplet 2 adj H Quartet 3 adj HC2H5+ = 29C2H5O+ = 45CH3+ = 15HCOO+ = 45H in CHO singlet 8 ppm (next to COO)H in CH2 - quartet 4 ppm (next to O)H in CH3 - triplet 1 ppm (next to CH2) Singlet No adj H

Ester groupUnknown X mass composition of 48.63% C, 8.18% H, 43.19% O Mass spectra, IR and NMR shown below. O H H HC O C C H H H

% composition massC15.4H3.24I81.36

m/eIB Questioni. Deduce EF and MF of compoundabundance

29127156

3 1 3Empirical formula = C2H5IMolecular ion peak, M+ = RMM = 156n(EF) = MFn(C2H5I)= 156n (12 x 2 + 1.01 x 5 + 127) = 156n = 1MF = C2H5I C H stretch(2840 3000) C H bend(1200)

MoleculeIndex H2 DeficiencyC2H5I0

M+ peakM+ = C2H5I+ = 156ii. Deduce IHD of compoundIHD = 0 (Saturated)iii. Deduce the molecular structureIR spec C-H stretch and C H bendNo C=C absorption at 1610No C =O/C-O/OH functional gp

HNMR spec triplet/quartet - CH3CH2 present 2 peak/ diff environment, ratio 3:2Molecular Formula

20 30 40 120 150 Unknown X have mass composition 15.4% C, 3.24% H, 81.36% IMass spectra, IR and NMR shown below.

C CI stretch(700-800) H H H - C- C - I H H C2H5+ = 29I+ = 127 Triplet 2 adj H Quartet 3 adj H2H in CH3 - triplet 1 ppm (next to CH2)H in CH2 - quartet 3 ppm (next to I) H H I - C- C - H H H

Tetramethyl Silane (TMS) as STD Strong peak upfield (shielded)Silicon has lower EN value < carbon Electron shift to carbon H in CH3 more shielded Experience lower EMF, absorb freq UPFIELD 0Click here for more complicated proton chemical shift 3 diff proton environment Ratio of 3:2:1CH3 chemical shift 1 integration = 3 H split to 3 CH2 chemical shift 3.8 integration = 2 H split to 4 OH chemical shift 4.8 integration = 1 H No split (Singlet)

321Upfield12Advantages 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 () are relative to this STD, ( zero)Nuclear Magnetic Resonance Spectroscopy (HNMR)HO-CH2-CH3 CH3 H3C Si CH3 CH3

Click here Spectra database (Ohio State)

Click here Spectra database (NIST) TMSDownfield

1H NMR Spectrum O HO-C-CH2-CH3

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 splitABCBAC

12323321 O CH3-C-O-CH2-CH3

O CH3-C-CH2-CH2-CH3

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 split4 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)ABC3BACD213223HO-CH2-CH31H NMR Spectrum

O H-C-CH3

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)ABCD2 diff environment, Ratio H - 3:1 Peak A split to 2 (1H on neighbour C) Peak B split to 4 (3H on neighbour C)

9.8AB322331 O CH3-C-O-CH2-CH2-CH31H NMR Spectrum

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 CH3 CH3-C-O-C-H CH3ABC

ABC3 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 symmetry611631 CH3 H-C-OH CH3Molecule with plane of symmetry1H NMR Spectrum

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)ABAB64912 diff environment, Ratio H 9:1 Peak A No split Peak B No splitMolecule with plane of symmetry O CH3-CH2-C-CH2-CH3Molecule with plane of symmetry O CH3 H-C-C-CH3 CH31H NMR Spectrum

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)ABDC

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)AB611261Molecule with plane of symmetry CH3 HO-CH2-CH CH3Molecule with plane of symmetryCH3-CH-CH3 CI1H NMR Spectrum

Acknowledgements

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Prepared by Lawrence Kok

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IB Chemistry on HNMR Spectroscopy and Spin spin coupling