ib chemistry on hnmr spectroscopy and spin spin coupling

• 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 108ms-1) • Radiation with high ↑ frequency – short ↓ wavelength • Electromagnetic radiation/photon carry a quantum of energy given by

E = hf

hcE

h = plank constant = 6.626 x 10-34 Js f = frequency λ = wavelength

Click here notes spectroscopy

http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared:_Interpretation

http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared:_Interpretation

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

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 other

Net spin

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 - Number of peak split is equal to number of hydrogen on neighbouring carbon+1 (n+1) peak

Chemical Shift NMR spectrum CH3CH2Br

Number of peaks

Area under peaks Chemical shift

Splitting pattern

Nuclear Magnetic Resonance Spectroscopy (NMR)

Click here khan NMR videos.

https://www.youtube.com/watch?v=jjcHZuTGWXk


NMR spectrum CH3CH2Br

Number of peaks

Splitting pattern

Click here khan NMR videos.

Number equi H Multiplicity Ratio

0 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 - 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)

EMF

EMF 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 H

MF MF

Split with relative intensity of 1 : 1

Downfield Upfield


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

(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

EMF

EMF 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 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 CH3 the same

Split with relative intensity of 1 : 2 : 1

• CH3 spilt to triplet by 2 adj H • CH3 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



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

EMF

EMF 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 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 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 ׀ 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




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 H

CH3

• 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)

3 2 1

Triplet split Quartet split

Singlet split

triplet quartet

- Equiv H in same chemical environment have no splitting effect on each other - All Equiv H produce same signal

O ‖ CH3-C-O-CH2-CH3

HO-CH2-CH3

O ‖ HO-C-CH2-CH3

O ‖ CH3-C-CH2-CH2-CH3

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



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

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.

CH3

| CH3 – C -CH3

| CH3

1 chemical environment • 1 peak

O ║ CH3-CH2-C-CH2- CH3

2 diff chemical environment • 2 peak ratio 3:2

CI ׀ CH3-C-CH3

׀ H

CH3

׀ HO-CH2- C- H

׀ CH3

2

4 2

3 2 12

6 1 2 1 6 1

12 equiv H


4 diff chemical environment • 4 peak ratio 6:1:1:2

1 equiv H

6 equiv H


3 equiv H 2 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

9 1 6 1 1

9 2 6 3 1





1 equiv H 6 equiv H




6 equiv H

3 equiv H

1 equiv H 9 equiv H

2 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

4.5 6.1

2 2 1

6 4





2 equiv H

1 equiv H 2 equiv H


4 equiv H


6 equiv H

O ‖ CH3-C-O-CH2CH3

HO-CH2CH3

O ‖ HO-C-CH2CH3

O ‖ CH3-C-CH2CH2CH3

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 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



CH3

׀ HO-CH2–C- H

׀ CH3

CH3

׀ CH3 – C – CH3

׀ CH3

O ‖ CH3CH2-C-CH2CH3

CI ׀ CH3- C – CH3

׀ 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



2 chemical environment • 2 peak ratio 6:1



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 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







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

Singlet

No 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 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 • Equiv H do not split each other


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

2

Singlet

No adj H

O ‖ HO-C-H

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 equiv 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 • OH in COOH • H in CHO

Singlet due to • OH in COOH • 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)

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


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






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




Unknown X have MF of C3H6O2 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-CH2-CH3

yx HC

1

2

)12262(

2

22

IHD

IHD

yxIHD

ii. Deduce molecular structure

Molecule Index H2 Deficiency

C3H6O2 1

IHD = 1 (1 double bond or ring)

H in COOH – singlet – 8 ppm (next to COO) - No adj H bond neighbour C H in CH3 - triplet – 1.3 ppm (next to CH2) - 2 adj H bond neighbour C H in CH2 - quartet – 4.3 ppm (next to C=O) - 3 adj H bond neighbour C

O ‖ HO-C-CH2CH3

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 = CH2

Molecular ion, M+ = RMM = 84 n(EF) = MF n(CH2)= 84 n (12 + 2.01) = 84 n = 6 MF = C6H12

C – H stretch (2840 – 3000)

C – H bend (1200)

Mass spec

IR spec

HNMR spec

yx HC

1

2

)12262(

2

22

IHD

IHD

yxIHD


C6H12 1

M+ peak

M+ = C6H12+

= 84

ii. Deduce IHD of compound


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 CH2- singlet – 1 ppm All 12 H in same chemical environment (symmetry)

O

‖

CH3C-OH

% composition mass

C 40

H 6.7

O 53.3

m/e

IB Question


28

45

60

2

Singlet

No adj H

3

Empirical formula = CH2O Molecular ion, M+ = RMM = 60 n(EF) = MF n(CH2O)= 60 n (12 + 2.01 + 16) = 60 n = 2 MF = C2H4O2

C – H stretch (2840 – 3000)

C – H bend (1200)

yx HC

1

2

)4222(

2

22

IHD

IHD

yxIHD


C2H4O2 1

M+ peak

M+ = C2H4O2+

= 60




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 ‖ CH3C-OH

H in COOH – singlet – 12ppm (next to COO) H in CH3 - singlet – 2 ppm (next to C=O)

m/e

IB Question

i. Deduce structural formula X

29

45

74

1

3

Molecular ion, M+ = RMM = 74 MF = C3H6O2

C – H stretch (2840 – 3000)

C – H bend (1200)

yx HC

1

2

)6232(

2

22

IHD

IHD

yxIHD


C3H6O2 1

M+ peak

M+ = C3H6O2+

= 74




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 – CH3CH2 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 C3H6O2

Mass spectra, IR and NMR shown below.

O

‖

CH3CH2-COH

2

2

Triplet

2 adj H

Quartet

3 adj H

C2H5+ = 29 COOH+ = 45

CH3+ = 15 OH+ = 17

O ‖ HO-C-CH2-CH3

H 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+ = 15

C2H5+ = 29

COOH+ = 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 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 = C2H4O Molecular ion peak, M+ = RMM = 88 n(EF) = MF n(C2H4O)= 88 n (24 + 1.01 x 4 + 16) = 88 n = 2 MF = C4H8O2

C – H stretch (2840 – 3000)

yx HC

1

2

)8242(

2

22

IHD

IHD

yxIHD


C4H8O2 1

M+ peak

M+ = C4H8O2+

= 88




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 – CH3CH2 present → singlet – at 2 ppm – CH3 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

C2H5+ = 29

C2H5O+ = 45

CH3+ = 15 CH3CO+ = 43

H 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 C2H4O Mass spectra, IR and NMR shown below.

43

Singlet

No adj H

Ester group

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 = C3H6O2

Molecular ion peak, M+ = RMM = 74 n(EF) = MF n(C3H6O2)= 74 n (12 x 3 + 1.01 x 6 + 16 x 2) = 74 n = 1 MF = C3H6O2

C – H stretch (2840 – 3000)

yx HC

1

2

)6232(

2

22

IHD

IHD

yxIHD


C3H6O2 1

M+ peak

M+ = C3H6O2+

= 74




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 – CH3CH2 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

C2H5+ = 29

C2H5O+ = 45

CH3+ = 15

HCOO+ = 45

H 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 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

% composition mass

C 15.4

H 3.24

I 81.36

m/e

IB Question


29

127 156

3 1

3

Empirical formula = C2H5I Molecular ion peak, M+ = RMM = 156 n(EF) = MF n(C2H5I)= 156 n (12 x 2 + 1.01 x 5 + 127) = 156 n = 1 MF = C2H5I

C – H stretch (2840 – 3000)

C – H bend (1200)

syx XHCMolecule Index H2 Deficiency

C2H5I 0

M+ peak

M+ = C2H5I+ = 156


IHD = 0 (Saturated)


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 - CH3CH2 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.

0

2

)15222(

2

22

IHD

IHD

syxIHD

C – CI stretch (700-800)

H H ׀ ׀ H - C- C - I

׀ ׀ H H

C2H5+ = 29

I+ = 127

Triplet

2 adj H

Quartet

3 adj H

2

H 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 ≈ 0

Click here for more complicated proton chemical shift

• 3 diff proton environment • Ratio of 3:2:1

CH3

• 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)

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-CH2-CH3

CH3

׀ H3C – Si – CH3

׀ CH3

Click here Spectra database (Ohio State) Click here Spectra database (NIST)

TMS

Downfield

http://www.chem.wisc.edu/areas/reich/handouts/chem343-345/345-nmr-handout.pdf

https://undergrad-ed.chemistry.ohio-state.edu/anim_spectra/index.html

https://undergrad-ed.chemistry.ohio-state.edu/anim_spectra/index.html

http://webbook.nist.gov/chemistry/name-ser.html

http://webbook.nist.gov/chemistry/name-ser.html

1H NMR Spectrum

O ‖

HO-C-CH2CH3

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 ‖ CH3-C-O-CH2CH3

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 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-CH2-CH3

1H 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)

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 ‖ CH3-C-O-CH2-CH2CH3

1H 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

׀ CH3

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

CH3

׀ H- C – OH

׀ CH3


1H 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)

A

B

A

B

6 4

9 1

2 diff environment, Ratio H – 9:1 • Peak A – No split • Peak B – No split


O ‖ CH3CH2-C-CH2CH3


O CH3

׀ ‖ H-C – C – CH3

׀ CH3

1H 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)

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


CH3

׀ HO-CH2-C-H

׀ CH3

Molecule with plane of symmetry CH3-CH-CH3

׀ CI

1H NMR Spectrum

ib chemistry on hnmr spectroscopy and spin spin coupling

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