IB Chemistry on HNMR Spectroscopy and Spin spin coupling

• Spectroscopymeasuresinteractionof moleculeswith electromagneticradiation
• Particles(molecule,ion, atom) can interact/absorba quantumof light
Spectroscopy
Electromagnetic
Radiation
Nuclear spin
High Energy Radiation
Gamma/X ray
Transitionof
innerelectrons
UV or visible
Transitionof outermost
valenceelectrons
Infrared
Molecularvibration
Microwave
Molecularrotation
Radiowaves
Low Energy Radiation
InfraredSpectroscopy Nuclear MagneticResonance
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 108
ms-1
)
• Radiation with high ↑ frequency – short ↓ wavelength
• Electromagnetic radiation/photon carry a quantum of energy given by
E = hf

hc
E 
h = plank constant = 6.626 x 10-34
Js
f = frequency
λ = wavelength
Click here notes spectroscopy

ElectromagneticRadiationand Spectroscopy
Radiowaves
Nuclear spin
Nuclear MagneticResonance
Spectroscopy
• Organic structure determination
• MRI and body scanning
Infrared
Molecularvibration
InfraredSpectroscopy
UV or visible
Transitionof outervalenceelectron
• Organic structure determination
• Functional gp determination
• Measure bond strength
• Measure degree unsaturationin fat
• Measure level of alcohol in breath
ElectromagneticRadiation
UV Spectroscopy Atomic A Spectroscopy
• Quantification of metal ions
• Detection of metal in various samples
ElectromagneticRadiation Interact with Matter (Atoms, Molecules)= Spectroscopy

Nuclear Magnetic Resonance Spectroscopy (NMR)
• Involvenucleus(proton+ neutron)NOT electron
• Proton+ neutron= Nucleons
• Nucleonslike electronhave spin and magneticmoment (acts like tiny magnet)
Nuclei with even number of nucleon (12C and 16O)
• Even numberof protonand neutron– NO net spin
• Nucleonspin cancel out each other –Nucleushave NO overallmagneticmoment – NOT absorb radiowave
Nuclei with odd number of nucleon (1H, 13C, 19F, 31P)
-Nucleonhave net spin – Nucleushave NET magneticmoment – Absorbradiowave
• Nuclei with net spin – magnetic momentwill interactwith radiowaves
• Nuclei have a “spin”associated with them (i.e., they act as if they were spinningaboutan axis) due to the spin
associatedwith theirprotonand neutron.
• Nuclei are positivelycharged,theirspin inducesa magnetic field
• NMR spectroscopy does not work for nuclei with even numberof protons and neutrons - nuclei have no net spin.
Spin cancel each other
Net spin

Main features of HNMR Spectra
1. Number of diff absorption peak
– Number of diff proton/chemicalenvironment
2. Area under peak
- Number of hydrogen in a particularproton/chemicalenvironment(Integration trace)
- Ratio of number of hydrogen in each environment
3. Chemicalshift
- Chemicalenvironment 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 ShiftNMR spectrum CH3CH2Br
Number of peaks
Area under peaks Chemical shift
Splitting pattern
Click here khan NMR videos.

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
SplittingPattern
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 chemicalenvironment have no splitting effect on each other
• Equiv H do not split each other
• All Equiv H in same chemicalenvironment will produce a same peak
• Spin spin coupling – occurwhen proton have diff chemicalshift
• Splitting not observed for proton that are chemicallyequivalent/samechemicalshift
High Resolution (NMR)
Main features of HNMR Spectra
1. Number of diff absorption peak
– Number of diff proton/chemicalenvironment
2. Area under peak
- Number of hydrogen in a particularproton/chemicalenvironment(Integration trace)
- Ratio of number of hydrogen in each environment
3. Chemicalshift
- Chemicalenvironment 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 experience2 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 experience2 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 occuras 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
• 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)
• 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 experience3 diff MF due to 2
adjacentH
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
Splitting peaks occur as effective MF experience by H nuclei

3H nuclei split CH2 methylene to quartet
• 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)
• 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 experience4 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
Splitting peaks occur as effective MF experience by H nuclei

Singlet peak
• H nuclei attachto electronegativeatom, O - NO splitting – Singlet
• H nuclei attachto 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 experience3 diff MF due to 2 adj H
• CH2 spilt to quartet by 3 adj H
• CH2 experience4 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
• chemicalshift ≈ 1
• integration = 3 H
• split into 3
CH2
• chemicalshift ≈ 3.8
• split to 4
OH
• No split (Singlet)
3
21
Triplet splitQuartet split
Singlet split
tripletquartet

- Equiv H in same chemicalenvironment 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 chemicalenvironment
• 4 peak ratio 3:2:3:2
• 3 peak ratio 3:2:1
12
33232
32132 3
2 equiv H
3 equiv H
3 equiv H
2 equiv H
21
3 equiv H2 equiv H1 equiv H
3 equiv H2 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 particularchemicalenvironment
• Equiv H in same environment have no splitting effect on each other
• H on neighbouring carboncan 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 chemicalenvironment
• 1 peak
O
║
CH3-CH2-C-CH2- CH3
• 2 peak ratio 3:2
CI
‫׀‬
CH3-C-CH3
‫׀‬
H
CH3
‫׀‬
HO-CH2- C- H
‫׀‬
CH3
2
4 2
3212
612 161
12 equiv H
1 equiv H
6 equiv H
6 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
91611
92631
1 equiv H
6 equiv H
1 equiv H
9 equiv H1 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
4.56.1
22 1
64
• 1 peak
2 equiv H
1 equiv H 2 equiv H
• 1 peak
4 equiv H
• 1 peak
6 equiv H

O
‖
CH3-C-O-CH2CH3
HO-CH2CH3
O
‖
HO-C-CH2CH3
O
‖
CH3-C-CH2CH2CH3
12
• 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 HSinglet
No adj H
Splitting Pattern by neighbouring H
3213232
321332

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
3212
611261
1 chemical environment
• 1 peak

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
91611
9261 3

Singlet Splitting Pattern
• All equiv H in the same environment will produce a same peak .
• Singlet can be due to presence of OH or no adjacentH
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
924
12 91
Singlet due to
• Equiv H in same environ
• No adj H
Singlet due to
• No adj H
Singlet due to
Singlet due to
• 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
2
Singlet
No adj H
O
‖
HO-C-H
Singlet
No adj H
10.6 8.3
Singlet
No adj H
3111
6 93
Singlet Splitting Pattern
• Singlet can be due to presence of OH or no adjacentH
Singlet due to
Singlet due to
• No adj H
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
7.38
All H in benzene consider
• as 1 proton environment
7.38
2
E
1
D
2
5
C
2
32
A
B
3 diff proton environment, Ratio H – 3: 2 : 5
• Peak A – split to 3 (2H on adj C)
• Peak E – split to 2 (1H on adj C)
Moleculewith benzenering

A
C
3
5
2 1 2
D
E
7.38
7.38
2
F
1
E
2
5
D
3
12
AB
4 diff environment, Ratio H – 1 : 2 : 2 : 5
• Peak A – No split for OH
• Peak F – split to 2 (1H on adj C)
2
B
3 diff proton environment, Ratio H – 3 : 2 : 5
3
4
C
2

A
C
6
5
2 1 2
D
E
7.38
1
B
3 diff environment, Ratio H – 6 : 1 : 5
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
32
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
yx
IHD
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
yx
IHD
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
MolecularFormula
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
abundance
28
45
60
2
Singlet
No adj H
3
Empirical formula = CH2O
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
yx
IHD
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
MolecularFormula
Unknown X mass composition of 40% C, 6.7%H, 53.3%O
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
MF = C3H6O2
C – H stretch
(2840 – 3000)
C – H bend
(1200)
yx HC
 
1
2
)6232(
2
22





IHD
IHD
yx
IHD
C3H6O2 1
M+ peak
M+ = C3H6O2
+ = 74
IR spec → C-H /O-H stretch (broad absorption)
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
MolecularFormula
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
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
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
yx
IHD
C4H8O2 1
M+ peak
M+ = C4H8O2
+ = 88
IR spec → No O-H stretch (No broad absorption)
C=O/C-O, functional gp
→ singlet – at 2 ppm – CH3 next to C=O
3 peak/diff environment, ratio 3:3:2
MolecularFormula
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 – singlet – 2 ppm (next to C=O)
H in CH2 - quartet – 4 ppm (next to O)
Unknown X have EF C2H4O
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
29
45
74
1
3
Empiricalformula = C3H6O2
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
yx
IHD
C3H6O2 1
M+ peak
M+ = C3H6O2
+ = 74
IR spec → No O-H stretch (No broad absorption)
C=O/C-O, functional gp
→ singlet – at 8 ppm – H next to COO
3 peak/diff environment, ratio 3: 2: 1
MolecularFormula
10 20 30 40 50 60 70 80 90
15
C – O stretch
(1000-1300)C = O stretch
(1680 -1740)
8
21
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)
Singlet
No adj H
Ester group
Unknown X mass composition of 48.63%C, 8.18% H, 43.19%O
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
abundance
29
127
156
3 1
3
Empiricalformula = C2H5I
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
MolecularFormula
20 30 40 120 150
Unknown X have mass composition 15.4%C, 3.24%H, 81.36%I
 
0
2
)15222(
2
22





IHD
IHD
syx
IHD
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 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
• chemicalshift ≈ 1
• split to 3
CH2
• split to 4
OH
• No split (Singlet)
321
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

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)
• Peak B – split to 4 (3H on neighbourC)
• Peak C – No split
A
B
C
B
A
C
12
32 3
321
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 neighbourC)
4 diff environment,Ratio H - 3:2:2:3
• Peak D – split to 3 (2H on neighbourC)
A
B
C
3
B
A
C
D
21
322 3
HO-CH2-CH3
1H NMR Spectrum

O
‖
H-C-CH3
4 diff environment, Ratio H – 3:2:2:3
• Peak B – split to 6 (5H on neighbour C)
• Peak D – split to 3 (2H on neighbour C)
ABCD
2 diff environment,Ratio H - 3:1
9.8
A
B
322 3
3
1
O
‖
CH3-C-O-CH2-CH2CH3
1H NMR Spectrum

• Peak B – No split
O CH3
‖ ‫׀‬
CH3-C-O-C-H
‫׀‬
CH3
A
B
C
A
B
C
Molecule with plane of symmetry
611
631
CH3
‫׀‬
H- C – OH
‫׀‬
CH3
1H NMR Spectrum

2 diff environment, Ratio H – 6:4
A
B
A
B
64
91
2 diff environment,Ratio H – 9:1
• Peak A – 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 D – split to 2 (1H on neighbour C)
A
B
D C
2 diff environment,Ratio H – 6:1
A
B
6112
61
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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