Atomic Structure Protons, Neutrons, and Electrons Section 4.3.
NMR SPECTROSCOPY The basis of MRI scanning. NUCLEAR SPIN Protons and neutrons can be regarded as...
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Transcript of NMR SPECTROSCOPY The basis of MRI scanning. NUCLEAR SPIN Protons and neutrons can be regarded as...
NMR SPECTROSCOPY
The basis of MRI scanning
NUCLEAR SPIN
bull Protons and neutrons can be regarded as spinning about their axis
bull In many atoms these spins are paired against each other and so the nucleus has no overall spin (eg 12C)
bull In some atoms (eg 1H and 13C ndash an odd number of subatomic particles) the nucleus has an overall spin
bull A nucleus that spins generates a magnetic field ie behaves like a small magnet
bull The direction of the magnetic field depends which way the nucleus spins
NUCLEAR SPIN
bull Usually the two possible spin states of the nucleus have the same amount of energy
bull However in a magnetic field the two spin states have different energies
Energy Applied magnetic
field
Magnetic field in same direction as
applied field
Magnetic field opposed to applied field
Energy gap corresponds to frequency of radiowaves
Different ldquotypes of hydrogensrdquo
bull The frequency absorbed is determined by the environment of the atoms giving the signal
bull Atoms in different environments are sometimes said to be of different types
bull To be of the same type hydrogen atoms must be indistinguishable
bull Example ethane ndash all hydrogens identical so one type
bull Propane (2 types)bull Propan-1-ol (4 types)bull Notes
Example
bull Draw a displayed formula of butanone and decide how many types of hydrogen there are
bull This determines the number of peaks seen in the spectrum
butanone
2
3
3
EQUIVALENT H ATOMS
bull In a spectrum there is one signal for each set of equivalent H atoms ie H atoms in an identical situationenvironment
bull The intensity of each signal being proportional to the number of equivalent H atoms it represents so the hydrogen atoms are counted (area under peak)
bull In a real spectrum the relative peak areas is given rather than actual numbers so
bull 21 = 42 = 63 etc ndash beware of this
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
NUCLEAR SPIN
bull Protons and neutrons can be regarded as spinning about their axis
bull In many atoms these spins are paired against each other and so the nucleus has no overall spin (eg 12C)
bull In some atoms (eg 1H and 13C ndash an odd number of subatomic particles) the nucleus has an overall spin
bull A nucleus that spins generates a magnetic field ie behaves like a small magnet
bull The direction of the magnetic field depends which way the nucleus spins
NUCLEAR SPIN
bull Usually the two possible spin states of the nucleus have the same amount of energy
bull However in a magnetic field the two spin states have different energies
Energy Applied magnetic
field
Magnetic field in same direction as
applied field
Magnetic field opposed to applied field
Energy gap corresponds to frequency of radiowaves
Different ldquotypes of hydrogensrdquo
bull The frequency absorbed is determined by the environment of the atoms giving the signal
bull Atoms in different environments are sometimes said to be of different types
bull To be of the same type hydrogen atoms must be indistinguishable
bull Example ethane ndash all hydrogens identical so one type
bull Propane (2 types)bull Propan-1-ol (4 types)bull Notes
Example
bull Draw a displayed formula of butanone and decide how many types of hydrogen there are
bull This determines the number of peaks seen in the spectrum
butanone
2
3
3
EQUIVALENT H ATOMS
bull In a spectrum there is one signal for each set of equivalent H atoms ie H atoms in an identical situationenvironment
bull The intensity of each signal being proportional to the number of equivalent H atoms it represents so the hydrogen atoms are counted (area under peak)
bull In a real spectrum the relative peak areas is given rather than actual numbers so
bull 21 = 42 = 63 etc ndash beware of this
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
NUCLEAR SPIN
bull Usually the two possible spin states of the nucleus have the same amount of energy
bull However in a magnetic field the two spin states have different energies
Energy Applied magnetic
field
Magnetic field in same direction as
applied field
Magnetic field opposed to applied field
Energy gap corresponds to frequency of radiowaves
Different ldquotypes of hydrogensrdquo
bull The frequency absorbed is determined by the environment of the atoms giving the signal
bull Atoms in different environments are sometimes said to be of different types
bull To be of the same type hydrogen atoms must be indistinguishable
bull Example ethane ndash all hydrogens identical so one type
bull Propane (2 types)bull Propan-1-ol (4 types)bull Notes
Example
bull Draw a displayed formula of butanone and decide how many types of hydrogen there are
bull This determines the number of peaks seen in the spectrum
butanone
2
3
3
EQUIVALENT H ATOMS
bull In a spectrum there is one signal for each set of equivalent H atoms ie H atoms in an identical situationenvironment
bull The intensity of each signal being proportional to the number of equivalent H atoms it represents so the hydrogen atoms are counted (area under peak)
bull In a real spectrum the relative peak areas is given rather than actual numbers so
bull 21 = 42 = 63 etc ndash beware of this
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Different ldquotypes of hydrogensrdquo
bull The frequency absorbed is determined by the environment of the atoms giving the signal
bull Atoms in different environments are sometimes said to be of different types
bull To be of the same type hydrogen atoms must be indistinguishable
bull Example ethane ndash all hydrogens identical so one type
bull Propane (2 types)bull Propan-1-ol (4 types)bull Notes
Example
bull Draw a displayed formula of butanone and decide how many types of hydrogen there are
bull This determines the number of peaks seen in the spectrum
butanone
2
3
3
EQUIVALENT H ATOMS
bull In a spectrum there is one signal for each set of equivalent H atoms ie H atoms in an identical situationenvironment
bull The intensity of each signal being proportional to the number of equivalent H atoms it represents so the hydrogen atoms are counted (area under peak)
bull In a real spectrum the relative peak areas is given rather than actual numbers so
bull 21 = 42 = 63 etc ndash beware of this
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Example
bull Draw a displayed formula of butanone and decide how many types of hydrogen there are
bull This determines the number of peaks seen in the spectrum
butanone
2
3
3
EQUIVALENT H ATOMS
bull In a spectrum there is one signal for each set of equivalent H atoms ie H atoms in an identical situationenvironment
bull The intensity of each signal being proportional to the number of equivalent H atoms it represents so the hydrogen atoms are counted (area under peak)
bull In a real spectrum the relative peak areas is given rather than actual numbers so
bull 21 = 42 = 63 etc ndash beware of this
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
butanone
2
3
3
EQUIVALENT H ATOMS
bull In a spectrum there is one signal for each set of equivalent H atoms ie H atoms in an identical situationenvironment
bull The intensity of each signal being proportional to the number of equivalent H atoms it represents so the hydrogen atoms are counted (area under peak)
bull In a real spectrum the relative peak areas is given rather than actual numbers so
bull 21 = 42 = 63 etc ndash beware of this
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
EQUIVALENT H ATOMS
bull In a spectrum there is one signal for each set of equivalent H atoms ie H atoms in an identical situationenvironment
bull The intensity of each signal being proportional to the number of equivalent H atoms it represents so the hydrogen atoms are counted (area under peak)
bull In a real spectrum the relative peak areas is given rather than actual numbers so
bull 21 = 42 = 63 etc ndash beware of this
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 2 sets of equivalent Hrsquos ratio 62 (31)
CH3 C
CH3
CH2
OH
CH3
4 sets of equivalent Hrsquos ratio 3123
5 sets of equivalent Hrsquos ratio 31223
4 sets of equivalent Hrsquos ratio 6123
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
For each of the following compounds predict the number of signals and the relative intensityarea of the signals
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
2 signals ratio 62 (31)
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 213
2 signals ratio 61
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 32
4 signals ratio 3223
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
3 signals ratio 213
2 signals ratio 62 (31)
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 329
CH CHCH3 CH3
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
SOLVENTS amp CALIBRATION
bull Samples are dissolved in solvents free of 1H atoms eg CCl4 CDCl3
CH3
Si CH3
CH3
CH3
bull A small amount of TMS (tetramethylsilane) is added to calibrate the spectrum
bull It is used because
bull its signal is away from all the othersbull it only gives one signalbull it is non-toxicbull it is inertbull it has a low boiling point so is easy to remove
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
CHEMICAL SHIFT
bull The d is a measure in parts per million (ppm) of how far the magnetic field required for absorption is shifted away from that for TMS
0 1 2 3 4 5 6 7 8 9 10
chemical shift (ppm)
bull The depends on what other atomsgroups are near the H ndash more electronegative groups gives a greater shift
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Butanone again ndash look at the chemical shift this time See data sheet
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a
ldquopeakrdquo)
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
SPIN-SPIN COUPLING
bull Each signal in the spectrum can be split into further lines due to inequivalent Hrsquos on neighbouring C atoms
bull NOTE This splitting effect is not observed if adjacent H atoms are in an identical situationenvironment
Number of lines that the peak is split into = number of inequivalent Hrsquos on neighbouring C atoms + 1
The n+1 rule
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of lines
1 2 3 4
number of neighbouring
inequivalent H atoms
0 1 2 3
relative size 11 121 1331
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
butanone CH3 CH2 C CH3
O
shift () assignmentrelative intensity
coupling coupled to
10 CH3CH2 3 triplet CH2
20 CH3CO 3 singlet
24 CH2 2 quartet CH3
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
butane
shift () Assignmentrelative intensity
coupling coupled to
13 CH2 2 quartet CH3
08 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
SUMMARY
Number of signals
-radiowave region
Position of signals
Relative intensities
Splitting
how many different sets of equivalent H atoms there are
information about chemical environment of H atom
gives ratio of numbers of H atoms for the peaks
how many H atoms on adjacent C atoms (alcohol OH [aldehyde H] no splitting)
IDENTICAL PROTONS DO NOT CAUSE SPLITTING
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
For each of the following compounds predict the number of signals the relative intensity of the signals and the multiplicity (singlet doublet etc) of each signal
a) methylpropene
b) propene
c) 2-chloropropane
d) propanone
e) methylamine
f) ethyl propanoate
g) 12-dibromopropane
h) dimethylethyl propanoate
i) but-2-ene
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
2 signals ratio 6 2 (3 1) s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals ratio 2 1 3 d m d
2 signals ratio 6 1 d m
1 signal
CH3 CH2 C
O
O CH2 CH3
CH3 NH2
2 signals ratio 3 2 t q
4 signals ratio 3 2 2 3 t q q t
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
3 signals ratio 2 1 3 t m d
2 signals ratio 6 2 (3 1) d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH33 signals ratio 3 2 9 t q s
CH CHCH3 CH3
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Use of deuterium based compounds in NMR
bull Deuterium is 12H 1e 1 proton and 1 neutron in
the nucleusbull Deuterium does not give nmr signal (even
number of sub-atomic particles)
bull CDCl3 is one of the standard nmr solvents (the solvent does not interfere with the spectrum)
bull D2O removes peaks due to O-H and N-H from the spectrum
bull As a result of rapid HD exchange
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Past paper example
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Mark scheme
bull (i) carboxylic acid COOH (protons) NOT just ldquoOH protonsrdquo [1]
bull (ii) D replaces protons on OH groups OH protons are labile 1048633
bull peak for (CO)OH protons disappears 1048633
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Further example
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Further example continued
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Protons on a benzene ring
bull Need to identify aromatic protons from chemical shift patterns
bull No need to worry about their splitting patterns
bull Signals can appear as one or two complex peaks
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Example (past paper)
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-
Pentan-3-one
Mass spec mol Ion peak at 86
- NMR SPECTROSCOPY
- Slide 2
- Slide 3
- Different ldquotypes of hydrogensrdquo
- Example
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Butanone again ndash look at the chemical shift this time See data sheet
- Note that the signal from each type of hydrogen is seen in this case as a group of lines (still called a ldquopeakrdquo)
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Use of deuterium based compounds in NMR
- Past paper example
- Mark scheme
- Further example
- Further example continued
- Protons on a benzene ring
- Example (past paper)
- Pentan-3-one
-