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