IAP 5.301
Spectroscopy
Chemists use spectroscopy (IR,UV-Vis,NMR) to determine molecular structure,
Spectroscopy,technique that measures the amount of radiation a substance absorbs at various
wavelengths (based on the quantization of molecular energy levels)
Molecules exist in quantized energy states (rotational,vibrational,and electronic quantum states),
Molecules absorb discrete amounts of energy (?E) and are excited to higher energy states,
We can use electromagnetic radiation to excite molecules,
E
1
E
2
E = hν
(ν = c/λ)
10
cm/s
ν ( )
λ
h = 6.62608 x 10
–34
J·s
c = 2.998 x 10
= frequency Hz
= wavelength (cm)
Molecules have unique absorption spectra that depend on their structure,
Specific absorptions provide useful structural information,
How do you know what type of electromagnetic radiation to use?
EM Radiation Wavelength (cm)
10
–10
10
–8
10
–8 –6
10
–6 –5
–5 –5
–5 –2
–2 2
2 5
Excitation
––
––
––
––
Approximate?E
––
––
––
/
/
––
cosmic rays
gamma rays
X-rays
ultraviolet
visible
infrared
microwave
radio*
–12
to 10
–10
to 10
to 10
to 3.8 x 10
3.8 x 10 to 7.8 x 10
7.8 x 10 to 3 x 10
3 x 10 to 3 x 10
3 x 10 to 3 x 10
electronic
electronic
vibrational
rotational
70–300 kcal mol
40–70 kcal mol
1–10 kcal/mol
~ 1 cal/mol
* Radio frequency EM radiation is used in NMR spectroscopy,
1
Infrared Spectroscopy (vibrational energy)
Chemical bonds are not rigid; they are constantly vibrating,
Different vibrational modes are of higher energy than others,and complex organic molecules have
a large number of vibrational modes (complicated!),
A nonlinear molecule containing n atoms has 3n–6 possible fundamental vibrational modes!
Luckily,different functional groups (parts of a molecule) exhibit unique chracteristic absorptions,
This allows us to look at an IR spectrum (% transmission versus wavenumber) and determine
which functional groups are present based on the absorption bands,
Wavenumber is the number of waves per centimeter (1/λ),
Characteristic IR Absorption Bands
for Common Functional Groups
Class Wavenumber,cm
–1
Intensity Assignment
Alkanes
2850–3000 s C–H stretch
1450–1470 s
1370–1380 s CH
2
and CH
3
bend
720–725 m
Alkenes
3080–3140
1650
809–990
m =C–H stretch
m C=C stretch
s
C–H out-of-plane bend
Alkynes
3300
2100–2140
600–700
s ≡C–H stretch
m C≡C stretch
s ≡C–H bend
var free O–H stretch
Alcohols
3600
3400 s bonded O–H stretch
Ethers
1070–1150 s C–O stretch
Carbonyls 1540–1870
s C=O stretch
Aromatic Hydrocarbons
675–900
1000–1300
1400–1600
s C–H out-of-plane-bend
m C–H in-plane bending
m
C–C stretching
2
Ultraviolet-Visible Spectroscopy (electronic energy)
It is easy to think of atoms as tiny little solar systems with the electrons orbitting the
nucleus in a similar fashion as planets orbiting the sun,This simplistic picture satisfies
our intuition,but it does not correspond with what we know about the atom,
It is more effective to treat electrons as waves rather than particles!
(This may sound complicated,but you'll have a chance to explore this in great depth
during your stay here at MIT.)
What it boils down to is that electrons occupy discrete orbitals that are best
described by mathematical functions called wavefunctions,Because these orbitals
have quantized energy,we can use spectroscopy to investigate the excitation of
electrons from lower energy to higher energy orbitals,
i
i
hν
E
Ant bonding
Bond ng
excitation
These excitations usually involve
an electron moving from a low
energy bonding orbital to a high
energy antibonding orbital,
UV-Vis spectroscopy is most frequently used to detect the presence of a specific
compound in solution or to measure concentrations very accurately,You will get a
chance to do this later this month,
3
Nuclear Magnetic Resonance (NMR) Spectroscopy
NMR is the most common spectroscopic tool used by organic chemists,
NMR differs from other spectroscopy because the different energy states (nuclear spin states) exist
only in the presence of a magnetic field.
The nuclei of most (but not all) atoms behave as if they are spinning on an axis (nuclear spin).
Because nuclei are positively charged,these spinning nuclei create small magnetic moments.
In the absence of a magnetic field,these magnetic moments are oriented in a random fashion.
For some nuclei (those having nuclear spin = ±1/2),the presence of an external magnetic field (H)
results in the alignment of each nucleus either with (α) or against (β) the magnetic field.
1
H,
13
C,and
19
F are the three most common nuclei observed using NMR spectroscopy,In this
class,we will focus on
1
H (proton) NMR.
α
α
α
α
α
β
β
β
β
H
)
Nuclear Magnetic Moments
(no magnetic field)
Nuclear Magnetic Moments
(in a magnetic field
The energy difference between the α and β states (?E) is proportional to the strength of the
magnetic field (H) at the nucleus.
E is relatively small (~10
–2
cal/mol); radio waves are used to excite (flip) nuclei from α to β.
β
α
ν
[(γ·H) π]}
γ
)
)
β /
2
)
α /
2
)
E
E
H
E = h
E = {9.54 x 10
–11
/2
= magnetogyric ratio
(specific to nucleus
H = magnetic field strength
(at the nucleus
(–1
(+1
4
If the energy difference between the α and β spin states of a proton nucleus
(?E) depends only on the
1
H NMR magnetogyric ratio (γ = 2.6753 x 10
4
radians·sec
–1
·gauss
–1
) and the magnetic field (H),then won't all of the protons
in a molecule absorb radiation of the exact same energy?
Luckily,no! We wouldn't be able to obtain any structural information from a
1
H NMR spectrum if that were the case,
Remember,nuclei are surrounded by clouds of negatively charged electrons,
In the presence of an applied magnetic field (H
0
),these electrons move in such a
way that their motion induces their own small magnetic field (H'),
At the nucleus,the induced magnetic field (H') opposes the applied magnetic
field (H
0
),Therefore,the nucleus experiences a magnetic field (H) slightly less
than the applied magnetic field (H
0
),
0
)
0
induced magnetic
field (H')
electrons
applied
magnetic
field (H
H = H – H'
This phenomenon is called diamagnetic shielding,Protons with a lot of electron
density around them are well-shielded and experience a reduced magnetic field,As
a result,there is a smaller energy difference between the two spin states,and lower
energy radiation is required to flip the spin from α to β,
In other words,protons in different electronic environments experience different
amounts of shielding and absorb radiation of different frequencies,These
differences are referred to as chemical shifts,
5
6
That's all well and good,but what does it mean practically?
The location of the absorbance peak
on this scale gives the chemical shift
(δ) of that particular proton,
The chemical shift of a proton gives us information about its
chemical environment,but there's more to be learned!
Let's look at a sample
1
H NMR to get a better idea of the information
we can take from it,,,
δ
δ
downfield region
(deshielded,electron poor)
upfield region
(shielded,electron rich)
All chemical shifts ( ) are reported relative to a
standard peak,Tetramethylsilane (TMS) is used as
the standard because the protons are very electron
rich,and the resulting absorption peak is upfield
from most other protons,The chemical shift of the
TMS peak is assigned a value of
TMS
= 0.0 ppm
9
8 7 6 5 4 3 2
1 0
-1
ppm
H
1
H
1
H
2
H
2
C C
Cl
C H
2
Cl Cl
The
1
H NMR spectrum of 1,2,2-trichloropropane has two peaks other than the TMS peak,
How can you tell which peak corresponds to the two H
1
protons and which is from the
three H
2
protons?
Here's a clue,Chlorine atoms are very electronegative and strongly electron withdrawing,This
means that they suck electron density away from the atoms close to them,
The two H
1
protons are attached to a carbon that is also attached to a chlorine atom,As a result,
these protons have less electron density around them,are less effectively shielded,and have a
chemical shift further downfield than the H
2
protons (H
1
= A,H
2
= B),
Integration
In addition to chemical shift,the relative size of the peaks provides information.
When you take a
1
H NMR,you will integrate the different peaks in your spectrum,The relative
values of the integrals are proportional to the number of protons absorbing energy at that frequency.
For example (from above):
B, A = 75, 50 = 3, 2
The spectrum tells you that there is a 3, 2 ratio of different protons in 1,2,2-trichloropropane,Of
course you know that there are three H
2
protons and two H
1
protons that fit this criteria,
7
Spin-Spin Splitting
Unlike in the
1
H NMR spectrum of 1,2,2-trichloropropane on the previous page,
most peaks are not single lines (singlets),
In fact,the multiplicity of a peak (singlet,doublet,triplet) provides even more
information about the structure of a molecule,This is because protons on adjacent
atoms communicate with eachother,
In simple systems,n adjacent protons cause splitting into n + 1 peaks,
For example,look at the spectrum of 1-chloropropane,
H
1
H
1
H
3
H
3
C C
Cl
C H
3
H
2
H
2
There are three inequivalent types of protons in this molecule (H
1
,H
2
,H
3
),
It is easy to assign chemical shifts to the appropriate protons because you
already know that chlorine is an electron withdrawing group,
The protons closest to the chlorine will have the least electron density
(deshielded,downfield),and the protons furthest from the chlorine will have the
most electron density (shielded,upfield),Using this reasoning,you can assign,
A = H
1
,B = H
2
,C = H
3
Now,let's look at the multiplicity of the three peaks.,,
8
Cl
C
C
C
H
3
H
1
H
1
H
2
H
2
H
3
H
3
triplet
sextet
triplet
Peak A,corresponding to the two H
1
protons,is a triplet (three lines),This is
because there are two adjacent H
2
protons (2 + 1 = 3),
Peak B,corresponding to the two H
2
protons,is a sextet (six lines),This is because
there are five adjacent protons,two H
1
and three H
3
(5 + 1 = 6),
Peak C,corresponding to the three H
3
protons,is a triplet (three lines),This is
because there are two adjacent H
2
protons (2 + 1 = 3),
Spin-spin splitting,also known as coupling,can get much more complicated than
this,but we'll stick to simple systems for now,
It is very important to remember that spin-spin splitting and peak integrals are not
related,The multiplicity of a peak does not have any correlation to the number of
protons absorbing energy at that frequency!
There is a useful table on p,301 of Zubrick that will help you assign the peaks in
the
1
H NMR spectra you will be obtaining this month,
9
Spectroscopy
Chemists use spectroscopy (IR,UV-Vis,NMR) to determine molecular structure,
Spectroscopy,technique that measures the amount of radiation a substance absorbs at various
wavelengths (based on the quantization of molecular energy levels)
Molecules exist in quantized energy states (rotational,vibrational,and electronic quantum states),
Molecules absorb discrete amounts of energy (?E) and are excited to higher energy states,
We can use electromagnetic radiation to excite molecules,
E
1
E
2
E = hν
(ν = c/λ)
10
cm/s
ν ( )
λ
h = 6.62608 x 10
–34
J·s
c = 2.998 x 10
= frequency Hz
= wavelength (cm)
Molecules have unique absorption spectra that depend on their structure,
Specific absorptions provide useful structural information,
How do you know what type of electromagnetic radiation to use?
EM Radiation Wavelength (cm)
10
–10
10
–8
10
–8 –6
10
–6 –5
–5 –5
–5 –2
–2 2
2 5
Excitation
––
––
––
––
Approximate?E
––
––
––
/
/
––
cosmic rays
gamma rays
X-rays
ultraviolet
visible
infrared
microwave
radio*
–12
to 10
–10
to 10
to 10
to 3.8 x 10
3.8 x 10 to 7.8 x 10
7.8 x 10 to 3 x 10
3 x 10 to 3 x 10
3 x 10 to 3 x 10
electronic
electronic
vibrational
rotational
70–300 kcal mol
40–70 kcal mol
1–10 kcal/mol
~ 1 cal/mol
* Radio frequency EM radiation is used in NMR spectroscopy,
1
Infrared Spectroscopy (vibrational energy)
Chemical bonds are not rigid; they are constantly vibrating,
Different vibrational modes are of higher energy than others,and complex organic molecules have
a large number of vibrational modes (complicated!),
A nonlinear molecule containing n atoms has 3n–6 possible fundamental vibrational modes!
Luckily,different functional groups (parts of a molecule) exhibit unique chracteristic absorptions,
This allows us to look at an IR spectrum (% transmission versus wavenumber) and determine
which functional groups are present based on the absorption bands,
Wavenumber is the number of waves per centimeter (1/λ),
Characteristic IR Absorption Bands
for Common Functional Groups
Class Wavenumber,cm
–1
Intensity Assignment
Alkanes
2850–3000 s C–H stretch
1450–1470 s
1370–1380 s CH
2
and CH
3
bend
720–725 m
Alkenes
3080–3140
1650
809–990
m =C–H stretch
m C=C stretch
s
C–H out-of-plane bend
Alkynes
3300
2100–2140
600–700
s ≡C–H stretch
m C≡C stretch
s ≡C–H bend
var free O–H stretch
Alcohols
3600
3400 s bonded O–H stretch
Ethers
1070–1150 s C–O stretch
Carbonyls 1540–1870
s C=O stretch
Aromatic Hydrocarbons
675–900
1000–1300
1400–1600
s C–H out-of-plane-bend
m C–H in-plane bending
m
C–C stretching
2
Ultraviolet-Visible Spectroscopy (electronic energy)
It is easy to think of atoms as tiny little solar systems with the electrons orbitting the
nucleus in a similar fashion as planets orbiting the sun,This simplistic picture satisfies
our intuition,but it does not correspond with what we know about the atom,
It is more effective to treat electrons as waves rather than particles!
(This may sound complicated,but you'll have a chance to explore this in great depth
during your stay here at MIT.)
What it boils down to is that electrons occupy discrete orbitals that are best
described by mathematical functions called wavefunctions,Because these orbitals
have quantized energy,we can use spectroscopy to investigate the excitation of
electrons from lower energy to higher energy orbitals,
i
i
hν
E
Ant bonding
Bond ng
excitation
These excitations usually involve
an electron moving from a low
energy bonding orbital to a high
energy antibonding orbital,
UV-Vis spectroscopy is most frequently used to detect the presence of a specific
compound in solution or to measure concentrations very accurately,You will get a
chance to do this later this month,
3
Nuclear Magnetic Resonance (NMR) Spectroscopy
NMR is the most common spectroscopic tool used by organic chemists,
NMR differs from other spectroscopy because the different energy states (nuclear spin states) exist
only in the presence of a magnetic field.
The nuclei of most (but not all) atoms behave as if they are spinning on an axis (nuclear spin).
Because nuclei are positively charged,these spinning nuclei create small magnetic moments.
In the absence of a magnetic field,these magnetic moments are oriented in a random fashion.
For some nuclei (those having nuclear spin = ±1/2),the presence of an external magnetic field (H)
results in the alignment of each nucleus either with (α) or against (β) the magnetic field.
1
H,
13
C,and
19
F are the three most common nuclei observed using NMR spectroscopy,In this
class,we will focus on
1
H (proton) NMR.
α
α
α
α
α
β
β
β
β
H
)
Nuclear Magnetic Moments
(no magnetic field)
Nuclear Magnetic Moments
(in a magnetic field
The energy difference between the α and β states (?E) is proportional to the strength of the
magnetic field (H) at the nucleus.
E is relatively small (~10
–2
cal/mol); radio waves are used to excite (flip) nuclei from α to β.
β
α
ν
[(γ·H) π]}
γ
)
)
β /
2
)
α /
2
)
E
E
H
E = h
E = {9.54 x 10
–11
/2
= magnetogyric ratio
(specific to nucleus
H = magnetic field strength
(at the nucleus
(–1
(+1
4
If the energy difference between the α and β spin states of a proton nucleus
(?E) depends only on the
1
H NMR magnetogyric ratio (γ = 2.6753 x 10
4
radians·sec
–1
·gauss
–1
) and the magnetic field (H),then won't all of the protons
in a molecule absorb radiation of the exact same energy?
Luckily,no! We wouldn't be able to obtain any structural information from a
1
H NMR spectrum if that were the case,
Remember,nuclei are surrounded by clouds of negatively charged electrons,
In the presence of an applied magnetic field (H
0
),these electrons move in such a
way that their motion induces their own small magnetic field (H'),
At the nucleus,the induced magnetic field (H') opposes the applied magnetic
field (H
0
),Therefore,the nucleus experiences a magnetic field (H) slightly less
than the applied magnetic field (H
0
),
0
)
0
induced magnetic
field (H')
electrons
applied
magnetic
field (H
H = H – H'
This phenomenon is called diamagnetic shielding,Protons with a lot of electron
density around them are well-shielded and experience a reduced magnetic field,As
a result,there is a smaller energy difference between the two spin states,and lower
energy radiation is required to flip the spin from α to β,
In other words,protons in different electronic environments experience different
amounts of shielding and absorb radiation of different frequencies,These
differences are referred to as chemical shifts,
5
6
That's all well and good,but what does it mean practically?
The location of the absorbance peak
on this scale gives the chemical shift
(δ) of that particular proton,
The chemical shift of a proton gives us information about its
chemical environment,but there's more to be learned!
Let's look at a sample
1
H NMR to get a better idea of the information
we can take from it,,,
δ
δ
downfield region
(deshielded,electron poor)
upfield region
(shielded,electron rich)
All chemical shifts ( ) are reported relative to a
standard peak,Tetramethylsilane (TMS) is used as
the standard because the protons are very electron
rich,and the resulting absorption peak is upfield
from most other protons,The chemical shift of the
TMS peak is assigned a value of
TMS
= 0.0 ppm
9
8 7 6 5 4 3 2
1 0
-1
ppm
H
1
H
1
H
2
H
2
C C
Cl
C H
2
Cl Cl
The
1
H NMR spectrum of 1,2,2-trichloropropane has two peaks other than the TMS peak,
How can you tell which peak corresponds to the two H
1
protons and which is from the
three H
2
protons?
Here's a clue,Chlorine atoms are very electronegative and strongly electron withdrawing,This
means that they suck electron density away from the atoms close to them,
The two H
1
protons are attached to a carbon that is also attached to a chlorine atom,As a result,
these protons have less electron density around them,are less effectively shielded,and have a
chemical shift further downfield than the H
2
protons (H
1
= A,H
2
= B),
Integration
In addition to chemical shift,the relative size of the peaks provides information.
When you take a
1
H NMR,you will integrate the different peaks in your spectrum,The relative
values of the integrals are proportional to the number of protons absorbing energy at that frequency.
For example (from above):
B, A = 75, 50 = 3, 2
The spectrum tells you that there is a 3, 2 ratio of different protons in 1,2,2-trichloropropane,Of
course you know that there are three H
2
protons and two H
1
protons that fit this criteria,
7
Spin-Spin Splitting
Unlike in the
1
H NMR spectrum of 1,2,2-trichloropropane on the previous page,
most peaks are not single lines (singlets),
In fact,the multiplicity of a peak (singlet,doublet,triplet) provides even more
information about the structure of a molecule,This is because protons on adjacent
atoms communicate with eachother,
In simple systems,n adjacent protons cause splitting into n + 1 peaks,
For example,look at the spectrum of 1-chloropropane,
H
1
H
1
H
3
H
3
C C
Cl
C H
3
H
2
H
2
There are three inequivalent types of protons in this molecule (H
1
,H
2
,H
3
),
It is easy to assign chemical shifts to the appropriate protons because you
already know that chlorine is an electron withdrawing group,
The protons closest to the chlorine will have the least electron density
(deshielded,downfield),and the protons furthest from the chlorine will have the
most electron density (shielded,upfield),Using this reasoning,you can assign,
A = H
1
,B = H
2
,C = H
3
Now,let's look at the multiplicity of the three peaks.,,
8
Cl
C
C
C
H
3
H
1
H
1
H
2
H
2
H
3
H
3
triplet
sextet
triplet
Peak A,corresponding to the two H
1
protons,is a triplet (three lines),This is
because there are two adjacent H
2
protons (2 + 1 = 3),
Peak B,corresponding to the two H
2
protons,is a sextet (six lines),This is because
there are five adjacent protons,two H
1
and three H
3
(5 + 1 = 6),
Peak C,corresponding to the three H
3
protons,is a triplet (three lines),This is
because there are two adjacent H
2
protons (2 + 1 = 3),
Spin-spin splitting,also known as coupling,can get much more complicated than
this,but we'll stick to simple systems for now,
It is very important to remember that spin-spin splitting and peak integrals are not
related,The multiplicity of a peak does not have any correlation to the number of
protons absorbing energy at that frequency!
There is a useful table on p,301 of Zubrick that will help you assign the peaks in
the
1
H NMR spectra you will be obtaining this month,
9
