structure determination how to determine what …biewerm/9h-analytical.pdfstructure determination!...

Structure Determination

How to determine what compound that you have?

One way to determine compound is to get an elemental analysis

-basically burn the compound to determine %C, %H, %O, etc.

from these percentages can determine the molecular formula

Still need to determine structure from molecular formula

We have learned various isomers can result from a given molecular formula

Consider C3H6O

O O

H OH

Different type of carbonyl

No carbonyl present

Could have a ketone

If we only know the molecular formula, would not know which structure is present

Structure Determination

Even if a pure sample is obtained, how do we know the actual structure of the compound?

The development and improvement of analytical instruments to determine structure has been one of the biggest advancements in organic chemistry during the past 60 years

Today almost any structure can be determined with these instruments

The important part is to recognize what information each instrument provides,

and if deciding between possible isomers which technique can be used to differentiate

Techniques to be learned:

Mass Spectrometry

UV-vis Spectroscopy

IR Spectroscopy

NMR Spectroscopy

-mass of compound

-isotopes present

-distinguish some atoms

-conjugation present

-functional groups

-bond connectivity of structure

-symmetry

-most important for

structure determination

Mass Spectrometry

Can determine the molecular weight of a sample and some information about the structure

A key part of a mass spectrometry is the need to create a charged species

The most common method to create the charged species is electron impact ionization

C C C

H

H

H

H

H

H

H

H

C C C

H

H

H

H

H

H

H

H

e

e

e

An electron is accelerated toward a gaseous sample of the compound under consideration

Due to the high energy of the electron moving at high speed, an electron is expelled

The sample thus is now positively charged, due to the loss of one electron,

and is a radical/cation structure

Mass Spectrometry

R

R(•+)

magnet

The compound (R) thus becomes a radical/cation when bombarded with electrons

The radical/cation continues along a path until it reaches a magnetic field

Charged species become deflected (are attracted to one magnet) in the presence of the magnetic field and hence the path direction is bent

The radius of curvature is dependent upon the mass of the species (m/z),

lighter mass species are deflected more and heavier species are deflected less

Only a certain mass can thus deflect the correct amount with the curvature of the instrument, heavier species will hit one wall while lighter species will hit the other wall

The magnet strength is changed and depending upon when species hit the detector the mass of the compound can be determined

e

detector

Mass Spectrometry

The parent ion is called the molecular ion peak (M+)

m/z 72

Can find molecular ion, but what are the other peaks?

The molecular ion peak can fragment

Due to the high energy of the radical/cation generated, this species can fragment

Remember only the charged species will be detected

(the radical species will not be affected by the magnetic field)

The probability of obtaining a given fragment is due to the

STABILITY of the cations produced

Mass Spectrometry

m/z 72

CH3CH2

m/z 43

CH3

m/z 57

Effect of Isotopes

Remember that an isotope has the same number of protons and electrons,

but a different number of neutrons

Since neutrons and protons are the “heavy” parts of an atom,

the extra number of neutrons will cause a greater mass

In a mass spectrometer we can see the effect of this by peaks above the molecular ion peak (M, M+1, M+2, etc.)

M

M+1

M+2

H

100%

C

98.9%

1.1%

S

95%

0.8%

4.2%

Cl

75.5%

24.5%

Br

50.5%

49.5%

I

100%

The ratio of these peaks is diagnostic for which atoms are present

The natural abundance of isotopes is well known

Can distinguish atoms by the ratio of peaks above the molecular ion

Especially useful to distinguish which halogen is present

Cl Br I

m/z 78

M/M+2 = 3

~ 3/1

m/z 122

M/M+2 = 1

M/M+2 = 1

m/z 170

Effect of Isotopes

Nitrogen

Nitrogen is also diagnostic in a mass spectrum due to the odd/even parity of the mass

Consider small molecules and their corresponding mass

CH4

m/z = 16

NH3

m/z = 17

The molecular ion peak for a molecule with one nitrogen is always odd,

all other common atoms in an organic compound yield an even mass

Mass Spectrometry

Fragmentation Behavior of Common Functional Groups

Alkenes

With an alkene the common fragmentation is to create an allylic carbocation

m/z 70 m/z 55

Alcohols

Two common effects

OHm/z 74 m/z 56

H2O

1) Loss of water

Alcohols

2) α-cleavage

m/z 74

OH OH OH

m/z 45

CH2CH3

McLafferty Rearrangement

Any ketone containing a γ-hydrogen can rearrange to the enol form in a MS

HO

m/z 100

O H

m/z 58

Ketones

Ketones can also do α-cleavage similar to alcohols

O

m/z 100

O OCH2CH2CH2CH3

m/z 43

High Resolution Mass Spectrometry (HRMS)

These high sensitivity mass spectrometers, called HRMS,

can be used to determine molecular formula

a HRMS can detect particle masses with an accuracy of 1/20,000

therefore > 0.0001 amu (atomic mass units)

Can use this to distinguish compounds with a similar rough mass

but with a different molecular formula

12C

12.0000 amu (by definition)

1H

1.0078 amu

16O

15.9949 amu

For example:

Many structures may have the same integer value molecular weight,

but different molecular formulas

O

C4H6O1

70.0418 amu

C5H10

70.0783 amu

HN NH

C3H6N2

70.0531 amu

Structure Determination Using Spectroscopy

Need methods to distinguish between possible structures

A nondestructive way is to use absorption spectroscopy

In a simplified picture:

The ability of the sample to absorb incident radiation is measured by the difference in absorbance at the detector versus the blank

Monochromatic light source

sample

blank

detector

Beam splitter

Electromagnetic Spectrum

All light travels at a constant speed

The difference is the wavelength of the light

(which also determines the energy of the light)

E = hν = (hc) / λ

UV-vis

IR

NMR

Infrared Region

Wavelength of infrared radiation is ~800 cm-1 to 4000 cm-1 wavenumbers

(wavenumbers correspond to number of wavelengths of light in 1 cm)

-common descriptor for IR frequencies by organic chemists

As the wavenumber becomes larger the energy increases

The energy level of infrared light corresponds to the energy required

to cause molecular vibrations

Depending upon what type of bond is present determines

the exact energy required to cause the vibration

The energy of light absorbed therefore indicates what functional group is present

Bond Vibration

The energy of the infrared light can interact

with the resonant vibrational frequency of the bond

Since different bonds have different energies,

they require different energy to cause vibration

H3C

O

CH3

consider acetone

The carbonyl has a strong dipole

H3C

O

CH3

When electric field aligns with dipole, bond shortens

E

The absorption of the infrared light thus changes the dipole for this bond as it vibrates

Active versus Inactive

IR only causes a vibration if there is a change in dipole during vibration

Therefore symmetric bonds are inactive

CH3-CH3

the carbon-carbon bond of ethane will not observe an IR stretch

Or any other symmetric bond

An IR “active” bond is therefore a bond that changes dipole during vibration,

While an IR “inactive” bond is a symmetric bond that doesn’t change dipole during vibration

Infrared Spectroscopy

Number of Vibrations

The number of possible vibrations for a given molecule

is determined by the number of atoms present

For nonlinear molecules obtain 3N-6 vibrations

(N equals number of atoms present)

3N-5 vibrations for linear molecule

For example consider acetone again (C3H6O1)

Acetone has 10 atoms and is nonlinear

Therefore expect 3(10)-6 = 24 vibrations

The other vibrations are due to different bonds besides the carbonyl stretching,

for example the hydrogens

Or bending motions

Intensity of Absorbance

Intensity of light absorbed by a molecule is related to the dipole of the bond

The greater the dipole, the greater the absorbance intensity

C-O bond stretches are therefore more intense than C-C stretches

Realize the intensity of absorbance is not related to the wavenumber

The wavenumber is related to the force constant for the bond vibrating

(the stiffness of the bond)

O

C O

Factors to be considered in an IR spectrum

1)  Position of absorbance (wavenumber)

Energy required for absorbance

2)  Intensity of absorbance

Related to the dipole of the bond

3)  Shape of absorbance

(broad or sharp peaks)

Tells information about the type of bond


Specific functional groups

As mentioned specific functional groups have characteristic absorbance frequencies

Consider carbon-carbon bonds

As the number of bonds increases between two atoms,

the stiffness of the bond increases which results in a harder bond to stretch


C C

Wavenumber (cm-1)

C C

C C

~1200 cm-1

~1660 cm-1

~2200 cm-1

Conjugation lowers the stretching frequency

(RESONANCE!!!)


Wavenumber (cm-1)

~1640-1680 cm-1

~1620-1640 cm-1

Whenever a functional group becomes more conjugated

(adjacent to double bonds for example) the stretching frequency lowers

C-H bond stretching

As the %s character increases in a bond, the bond becomes stiffer

(already saw that sp hybridized C-C bonds are stiffer than sp3 hybridized C-C bonds)

Same is true for carbon-hydrogen bonds

sp3 hybridized

2800-3000 cm-1

sp2 hybridized

3000-3100 cm-1

sp hybridized

~3300 cm-1

Key point: only sp3 hybridized C-H bond stretches are below 3000 cm-1


Alcohols and amines

Both O-H and N-H bonds are “stiff” bonds

Therefore they have a higher vibrational frequencies

Alcohol

RO-H

~3300 cm-1

Acid

RCO2-H

~3000 cm-1

Amine

RN-H

~3300 cm-1


In addition, both N-H and O-H bonds are involved in hydrogen bonding

therefore each bond will experience a slightly different vibrational frequency

Therefore this causes the appearance of a broad peak

Amine peaks show the same broad features

(N-H bonds are also involved in hydrogen bonding)

Difference is that often observe a sharp peak in the midst of the broad peak

(due to one conformation of hydrogen bonding having a preferential formation)


Carbonyl Compounds

One of the best diagnostic features of IR is for carbonyl compounds

Remember there are many types of carbonyl groups

(each can be differentiated only with an IR spectrum)

R

O

R R

O

H R

O

OH R

O

ORR

O

NH2 R

O

Cl

Ketone

Aldehyde

Amide

Acid

Ester

Acid chloride

ν (cm-1)

1700-1730

1700-1730

1700-1730

1620-1680

1735-1750

1770-1820

In addition to the carbonyl stretch, other characteristic peaks can distinguish carbonyl groups that display similar C=O stretching frequencies

2700-2800

Two peaks

3000

Broad peak

Carbonyl Compounds

Due to the large dipole of carbonyl bonds, all carbonyl groups display strong,

relatively sharp peaks

C=O

C=C

large dipole

small dipole

Most carbonyl stretching frequencies are centered around 1700-1730 cm-1 and can be distinguished easily from alkene stretches (~low 1600’s cm-1) due to both the higher

frequency and the more intense absorbance

Carbonyl Compounds

Some carbonyl stretching frequencies are noticeably different than 1700-1730 cm-1

Esters are one type

Esters have an appreciably higher stretching frequency

Higher frequency means a “stiffer” bond

O

O ν (cm-1) = 1742

Carbonyl Compounds

What causes a “stiffer” carbonyl bond?

Substituents on the carbonyl carbon can affect the C=O bond stretch in two ways:

Inductive effect

O

R Y

More electronegative Y pulls electron density from carbon, which then pulls electrons from

oxygen to create a stiffer bond

Resonance effect

O

R Y

Lone pair of electrons on Y atom can resonate to create a C=Y double bond and thus a C-O single bond – therefore a weaker C-O bond

O

R Y

The question is which effect is larger

Generally the greater difference in electronegativity between C and Y

causes inductive effect to become dominant

Y

ν (cm-1)

Stronger effect

Cl

1810

inductive

OR

1735

inductive

NH2

1660

resonance

Carbonyl Compounds

Amide group lowers the frequency due to the resonance effect

If a nitrogen is attached to the carbonyl carbon then the lone pair of electrons

on nitrogen can stabilize the resonance form

Due to this lower energy resonance form the carbonyl carbon-oxygen bond is less “stiff”,

therefore the stretching frequency is LOWER

O

NH2

O

NH2

Carbonyl Compounds

Resonance with extra conjugation will also lower the stretching frequency for a carbonyl

Resonance allows delocalization of π electrons,

therefore carbonyl is less “stiff”

O

H

O

H

O

H

O

H

ν = 1721 cm-1

ν = 1699 cm-1

Carbonyl Compounds

As already observed many carbonyl groups are ~1700-1730 cm-1

How to distinguish?

R

O

R

Ketone

~1715 cm-1 for carbonyl

R

O

H

Aldehyde

observe aldehyde C-H stretch

Two peaks between 2700-2900 cm-1

R

O

OH

Acid

observe broad O-H stretch

~3000 cm-1

Small Rings

Small rings also have a shift in vibrational frequency to higher energy,

Therefore 5,4, or 3-membered rings have the carbonyl stretching frequency shifted

O

1785 cm-1

O

1745 cm-1

O

1715 cm-1

Angle strain in these rings causes the carbonyl group to have more electron density,

Therefore a “stiffer” bond

C-N bonds

C-N bonds are in similar regions to C-C bonds

The intensity of absorbance, however is higher

due to greater dipole of C-N bond compared to C-C


Wavenumber (cm-1)

~1200 cm-1

~1600 cm-1

>2200 cm-1

C N

C N

C N

Fingerprint Region

The so-called “fingerprint” region is below ~1500 cm-1

Vibrations in this region are often complex and hard to assign

to a specific functional group of the molecule

-a given molecule, though, has a DISTINCT pattern in this region

(reason for this region being called the “fingerprint” region)

One common pattern – differentiating substitution isomers

One example:

Aromatic isomers

Ortho

one peak

770-735 cm-1

Meta

three peaks

900-860, 810-750, 725-680 cm-1

Para

one peak

860-800 cm-1

CH3Cl

CH3

Cl

CH3

Cl

Fingerprint Region

CH3Cl

CH3

Cl

CH3

Cl

Strong peak

747 cm-1

Strong peaks

863, 773, 682 cm-1

Strong peak

806 cm-1

Overtone and Combination Bands

Overtone

-when assigning IR spectra be careful to note overtone bands

(an intense peak will display a smaller peak at a multiple [2x, 3x, etc.] of that peak)

Combination Bands

Two or more vibrations can couple to cause a vibration at a different position

(vibrations must be “coupled” to observe these combination bands)

Strong carbonyl stretch

~1715 cm-1

2nd Overtone

~3430 cm-1

Ultraviolet-Visible Spectroscopy

(UV-vis)

Another analytical tool to determine organic structures is UV-vis spectroscopy

Similar to IR, this is another spectroscopic technique,

therefore a sample is irradiated with light of a particular wavelength

If the compound absorbs the light, the detector will record the intensity of absorbance

In an UV-vis spectrum the light used is between ~200 nm to ~700 nm

(UV range is ~200 nm to ~370 nm, while visible light is ~370 nm to 700 nm)

Remember that in IR the wavelength of light used was 2.5 – 25 µm

(between one to two orders of magnitude larger than UV light)

UV light is thus much higher in energy than IR light

Instead of causing molecular vibrations,

UV-vis light causes electronic excitations

An electron is excited from the HOMO to the LUMO

E h!

Ethylene HOMO

Ethylene LUMO

If the correct amount of energy is applied (i.e. the correct wavelength of light),

the excitation of one electron from the HOMO to the LUMO will occur


Required energy

The amount of energy required is thus the energy gap between the HOMO and LUMO

As the HOMO-LUMO gap changes, the wavelength required for excitation changes

(remember that a lower wavelength is higher in energy)

The HOMO-LUMO energy gap is affected by the amount of conjugation

A conjugated diene system has a lower HOMO-LUMO energy gap

than an isolated double bond

Therefore a more conjugated system has a higher wavelength of absorbance

(higher wavelength is lower in energy, therefore smaller energy gap)


Information Obtained from a UV-vis

The major piece of information is the point of maximum absorbance

(called λmax)

In addition an absorbance is characterized by how strongly the molecule absorbs

Beer’s law:

A = ε•c•l

c = concentration of sample

l = path length of sample

ε = molar absorbtivity (extinction coefficient)

*characteristic of sample

If the molecule absorbs more strongly, it has a higher ε


As seen earlier, the position of the λmax indicates the HOMO-LUMO energy gap

A more conjugated system has a lower HOMO-LUMO energy gap,

therefore the λmax will be of a longer wavelength

Only conjugated alkenes will cause a shift in λmax

As the conjugation increases, the shift will increase

Compounds that are colored to our eye therefore must be very conjugated

(the lowest wavelength human eyes detect is ~370 nm)


Alkyl substitution causes a shift of ~5 nm, but conjugation causes shift of ~30 nm

λmax

171 nm

180 nm

227 nm


Organic colored compounds:

β-carotene (λmax = 453 and 483 nm)

OBr

OBr

O

BrBrCO2

Eosin Y (λmax = 517 nm)

structure determination how to determine what …biewerm/9h-analytical.pdfstructure determination!...

Documents