11.3 Spectroscopic Identification of Organic compounds - PowerPoint Presentation

Slide1

11.3 Spectroscopic Identification of Organic compounds

IHD

IR Spectroscopy

Mass Spectroscopy

NMRSlide2

Chapter 11.3-21.1:Spectroscopic identification of organic compounds

11.3: Analytical techniques can be used to determine the structure of a compound, analyze the composition of a substance, or determine the purity of a compound. Spectroscopic techniques are used in the structural identification of organic and inorganic compounds.

20.1: Although spectroscopic characterization techniques form the backbone of structural identification of compounds, typically no one techniques results in a full structural identification of a moleculeSlide3

Analytical techniquesQualitative analysis: detection of presence but not quantity of substance in mixture

Quantitative analysis: measurement of the quantity of a particular substance in a mixture

Structural analysis: description of how atoms are arranged in molecular structuresSlide4

Index of Hydrogen Deficiency Learning outcomes

Understand how the number of double bonds/rings can be worked out from molecular formula Slide5

Index of Hydrogen Deficiency Alkanes have maximum number of hydrogen atoms

For every two hydrogen atoms fewer than in alkane with same number of carbon atoms, there is one double bond or ring present. (double bond equivalent)

The number of double bond equivalents sometimes called degree of unsaturation or the Index of hydrogen deficiency. Slide6

How the index of hydrogen deficiency works.

1

 A double bond and ring each counts as one IHD.

2 A triple bond counts as two IHD

Hydrocarbons (

C

x

H

y

):

IHD=1/2[2C+2-H]

(where x = number of carbon, Y= number of hydrogen) Example C4H8 ½(8+2-8)= 1Slide7

Compounds Containing Elements Other than C and H

O and S atoms do not affect the IHD.

Halogens (F,

Cl

, Br, I) are treated like H atoms (CH

2

Cl

2

 has the same IHD as CH4).

For each N, add one to the number of C and one to the number H (CH

5

N is treated as C2H6. CH4N2O is treated as C3H6 by adding 2 to # of C and 2 to # of H).Slide8

Calculate IHD forC

3

H

5

N will be treated as C

4

H

6

IHD = 1/2(4x2 + 2- (

6

) = 4/2=2

Possible structuresSlide9

Calculate IHD forC6

H9Cl will be treated

like

C

6

H

10

IHD=1/2[2C+2-H]

IHD =1/2[12+2-10]=4/2=2Slide10

Practice problems

CH3CHCHCH2CHCH2

CH3C≡

CCOCH3

I

HD

=

3

IHD

=

2

IHD

=

5

IHD

=

1

IHD

= 3Slide11

Learning outcomes

Understanding

Mass

spectrometry (MS), proton nuclear magnetic resonance spectroscopy (1H NMR) and infrared spectroscopy (IR) are techniques that can be used to help identify compounds and to determine their structure.Slide12

Learning outcomes Deduction of information about the structural features of a compound from percentage composition data, MS, 1H NMR or IR.Slide13
Slide14

Mass spectrometer

Picture Source http://chemistry.umeche.maine.edu/CHY251/Ch13-Overhead4.htmlSlide15

Using a mass spectrum to find relative formula mass

The formation of molecular ions

When

the

vaporized

organic sample passes into the

ionization

chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion.Slide16

The molecular ion is often given the symbol M+ 

or M•

-

the dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the

ionization

process.

The molecular ions tend to be unstable and some of them break into smaller fragments. These fragments produce the familiar stick diagram. Fragmentation is

irrelevant at this stage. Slide17

Using the molecular ion to find the relative formula mass

In the mass spectrum, the heaviest ion (the one with the greatest m/z value) is likely to be the molecular

ion.

For example, in the mass spectrum of pentane, the heaviest ion has an m/z value of 72.Slide18

Analyzing mass spectrometer dataC3H

8Slide19

MS data of N,N-diethylmethylamineSlide20

WHAT IS AN INFRA-RED SPECTRUM?

If a range of infra-red frequencies shine at an organic sample, some of the frequencies get absorbed by the compound. A detector attached to the other end of spectrum measured frequencies absorbed or transmitted.

How much of a particular frequency gets through the compound is measured as 

percentage transmittance.Slide21

What an infra-red spectrum looks like

A graph is produced showing how the percentage transmittance varies with the frequency of the infra-red radiation.Slide22

Analyzing IR spectrum IR spectra are not particularly easy to

analyse

, nor do they give definitive information about structure. There are however, two different stages in an analysis.

1 Identification of absorptions

2 Fingerprinting

The first stage involves looking for characteristic absorptions and attempting to ascribe them to specific structural features. The second stage is usually carried out after a series of analyses leads to a possible conclusion.Slide23

IDENTIFICATION OF PARTICULAR BONDS

IN A MOLECULE

INFRA RED SPECTRA -

USES

The presence of bonds such as O-H and C=O within a molecule can be confirmed because they have characteristic peaks in identifiable parts of the spectrum.

IDENTIFICATION OF COMPOUNDS BY DIRECT COMPARISON OF SPECTRA

The only way to completely identify a compound using IR is to compare its spectrum with a known sample. The part of the spectrum known as the ‘Fingerprint Region’ is unique to each compound.Slide24

Infra-red spectra are complex due to the many vibrations in each molecule.

Total characterisation of a substance based only on its IR spectrum is almost impossible unless one has computerised data handling facilities for comparison of the obtained spectrum with one in memory.

However, the technique is useful when used in conjunction with other methods such as nuclear magnetic resonance (nmr) spectroscopy and mass spectroscopy.

Peak position depends on

bond strength

masses of the atoms joined by the bond

strong bonds

and

light atoms

absorb at

lower wavenumbers

weak bonds

and

heavy atoms

absorb at

high wavenumbers

INFRA RED SPECTRA -

INTERPRETATIONSlide25

Vertical axis Absorbance the stronger the absorbance the larger the peak

Horizontal axis Frequency wavenumber (waves per centimetre) / cm

-1

Wavelength microns (m); 1 micron = 1000 nanometres

INFRA RED SPECTRA -

INTERPRETATIONSlide26

FINGERPRINT REGION

• organic molecules have a lot of C-C and C-H bonds within their structure

• spectra obtained will have peaks in the 1400 cm

-1

to 800 cm

-1

range

• this is referred to as the

“fingerprint”

region

• the pattern obtained is characteristic of a particular compound the frequency

of any absorption is also affected by adjoining atoms or groups.Slide27

IR SPECTRUM OF A CARBONYL COMPOUND

• carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm

-1

• this is due to the presence of the C=O bondSlide28

IR SPECTRUM OF AN ALCOHOL

• alcohols show a broad absorption between 3200 and 3600 cm

-1

• this is due to the presence of the O-H bondSlide29

IR SPECTRUM OF A CARBOXYLIC ACID

• carboxylic acids show a broad absorption between 3200 and 3600 cm

-1

• this is due to the presence of the O-H bond

• they also show a strong absorption around 1700 cm

-1

• this is due to the presence of the C=O bondSlide30

IR SPECTRUM OF AN ESTER

• esters show a strong absorption between 1750 cm

-1

and 1730 cm

-1

• this is due to the presence of the C=O bondSlide31

WHAT IS IT!

O-H STRETCH

C=O STRETCH

O-H STRETCH

C=O STRETCH

AND

ALCOHOL

ALDEHYDE

CARBOXYLIC ACID

One can tell the difference between alcohols, aldehydes and carboxylic acids by comparison of their spectra.Slide32

O-H

C=O

C-O

N-H

Aromatic C-C

C-H

C=C

C-C alkanes

C



N

C-Cl

CHARACTERISTIC FREQUENCIESSlide33

Bond Class of compound Range / cm

-1

Intensity

C-H Alkane 2965 - 2850 strong

C-C Alkane 1200 - 700 weak

C=C Alkene 1680 - 1620 variable

C=O Ketone 1725 - 1705 strong

Aldehyde 1740 - 1720 strong

Carboxylic acid 1725 - 1700 strong

Ester 1750 - 1730 strong

Amide 1700 - 1630 strong

C-O Alcohol, ester, acid, ether 1300 - 1000 strong

O-H Alcohol (monomer) 3650 - 3590 variable, sharp

Alcohol (H-bonded) 3420 - 3200 strong, broad

Carboxylic acid (H-bonded) 3300 - 3250 variable, broad

N-H Amine, Amide 3500 (approx) medium

C



N Nitrile 2260 - 2240 medium

C-X Chloride 800 - 600 strong

Bromide 600 - 500 strong

Iodide 500 (approx) strong

CHARACTERISTIC ABSORPTION FREQUENCIESSlide34

H-NMRNuclear Magnetic Resonance Slide35

NMR Nuclear magnetic resonance relies on the magnetic field produced by a spinning nucleus containing an odd number of nucleons (protons or neutrons). In the presence of an external magnetic field the nucleus can exhibit more than one spin state and can move between these states by the absorption of electromagnetic radiation of a specific frequency (energy).

The energy absorbed can be detected and from this information about the location (environment) of the nucleus can be deduced.Slide36
Slide37

https://www.youtube.com/watch?v=k0eR8YqcA8c

https://

www.youtube.com/watch?v=OrvAUDgVoT8&feature=iv&src_vid=k0eR8YqcA8c&annotation_id=annotation_877622Slide38

NMR is probably the most useful tool in the organic chemists arsenal for structural determination.As organisms are mainly water (containing H atoms with an odd number of nucleons), NMR has developed into an invaluable medical

diagnostics

tool, called an MRI (

magnetic

resonance instrument) scan

.Slide39

Nuclear magnetic resonance

This tells us the number of hydrogen atoms in different environments within the molecule. As hydrogen is present in (almost) all organic compounds this technique is very useful. The pattern produced by the

hydrogen atoms

is often split into finer structure, that also gives information about the number of hydrogen adjacent to the absorbing atoms.Slide40

Nuclear energy levels

Nuclei with odd numbers of nucleons (protons and neutrons) can have different energy levels. To help us differentiate these energy levels we say that they have 'spin'.

A hydrogen nucleus can have two different 'spin' states. These are designated as spin = +1/2 and spin = -1/2.

For a hydrogen nucleus to change spin states it must absorb energy (promotion) or lose energy (relaxation).

Animation Slide41

Low resolution NMRA low resolution spectrum looks much simpler because it can't distinguish between the individual peaks in the various groups of peaks

The numbers against the peaks represent the relative areas under each peak. That information is extremely important in interpreting the spectraSlide42

Interpreting L R NMR spectrum

Different sizes of peak give valuable information

Are underneath a peak is proportional to number of hydrogen atoms in that environment.

Area underneath peaks can be worked out by integration trace. The vertical heights of the steps in Integration trace are proportional to the number of hydrogen atoms in each

envirnoment

.

Three peaks in the spectrum corresponds to different chemical environments of H atomsSlide43

Interpreting L R NMR spectrum

The Chemical Shift gives information about the environment of protons( Hydrogen atoms). The protons in different chemical environment give different chemical shift

Detail about chemical shift will be covered in HL syllabus.

Chemical

Shift

: The Horizontal scale on NMR, is given by the symbol

δ

has a unit parts per million ppm. Slide44

NMR Spectrum of Pentan-3-oneSymmetrical molecule

Two peaks show two different chemical environment

Heights of peaks as ratio of 2;3 in integration trace, show four H atom in one environment and 6 in other. Slide45

Identify number of different chemical environments and ratio of H atoms in each environment

3

2

3Slide46

Chemical Shift ( HL only)The horizontal scale on a nuclear magnetic resonance spectrum is called chemical shift.

The symbol for chemical shift is

δ

.

It is measured as parts per million

The Chemical Shift gives information about the environment of protons( Hydrogen atoms). The protons in different chemical environment give different chemical shift

Chemical shift are measured relative to TMS

Chemical shift for TMS is ZeroSlide47

Tetramethylsilane is the standard

All

Hs

are the same = 1 signal

Si

Has lower EN than Carbon

Si absorbs in a different part of the spectrum than C when bonded to H

Si(CH

3

)

4

Has low boiling point

Is chemically inert (non-reactive)

Is soluble in most organic solventsSlide48

Chemical Shift values relative to TMS

Values are given in data book let page 26, table 27Slide49

Using Chemical ShiftSlide50

High-resolution 1H NMR

Can show the difference in the spins of nuclei

Right: Hi-res

1

H NMR

Below:

1

H NMR Slide51

What a low resolution NMR spectrum tells youLow Resolution

High Resolution

Remember

:

The number of peaks tells you the number of different environments the hydrogen atoms are in.

The ratio of the areas under the peaks tells you the ratio of the numbers of hydrogen atoms in each of these environments.

The chemical shifts give you important information about the sort of environment the hydrogen atoms are in.

In a high resolution spectrum,

the

low resolution spectrum are split into clusters of peaks

.

1

peak

a

singlet

2 peaks in the

cluster a doublet3 peaks in the

cluster

a

triplet

4 peaks in the

cluster

a

quartet

T

he

amount of splitting of the peaks gives you important extra information.Slide52

Interpreting a high resolution spectrum

The n+1

rule

The amount of splitting tells you about the number of

hydrogens

attached to the carbon atom or atoms

next door

to the one you are currently interested in

.

The number of sub-peaks in a cluster is one more than the number of

hydrogens

attached to the next door carbon(s).Singlet next door to carbon with no hydrogens attacheddoublet next door to a CH grouptriplet next door to a CH2 groupquartet next door to a CH3 groupSlide53

Multiplicity (Splitting)

NEIGHBOR

hydrogens

= number of peaks -1Slide54

Multiplicity (Splitting)

3 peaks (triplet)

2 neighbor H’s

(probably CH

2

)

6 peaks (hextet)

5 H’s

(CH

3

& CH

2

)

3 peaks (triplet)

2 neighbor H’s

(CH

2

)Slide55
Slide56

Practice Example 11.6 page 540