ir spectroscopy - university of york€¦ · web view1) predict the ir, nmr and mass spectra of...

A-level Spectroscopy

An image of a human brain from a live patient recorded using magnetic resonance imaging - a 3D form of n.m.r. spectroscopy

IntroductionSpectroscopy is a collective name for the various techniques that use the interaction between molecules and electromagnetic radiation to elucidate

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the structure of molecules. Spectroscopic methods are fundamental to the study of Chemistry, Molecular Biology, Medicine and Astrophysics.

This booklet covers the following techniques:-

A)Infrared Spectroscopy - from ‘What’s in a Medicine?’

Describe how infrared spectroscopy (i.r.) can be used for the elucidation of molecular structure;

Interpret infrared spectra for salicylic acid and simple compounds containing a limited range of functional groups (hydroxyl, carbonyl, carboxylic acid and ester) given relevant information.

B)Mass Spectrometry - from ‘What’s in a Medicine?’

Describe how mass spectrometry (m.s.) can be used for the elucidation of molecular structure;

Interpret mass spectra (molecular ion and significance of the fragmentation pattern) for salicylic acid and simple compounds containing a limited range of functional groups (hydroxyl, carbonyl, carboxylic acid and ester) given relevant information.

C)Nuclear Magnetic Spectroscopy - from ‘Engineering Proteins’

Describe how nuclear magnetic resonance spectroscopy (n.m.r.) can be used for the elucidation of molecular structure;

Interpret nuclear magnetic resonance spectra for simple compounds given relevant information (reference to splitting on the resonances is not required)

This work builds on AS topics of:-

Interaction of radiation with matter (‘Elements of Life’ and ‘Atmosphere’); Mass spectrometry (‘Elements of Life’).

A) Infrared (i.r.) spectroscopy

Used to identify bonds / functional groups Can only identify the exact molecule by comparison with

library spectra

Experiment 1 RSC Video

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c.f.springoscillations

m

Infrared radiation is passed simultaneously through the sample and a reference cell.

The reference ensures that peaks due to water or carbon dioxide in the air can be cancelled out.

The frequencies of i.r. radiation absorbed are determined by passing through a rotating prism to focus one frequency at a time onto the detector.

The spectrum shows the ________________ (cm-1) on the x axis (which is 1/) and the ____________________ on the y-axis.

Calculations

1) c = 2) Wavenumber = 1/(cm)

e.g. What wavenumber would appear on an i.r. spectrum if the frequency of radiation absorbed by a molecule was 2.5 x 1013 Hz?

Theory

IR radiation corresponds to the energy required to make chemical bonds vibrate more / move to a higher vibrational energy level.

Therefore, energy of certain wavelengths is absorbed by molecules. The actual energy depends on the mass of the atoms and the

strength of the bond, so different bonds will absorb at different frequencies.

Stronger bonds need more energy to make them vibrate, so absorb a higher frequency of i.r. radiation (higher wavenumber)

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e.g. hydrogen halides

Molecules with more than 2 atoms can vibrate in different wayse.g. sulphur dioxide

So these spectra will contain more absorptions

Most organic molecules contain a number of types of bond, so characteristic absorptions will be seen for each bond.

e.g. ethanol

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The following types of bond need to be recognised:-

Bond Functional group Absorbance (cm-1)O – H Alcohols 3200 – 3600 / strong and broad*

O – H Carboxylic acids 2500 – 3200 / medium and very broad*

C=O Aldehydes / ketones / carboxylic acids/ esters

1680 – 1750 / strong and sharp

C-O Alcohols / esters / ethers 1050 – 1300 / medium

C-H Alkanes / alkenes etc 2850 – 3100 / medium

*Broad due to Hydrogen Bonding between O-H groups

i.r. bands.ppt

Examples of infrared spectra

1) ethanol (CH3CH2OH)

displayed formula

i.r. spectrum

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Bond / (Functional group) Absorption / cm-1

2) ethanoic acid (CH3COOH)

displayed formula

i.r. spectrum

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3) Ethyl Ethanoate (CH3COOCH2CH3)

i.r. spectrum

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H C C CC

H

O

O H

H

H

HH

H


1750

1250

3000

4) a) i.r. spectrum of an alcohol with molecular formula C3H8O.

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NB: This Alcohol is oxidised to compound 4)b) when heated under distillation with acidified potassium dichromate and 4)c) when heated to reflux with acidified potassium dichromate.

Clue?


Displayed Formula of 4a

4) b) i.r. spectrum of the compound with molecular formula C3H6O obtained by distilling compound 4)a) with acidified potassium dichromate

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Displayed Formula of compound 4b

4)c) i.r. spectrum of the compound with molecular formula C3H6O2 formed when compound 4a is heated to reflux with acidified potassium dichromate

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Displayed Formula of Compound 4c

5)a) i.r. spectrum of an isomer of 4a which forms the same product 5)b) whether it is heated to distil or reflux with acidified potassium dichromate

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Displayed Formula of Compound 5a

5)b) i.r. spectrum of the product of the reaction of 5a with acidified potassium dichromate when heated to reflux or distillation.

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Displayed Formula of Compound 5b

6) Salicylic Acid (2-hydroxybenzoic acid - (HOC6H4COOH))

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displayed formula

i.r. spectrum


7) Aspirin (CH3COOC6H4COOH)

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i.r. spectrum

Bond / (Functional group) Absorption / cm-12900 v. broad

1750

1700

1200

B) Mass Spectrometry

Use M+ (molecular ion) to measure Mr Use M+2 isotope peaks to identify Cl or Br Use fragmentation pattern to confirm structure of molecule

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Experiment 2 – RSC video

AS-level

Vaporisation of atoms or molecules; Ionisation of atoms or molecules; Acceleration of ions; Deflection of ions; Detection of ions.

A2-level

The atoms or molecules are ionised by bombarding with high energy electrons:-

e.g. CH3COCH3 + e- [CH3COCH3] + + 2 e-

M+

Usually, the resulting molecular ion has such high energy that it splits up into a smaller ion and an uncharged molecule (fragmentation)

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e.g. [CH3COCH3] + [CH3CO] + + CH3

M+m/e = 58 43

or [CH3COCH3] + CH3CO + [CH3] +58 15

NB The first fragmentation route is more likely because fragments containing the [R-C=O] + group (acylium cations) are particularly stable.

The following peaks are often seen in the fragmentation patterns of mass spectra – the highlighted peaks usually provide very useful clues in determining the structure of a molecule

fragment m/eCH3 15CH3CH2 or CHO 29CH2NH2 30CH2OH 31CH3CO or C3H7 43CONH2 44COOH 45C6H5 77C6H5CH2 91C6H5CO 105

Examples of fragmentation and the interpretation of mass spectra

1) Propanone (CH3COCH3)

displayed formula

mass spectrum

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m/z Formula m/z lost Group lost

58

43

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2) Propanal (CH3CH2CHO)

displayed formula

mass spectrum

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58

57

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3) Methyl Benzoate (C6H5COOCH3)

mass spectrum

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C C H

H

O

O

H


136

105

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4) Ethyl Ethanoate (CH3COOCH2CH3)

mass spectrum

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H C C CC

H

O

O H

H

H

HH

H


88

73

43

29

15


displayed formula

mass spectrum

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138

120*

92

* NB 3- or 4- hydroxybenzoic acid isomers cannot eliminate water –why not?


mass spectrum

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180

138

120

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7) ethanamide (CH3CONH2)

displayed formula

mass spectrum

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59

44

43

8) paracetamol (4-hydroxyphenylethanamide) (HOC6H4NHCOCH3)

displayed formula

mass spectrum

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151

109

108

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C) Nuclear Magnetic Resonance (n.m.r.) spectroscopy

The number of peaks – number of proton types The chemical shift (δ) – what are the proton types The integration – how many protons of each type

Experiment 3 – RSC video

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Sample is placed in a very strong magnetic field A pulse of radiofrequency radiation is applied Radiofrequency signal emitted from sample is detected

Theory Nuclei have a property called nuclear spin which generates a tiny

magnetic field. The nuclei therefore behave like tiny bar magnets.

When such nuclei are placed in a large magnetic field they will become aligned with or against the direction of the external field.

The nuclei lined up with the field are slightly more stable (lower energy) than those that oppose the external field.

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The energy gap between these two states corresponds to radiofrequency radiation.

If the sample is irradiated with a pulse of radio waves, the nuclei in the lower energy state may be promoted to the higher energy state (the tiny bar magnets ‘flip’ from being aligned with to against the external field).

The excited nuclei will then return to the ground state releasing fixed quanta of energy which will be detected.

The energy gap depends on the chemical environment of the nuclei and can be used to deduce the exact structure of the molecule.

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Ethanal has two proton types, so produces 2 signals in the n.m.r. spectrum.

The important features of the spectrum are:- The number of peaks – number of proton types The integration – how many protons of each type The chemical shift (δ) – what are the proton types

The following table can be used to link the chemical shift to the proton type (chemical environment of H atom):-

type of proton chemical shift δ / ppmRCH3 / RCH2R (alkane) 0.8 - 1.4RCOCH3 (carbonyls, esters) 1.8 - 2.2RCH2Hal 3.2 - 4.6ROCH3 (esters, ethers) 3.2 - 3.5ROH (alcohol) 1.0 - 6.0RC6H4H (arenes) 6.0 - 9.0RC6H4CH3 (methylarene) 2.2 - 2.4RCONHR (amides) 7.0 - 10.0RCHO (aldehydes) 9.7 - 9.8RCOOH (carboxylic acids) 9.0 - 12.0RC6H4OH (phenols) variableRNHR (amines) variable

[R represents an alkyl group]

Examples of the interpretation of n.m.r spectra1) propanone (CH3COCH3)

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displayed formula

nmr spectrum

Proton integration inference δ / ppm inference

Ha

2) ethanoic acid (CH3COOH)

nmr spectrum

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Ha 11.4

Hb 2.1

3) propanal (CH3CH2CHO)

displayed formula

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nmr spectrum


Ha 9.7

Hb 2.4

Hc 1.1

High Resolution nmr Spectra

Most nmr spectra look more complicated than the first three examples.

The signal for each hydrogen atom may be split into a number of peaks.

The pattern of the splitting tells us how many hydrogen atoms are bonded to the adjacent carbon atom.

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The n+1 rule The no. peaks = the no. H atoms on the adjacent carbon + 1

e.g. ethanal

The Ha protons have one adjacent H atom (Hb) The signal will be split into ____ peaks – a doublet.

The Hb proton has three adjacent H atoms (Ha) The signal will be split into ____ peaks – a quartet.

High resolution spectra may be analysed as follows:-

The number of peaks – number of proton types The integration – how many protons of each type The chemical shift (δ) – what are the proton types The splitting pattern – the number of H atoms on the adjacent C

atom

NB This level of analysis is not required for the A-level examinations, but the nmr spectra shown are usually high resolution and the information from splitting patterns is extremely useful in working out the structure of complex molecules.

Why are the signals split by adjacent protons?

Each H nucleus generates its own tiny magnetic field, which may be aligned with or against the external magnetic field. This will affect the magnetic environment experienced by H nuclei bonded to adjacent C atoms.

e.g. ethanal

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HaC C

O Ha

Hb Ha

HaC C

O Ha

Hb Ha

The Ha protons have one adjacent H atom (Hb) Hb may be aligned with or against the field

This means that there are two possible environments for the Ha protons, of equal probability.

The signal for Ha will be split into 2 peaks – a 1:1 doublet.

The situation is slightly more complicated for the Hb proton, which has three adjacent H atoms (Ha).

Each of the three Ha protons may be aligned with or against the field.

This means that there are four possible orientations of the Ha nuclei:-

a. All nuclei are aligned with the field (1);b. Two nuclei with and one against the field (3);c. One nucleus with and two nuclei against the field (3);d. All nuclei aligned against the external magnetic field (1).

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1 : 1

HaC C

O Ha

Hb Ha

This means that there are four possible environments for the Hb proton, with a relative probability of 1:3:3:1

The signal will be split into 4 peaks – a 1:3:3:1 quartet.

4) Ethanal (CH3CHO)

High Resolution nmr Spectrum of ethanal

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/ ppm10 8 6 4 2 0

Ha

Hb 1

3

HaC C

O Ha

Hb Ha

1 : 3 : 3 : 1

Proton

integration

inference δ / ppm inference

splitting inference

Ha 2.1

Hb 9.7

5) ethyl ethanoate (CH3COOCH2CH3)

nmr spectrum

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Ha C C CC

Hb

O

O Hc

Hc

Ha

HbHa

Hc

Proton

integration


splitting inference

Ha 2.1

Hb 4.1

Hc 1.2

6) propan-2-ol (CH3CH(OH)CH3)

nmr spectrum

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Hc C C C

Hc

Hc

HbHc

Hc

Hc O

Ha

Proton

integration


splitting inference

Ha 2.1 singlet

Hb 3.9 septet

Hc 1.2 doublet


displayed formula

nmr spectrum

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Ha 8.0

Hb 7.6

Hc 7.0

Where are the O-H groups?


nmr spectrum

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HH

HH


Ha 1 11.3

Hb 4 x 1 7 - 8

Hc 3 2.1

9) Mystery compound – “Why are there no aspirin in the jungle?”

n.m.r. spectrum

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Ha 1 9.7

Hb 1 9.1

Hc 2 7.4

Hd 2 6.7

He 3 2.0

Structure

Combined Spectral Techniques

1) Predict the ir, nmr and mass spectra of propanoic acid

a) IR spectroscopy

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b) Nmr spectroscopy

Proton

integration


splitting inference

Ha

Hb

Hc

c) Mass Spectrometry


2) Deduce the structure of the molecule from these spectra

a. ir spectrum

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b) nmr spectrum

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Proton

integration


splitting inference

Ha 2.2 singlet

Hb 3.6 triplet

Hc 1.5overlappi

ngquartet

of triplets

Hd 0.9 triplet

c) mass spectrum

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60

31

44

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Structure of Unknown Molecule

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ir spectroscopy - university of york€¦ · web view1) predict the ir, nmr and mass spectra of...

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