slide1
Download
Skip this Video
Download Presentation
Structural Determination of Organic Compounds

Loading in 2 Seconds...

play fullscreen
1 / 107

Structural Determination of Organic Compounds - PowerPoint PPT Presentation


  • 196 Views
  • Uploaded on

Chapter 38. Structural Determination of Organic Compounds. 38.1 Introduction 38.2 Isolation and Purification of Organic Compounds 38.3 Qualitative Analysis of Elements in an Organic Compound 38.4 Determination of Empirical and Molecular Formulae from Analytical Data

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Structural Determination of Organic Compounds' - zudora


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Chapter 38

Structural Determination of Organic Compounds

38.1Introduction

38.2Isolation and Purification of Organic Compounds

38.3Qualitative Analysis of Elements in an Organic Compound

38.4Determination of Empirical and Molecular Formulae from Analytical Data

38.5Chemical Tests for Functional Groups

38.6Use of Infra-red Spectroscopy in the Identification of Functional Groups

slide2

38.1 Introduction (SB p.137)

The determination of the structure of an organic compound involves four steps:

  • Isolation and purification of the compound
  • Qualitative analysis of the elements present in the compound
  • Determination of the molecular formula of the compound
  • Determination of the functional group present in the compound
slide3

38.2 Isolation and Purification of Organic Compounds (SB p.138)

Filtration

  • The solid/ liquid mixture to be filtered is poured onto the folded filter paper in the filter funnel
  • Liquids are allowed to pass through the filter paper (known as filtrate) while insoluble solids (known as residue) are retained on the filter paper
slide4

38.2 Isolation and Purification of Organic Compounds (SB p.139)

Centrifugation

  • The tubes containing undissolved solids are spun around very rapidly and are thrown outwards in the centrifuge
  • The denser solid collects as a lump at the bottom of the tube with the clear liquid above it
  • The liquid can be removed by decantation
slide5

38.2 Isolation and Purification of Organic Compounds (SB p.139)

Crystallization

Crystallization by Making and Cooling a Hot Concentrated Solution

  • The solubility of most solids increase with a rise of temperature
  • As a hot concentrated solution cools, it cannot hold all of its dissolve solute. The excess solute separates out as crystals
slide6

38.2 Isolation and Purification of Organic Compounds (SB p.140)

Crystallization by Evaporating a Solution at Room Temperature

  • When the solution becomes saturated after evaporation, further evaporation causes crystallization to occur
  • The crystallization process will be slow if it occurs at room temperature
slide7

38.2 Isolation and Purification of Organic Compounds (SB p.141)

Solvent Extraction

  • To dissolve out a component from a mixture with a suitable solvent
  • An aqueous solution containing the organic product is usually shaken with diethyl ether in a separating funnel. The ether layer is then run off
  • Repeated extraction with fresh ether to extract any organic products remaining
  • The ether portions are combined and dried, and distilled away to obtain the organic products
slide8

38.2 Isolation and Purification of Organic Compounds (SB p.141)

Simple Distillation

  • To separate a liquid from a solution of a liquid and a non-volatile solid or liquid
  • Only the liquid vapourizes to form vapours which condense to liquid on cold surface which collected as distillate
  • Anti-bumping granules are added to prevent bumping
slide9

38.2 Isolation and Purification of Organic Compounds (SB p.142)

Fractional Distillation

  • To separate a liquid from a mixture of two or more miscible liquids which have large difference in boiling point
  • Fractionating column provides a large surface area for condensation and vapourization of the mixture to occur
slide10

38.2 Isolation and Purification of Organic Compounds (SB p.143)

Steam Distillation

  • Many liquid or even solid compounds which are immiscible with water can be purified by distillation in a current of steam
slide11

38.2 Isolation and Purification of Organic Compounds (SB p.143)

  • Principle of steam distillation:
  • the total vapour pressure of an immiscible mixture is equal to the sum of the vapour pressures of individual components the water and the compound will distil at a lower temperature than the boiling point of either one of them
slide12

38.2 Isolation and Purification of Organic Compounds (SB p.144)

Sublimation

  • Direct change of a solid to vapour on heating or vapour to solid on cooling
  • A mixture of two compounds is heated in an evaporating dish, one compound sublimes. The vapour changes back to solid on a cool surface while the other compound is not affected and remains in the evaporating dish
slide13

38.2 Isolation and Purification of Organic Compounds (SB p.144)

Chromatography

  • To separate a complex mixture of substances
  • As the various components are being absorbed or partitioned at different rates, they move upwards to different extent
  • The ratio of the distance travelled by the substance to the distance travelled by the solvent is known as Rf value, which is the characteristic of the substance
slide14

38.2 Isolation and Purification of Organic Compounds (SB p.146)

A summary of different techniques of isolation and purification

slide15

38.2 Isolation and Purification of Organic Compounds (SB p.146)

Tests for Purity

Determination of Melting Point

  • Some of the dry solid is placed in a thin-walled glass melting point tube. The tube is attached to a thermometer
  • The temperature at which the solid melts is its melting point
  • Pure solid has a sharp melting point
  • Impure solid melts gradually over a wide temperature range
slide16

38.2 Isolation and Purification of Organic Compounds (SB p.147)

Determination of Boiling Point

  • The boiling point of a liquid can be determined by using the simple distillation apparatus
  • The temperature at which the liquid boils steadily is its boiling point
  • The boiling point of a pure liquid is quite sharp, but varies with changes in external pressure
  • The presence of non-volatile solutes such as salts raises the boiling point of a liquid
slide17

38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.147)

Carbon and Hydrogen

  • Carbon and hydrogen can be detected by heating a small amount of the substance with Cu2O
  • Carbon and hydrogen would be oxidized to CO2 and H2O respectively
  • CO2 turns lime water milky, and H2O turns anhydrous CoCl2 paper pink
slide18

38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.147)

Halogens, Nitrogen and Sulphur

  • Sodium fusion test is performed to test for the presence of halogens, nitrogen and sulphur
  • The compound under test is fused with a small piece of sodium metal in a small combustion tube and then heated strongly
  • The product of the test are extracted with water and analyzed
  • During sodium fusion, halogens, nitrogens and sulphur in the organic compound are converted to sodium halide, sodium cyanide and sodium sulphide respectively
slide20

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.148)

  • Quantitative analysis is to find out the percentage composition by mass of the compound
  • Methods to determine the mass of the elements:
  • 1. Carbon and hydrogen
  • The organic compound is burnt in oxygen.
  • The carbon dioxide and water vapour formed are respectively absorbed by KOH solution and anhydrous CaCl2.
  • The increases in mass in KOH solution and CaCl2 represent the masses of carbon dioxide and water vapour formed respectively
slide21

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149)

  • 2. Nitrogen
  • The organic compound is heated with excess Cu2O
  • The NO and NO2 formed are passed over hot Cu and the volume of N2 formed is measured
slide22

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149)

  • 3. Halogens
  • The organic compound is heated with fuming HNO3 and excess AgNO3 solution
  • The mixture is allowed to cool and then water is added
  • The dry AgX formed is weighed
slide23

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149)

  • 4. Sulphur
  • The organic compound is heated with fuming HNO3
  • After cooling, barium nitrate solution is added
  • The dry barium sulphate formed is weighed
slide24

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149)

  • The empirical formula of the compound can be calculated after the determination of percentage composition by mass of a compound
  • The molecular formula can be calculated after knowing relative molecular mass and the empirical formula
slide25

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149)

  • The empirical formula of a compound is the formula which shows the simplest whole number ratio of the atoms present in the compound
  • The molecular formula of a compound is the formula which shows the actual numberof each kind of atoms present in a molecule of the compound
slide26

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149)

Example 38-1

On quantitative analysis, an organic compound was found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Calculate the empirical formula of the compound.(R.a.m.: H = 1.0, C = 12.0, O = 16.0)

Answer

slide27

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149)

Solution:

Let the mass of the compound be 100 g. Then,mass of C in the compound = 40.0 gmass of H in the compound = 6.7 gmass of O in the compound = 53.3 g

∴ The empirical formula of the organic compound is CH2O.

slide28

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150)

Example 38-2

An organic compound Z has the following composition by mass:

(a) Calculate the empirical formula of compound Z.

(b) If compound Z is found to have a relative molecular mass of 60, determine the molecular formula of compound Z.

(R.a.m.: H = 1.0, C = 12.0, O = 16.0)

Answer

slide29

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150)

Solution:

Let the mass of the compound be 100 g. Then,mass of C in the compound = 60.0 gmass of H in the compound = 13.33 gmass of O in the compound = 26.67 g

∴ The empirical formula of compound Z is C3H8O.

slide30

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150)

Solution:

The molecular formula of the compound is (C3H8O)n.

Relative molecular mass of (C3H8O)n = 60

n  (12.0 3 + 1.0 8 + 16.0) = 60

n = 1

∴ The molecular formula of compound Z is C3H8O.

slide31

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150)

Example 38-3

An organic compound is found to contain carbon, hydrogen and oxygen only. On complete combustion, 0.15 g of this compound gives 0.22 g of carbon dioxide and 0.09 g of water. If the relative molecular mass of this compound is found to be 60, determine the molecular formula of this compound.(R.a.m.: H = 1.0, C = 12.0, O = 16.0)

Answer

slide32

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.151)

Solution:

Let the empirical formula of the compound be CxHyOz.Mass of C in CxHyOz = Mass of C in CO2Mass of H in CxHyOz = Mass of H in H2OMass of O in CxHyOz = Mass of CxHyOz– mass of C in CxHyOz– mass of H in CxHyOz

Relative molecular mass of CO2 = 12.0 + 16.0  2 = 44.0

Mass of C in 0.22 g of CO2 = 0.22 g  = 0.06 g

Mass of H in 0.09 g of H2O = 0.09 g  = 0.01 g

slide33

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.151)

Solution:

Mass of O in the compound = (0.15 – 0.06 – 0.01) g = 0.08 g

The simplest whole number ratio of x, y and z can be determined by following the steps in the table below.

∴ The empirical formula of the organic compound is CH2O.

slide34

38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.151)

Solution:

The molecular formula of the compound is (CH2O)n.

Relative molecular mass of (CH2O)n = 60 n  (12.0 + 1.0  2 +16.0) = 60 n = 2

∴ The molecular formula of the compound is C2H4O2.

slide35

38.5 Chemical Tests for Functional Groups (SB p.152)

  • The molecular formula does not give enough information on the structure of the compound

∵ compounds having the same molecular formula may have widely different arrangement of atoms and even different functional groups

e.g.

slide36

38.5 Chemical Tests for Functional Groups (SB p.152)

  • ∴the following step is to find out the functional group(s) present so as to deduce the actual arrangement of atoms in the molecule
  • The presence of certain functional groups in an organic compound can be detected by simple chemical tests
slide45

38.5 Chemical Tests for Functional Groups (SB p.156)

Example 38-4

An organic compound is found to have an empirical formula of CH2O and a relative molecular mass of 60. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.

(a) Calculate the molecular formula of the compound.

Solution:

(a) Let the molecular formula of the compound be (CH2O)n.

Relative molecular mass of (CH2O)n = 60 n  (12.0 + 1.0  2 + 16.0) = 60 n = 2

∴ The molecular formula of the compound is C2H4O2.

Answer

slide46

38.5 Chemical Tests for Functional Groups (SB p.156)

Solution:

(b) The compound reacts with sodium hydrogencarbonate to give carbon dioxide gas (which turns lime water milky), indicating that the compound contains a carboxyl group (–COOH). Eliminating the –COOH group from the molecular formula of C2H4O2, the atoms left are one carbon and three hydrogen atoms. This obviously shows that a methyl group (–CH3) is present. The structural formula of the compound is therefore CH3COOH.

(c) The IUPAC name for the compound is ethanoic acid.

Example 38-4 (cont’d)

(b) Deduce the structural formula of the compound.

(c) Give the IUPAC name for the compound.

(R.a.m.: H = 1.0, C = 12.0, O = 16.0)

Answer

slide47

38.5 Chemical Tests for Functional Groups (SB p.156)

Example 38-5

15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3 of oxygen which was in excess. The mixture was exploded and after cooling, the residual volume was 105 cm3. On adding concentrated potassium hydroxide solution, the volume decreased to 75 cm3.

(a) Calculate the molecular formula of the compound, assuming all volumes were measured under room temperature and pressure.

Answer

slide48

38.5 Chemical Tests for Functional Groups (SB p.156)

Solution:

(a) Let the molecular formula of the compound be CxHy.

Volume of hydrocarbon reacted = 15 cm3 Volume of unreacted oxygen = 75 cm3 Volume of oxygen reacted = (120 – 75) cm3 = 45 cm3 Volume of carbon dioxide formed = (105 – 75) cm3 = 30 cm3

Volume of CxHy reacted : volume of CO2 formed = 1 : x = 15 : 30 ∵

∴ x = 2

slide49

38.5 Chemical Tests for Functional Groups (SB p.156)

Solution:

Volume of CxHy reacted : volume of O2 reacted = 1 : = 15 : 45 ∵

∴ y = 4 ∴ The molecular formula of the compound is C2H4

slide50

38.5 Chemical Tests for Functional Groups (SB p.156)

Solution:

(b) C2H4 belongs to the alkene series.

(c) The structural formula of the hydrocarbon is:

Example 38-5 (cont’d)

(b) To which homologous series does the hydrocarbon belong?

(c) Give the structural formula of the hydrocarbon.

Answer

slide51

38.5 Chemical Tests for Functional Groups (SB p.157)

Example 38-6

20 cm3 of a gaseous organic compound containing only carbon, hydrogen and oxygen were mixed with 110 cm3 of oxygen which was in excess. The mixture was exploded at 105°C and the volume of the gaseous mixture was 150 cm3. After cooling to room temperature, the residual volume was reduced to 90 cm3. On adding concentrated aqueous potassium hydroxide, the volume further decreased to 50 cm3.

(a) Calculate the molecular formula of the compound, assuming all volumes were measured under room temperature and pressure.

Answer

slide52

38.5 Chemical Tests for Functional Groups (SB p.157)

Solution:

(a) Let the molecular formula of the compound be CxHyOz.

Volume of CxHyOz reacted = 20 cm3 Volume of unreacted oxygen = 50 cm3 Volume of oxygen reacted = (110 – 50) cm3 = 60 cm3 Volume of carbon dioxide formed = (90 – 50) cm3 = 40 cm3 Volume of water (in the form of steam) formed = (150 – 90) cm3 = 60 cm3

Volume of CxHyOz reacted : volume of CO2 formed = 1 : x = 20 : 40 ∵

∴ x = 2

slide53

38.5 Chemical Tests for Functional Groups (SB p.157)

Solution:

(a) Volume of CxHyOz reacted : volume of H2O formed = 1 : = 20 : 60 ∵

∴ y = 6

Volume of CxHyOz reacted : volume of O2 formed = 1 : = 20 : 60

∴ z = 1 ∴ The molecular formula of the compound is C2H6O.

slide54

38.5 Chemical Tests for Functional Groups (SB p.157)

Solution:

(b) As the compound contains a –OH group in its structure, the hydrocarbon skeleton of the compound becomes C2H5 after eliminating the –OH group from C2H6O. The structural formula of the compound is CH3CH2OH.

(c) The compound is optically inactive as both carbon atoms in the compound are not asymmetric, i.e. both of them do not attach to four different groups of atoms.

Example 38-6 (cont’d)

(b) The compound is found to contain an –OH functional group in its structure. Deduce its structural formula.

(c) Is the compound optically active? Explain your answer.

Answer

slide55

38.5 Chemical Tests for Functional Groups (SB p.159)

Check Point 38-1

(a) A substance contains 42.8% carbon, 2.38% hydrogen, 16.67% nitrogen by mass and the remainder consists of oxygen.

(i) Given the fact that the relative molecular mass of the substance is 168, deduce the molecular formula of the substance.

(R.a.m.: H = 1.0, C = 12.0, N = 14.0, O = 16.0)

Answer

slide56

38.5 Chemical Tests for Functional Groups (SB p.159)

(a) (i) Let the mass of the compound be 100 g.

∴ The empirical formula of the compound is C3H2NO2.

Let the molecular formula of the compound be (C3H2NO2)n. Molecular mass of (C3H2NO2)n = 168n (12.0  3 + 1.0  2 + 14.0 + 16.0  2) = 168∴ n = 2

∴ The molecular formula of the compound is C6H4N2O4.

slide57

38.5 Chemical Tests for Functional Groups (SB p.159)

(a) (ii) 1,2-Dinitrobenzene 1,3-Dinitrobenzene

1,4-Dinitrobenzene

Check Point 38-1 (cont’d)

(ii) The substance is proved to be an aromatic compound with only one type of functional group. Give the structural formulae for all isomers of the substance (the name of each isomer should also be included).

Answer

slide58

38.5 Chemical Tests for Functional Groups (SB p.159)

Check Point 38-1 (cont’d)

(b) 30cm3 of a gaseous hydrocarbon were mixed with 140 cm3 of oxygen which was in excess, and the mixture was then exploded. After cooling to room temperature, the residual gases occupied 95 cm3 by volume. Be adding potassium hydroxide solution, the volume was reduced by 60 cm3. The remaining gas was proved to be oxygen.

(i) Determine the molecular formula of the hydrocarbon.

Answer

slide59

38.5 Chemical Tests for Functional Groups (SB p.159)

(b) (i) Volume of hydrocarbon reacted = 30 cm3 Volume of unreacted oxygen = (95 – 60) cm3 = 35 cm3 Volume of oxygen reacted = (140 – 35) cm3 = 105 cm3 Volume of carbon dioxide formed = 60 cm3

Volume of CxHy reacted : Volume of CO2 formed

= 1 : x= 30 : 60

∴ x = 2

slide60

38.5 Chemical Tests for Functional Groups (SB p.159)

Volume of CxHy reacted : Volume of O2 reacted

= 1 :

= 30 : 105

∴ y = 6

∴ The molecular formula of the hydrocarbon is C2H6.

slide61

38.5 Chemical Tests for Functional Groups (SB p.159)

(b) (ii) From the molecular formula of the hydrocarbon, it can be deduced that the hydrocarbon is saturated because it fulfils the general formula of alkanes CnH2n+2.

Check Point 38-1 (cont’d)

(ii) Is the hydrocarbon a saturated, unsaturated or aromatic hydrocarbon?

Answer

slide62

38.5 Chemical Tests for Functional Groups (SB p.159)

Check Point 38-1

(c) A hydrocarbon having a relative molecular mass of 56 contains 85.5% carbon and 14.5% hydrogen by mass. Detailed analysis shows that it has two geometrical isomers.

(i) Deduce the molecular formula of the hydrocarbon.

(R.a.m.: H = 1.0, C = 12.0)

Answer

slide63

38.5 Chemical Tests for Functional Groups (SB p.159)

(c) (i) Let the mass of the compound be 100 g.

∴ The empirical formula of the hydrocarbon is CH2.

Let the molecular formula of the hydrocarbon be (CH2)n. Molecular mass of (CH2)n = 56n (12.0 + 1.0  2) = 56n = 4

∴ The molecular formula of the compound is C4H8.

slide64

38.5 Chemical Tests for Functional Groups (SB p.159)

(c) (ii)

(iii) Since the but-2-ene molecule is unsymmetrical and free rotation of the but-2-ene molecule is restricted by the presence of the carbon-carbon double bond, geometrical isomerism exists.

Check Point 38-1 (cont’d)

(ii) Name the two geometrical isomers of the hydrocarbon.

(iii) Explain the existence of geometrical isomerism.

Answer

slide65

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.159)

  • The energy of a quantum of electromagnetic radiation is:
  • E = h where h is the Planck constant (i.e. 6.626  10–34 J s)
  • or

The Electromagnetic Spectrum

  • Electromagnetic radiation has dual property: as wave or as photons
  • The relationship of the frequency () of an electromagnetic radiation, its wavelength () and velocity (c) is shown as follows:
slide66

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.160)

  • Visible light has wavelength between 400 nm and 800 nm
  • Infra-red radiation has wavelength between 800 nm and 300 m
slide67

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.161)

Infra-red Spectroscopy

  • Organic compounds absorb electromagnetic energy in the infra-red (IR) region of the spectrum
  • IR radiation causes atoms and groups of atoms of organic compounds to vibrate with increased amplitude about the covalent bonds that connect them
  • The vibration is quantized, therefore, the IR radiation absorbed have particular energy
slide68

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.161)

  • Infra-red spectrometer:Instrument used to measure the amount of light absorbed at each wavelength of the IR region
  • Infra-red spectrum:
  • shows the absorption of IR radiation by a sample at different frequencies
  • The IR radiation is specified by its wavenumber (unit: cm–1) which is the reciprocal of wavelength
slide69

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.161)

  • Covalent bonds behave as if they were tiny springs connecting the atoms in the vibrations
  • The atoms only vibrate at certain frequencies, therefore covalently bonded atoms have only particular vibrational energy levels
  • The excitation of a molecule from one vibrational energy level to another occurs only when the compound absorbs IR radiation of the exact energy required
slide70

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.161)

Molecules vibrate in a variety of ways:Stretching vibration: two atoms joined by a covalent bond move back and forth as if they were joined by a spring

slide71

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.162)

A variety of stretching and bending vibrations:

slide72

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.162)

  • The frequency of a given stretching vibration of a covalent bond depends on:
  • 1. The masses of the bonded atoms
  • 2. The strength of the bond
  • Light atoms vibrate at higher frequencies than heavier ones
  • The stretching frequencies of groups involving hydrogen such as C – H, N – H and O – H occur at relatively high frequencies
slide73

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.162)

  • Triple bonds are stronger (and vibrate at higher frequencies) than double bonds
  • The IR spectra of simple compounds contain many absorption peaks possibility of two different compounds having the same IR spectrum is very small
  •  IR spectrum has been called the ‘fingerprint’ of a molecule
slide74

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.163)

Use of IR Spectrum in the Identification of Functional Groups

  • A 100% transmittance in the spectrum implies no absorption of IR radiation
  • Absorption peak (or band): Due to the absorption of IR radiation, a decrease in percent transmittance is resulted and hence a dip in the spectrum
slide75

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.163)

The four regions of an IR spectrum

slide76

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.164)

  • The region between 4 000 cm–1 and 1 500 cm–1 is often used for identification of functional groups from their characteristic absorption wavenumbers
slide77

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.164)

Interpretation of the IR spectrum of butane

Wavenumber (cm–1)

Intensity

Indication

2 968

2 890

vs

m

C – H stretching

1 468

s

C – H bending

Interpretation of IR Spectra

slide78

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.166)

Interpretation of the IR spectrum of cis-but-2-ene

Wavenumber (cm–1)

Intensity

Indication

3 044

3 028

vs

vs

C – H stretching

2 952

vs

C – H stretching

1 677

1 657

m

m

C = C stretching

1 411

s

C – H bending

slide79

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.167)

The IR spectrum of hex-1-yne

slide80

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.167)

Interpretation of the IR spectrum of hex-1-yne

slide81

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.168)

Interpretation of the IR spectrum of butanone

Wavenumber (cm–1)

Intensity

Indication

2 983

2 925

s

s

C – H stretching

1 720

vs

C = O stretching

1 416

m

C – H bending

slide82

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.169)

Interpretation of the IR spectrum of butan-1-ol

Wavenumber (cm–1)

Intensity

Indication

3 330

Broad band

O – H stretching

2 960

2 935

2 875

m

m

m

C – H stretching

slide83

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.170)

Interpretation of the IR spectrum of butanoic acid

Wavenumber (cm–1)

Intensity

Indication

3 100

Broad band

O – H stretching

1 708

s

C = O stretching

slide84

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.170)

  • The absorption of the O – H group in alcohols and carboxylic acids usually appear as broad band instead of sharp peak

∵ the vibration of the O – H group is complicated by the hydrogen bonding formed between the molecules

slide85

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.171)

Interpretation of the IR spectrum of butylamine

Wavenumber (cm–1)

Intensity

Indication

3 371

3 280

s

s

N – H stretching

2 960 – 2 875

w

C – H stretching

1 610

m

N – H bending

1 475

m

C – H bending

slide86

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.171)

Interpretation of the IR spectrum of butanenitrile

Wavenumber (cm–1)

Intensity

Indication

2 990 – 2 895

s

C – H stretching

2 246

vs

C  N stretching

1 420

s

C – H bending

1 480

s

slide87

38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.172)

Summary of strategies for the use of IR spectra in the identification of functional groups in an organic compound:

  • Focus at the IR absorption peak at or above 1500 cm–1. Concentrate initially on the major absorption peaks
  • For each absorption peak, try to list out all the possibilities using a correlation table or chart. Note that not all absorption peaks in the spectrum can be assigned
  • The absence and presence of absorption peaks at some characteristic ranges of wavenumbers are equally important. It is because the absence of particular absorption peaks can be used to identify certain functional groups or bonds do not exist in the molecule
slide88

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.173)

Example 38-7

An organic compound with a relative molecular mass of 72 was found to contain 66.66% carbon, 22.23% oxygen and 11.11% hydrogen by mass. A portion of its infra-red spectrum is shown below.

(a) Determine the molecular formula of the compound.(R.a.m.: H = 1.0, C = 12.0, O = 16.0)

Answer

slide89

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.173)

Solution:

(a) Let the mass of the compound be 100 g. Then, mass of carbon in the compound = 66.66 g mass of hydrogen in the compound = 11.11 g mass of oxygen in the compound = 22.23 g

∴ The empirical formula of the organic compound is C4H8O.

slide90

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.173)

Solution:

Let the molecular formula of the compound is (C4H8O)n.

Relative molecular mass of (C4H8O)n = 72

n  (12.0  4 + 1.0 8 + 16.0) = 72

∴ n = 1

∴ The molecular formula of the compound is C4H8O.

slide91

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.173)

  • Example 38-7 (cont’d)
  • Deduce two possible structures of the compound, each of which belongs to a different homologous series.

Answer

slide92

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.173)

Solution:

(b)From the IR spectrum, it can be observed that there are absorption peaks at 2950 cm–1 and 1700 cm–1. The absorption peak at 2950 cm–1 corresponds to the stretching vibration of the C – H bond, and the absorption peak at 1700 cm–1 corresponds to the stretching vibration of the C = O bond. As there is only one oxygen atom in the molecule of the compound, it is either an aldehyde or a ketone.If it is an aldehyde, its possible structure will be:

slide93

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

Solution:

(b)If it is a ketone, its possible structure will be:

slide94

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

Check Point 38-2

(a) An organic compound X forms a silver mirror with ammoniacal silver nitrate solution. Another organic compound Y reacts with ethanoic acid to give a product with a fruity smell. The portions of infra-red spectra of X and Y are shown below.

slide95

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

Check Point 38-2 (cont’d)

(a)

Sketch the infra-red spectrum of a carboxylic acid based on the IR spectra of X and Y.

Answer

slide96

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(a) From the information given, X would be an aldehyde and Y would be an alcohol.

Comparing the structures of an aldehyde and an alcohol with that of a carboxylic acid, some common features are found between the two.

In the IR spectrum of a carboxylic acid, it is expected that it contains the characteristic O – H (similar to the alcohol) and C = O (similar to the aldehyde) absorption peaks. Thus peak values at around 3300 cm–1 and 1720 cm–1 are predicted. A wide band at around 3300 cm–1 is observed due to the complication of the stretching vibration of the O – H group by hydrogen bonding and it overlaps with the absorption of the C – H bond in the 2950 – 2875 cm–1 region. The infra-red spectrum of a carboxylic acid is as follows:

slide98

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

Check Point 38-2 (cont’d)

(b) The infra-red spectra of two organic compounds A and B are shown below.

slide99

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(b) Compound B could be an alcohol. From the two spectra given, compound B shows a broad band at 3300 cm–1 and several peaks at 2960 – 2875 cm–1. This broad band corresponds to the complication of the stretching vibration of the O – H bond by hydrogen bonding occurring among alcohol molecules.

Check Point 38-2 (cont’d)

(b)

State which compound could be an alcohol. Explain your answer briefly.

Answer

slide100

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(c) The infra-red spectrum of the amino acid is shown as follows:

Check Point 38-2 (cont’d)

(c) Given the following characteristic absorption wavenumbers of some covalent bonds in infra-red spectra:

Sketch the expected infra-red spectrum for an amino acid with the following structure:

Answer

slide101

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(d) In the IR spectrum of compound X, the wide absorption band at 3500 – 3000 cm–1 corresponds to the stretching vibration of the O – H bond. Besides, the absorption peak at 1760 – 1720 cm–1 corresponds to the stretching vibration of the C = O bond. Therefore, compound X is a carboxylic acid.

Check Point 38-2 (cont’d)

(d) Below is a portion of the infra-red spectrum of an organic compound X. To which homologous series does it belong?

Answer

slide102

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(e) From the IR spectrum of compound Y, the two peaks in the 3300 – 3180 cm–1 region show that the compound contains the –NH2 group. Besides, the sharp peak at 1680 cm–1 implies that the compound also contains the C = O bond.

Check Point 38-2 (cont’d)

(e) Below is a portion of the infra-red spectrum of an organic compound Y. Identify the functional groups that it contains.

Answer

slide103

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

Check Point 38-2 (cont’d)

(f) An organic compound Z with a relative molecular mass of 88 was found to contain 54.54% carbon, 36.36% oxygen and 9.10% hydrogen by mass. A portion of its infra-red spectrum is shown below:

(i) Determine the molecular formula of compound Z. (R.a.m.: H = 1.0, C = 12.0, O = 16.0)

Answer

slide104

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(f) (i) Let the mass of the compound be 100 g.

∴ The empirical formula of the compound is C2H4O.

Let the molecular formula of the compound be (C2H4O)n. Molecular mass of (C2H4O)n = 88n (12.0  2 + 1.0  4 + 16.0) = 88n = 2∴ The molecular formula of the compound is C4H8O2.

slide105

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(f) (ii)

Check Point 38-2 (cont’d)

(f) (ii) Based on the result from (i), draw two possible structures of the compound, each of which belongs to a different homologous series.

Answer

slide106

38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.174)

(f) (iii) From the IR spectrum of compound Z, the absorption peak at 3200 – 2800 cm–1 corresponds to the stretching vibration of the C – H bond. Besides, the absorption peak at 1800 – 1600 cm–1 corresponds to the stretching vibration of the C = O bond. The absence of the characteristic peak of the O – H bond in the 3230 – 3670 cm–1 region indicates that compound Z is an ester.

Check Point 38-2 (cont’d)

(f) (iii) Using the information from the IR spectrum, name the homologous series that compound Z belongs to .

Answer

ad