new way chemistry for hong kong a-level 3b 1 structural determination of organic compounds...
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New Way Chemistry for Hong Kong A-Level 3B1
Structural Determination Structural Determination of Organic Compoundsof Organic Compounds
34.134.1 IntroductionIntroduction
34.234.2 Isolation and Purification of Organic CompoundsIsolation and Purification of Organic Compounds
34.334.3 Tests for PurityTests for Purity
34.434.4 Qualitative Analysis of Elements in an Organic CompoundQualitative Analysis of Elements in an Organic Compound
34.534.5 Determination of Empirical Formula and Molecular Determination of Empirical Formula and Molecular
Formula from Analytical DataFormula from Analytical Data
34.634.6 Structural Information from Physical PropertiesStructural Information from Physical Properties
34.734.7 Structural Information from Chemical PropertiesStructural Information from Chemical Properties
34.834.8 Use of Infra-red Spectrocopy in the Identification of Use of Infra-red Spectrocopy in the Identification of
Functional GroupsFunctional Groups
34.934.9 Use of Mass Spectra to Obtain Structural InformationUse of Mass Spectra to Obtain Structural Information
3434
New Way Chemistry for Hong Kong A-Level 3B2
34.134.1IntroductioIntroductio
nn
New Way Chemistry for Hong Kong A-Level 3B3
34.1 Introduction (SB p.77)
IntroductionIntroduction• The determination of the structure of an
organic compound involves:
1. Isolation and purification of the compound
2. Qualitative analysis of the elements present in the compound
3. Determination of the molecular formula of the compound
4. Determination of the functional group present in the compound
New Way Chemistry for Hong Kong A-Level 3B4
34.1 Introduction (SB p.77)
IntroductionIntroduction
The general steps to determine the structure of an organic compound
Check Point 34-1Check Point 34-1
New Way Chemistry for Hong Kong A-Level 3B5
34.234.2Isolation and Isolation and Purification Purification of Organic of Organic
CompoundsCompounds
New Way Chemistry for Hong Kong A-Level 3B6
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Isolation and Purification Isolation and Purification of Organic Compoundsof Organic Compounds
• These techniques include:
1. Filtration
2. Centrifugation
3. Crystallization
4. Solvent extraction
5. Distillation
New Way Chemistry for Hong Kong A-Level 3B7
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Isolation and Purification Isolation and Purification of Organic Compoundsof Organic Compounds
• These techniques include:
5. Fractional distillation
6. Sublimation
7. Chromatography
New Way Chemistry for Hong Kong A-Level 3B8
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Isolation and Purification Isolation and Purification of Organic Compoundsof Organic Compounds
• The selection of a proper technique
depends on the particular differences in physical properties of the substances present in the mixture
New Way Chemistry for Hong Kong A-Level 3B9
34.2 Isolation and Purification of Organic Compounds (SB p.78)
FiltrationFiltration
• To separate an insoluble solid from a liquid particularly when the solid is suspended throughout the liquid
• The solid/liquid mixture is called a suspension
New Way Chemistry for Hong Kong A-Level 3B10
34.2 Isolation and Purification of Organic Compounds (SB p.78)
FiltrationFiltration
The laboratory set-up of filtration
New Way Chemistry for Hong Kong A-Level 3B11
34.2 Isolation and Purification of Organic Compounds (SB p.78)
FiltrationFiltration
• There are many small holes in the filter paper
allow very small particles of solvent and dissolved solutes to pass
through as filtrate
• Larger insoluble particles are retained on the filter paper as residue
New Way Chemistry for Hong Kong A-Level 3B12
34.2 Isolation and Purification of Organic Compounds (SB p.79)
CentrifugatioCentrifugationn• When there is only a small amount of
suspension, or when much faster separation is required
Centrifugation is often used instead of filtration
New Way Chemistry for Hong Kong A-Level 3B13
34.2 Isolation and Purification of Organic Compounds (SB p.79)
CentrifugatioCentrifugationn
• The liquid containing undissolved solids is put in a centrifuge tube
• The tubes are then put into the tube holders in a centrifuge
A centrifuge
New Way Chemistry for Hong Kong A-Level 3B14
34.2 Isolation and Purification of Organic Compounds (SB p.79)CentrifugatioCentrifugationn• The holders and tubes are spun around at a
very high rate and are thrown outwards
• The denser solid is collected as a lump at the bottom of the tube with the clear liquid above
New Way Chemistry for Hong Kong A-Level 3B15
34.2 Isolation and Purification of Organic Compounds (SB p.79)
CrystallizatioCrystallizationn
• Crystals are solids that have
a definite regular shape
smooth flat faces and straight edges
• Crystallization is the process of forming crystals
New Way Chemistry for Hong Kong A-Level 3B16
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot Conce1. Crystallization by Cooling a Hot Concentrated Solutionntrated Solution• To obtain crystals from an unsaturated
aqueous solution
the solution is gently heated to make it more concentrated
• After, the solution is allowed to cool at room conditions
New Way Chemistry for Hong Kong A-Level 3B17
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot Conce1. Crystallization by Cooling a Hot Concentrated Solutionntrated Solution
• The solubilities of most solids increase with temperature
• When a hot concentrated solution is cooled
the solution cannot hold all of the dissolved solutes
• The “excess” solute separates out as crystals
New Way Chemistry for Hong Kong A-Level 3B18
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot Conce1. Crystallization by Cooling a Hot Concentrated Solutionntrated Solution
Crystallization by cooling a hot concentrated solution
New Way Chemistry for Hong Kong A-Level 3B19
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a Cold S2. Crystallization by Evaporating a Cold Solution at Room Temperatureolution at Room Temperature
• As the solvent in a solution evaporates,
the remaining solution becomes more and more concentrated
eventually the solution becomes saturated
further evaporation causes crystallization to occur
New Way Chemistry for Hong Kong A-Level 3B20
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a Cold S2. Crystallization by Evaporating a Cold Solution at Room Temperatureolution at Room Temperature
• If a solution is allowed to stand at room temperature,
evaporation will be slow
• It may take days or even weeks for crystals to form
New Way Chemistry for Hong Kong A-Level 3B21
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a Cold S2. Crystallization by Evaporating a Cold Solution at Room Temperatureolution at Room Temperature
Crystallization by slow evaporation of a solution (preferably saturated) at room
temperature
New Way Chemistry for Hong Kong A-Level 3B22
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• Involves extracting a component from a mixture with a suitable solvent
• Water is the solvent used to extract salts from a mixture containing salts and sand
• Non-aqueous solvents (e.g. 1,1,1-trichloroethane and diethyl ether) can be used to extract organic products
New Way Chemistry for Hong Kong A-Level 3B23
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• Often involves the use of a separating funnel
• When an aqueous solution containing the organic product is shaken with diethyl ether in a separating funnel,
the organic product dissolves into the ether layer
New Way Chemistry for Hong Kong A-Level 3B24
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
The organic product in an aqueous solution can be extracted by solvent extraction using diethyl
ether
New Way Chemistry for Hong Kong A-Level 3B25
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• The ether layer can be run off from the separating funnel and saved
• Another fresh portion of ether is shaken with the aqueous solution to extract any organic products remaining
• Repeated extraction will extract most of the organic product into the several portions of ether
New Way Chemistry for Hong Kong A-Level 3B26
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• Conducting the extraction with several small portions of ether is more efficient than extracting in a single batch with the whole volume of ether
• These several ether portions are combined and dried
the ether is distilled off
leaving behind the organic product
New Way Chemistry for Hong Kong A-Level 3B27
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
• A method used to separate a solvent from a solution containing non-volatile solutes
• When a solution is boiled,
only the solvent vaporizes
the hot vapour formed condenses to liquid again on a cold surface
• The liquid collected is the distillate
New Way Chemistry for Hong Kong A-Level 3B28
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
The laboratory set-up of distillation
New Way Chemistry for Hong Kong A-Level 3B29
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
• Before the solution is heated,
several pieces of anti-bumping granules are added into the flask
prevent vigorous movement of the liquid called bumping to occur
during heating
make boiling smooth
New Way Chemistry for Hong Kong A-Level 3B30
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
• If bumping occurs during distillation,
some solution (not yet vaporized) may spurt out into the collecting vessel
New Way Chemistry for Hong Kong A-Level 3B31
34.2 Isolation and Purification of Organic Compounds (SB p.81)
Fractional DistillationFractional Distillation
• A method used to separate a mixture of two or more miscible liquids
New Way Chemistry for Hong Kong A-Level 3B32
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional DistillationFractional Distillation
The laboratory set-up of fractional distillation
New Way Chemistry for Hong Kong A-Level 3B33
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional DistillationFractional Distillation
• A fractionating column is attached vertically between the flask and the condenser
a column packed with glass beads
provide a large surface area for the repeated condensation and vaporization of the mixture to occur
New Way Chemistry for Hong Kong A-Level 3B34
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Fractional DistillationDistillation• The temperature of the escaping vapour
is measured using a thermometer
• When the temperature reading becomes steady,
the vapour with the lowest boiling point firstly comes out from the top
of the column
New Way Chemistry for Hong Kong A-Level 3B35
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Fractional DistillationDistillation• When all of that liquid has distilled off,
the temperature reading rises and becomes steady later on
another liquid with a higher boiling point distils out
• Fractions with different boiling points can be collected separately
New Way Chemistry for Hong Kong A-Level 3B36
34.2 Isolation and Purification of Organic Compounds (SB p.82)
SublimatioSublimationn
• Sublimation is the direct change of
a solid to vapour on heating, or
a vapour to solid on cooling
without going through the liquid state
New Way Chemistry for Hong Kong A-Level 3B37
34.2 Isolation and Purification of Organic Compounds (SB p.82)
SublimatioSublimationn• A mixture of two compounds is heated in an
evaporating dish
• One compound changes from solid to vapour directly
The vapour changes back to solid on a cold surface
• The other compound is not affected by heating and remains in the evaporating dish
New Way Chemistry for Hong Kong A-Level 3B38
34.2 Isolation and Purification of Organic Compounds (SB p.83)
SublimatioSublimationn
A mixture of two compounds can be separated by sublimation
New Way Chemistry for Hong Kong A-Level 3B39
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• An effective method of separating a complex mixture of substances
• Paper chromatography is a common type of chromatography
New Way Chemistry for Hong Kong A-Level 3B40
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
The laboratory set-up of paper chromatography
New Way Chemistry for Hong Kong A-Level 3B41
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• A solution of the mixture is dropped at one end of the filter paper
New Way Chemistry for Hong Kong A-Level 3B42
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• The thin film of water adhered onto the surface of the filter paper forms the stationary phase
• The solvent is called the mobile phase or eluent
New Way Chemistry for Hong Kong A-Level 3B43
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• When the solvent moves across the sample spot of the mixture,
partition of the components between the stationary phase and the mobile phase
occurs
New Way Chemistry for Hong Kong A-Level 3B44
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy• As the various components are being
adsorbed or partitioned at different rates,
they move upwards at different rates
• The ratio of the distance travelled by the substance to the distance travelled by the solvent
known as the Rf value
a characteristic of the substance
New Way Chemistry for Hong Kong A-Level 3B45
34.2 Isolation and Purification of Organic Compounds (SB p.84)
Technique Aim
(a) Filtration To separate an insoluble solid from a liquid (slow)
(b) Centrifugation To separate an insoluble solid from a liquid (fast)
(c) Crystallization To separate a dissolved solute from its solution
(d) Solvent extraction
To separate a component from a mixture with a suitable solvent
(e) Distillation To separate a liquid from a solution containing non-volatile solutes
A summary of different techniques of isolation and purification
New Way Chemistry for Hong Kong A-Level 3B46
34.2 Isolation and Purification of Organic Compounds (SB p.84)
Technique Aim
(f) Fractional distillation
To separate miscible liquids with widely different boiling points
(g) Sublimation To separate a mixture of solids in which only one can sublime
(h) Chromatography
To separate a complex mixture of substances
A summary of different techniques of isolation and purification
Check Point 34-2Check Point 34-2
New Way Chemistry for Hong Kong A-Level 3B47
34.334.3Tests for Tests for
PurityPurity
New Way Chemistry for Hong Kong A-Level 3B48
34.3 Tests for Purity (SB p.84)
• If the substance is a solid,
its purity can be checked by determining its melting point
• If it is a liquid,
its purity can be checked by determining its boiling point
Tests for PurityTests for Purity
New Way Chemistry for Hong Kong A-Level 3B49
34.3 Tests for Purity (SB p.85)
Determination of Melting Determination of Melting PointPoint
• To determine the melting point of a solid,
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
New Way Chemistry for Hong Kong A-Level 3B50
34.3 Tests for Purity (SB p.85)
Determination of Melting Determination of Melting PointPoint
Determination of the melting point of a solid using an oil bath
New Way Chemistry for Hong Kong A-Level 3B51
34.3 Tests for Purity (SB p.85)
Determination of Melting Determination of Melting PointPoint
• A pure solid has a sharp melting point
melting occurs within a narrow temperature range (usually less than 0.5°C)
• An impure solid does not have a sharp melting point
melts gradually over a wide temperature range
New Way Chemistry for Hong Kong A-Level 3B52
34.3 Tests for Purity (SB p.85)
Determination of Melting Determination of Melting PointPoint
• The presence of impurities lowers the melting point of a solid
• Melting point is a useful indication of the purity of a substance
New Way Chemistry for Hong Kong A-Level 3B53
34.3 Tests for Purity (SB p.85)
Determination of Boiling Determination of Boiling PointPoint• The boiling point of a liquid can be
determined by using the distillation apparatus
• The temperature at which the liquid boils steadily is its boiling point
• A flammable liquid should be heated in a water bath, instead of heated with a naked flame
New Way Chemistry for Hong Kong A-Level 3B54
34.3 Tests for Purity (SB p.85)
Determination of Boiling Determination of Boiling PointPoint
• The boiling point of a pure liquid is quite sharp
• The presence of non-volatile solutes such as salts raises the boiling point of a liquid
New Way Chemistry for Hong Kong A-Level 3B55
34.434.4Qualitative Qualitative Analysis of Analysis of
Elements in an Elements in an Organic Organic
CompoundCompound
New Way Chemistry for Hong Kong A-Level 3B56
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
• Qualitative analysis of an organic compound is
to determine what elements are present in the compound
Qualitative Analysis of Qualitative Analysis of an Organic Compoundan Organic Compound
New Way Chemistry for Hong Kong A-Level 3B57
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Carbon and Carbon and HydrogenHydrogen• Tests for carbon and hydrogen in an
organic compound are usually unnecessary
an organic compound must contain carbon and hydrogen
New Way Chemistry for Hong Kong A-Level 3B58
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Carbon and Carbon and HydrogenHydrogen• Carbon and hydrogen can be detected by
heating a small amount of the substance with copper(II) oxide
• Carbon and hydrogen would be oxidized to carbon dioxide and water respectively
• Carbon dioxide turns lime water milky
• Water turns anhydrous cobalt(II) chloride paper pink
New Way Chemistry for Hong Kong A-Level 3B59
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur• Halogens, nitrogen and sulphur in organic
compounds can be detected
by performing the sodium fusion test
New Way Chemistry for Hong Kong A-Level 3B60
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur• The compound under test is
fused with a small piece of sodium metal in a small combustion tube
heated strongly
• The products of the test are extracted with water and then analyzed
New Way Chemistry for Hong Kong A-Level 3B61
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur
• During sodium fusion,
halogens in the organic compound is converted to sodium halides
nitrogen in the organic compound is converted to sodium cyanide
sulphur in the organic compound is converted to sodium sulphide
New Way Chemistry for Hong Kong A-Level 3B62
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Element Material used Observation
Halogens, as Acidified silver nitrate solution
chloride ion (Cl-) A white precipitate is formed. It is soluble in excess NH3(aq).
bromide ion (Br-) A pale yellow precipitate is formed. It is sparingly soluble in excess NH3(aq).
iodide ion (I-) A creamy yellow precipitate is formed. It is insoluble in excess NH3(aq).
Results for halogens, nitrogen and sulphur in the sodium fusion test
New Way Chemistry for Hong Kong A-Level 3B63
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Results for halogens, nitrogen and sulphur in the sodium fusion test
Element Material used Observation
Nitrogen,as
cyanide ion (CN-)
A mixture of iron(II) sulphate and iron(III) sulphate solutions
A blue-green colour is observed.
Sulphur, assulphide ion (S2-)
Sodium pentacyanonitrosylferrate(II) solution
A black precipitate is formed
Check Point 34-4Check Point 34-4
New Way Chemistry for Hong Kong A-Level 3B64
34.534.5Determination of Determination of
Empirical Empirical Formula and Formula and
Molecular Molecular Formula from Formula from
Analytical DataAnalytical Data
New Way Chemistry for Hong Kong A-Level 3B65
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
• After determining the constituent elements of a particular organic compound
perform quantitative analysis to find the percentage composition by mass of the compound
the masses of different elements in an organic compound are determined
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound
New Way Chemistry for Hong Kong A-Level 3B66
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
1. Carbon and Hydrog1. Carbon and Hydrogenen• The organic compound is burnt in excess
oxygen
• The carbon dioxide and water vapour formed are respectively absorbed by
potassium hydroxide solution and anhydrous calcium chloride
New Way Chemistry for Hong Kong A-Level 3B67
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
1. Carbon and Hydrog1. Carbon and Hydrogenen• The increases in mass in potassium
hydroxide solution and calcium chloride represent
the masses of carbon dioxide and water vapour formed respectively
New Way Chemistry for Hong Kong A-Level 3B68
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
2. Nitroge2. Nitrogenn • The organic compound is heated with
excess copper(II) oxide
• The nitrogen monoxide and nitrogen dioxide formed are passed over hot copper
the volume of nitrogen formed is measured
New Way Chemistry for Hong Kong A-Level 3B69
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
3. Haloge3. Halogensns• The organic compound is heated with
fuming nitric(V) acid and excess silver nitrate solution
• The mixture is allowed to cool
then water is added
the dry silver halide formed is weighed
New Way Chemistry for Hong Kong A-Level 3B70
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
4. Sulphu4. Sulphurr
• The organic compound is heated with fuming nitric(V) acid
• After cooling,
barium nitrate solution is added
the dry barium sulphate formed is weighed
New Way Chemistry for Hong Kong A-Level 3B71
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound
• After determining the percentage composition by mass of a compound,
the empirical formula of the compound can be calculated
New Way Chemistry for Hong Kong A-Level 3B72
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound
The empirical formula of a compound is the formula which shows the simplest whole number ratio of the atoms present in the compound
New Way Chemistry for Hong Kong A-Level 3B73
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound
• When the relative molecular mass and the empirical formula of the compound are known,
the molecular formula of the compound can be calculated
New Way Chemistry for Hong Kong A-Level 3B74
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound
The molecular formula of a compound is the formula which shows the actual number of each kind of atoms present in a molecule of the compound
New Way Chemistry for Hong Kong A-Level 3B75
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
Example 34-5AExample 34-5A Example 34-5BExample 34-5B
Check Point 34-5Check Point 34-5
New Way Chemistry for Hong Kong A-Level 3B76
34.634.6Structural Structural
Information Information from Physical from Physical
PropertiesProperties
New Way Chemistry for Hong Kong A-Level 3B77
34.6 Structural Information from Physical Properties (SB p.89)
• The physical properties of a compound include its colour, odour, density, solubility, melting point and boiling point
• The physical properties of a compound depend on its molecular structure
Structural Information from Structural Information from Physical PropertiesPhysical Properties
New Way Chemistry for Hong Kong A-Level 3B78
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from Structural Information from Physical PropertiesPhysical Properties
• From the physical properties of a compound,
obtain preliminary information about the structure of the compound
New Way Chemistry for Hong Kong A-Level 3B79
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from Structural Information from Physical PropertiesPhysical Properties• e.g.
Hydrocarbons have low densities, often about 0.8 g cm–3
Compounds with functional groups have higher densities
New Way Chemistry for Hong Kong A-Level 3B80
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from Structural Information from Physical PropertiesPhysical Properties
• The densities of most organic compounds are < 1.2 g cm–3
• Compounds having densities > 1.2 g cm–3 must contain multiple halogen atoms
New Way Chemistry for Hong Kong A-Level 3B81
34.6 Structural Information from Physical Properties (SB p.90)
Organic compound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly
polar solvents
In non-polar
organic solvents
Hydrocarbons (saturated and unsaturated)
All
have densities < 0.8 g cm–3
• Generally low but increases with number of carbon atoms in the molecule
• Branched-chain hydrocarbons have lower boiling points but higher melting points than the corresponding straight-chain isomers
Insoluble Soluble
Physical properties of some common organic compounds
New Way Chemistry for Hong Kong A-Level 3B82
34.6 Structural Information from Physical Properties (SB p.90)
Organic compound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly
polar solvents
In non-polar
organic solvents
Aromatic hydrocarbons
Between 0.8 and 1.0 g cm–3
Generally low Insoluble Soluble
Physical properties of some common organic compounds
New Way Chemistry for Hong Kong A-Level 3B83
34.6 Structural Information from Physical Properties (SB p.90)
Organic compound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly
polar solvents
In non-polar
organic solvents
Halo-alkanes
• 0.9 - 1.1 g cm–3 for chloro-alkanes
• >1.0 g cm–3 for bromo-alkanes and iodo-alkanes
• Higher than alkanes of similar relative molecular masses ( haloalkane molecules are polar)
• All haloalkanes are liquids except halomethanes
• Both the m.p. and b.p. increase in the order: RCH2F < RCH2Cl < RCH2Br < RCH2I
Insoluble Soluble
Physical properties of some common organic compounds
New Way Chemistry for Hong Kong A-Level 3B84
34.6 Structural Information from Physical Properties (SB p.90)
Organic comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Alcohols • Simple alcohols are liquids and alcohols with > 12 carbons are waxy solids
• Much higher than hydrocarbons of similar relative molecular masses ( formation of hydrogen bonds between alcohol molecules)
• Lower members: Completely miscible with water ( formation of hydrogen bonds between alcohol molecules and water molecules)
Soluble
Physical properties of some common organic compounds
New Way Chemistry for Hong Kong A-Level 3B85
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Alcohols • All simple alcohols have densities < 1.0 g cm–3
• Straight-chain alcohols have higher b.p. than the corresponding branched-chain alcohols
• Solubility decreases gradually as the hydrocarbon chain lengthens
Soluble
New Way Chemistry for Hong Kong A-Level 3B86
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Carbonyl
comp-ounds
(alde-hydes
and ketones)
• <1.0 g cm–3 for aliphatic carbonyl compounds
Higher than alkanes but lower than alcohols of similar relative molecular masses (Molecules of aldehydes or ketones are held together by strong dipole-dipole interactions but not hydrogen bonds)
• Lower members:Soluble in water ( the formation of hydrogen bonds between molecules of aldehydes or ketones and water molecules)
Soluble
New Way Chemistry for Hong Kong A-Level 3B87
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Carbonyl
comp-ounds
(alde-hydes
and ketones)
• > 1.0 g cm–3 for aromatic carbonyl compounds
• Solubility decreases gradually as the hydrocarbon chain lengthens
Soluble
New Way Chemistry for Hong Kong A-Level 3B88
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Carbo-xylic
acids
• Lower members have densities similar to water
• Methanoic acid has a density of 1.22 g cm–3
Higher than alcohols of similar relative molecular masses ( the formation of more extensive intermolecular hydrogen bonds)
• First four members are
miscible with water in all proportions
• Solubility decreases gradually as the hydrocarbon chain lengthens
Soluble
New Way Chemistry for Hong Kong A-Level 3B89
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Esters Lower members
have densities less than water
Slightly higher than hydrocarbons but lower than carbonyl compounds and
alcohols of similar relative molecular masses
Insoluble Soluble
New Way Chemistry for Hong Kong A-Level 3B90
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Amines Most amines have densities less than water
• Higher than alkanes but lower than alcohols of similar relative molecular masses
• Generally soluble
• Solubility decreases in the order:
1o amines > 2o amines > 3o amines
Soluble
New Way Chemistry for Hong Kong A-Level 3B91
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Amines • 1o and 2o amines are able to form hydrogen bonds with each other but the strength is less than that between alcohol molecules (NH bond is less polar than O H bond)
New Way Chemistry for Hong Kong A-Level 3B92
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Amines • 3o amines have lower m.p. and b.p. than the isomers of 1o and 2o amines ( molecules of 3o amines cannot form intermolecular hydrogen bonds)
New Way Chemistry for Hong Kong A-Level 3B93
34.6 Structural Information from Physical Properties (SB p.92)
Example 34-6Example 34-6 Check Point 34-6Check Point 34-6
New Way Chemistry for Hong Kong A-Level 3B94
34.734.7Structural Structural
Information Information from Chemical from Chemical
PropertiesProperties
New Way Chemistry for Hong Kong A-Level 3B95
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from Structural Information from Chemical PropertiesChemical Properties
• The molecular formula of a compound
does not give enough clue to the structure of the compound
• Compounds having the same molecular formula
may have different arrangements of atoms and even different functional groups
New Way Chemistry for Hong Kong A-Level 3B96
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from Structural Information from Chemical PropertiesChemical Properties
• e.g.The molecular formula of C2H4O2 may represent a carboxylic acid or an ester:
New Way Chemistry for Hong Kong A-Level 3B97
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from Structural Information from Chemical PropertiesChemical Properties
• The next stage is
to find out the functional group(s) present
to deduce the actual arrangement of atoms in the molecule
New Way Chemistry for Hong Kong A-Level 3B98
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Saturated
hydrocarbons
• Burn the saturated hydrocarbon in a non-luminous Bunsen flame
• A blue or clear yellow flame is observed
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B99
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Unsaturated hydrocarbons (C = C, C C)
• Burn the unsaturated hydrocarbon in a non-luminous Bunsen flame
• A smoky flame is observed
• Add bromine in 1,1,1-trichloroethane at room temperature and in the absence of light
• Bromine decolourizes rapidly
• Add 1% (dilute) acidified potassium manganate(VII) solution
• Potassium manganate(VII) solution decolourizes rapidly
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B100
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Haloalkanes
(1°, 2° or 3°)
• Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution
• For chloroalkanes, a white precipitate is formed
• For bromoalkanes, a pale yellow precipitate is formed
• For iodoalkanes, a creamy yellow precipitate is formed
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B101
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Halobenzenes • Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution
• No precipitate is formed
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B102
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• Add a small piece of sodium metal
• A colourless gas is evolved
• Esterification: Add ethanoyl chloride
• The temperature of the reaction mixture rises
• A colourless gas is evolved
New Way Chemistry for Hong Kong A-Level 3B103
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• Add acidified potassium dichromate(VI) solution
• For 1° and 2° alcohols, the clear orange solution becomes opaque and turns green almost immediately
• For 3° alcohols, there are no observable changes
New Way Chemistry for Hong Kong A-Level 3B104
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• Iodoform test for:
Add iodine in sodium hydroxide solution
• A yellow precipitate is formed
New Way Chemistry for Hong Kong A-Level 3B105
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• Lucas test: add a solution of zinc chloride in concentrated hydrochloric acid
• For 1° alcohols, the aqueous phase remains clear
• For 2° alcohols, the clear solution becomes cloudy within 5 minutes
• For 3° alcohols, the aqueous phase appears cloudy immediately
34.7 Structural Information from Chemical Properties (SB p.94)
New Way Chemistry for Hong Kong A-Level 3B106
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Ethers ( O )
• No specific test for ethers but they are soluble in concentrated sulphuric(VI) acid
34.7 Structural Information from Chemical Properties (SB p.94)
New Way Chemistry for Hong Kong A-Level 3B107
34.7 Structural Information from Chemical Properties (SB p.94)
Organic compound
Test Observation
Aldehydes
( )
• Add aqueous sodium hydrogensulphate(IV)
• Crystalline salts are formed
• Add 2,4-dinitrophenylhydrazine
• A yellow, orange or red precipitate is formed
• Silver mirror test: add Tollens’ reagent (a solution of aqueous silver nitrate in aqueous ammonia)
• A silver mirror is deposited on the inner wall of the test tube
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B108
34.7 Structural Information from Chemical Properties (SB p.94)
Organic compound
Test Observation
Ketones
( )
• Add aqueous sodium hydrogensulphate(IV)
• Crystalline salts are formed (for unhindered ketones only)
• Add 2,4-dinitrophenylhydrazine
• A yellow, orange or red precipitate is formed
• Iodoform test for:
Add iodine in sodium hydroxide solution
• A yellow precipitate is formed
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B109
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Carboxylic acids
( )
• Esterification: warm the carboxylic acid with an alcohol in the presence of concentrated sulphuric(VI) acid, followed by adding sodium carbonate solution
• A sweet and fruity smell is detected
• Add sodium hydrogencarbonate
• The colourless gas produced turns lime water milky
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B110
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Esters
( )
• No specific test for esters but they can be distinguished by its characteristic smell
• A sweet and fruity smell is detected
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B111
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Acyl halides
( )
• Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution
• For acyl chlorides, a white precipitate is formed
• For acyl bromides, a pale yellow precipitate is formed
• For acyl iodides, a creamy yellow precipitate is formed
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B112
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Amides
( )
• Boil with sodium hydroxide solution
• The colourless gas produced turns moist red litmus paper or pH paper blue
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B113
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Amines
(NH2)
• 1o aliphatic amines: dissolve the amine in dilute hydrochloric acid at 0 – 5 oC, then add cold sodium nitrate(III) solution slowly
• Steady evolution of N2(g) is observed
• 1o aromatic amines: add naphthalen-2-ol in dilute sodium hydroxide solution
• An orange or red precipitate is formed
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B114
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Aromatic compounds
( )
• Burn the aromatic compound in a non-luminous Bunsen flame
• A smoky yellow flame with black soot is produced
• Add fuming sulphuric(VI) acid
• The aromatic compound dissolves
• The temperature of the reaction mixture rises
Chemical tests for different groups of organic compounds
New Way Chemistry for Hong Kong A-Level 3B115
34.7 Structural Information from Chemical Properties (SB p.96)
Example 34-7AExample 34-7A Example 34-7BExample 34-7B
Example 34-7CExample 34-7C Check Point 34-7Check Point 34-7
New Way Chemistry for Hong Kong A-Level 3B116
34.834.8Use of Infra-red Use of Infra-red Spectrocopy in Spectrocopy in
the the Identification of Identification of
Functional Functional GroupsGroups
New Way Chemistry for Hong Kong A-Level 3B117
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.99)
The Electromagnetic The Electromagnetic SpectrumSpectrum• Electromagnetic radiation has dual property
i.e. the properties of both wave and particle
• Can be described as a wave occurring simultaneously in electrical and magnetic fields
• Can also be described as consisting of particles called quanta or photons
New Way Chemistry for Hong Kong A-Level 3B118
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.99)
The Electromagnetic The Electromagnetic SpectrumSpectrum• All electromagnetic radiation travels through
vacuum at the same velocity, 3 108 m s-1
• The relationship between the frequency () of an electromagnetic radiation, its wavelength () and velocity (c) is:
λcν
New Way Chemistry for Hong Kong A-Level 3B119
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum• The energy of a quantum of electromagnetic
radiation is directly related to its frequency:
E = h
where h is the Planck constant (i.e. 6.626 10-34 J s).
New Way Chemistry for Hong Kong A-Level 3B120
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum
λcν
the energy of a quantum of electromagnetic radiation is inversely proportional to its wavelength:
λhcE
New Way Chemistry for Hong Kong A-Level 3B121
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum
• Electromagnetic radiation of long wavelength has low energy
• Electromagnetic radiation of short wavelength has high energy
New Way Chemistry for Hong Kong A-Level 3B122
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum
• Visible light has wavelength between 400 nm and 800 nm
• Infra-red radiation has wavelength between 800 nm and 300 m
New Way Chemistry for Hong Kong A-Level 3B123
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum
Regions of the electromagnetic spectrum
New Way Chemistry for Hong Kong A-Level 3B124
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum
• When electromagnetic radiation falls onto a hydrogen atom,
the electron in the hydrogen atom will absorb a definite amount of energy
• The electron is excited from the ground state to a higher energy level
New Way Chemistry for Hong Kong A-Level 3B125
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum
The electron is unstable at a higher energy level
it will fall back to a lower energy level
• Excess energy is given out in the form of electromagnetic radiation
New Way Chemistry for Hong Kong A-Level 3B126
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum• The radiation emitted has the frequency
as shown by the following relationship:
λhcE = E2 – E1 = h =
New Way Chemistry for Hong Kong A-Level 3B127
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100)
The Electromagnetic The Electromagnetic SpectrumSpectrum
• The atomic spectrum of hydrogen is originated from
electron transitions between energy levels in a hydrogen atom
New Way Chemistry for Hong Kong A-Level 3B128
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
The Electromagnetic The Electromagnetic SpectrumSpectrum• In the case of molecules, the absorption
of energy can
cause the excitation of electrons
increase the extent of vibration of the bonds and the speed of rotation of the molecule
• This is the basis of infra-red spectroscopy
New Way Chemistry for Hong Kong A-Level 3B129
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy
• Organic compounds absorb electromagnetic radiation in the IR region of the spectrum
IR radiation does not have sufficient energy to cause the excitation of electrons
New Way Chemistry for Hong Kong A-Level 3B130
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy
• IR radiation causes
atoms and groups of atoms of organic compounds to vibrate with increased amplitude about the covalent bonds that connect them
New Way Chemistry for Hong Kong A-Level 3B131
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy
• These vibrations are quantized
the compounds absorb IR radiation of a particular amount of energy
only
New Way Chemistry for Hong Kong A-Level 3B132
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy
Effect of absorption of IR radiation on vibration of atoms in a molecule
New Way Chemistry for Hong Kong A-Level 3B133
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy• Infra-red spectrometer is used to
measure the amount of energy absorbed at each wavelength of the IR region
An infra-red spectrometer
New Way Chemistry for Hong Kong A-Level 3B134
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy• A beam of IR radiation is passed through
the sample
the intensity of the emergent radiation is carefully measured
• The spectrometer plots the results as a graph called infra-red spectrum
shows the absorption of IR radiation by a sample at different frequencies
New Way Chemistry for Hong Kong A-Level 3B135
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy
• The IR radiation is usually specified by its wavenumber (unit: cm-1)
the reciprocal of wavelength
Frequency and wavelength are related by the equation c =
Wavenumber is a direct measure of frequency
New Way Chemistry for Hong Kong A-Level 3B136
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy
• Covalently bonded atoms have only particular vibrational energy levels
the levels are quantized
New Way Chemistry for Hong Kong A-Level 3B137
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy• When the compound absorbs IR radiation of
the exact energy required (or a particular wavelength or a particular frequency)
the excitation of a molecule from one vibrational energy level to another occ
urs only
New Way Chemistry for Hong Kong A-Level 3B138
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101)
Infra-red Infra-red SpectroscopySpectroscopy• Molecules can vibrate in a variety of ways
• Two atoms joined by a covalent bond can undergo a stretching vibration where the atoms move back and forth as if they were joined by a spring
A stretching vibration of two atoms
New Way Chemistry for Hong Kong A-Level 3B139
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)
Infra-red Infra-red SpectroscopySpectroscopy
A variety of stretching and bending vibrations
New Way Chemistry for Hong Kong A-Level 3B140
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)
Infra-red Infra-red SpectroscopySpectroscopy• The frequency of a given stretching
vibration of a covalent bond
depends on the masses of the bonded atoms and the strength of the bond
• Lighter atoms vibrate at higher frequencies than heavier ones
New Way Chemistry for Hong Kong A-Level 3B141
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)
Infra-red Infra-red SpectroscopySpectroscopy• The stretching vibrations of single bonds
involving hydrogen (C H, O H and N H) occur at relatively high frequencies
3350 – 3500N H
3230 – 3670O H
2840 – 3095C H
Range of wavenumber (cm-1)Bond
Characteristic absorption wavenumbers of some single bonds in infra-red spectra
New Way Chemistry for Hong Kong A-Level 3B142
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)
Infra-red Infra-red SpectroscopySpectroscopy
• Triple bonds are stronger and vibrate at higher frequencies than double bonds
1680 – 1750C = O
1610 – 1680C = C
2200 – 2280C N
2070 – 2250C C
Range of wavenumber (cm-1)Bond
Characteristic absorption wavenumbers of some double bonds and triple bonds in infra-
red spectra
New Way Chemistry for Hong Kong A-Level 3B143
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)
Infra-red Infra-red SpectroscopySpectroscopy• The IR spectra of even relatively simple
compounds contain many absorption peaks
• The possibility of two different compounds having the same IR spectrum is very small
• An IR spectrum has been called the “fingerprint” of a compound
New Way Chemistry for Hong Kong A-Level 3B144
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103)
• An IR spectrum is a plot of percentage of transmittance against wavenumber of IR radiation
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups
New Way Chemistry for Hong Kong A-Level 3B145
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103)
The IR spectrum of hex-1-yne
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups
New Way Chemistry for Hong Kong A-Level 3B146
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103)
• 100% transmittance in the spectrum
implies no absorption of IR radiation
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups
New Way Chemistry for Hong Kong A-Level 3B147
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103)
• When a compound absorbs IR radiation,
the intensity of transmitted radiation decreases
results in a decrease in percentage of transmittance
a dip in the spectrum
often called an absorption peak or absorption band
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups
New Way Chemistry for Hong Kong A-Level 3B148
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103)
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional Identification of Functional GroupsGroups• In general, an IR spectrum can be split
into four regions for interpretation purpose
New Way Chemistry for Hong Kong A-Level 3B149
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103)
Range of wavenumber (cm-1)
Interpretation
400 – 1500 • Often consists of many complicated bands
• Unique to each compound
• Often called the fingerprint region
• Not used for identification of particular functional groups
1500 – 2000 Absorption of double bonds,
e.g. C = C, C = O
2000 – 2500 Absorption of triple bonds, e.g. C C, C N
2500 – 4000 Absorption of single bonds involving hydrogen, e.g. C H, O H, N H
The four regions of an IR spectrum
New Way Chemistry for Hong Kong A-Level 3B150
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104)
• 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
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional Identification of Functional GroupsGroups
New Way Chemistry for Hong Kong A-Level 3B151
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104)
Compound Bond Characteristic range of wavenumber (cm-1)
Alkenes C = C 1610 – 1680
Aldehydes, ketones, acids, esters
C = O 1680 – 1750
Alkynes C C 2070 – 2250
Nitriles C N 2200 – 2280
Acids (hydrogen-bonded) O H 2500 – 3300
Alkanes, alkenes, arenes C H 2840 – 3095
Alcohols, phenols (hydrogen-bonded)
O H 3230 – 3670
Primary amines N H 3350 – 3500
Characteristic range of wavenumbers of covalent bonds in IR spectra
New Way Chemistry for Hong Kong A-Level 3B152
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104)
Interpretation of IR SpectraInterpretation of IR Spectra
1. Buta1. Butanene
The IR spectrum of butane
New Way Chemistry for Hong Kong A-Level 3B153
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104)
1. Buta1. ButaneneWavenumber (cm-1) Intensity Indication
2968 Very strong
C H stretching
2890 Medium
1468 Strong C H bending
Interpretation of the IR spectrum of butane
New Way Chemistry for Hong Kong A-Level 3B154
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105)
2. 2. ciscis-But-2-e-But-2-enene
The IR spectrum of cis-but-2-ene
New Way Chemistry for Hong Kong A-Level 3B155
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105)
2. 2. ciscis-But-2-e-But-2-eneneWavenumber (cm-1) Intensity Indication
3044 Very strong C H stretching (sp2 C H)3028 Very strong
2952 Very strong C H stretching (sp3 C H)
1677 Medium C = C stretchinh
1657 Medium
1411 Strong C H bending
Interpretation of the IR spectrum of cis-but-2-ene
New Way Chemistry for Hong Kong A-Level 3B156
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105)
3. Hex-1-y3. Hex-1-ynene
The IR spectrum of hex-1-yne
New Way Chemistry for Hong Kong A-Level 3B157
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105)
3. Hex-1-y3. Hex-1-yneneWavenumber (cm-1) Intensity Indication
3313 Very strong C H stretching (sp C H)
2963 Very strong C H stretching (sp3 C H)2938 Very strong
2874 Strong
2119 Strong C C stretching
1468 Strong C H bending (sp C H)
1445 Medium C H bending (sp3 C H)
Interpretation of the IR spectrum of hex-1-yne
New Way Chemistry for Hong Kong A-Level 3B158
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.106)
4. Butano4. Butanonene
The IR spectrum of butanone
New Way Chemistry for Hong Kong A-Level 3B159
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.106)
4. Butano4. ButanoneneWavenumber (cm-1) Intensity Indication
2983 Strong C H stretching
2925 Strong
1720 Very strong C = O stretching
1416 Medium C H bending (shifted as adjacent to C = O)
Interpretation of the IR spectrum of butanone
New Way Chemistry for Hong Kong A-Level 3B160
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107)
5. Butan-1-ol5. Butan-1-ol
The IR spectrum of butan-1-ol
New Way Chemistry for Hong Kong A-Level 3B161
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107)
5. Butan-1-ol5. Butan-1-olWavenumber
(cm-1)Intensity Indication
3330 Broad band O H stretching
2960 Medium C H stretching
2935 Medium
2875 Medium
Interpretation of the IR spectrum of butan-1-ol
New Way Chemistry for Hong Kong A-Level 3B162
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107)
6. Butanoic Ac6. Butanoic Acidid
The IR spectrum of butanoic acid
New Way Chemistry for Hong Kong A-Level 3B163
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107)
6. Butanoic Ac6. Butanoic AcididWavenumber (cm-
1)Intensity Indication
3100 Broad band O H stretching
1708 Strong C = O stretching
Interpretation of the IR spectrum of butanoic acid
New Way Chemistry for Hong Kong A-Level 3B164
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108)
6. Butanoic Ac6. Butanoic Acidid
• The absorption of the O H group in alcohols and carboxylic acids does not usually appear as a sharp peak
a broad band is observed
the vibration of the O H group is complicated by the hydrogen
bonding formed between the molecules
New Way Chemistry for Hong Kong A-Level 3B165
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108)
7. Butylami7. Butylaminene
The IR spectrum of butylamine
New Way Chemistry for Hong Kong A-Level 3B166
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108)
7. Butylami7. ButylamineneWavenumber (cm-
1)Intensity Indication
3371 Strong N H stretching
3280 Strong
2960 – 2875 Weak C H stretching
1610 Medium N H bending
1475 Medium C H bending
Interpretation of the IR spectrum of butylamine
New Way Chemistry for Hong Kong A-Level 3B167
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108)
8. Butanenitri8. Butanenitrilele
The IR spectrum of butanenitrile
New Way Chemistry for Hong Kong A-Level 3B168
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109)
8. Butanenitri8. Butanenitrilele
Wavenumber (cm-1)
Intensity Indication
2990 – 2895 Strong C H stretching
2246 Very strong C N stretching
1420 Strong C H bending
1480 Strong
Interpretation of the IR spectrum of butanenitrile
New Way Chemistry for Hong Kong A-Level 3B169
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109)
Strategies for the Use of IR Spectra in Strategies for the Use of IR Spectra in the Identification of Functional the Identification of Functional GroupsGroups
1. Focus at the IR absorption peak at or above 1500 cm–1
Concentrate initially on the major absorption peaks
New Way Chemistry for Hong Kong A-Level 3B170
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109)
Strategies for the Use of IR Spectra in Strategies for the Use of IR Spectra in the Identification of Functional the Identification of Functional GroupsGroups
2. For each absorption peak, try to list out all the possibilities using a table or chart
Not all absorption peaks in the spectrum can be assigned
New Way Chemistry for Hong Kong A-Level 3B171
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109)
Strategies for the Use of IR Spectra in Strategies for the Use of IR Spectra in the Identification of Functional the Identification of Functional GroupsGroups3. The absence and presence of absorption
peaks at some characteristic ranges of wavenumbers are equally important
the absence of particular absorption peaks can be used to eliminate the presence of certain functional group
s or bonds in the molecule
New Way Chemistry for Hong Kong A-Level 3B172
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109)
Limitation of the Use of IR Spectroscopy Limitation of the Use of IR Spectroscopy in the Identification of Organic in the Identification of Organic CompoundsCompounds
1. Some IR absorption peaks have very close wavenumbers and the peaks always coalesce
2. Not all vibrations give rise to strong absorption peaks
New Way Chemistry for Hong Kong A-Level 3B173
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109)
Limitation of the Use of IR Spectroscopy Limitation of the Use of IR Spectroscopy in the Identification of Organic in the Identification of Organic CompoundsCompounds
3. Not all absorption peaks in a spectrum can be associated with a particular bond or part of the molecule
4. Intermolecular interactions in molecules can result in complicated infra-red spectra
New Way Chemistry for Hong Kong A-Level 3B174
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
Example 34-8Example 34-8 Check Point 34-8Check Point 34-8
New Way Chemistry for Hong Kong A-Level 3B175
34.934.9Use of Mass Use of Mass Spectra to Spectra to
Obtain Obtain Structural Structural
InformationInformation
New Way Chemistry for Hong Kong A-Level 3B176
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Mass SpectrometrySpectrometry
• One of the most sensitive and versatile analytical tools
• More sensitive than other spectroscopic methods (e.g. IR spectroscopy)
• Only a microgram or less of materials is required for the analysis
New Way Chemistry for Hong Kong A-Level 3B177
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Mass SpectrometrySpectrometryIn a mass spectrometric analysis, it involves:
1. the conversion of molecules to ions
2. separation of the ions formed according to their mass-to-charge (m/e) ratio
m is the mass of the ion in atomic mass units and e is its charge
New Way Chemistry for Hong Kong A-Level 3B178
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Mass SpectrometrySpectrometry
• Finally, the number of ions of each type (i.e. the relative abundance of ions of each type) is determined
• The analysis is carried out using a mass spectrometer
New Way Chemistry for Hong Kong A-Level 3B179
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
Components of a mass spectrometer
New Way Chemistry for Hong Kong A-Level 3B180
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
In the vaporization chamber,
• the sample is heated until it vaporizes
changes to the gaseous state
New Way Chemistry for Hong Kong A-Level 3B181
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Mass
SpectrometrySpectrometry• The molecules in the gaseous state are
bombarded with a beam of fast-moving electrons
Positively-charged ions called the molecular ions are formed
One of the electrons of the molecule is knocked off
New Way Chemistry for Hong Kong A-Level 3B182
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Mass
SpectrometrySpectrometry
• Molecular ions are sometimes referred to as the parent ion
New Way Chemistry for Hong Kong A-Level 3B183
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Mass
SpectrometrySpectrometry one of the electrons is removed
from the molecules during the ionization process
the molecular ion contains a single unpaired electron
the molecular ion is not only a cation, it is also a free radical
New Way Chemistry for Hong Kong A-Level 3B184
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Mass
SpectrometrySpectrometry• e.g.
if a molecule of methanol (CH3OH) is bombarded with a beam of fast-moving electrons
the following reaction will take place:
New Way Chemistry for Hong Kong A-Level 3B185
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
• The molecular ions formed in the ionization chamber are energetically unstable
undergo fragmentation
• Fragmentation can take place in a variety of ways
depend on the nature of the particular molecular ion
New Way Chemistry for Hong Kong A-Level 3B186
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
• The way that a molecular ion fragments
give us highly useful information about the structure of a complex molecule
New Way Chemistry for Hong Kong A-Level 3B187
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry• The positively charged ions formed are
then accelerated by electric field and deflected by magnetic field
causes the ions to arrive the ion detector
• The lighter the ions, the greater the deflection
New Way Chemistry for Hong Kong A-Level 3B188
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
• Positively charged ions of higher charge have greater deflection
• Ions with a high m/e ratio are deflected to smaller extent than ions with a low m/e ratio
New Way Chemistry for Hong Kong A-Level 3B189
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Mass SpectrometrySpectrometry• In the ion detector,
the number of ions collected is measured electronically
• The intensity of the signal is
a measure of the relative abundance of the ions with a particular m/e ratio
New Way Chemistry for Hong Kong A-Level 3B190
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Mass SpectrometrySpectrometry
• The spectrometer shows the results by
plotting a series of peaks of varying intensity
each peak corresponds to ions of a particular m/e ratio
• The graph obtained is known as a mass spectrum
New Way Chemistry for Hong Kong A-Level 3B191
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Mass
SpectrumSpectrum• Generally published as bar graphs.
Mass spectrum of methanol
New Way Chemistry for Hong Kong A-Level 3B192
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Mass
SpectrumSpectrum
Corresponding ion m/e ratio
H3C+ 15
H CO+ 29
H2C = OH+ 31
CH3OH 32
Interpretation of the mass spectrum of methanol
New Way Chemistry for Hong Kong A-Level 3B193
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Formation of Formation of
FragmentsFragments
• The molecular ions formed in the ionization chamber are energetically unstable
Some of them may break up into smaller fragments
Called the daughter ions
New Way Chemistry for Hong Kong A-Level 3B194
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Formation of Formation of
FragmentsFragments• These ionized fragments are accelerated
and deflected by the electric field and magnetic field
• Finally, they are detected by the ion detector and
their m/e ratios are measured
explains why there are so many peaks appeared in mass spectra
New Way Chemistry for Hong Kong A-Level 3B195
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments
Mass spectrum of methanol
• The peak at m/e 31
the most intense peak
• Arbitrarily assigned an intensity of 100%
Called the base peak
Corresponds to the most
common ion formed
New Way Chemistry for Hong Kong A-Level 3B196
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments• The peak at m/e 31
corresponds to the ion H2C = OH+
formed by losing one hydrogen atom from the molecular ion
New Way Chemistry for Hong Kong A-Level 3B197
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments• The ion H2C = OH+ is a relatively stable ion
the positive charge is not localized on a particular atom
it spreads around the carbon and the oxygen atoms to form a delocalized system
New Way Chemistry for Hong Kong A-Level 3B198
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments• The peak at m/e 29 corresponds to the ion
HC O+
formed by losing two hydrogen atoms from the ion H2C = OH+
New Way Chemistry for Hong Kong A-Level 3B199
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments• The ion HC O+ has two resonance
structures:
New Way Chemistry for Hong Kong A-Level 3B200
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments• The peak at m/e 15 corresponds to the
ion H3C+
formed by the breaking of the C O bond in the molecular ion
New Way Chemistry for Hong Kong A-Level 3B201
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments
Mass spectrum of pentan-3-one
New Way Chemistry for Hong Kong A-Level 3B202
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments
Corresponding ion m/e ratio
CH3CH2+ 29
CH3CH2CO+ 57
CH3CH2COCH2CH3 86
Interpretation of the mass spectrum of pentan-3-one
New Way Chemistry for Hong Kong A-Level 3B203
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Formation of
FragmentsFragments• The fragmentation pattern of pentan-3-one
is summarized below:
New Way Chemistry for Hong Kong A-Level 3B204
Example 34-9AExample 34-9A Example 34-9BExample 34-9B
Example 34-9CExample 34-9C
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
New Way Chemistry for Hong Kong A-Level 3B205
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) Fragmentation Fragmentation
PatternPattern1. Straight-chain Alka1. Straight-chain Alkanesnes• Simple alkanes tend to undergo
fragmentation by
This carbocation can then undergo stepwise cleavage down the alkyl chain
the initial loss of a • CH3 to give a peak at M – 15
New Way Chemistry for Hong Kong A-Level 3B206
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 1. Straight-chain Alka1. Straight-chain Alka
nesnes• Take hexane as an example:
New Way Chemistry for Hong Kong A-Level 3B207
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)
2. Branched-chain Alka2. Branched-chain Alkanesnes
• Tend to cleave at the “branch point”
more stable carbocations are formed
New Way Chemistry for Hong Kong A-Level 3B208
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 2. Branched-chain Alka2. Branched-chain Alka
nesnes• e.g.
New Way Chemistry for Hong Kong A-Level 3B209
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
3. Alkyl-substituted Aromatic Hydrocar3. Alkyl-substituted Aromatic Hydrocarbonsbons• Undergo loss of a hydrogen atom or alkyl
group
yield the relatively stable tropylium ion
• Gives a prominent peak at m/e 91
New Way Chemistry for Hong Kong A-Level 3B210
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
3. Alkyl-substituted Aromatic Hydrocar3. Alkyl-substituted Aromatic Hydrocarbonsbons
• e.g.
New Way Chemistry for Hong Kong A-Level 3B211
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
4. Aldehydes and Keto4. Aldehydes and Ketonesnes• Frequently undergo fragmentation by losing
one of the side chains
generate the substituted oxonium ion
often represents the base peak in the mass spectra
New Way Chemistry for Hong Kong A-Level 3B212
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
4. Aldehydes and Keto4. Aldehydes and Ketonesnes
New Way Chemistry for Hong Kong A-Level 3B213
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 5. Esters, Carboxylic Acids and A5. Esters, Carboxylic Acids and A
midesmides• Often undergo cleavage that involves the
breaking of the C X bond
form substituted oxonium ions as shown below:
(where X = OH, OR, NH2, NHR, NR2)
New Way Chemistry for Hong Kong A-Level 3B214
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 5. Esters, Carboxylic Acids and A5. Esters, Carboxylic Acids and A
midesmides• For carboxylic acids and unsubstituted
amides,
characteristic peaks at m/e 45 and 44 are observed respectively
New Way Chemistry for Hong Kong A-Level 3B215
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 6. Alcoho6. Alcoho
lsls• In addition to the loss of a proton and the
hydroxyl radical,
alcohols tend to lose one of the alkyl groups (or hydrogen atoms)
form oxonium ions
New Way Chemistry for Hong Kong A-Level 3B216
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 6. Alcoho6. Alcoho
lsls• For primary alcohols,
the peak at m/e 31, 45, 59 or 73 often appears
depends on what the R1 group is
New Way Chemistry for Hong Kong A-Level 3B217
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 7. Haloalkan7. Haloalkan
eses• Haloalkanes simply break at the C X b
ond
New Way Chemistry for Hong Kong A-Level 3B218
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
7. Haloalkan7. Haloalkaneses
• In the mass spectra of chloroalkanes,
two peaks, separated by two mass units, in the ratio 3 : 1 will be appeared
New Way Chemistry for Hong Kong A-Level 3B219
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
7. Haloalkan7. Haloalkaneses
• In the mass spectra of bromoalkanes,
two peaks, separated by two mass units, having approximately equal intensities will be appeared
New Way Chemistry for Hong Kong A-Level 3B220
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
Check Point 34-9Check Point 34-9
New Way Chemistry for Hong Kong A-Level 3B221
The END
New Way Chemistry for Hong Kong A-Level 3B222
34.1 Introduction (SB p.77)
What are the necessary information to determine the structure of an organic compound? AnswerMolecular formula from analytical data,
functional group present from physical and
chemical properties, structural information from
infra-red spectroscopy and mass spectrometry
Back
New Way Chemistry for Hong Kong A-Level 3B223
34.2 Isolation and Purification of Organic Compounds (SB p.84)
For each of the following, suggest a separation technique.
(a) To obtain blood cells from blood
(b) To separate different pigments in black ink
(c) To obtain ethanol from beer
(d) To separate a mixture of two solids, but only one sublimes
(e) To separate an insoluble solid from a liquidAnswer(a) Centrifugation
(b) Chromatography
(c) Fractional distillation
(d) Sublimation
(e) Filtration
Back
New Way Chemistry for Hong Kong A-Level 3B224
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.87)
(a) Why is detection of carbon and hydrogen in organic compounds not necessary?
(b) What elements can be detected by sodium fusion test? Answer
(a) All organic compounds contain carbon
and hydrogen.
(b) Halogens, nitrogen and sulphur
Back
New Way Chemistry for Hong Kong A-Level 3B225
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
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. Answer
New Way Chemistry for Hong Kong A-Level 3B226
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
Let the mass of the compound be 100 g. Then,
mass of carbon in the compound = 40.0 g
mass of hydrogen in the compound = 6.7 g
mass of oxygen in the compound = 53.3 g
∴ The empirical formula of the organic compound is CH2O.
Carbon Hydrogen Oxygen
Mass (g) 40.0 6.7 53.3
Number of moles (mol)
Relative number of moles
Simplest mole ratio
1 2 1
33.312.040.0 7.6
1.06.7 33.3
16.053.3
13.333.33 2
3.336.7 1
3.333.33
Back
New Way Chemistry for Hong Kong A-Level 3B227
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
An organic compound Z has the following composition by mass:
(a) Calculate the empirical formula of compound Z.
(b) If the relative molecular mass of compound Z is 60.0, determine the molecular formula of compound Z. Answer
Element Carbon Hydrogen
Oxygen
Percentage by mass (%)
60.00 13.33 26.67
New Way Chemistry for Hong Kong A-Level 3B228
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89)
(a) Let the mass of the compound be 100 g. Then,
mass of carbon in the compound = 60.00 g
mass of hydrogen in the compound = 13.33 g
mass of oxygen in the compound = 26.67 g
∴ The empirical formula of the organic compound is C3H8O.
Carbon Hydrogen Oxygen
Mass (g) 60.00 13.33 26.67
Number of moles (mol)
Relative number of moles
Simplest mole ratio
3 8 1
512.060.00 33.13
1.013.33 67.1
16.026.67
81.67
13.33 11.671.67 3
1.675
New Way Chemistry for Hong Kong A-Level 3B229
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89)
(b) The molecular formula of the compound is (C3H8O)n.
Relative molecular mass of (C3H8O)n = 60.0
n × (12.0 × 3 + 1.0 × 8 + 16.0) = 60.0
n = 1
∴ The molecular formula of compound Z is C3H8O.
Back
New Way Chemistry for Hong Kong A-Level 3B230
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89)
An organic compound was found to contain carbon, hydrogen and oxygen only. On complete combustion, 0.15 g of this compound gave 0.22 g of carbon dioxide and 0.09 g of water. If the relative molecular mass of this compound is 60.0, determine the molecular formula of this compound.
Answer
New Way Chemistry for Hong Kong A-Level 3B231
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89)
Relative molecular mass of CO2 = 12.0 + 16.0 × 2 = 44.0
Mass of carbon in 0.22 g of CO2 = 0.22 g ×
= 0.06 g
Relative molecular mass of H2O = 1.0 × 2 + 16.0
= 18.0
Mass of hydrogen in 0.09 g of H2O = 0.09 g ×
= 0.01 g
Mass of oxygen in the compound = (0.15 – 0.06 – 0.01) g
= 0.08 g
44.012.0
18.02.0
New Way Chemistry for Hong Kong A-Level 3B232
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89)
∴ The empirical formula of the organic compound is CH2O.
Carbon Hydrogen Oxygen
Mass (g) 0.06 0.01 0.08
Number of moles (mol)
Relative number of moles
Simplest mole ratio
1 2 1
005.012.00.06 01.0
1.00.01 005.0
16.00.08
20.0050.01 1
0.0050.005 1
0.0050.005
New Way Chemistry for Hong Kong A-Level 3B233
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89)
Let the molecular formula of the compound be (CH2O)n.
Relative molecular mass of (CH2O)n= 60.0
n × (12.0 + 1.0 × 2 + 16.0) = 60.0
n = 2
∴ The molecular formula of the compound is C2H4O2.
Back
New Way Chemistry for Hong Kong A-Level 3B234
34.6 Structural Information from Physical Properties (SB p.92)
Why do branched-chain hydrocarbons have lower boiling points but higher melting points than the
corresponding straight-chain isomers?
AnswerBranched-chain hydrocarbons have lower boiling points than the
corresponding straight-chain isomers because the straight-chain
isomers are being flattened in shape. They have greater surface
area in contact with each other. Hence, molecules of the straight-
chain isomer are held together by greater attractive forces. On the
other hand, branched-chain hydrocarbons have higher melting
points than the corresponding straight-chain isomers because
branched-chain isomers are more spherical in shape and are
packed more efficiently in solid state. Extra energy is required to
break down the efficient packing in the process of melting.
Back
New Way Chemistry for Hong Kong A-Level 3B235
34.6 Structural Information from Physical Properties (SB p.92)
Why does the solubility of amines in water decrease in the order:
1o amines > 2o amines > 3o amines?
AnswerThe solubility of primary and secondary amines is
higher than that of tertiary amines because tertiary
amines cannot form hydrogen bonds between
water molecules. On the other hand, the solubility
of primary amines is higher than that of secondary
amines because primary amines form a greater
number of hydrogen bonds with water molecules
than secondary amines.
Back
New Way Chemistry for Hong Kong A-Level 3B236
34.6 Structural Information from Physical Properties (SB p.92)
Match the boiling points 65oC, –6oC and –88oC with the compounds CH3CH3, CH3NH2 and CH3OH. Explain your answer briefly.
Answer
New Way Chemistry for Hong Kong A-Level 3B237
34.6 Structural Information from Physical Properties (SB p.92)
Compounds Boiling point (°C)
CH3CH3 –88
CH3NH2 –6
CH3OH 65
Ethane (CH3CH3) is a non-polar compound. In pure liquid form, ethane
molecules are held together by weak van der Waals’ forces. However,
both methylamine (CH3NH2) and methanol (CH3OH) are polar
substances. In pure liquid form, their molecules are held together by
intermolecular hydrogen bonds. As van der Waals’ forces are much
weaker than hydrogen bonds, ethane has the lowest boiling point
among the three. Besides, as the O H bond in alcohols is more
polar than the N H bond in amines, the hydrogen bonds formed
between methylamine molecules are weaker than those formed
between methanol molecules. Thus, methylamine has a lower boiling
point than methanol.
Back
New Way Chemistry for Hong Kong A-Level 3B238
34.6 Structural Information from Physical Properties (SB p.92)
(a) Butan-1-ol boils at 118°C and butanal boils at 76°C.
(i) What are the relative molecular masses of butan-1- ol and butanal?
(ii) Account for the higher boiling point of butan-1-ol. Answer
(a) (i) The relative molecular masses of butan-
1-ol and butanal are 74.0 and 72.0
respectively.
(ii) Butan-1-ol has a higher boiling point
because it is able to form extensive hydrogen
bonds with each other, but the forces holding
the butanal molecules together are dipole-
dipole interactions only.
New Way Chemistry for Hong Kong A-Level 3B239
34.6 Structural Information from Physical Properties (SB p.92)
(b) Arrange the following compounds in order of increasing solubility in water. Explain your answer.
Ethanol, chloroethane, hexan-1-olAnswer(b) The solubility increases in the order: chloroethane <
hexan-1-ol < ethanol. Both hexan-1-ol and ethanol are
more soluble in water than chloroethane because
molecules of the alcohols are able to form extensive
hydrogen bonds with water molecules. Molecules of
chloroethane are not able to form hydrogen bonds with
water molecules and that is why it is insoluble in water.
Hexan-1-ol has a longer carbon chain than ethanol and
this explains why it is less soluble in water than ethanol.
New Way Chemistry for Hong Kong A-Level 3B240
34.6 Structural Information from Physical Properties (SB p.92)
(c) Explain why (CH3)3N (b.p.: 2.9°C) boils so much lower than CH3CH2CH2NH2 (b.p.: 48.7°C) despite they have the same molecular mass. Answer
(c) They are isomers. The primary amine is able to form
hydrogen bonds with the oxygen atom of water
molecules, but there is no hydrogen atoms directly
attached to the nitrogen atom in the tertiary amine.
New Way Chemistry for Hong Kong A-Level 3B241
34.6 Structural Information from Physical Properties (SB p.92)
(d) Match the boiling points with the isomeric carbonyl compounds.
Compounds: Heptanal, heptan-4-one, 2,4-dimethylpentan-3-one
Boiling points: 124°C, 144°C, 155°C Answer(d)
1252,4-Dimethylpentan-3-one
144Heptan-4-one
155Heptanal
Boiling point (oC)Compound
Back
New Way Chemistry for Hong Kong A-Level 3B242
34.7 Structural Information from Chemical Properties (SB p.96)
The empirical formula of an organic compound is CH2O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.
(a) Calculate the molecular formula of the compound.
Answer(a) Let the molecular formula of the compound be (CH2O)n.
Relative molecular mass of (CH2O)n= 60.0
n (12.0 + 1.0 2 + 16.0) = 60.0
n = 2
∴ The molecular formula of the compound is C2H4O2.
New Way Chemistry for Hong Kong A-Level 3B243
34.7 Structural Information from Chemical Properties (SB p.96)
The empirical formula of an organic compound is CH2O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.
(b) Deduce the structural formula of the compound.
Answer
New Way Chemistry for Hong Kong A-Level 3B244
34.7 Structural Information from Chemical Properties (SB p.96)
(b) The compound reacts with sodium hydrogencarbonate to give a
colourless gas which turns lime water milky. This indicates 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.
Therefore, the structural formula of the compound is:
New Way Chemistry for Hong Kong A-Level 3B245
34.7 Structural Information from Chemical Properties (SB p.96)
The empirical formula of an organic compound is CH2O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.
(c) Give the IUPAC name for the compound.Answer(c) The IUPAC name for the compound
is ethanoic acid.
Back
New Way Chemistry for Hong Kong A-Level 3B246
34.7 Structural Information from Chemical Properties (SB p.96)
15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3 of oxygen which was in excess. The mixture was exploded. 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 the volumes were measured under room temperature and pressure.
(b) To which homologous series does the hydrocarbon belong?
(c) Give the structural formula of the hydrocarbon.
Answer
New Way Chemistry for Hong Kong A-Level 3B247
34.7 Structural Information from Chemical Properties (SB p.97)
(a) Let the molecular formula of the compound be CxHy.
Volume of CxHy 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
CxHy + (x + )O2 xCO2 + H2O
Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 15 : 30
x = 2
2y
4y
3015
x1
New Way Chemistry for Hong Kong A-Level 3B248
34.7 Structural Information from Chemical Properties (SB p.97)
(a) Volume of CxHy reacted : Volume of O2 reacted = 1 : ( )
= 15 : 45
y = 4
The molecular formula of the compound is C2H4.
(b) C2H4 belongs to alkenes.
(c) The structural formula of the hydrocarbon is:
4y
x
4515
)4y
2(
1
34y
2
Back
New Way Chemistry for Hong Kong A-Level 3B249
34.7 Structural Information from Chemical Properties (SB p.97)
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 105oC 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 potassium hydroxide solution, the volume further decreased to 50 cm3.
(a) Calculate the molecular formula of the compound, assuming that all the volumes were measured under room temperature and pressure.
(b) The compound is found to contain a hydroxyl group ( OH) in its structure. Deduce its structural formula.
(c) Is the compound optically active? Explain your answer.
Answer
New Way Chemistry for Hong Kong A-Level 3B250
34.7 Structural Information from Chemical Properties (SB p.97)
(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
= (90 - 50) cm3 = 40 cm3
CxHyOz + (x + - )O2 xCO2 + H2O
Volume of CxHyOz reacted : Volume of CO2 formed = 1 : x = 20 : 40
x = 2
2y
4y
2z
4020
x1
New Way Chemistry for Hong Kong A-Level 3B251
34.7 Structural Information from Chemical Properties (SB p.98)
(a) Volume of CxHyOz reacted : Volume of H2O formed = 1 : = 20 : 60
y = 6
Volume of CxHyOz reacted : Volume of O2 reacted = 1 :
= 20 : 60
z = 1
The molecular formula of the compound is C2H6O.
2y
6020
y2
)2z
4y
x(
6020
)2z
-4y
(x
1
32z
46
2
New Way Chemistry for Hong Kong A-Level 3B252
34.7 Structural Information from Chemical Properties (SB p.98)
(b) As the compound contains a OH group, the hydrocarbon skeleto
n of the compound becomes C2H5 after eliminating the
OH group from the molecular formula of C2H6O. The structural f
ormula of the compound is:
(c) The compound is optically inactive as both carbon atoms in the co
mpound are not asymmetric, i.e. both of them do not attach to four
different atoms or groups of atoms.
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New Way Chemistry for Hong Kong A-Level 3B253
34.7 Structural Information from Chemical Properties (SB p.99)
(a) A substance contains 42.8% carbon, 2.38% hydrogen, 16.67% nitrogen by mass and the remainder consists of oxygen.
(i) Given that the relative molecular mass of the substance is 168.0, deduce the molecular formula of the substance.
(ii) The substance is proved to be an aromatic compound with only one type of functional group. Give the names and structural formulae for all isomers of the substance. Answer
New Way Chemistry for Hong Kong A-Level 3B254
34.7 Structural Information from Chemical Properties (SB p.99)
(a) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the compound is C3H2NO2.
Carbon Hydrogen Nitrogen Oxygen
Mass (g) 42.8 2.38 16.67 38.15
Number of moles (mol)
Relative number of moles
Simplest mole ratio
3 2 1 2
57.30.128.42 38.2
0.138.2 19.1
0.1467.16 38.2
0.1615.38
319.157.3 2
19.138.2 1
19.119.1 2
19.138.2
New Way Chemistry for Hong Kong A-Level 3B255
34.7 Structural Information from Chemical Properties (SB p.99)
(a) (i) Let the molecular formula of the compound be (C3H2NO2)n.
Molecular mass of (C3H2NO2)n = 168.0
n × (12.0 × 3 + 1.0 × 2 + 14.0 + 16.0 × 2) = 168.0
∴ n = 2
∴ The molecular formula of the compound is C6H4N2O4.
(ii)
New Way Chemistry for Hong Kong A-Level 3B256
34.7 Structural Information from Chemical Properties (SB p.99)
(b) 30 cm3 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. By 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.
(ii) Is the hydrocarbon a saturated, an unsaturated or an aromatic hydrocarbon?Answer
New Way Chemistry for Hong Kong A-Level 3B257
34.7 Structural Information from Chemical Properties (SB p.99)
(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
CxHy + (x + )O2 xCO2 + H2O
Volume of CxHy reacted : Volume of CO2 formed
= 1 : x = 30 : 60
x = 2
2y
4y
6030
x1
New Way Chemistry for Hong Kong A-Level 3B258
34.7 Structural Information from Chemical Properties (SB p.99)
(b) (i) Volume of CxHy reacted : Volume of O2 reacted
= 1 : ( ) = 30 : 105
y = 6
The molecular formula of the compound is C2H6.
(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.
4y
2
10530
)4y
x(
1
105)4y
2(30
New Way Chemistry for Hong Kong A-Level 3B259
(c) A hydrocarbon having a relative molecular mass of 56.0 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.
(ii) Name the two geometrical isomers of the hydrocarbon.
(iii) Explain the existence of geometrical isomerism in the hydrocarbon. Answer
34.7 Structural Information from Chemical Properties (SB p.99)
New Way Chemistry for Hong Kong A-Level 3B260
34.7 Structural Information from Chemical Properties (SB p.99)
(c) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the compound is CH2.
Carbon Hydrogen
Mass (g) 85.5 14.5
Number of moles (mol)
Relative number of moles
Simplest mole ratio
1 2
125.70.125.85 5.14
0.15.14
1125.7125.7 2
125.75.14
New Way Chemistry for Hong Kong A-Level 3B261
34.7 Structural Information from Chemical Properties (SB p.99)
(c) (i) Let the molecular formula of the hydrocarbon be (CH2)n.
Molecular mass of (CH2)n = 56.0
n × (12.0 + 1.0 × 2) = 56.0
n = 4
∴ The molecular formula of the hydrocarbon is C4H8.
(ii)
(iii) Since but-2-ene is unsymmetrical and free rotation of but-
2-ene is restricted by the presence of the carbon-carbon double
bond, geometrical isomerism exists.
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New Way Chemistry for Hong Kong A-Level 3B262
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)
What is the relationship between frequency andwavenumber?
AnswerThe higher the frequency, the higher the
wavenumber.
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New Way Chemistry for Hong Kong A-Level 3B263
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
An organic compound with a relative molecular mass of 72.0 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.
New Way Chemistry for Hong Kong A-Level 3B264
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
(a) Determine the molecular formula of the compound.
(b) Deduce two possible structures of the compound, each of which belongs to a different homologous series. Answer
New Way Chemistry for Hong Kong A-Level 3B265
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
(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 compound is C4H8O.
Carbon Hydrogen Oxygen
Mass (g) 66.66 11.11 22.23
Number of moles (mol)
Relative number of moles
Simplest mole ratio
4 8 1
56.50.1266.66 11.11
0.111.11 39.1
0.1623.22
439.156.5 8
39.111.11 1
39.139.1
New Way Chemistry for Hong Kong A-Level 3B266
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
Let the molecular formula of the compound be (C4H8O)n.
Relative molecular mass of (C4H8O)n = 72.0
n × (12.0 × 4 + 1.0 × 8 + 16.0) = 72.0
∴ n = 1
∴ The molecular formula of the compound is C4H8O.
New Way Chemistry for Hong Kong A-Level 3B267
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
(b) From the IR spectrum, it can be observed that there are absorption
peaks at 2 950 cm–1 and 1 700 cm–1. The absorption peak at 2 950 c
m–1 corresponds to the stretching vibration of the C H bond, and t
he absorption peak at 1 700 cm–1 corresponds to the stretching vibra
tion of the C = O bond. Since there is only one oxygen atom in the
molecule of the compound, we can deduce that the compound is eit
her an aldehyde or a ketone.
If it is an aldehyde, its possible structure will be:
New Way Chemistry for Hong Kong A-Level 3B268
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
(b) If it is a ketone, its possible structure will be:
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New Way Chemistry for Hong Kong A-Level 3B269
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111)
(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.
New Way Chemistry for Hong Kong A-Level 3B270
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111)
Sketch the infra-red spectrum of a carboxylic acid based on the IR spectra of X and Y.
Answer
New Way Chemistry for Hong Kong A-Level 3B271
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111)
(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 expected. A
broad 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.
New Way Chemistry for Hong Kong A-Level 3B272
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111)
(a) The infrared spectrum of a carboxylic acid is as follows:
New Way Chemistry for Hong Kong A-Level 3B273
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(b) The infra-red spectra of two organic compounds A and B are shown below.
Decide which compound could be an alcohol. Explain your answer briefly.
Answer
New Way Chemistry for Hong Kong A-Level 3B274
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(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.
New Way Chemistry for Hong Kong A-Level 3B275
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(c) The table below shows the characteristic absorption wavenumbers of some covalent bonds in infra-red spectra.
Bond Range of wavenumber (cm-1)
C = O 1680 – 1750O H 2500 – 3300C H 2840 – 3095N H 3350 – 3500
Answer
New Way Chemistry for Hong Kong A-Level 3B276
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
Sketch the expected infra-red spectrum for an amino acid with the following structure:
New Way Chemistry for Hong Kong A-Level 3B277
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(c) The infra-red spectrum of the amino acid is shown as follows:
New Way Chemistry for Hong Kong A-Level 3B278
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(d) A portion of the infra-red spectrum of an organic compound X is shown below. To which homologous series does it belong?
Answer
New Way Chemistry for Hong Kong A-Level 3B279
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(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.
New Way Chemistry for Hong Kong A-Level 3B280
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(e) A portion of the infra-red spectrum of an organic compound Y is shown on the right. Identify the functional groups that it contains.
Answer
New Way Chemistry for Hong Kong A-Level 3B281
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(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.
New Way Chemistry for Hong Kong A-Level 3B282
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113)
(f) An organic compound Z with a relative molecular mass of 88.0 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:
New Way Chemistry for Hong Kong A-Level 3B283
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113)
(i) Determine the molecular formula of compound Z.
(ii) Based on the result from (i), draw two possible structures of the compound, each of which belongs to a different homologous series.
(iii) Using the information from the IR spectrum, name the homologous series that compound Z belongs to. Explain your answer.
Answer
New Way Chemistry for Hong Kong A-Level 3B284
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113)
(f) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the compound is C2H4O.
Carbon Hydrogen Oxygen
Mass (g) 54.54 9.10 36.36
Number of moles (mol)
Relative number of moles
Simplest mole ratio
2 4 1
55.40.1254.54 10.9
0.110.9 27.2
0.1636.36
227.255.4 4
27.210.9 1
27.227.2
New Way Chemistry for Hong Kong A-Level 3B285
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113)
(f) (i) Let the molecular formula of the compound be (C2H4O)n.
Relative molecular mass of (C2H4O)n = 88.0
n × (12.0 × 2 + 1.0 × 4 + 16.0) = 88.0
n = 2
∴ The molecular formula of the compound is C4H8O2.
(ii)
New Way Chemistry for Hong Kong A-Level 3B286
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113)
(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.
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New Way Chemistry for Hong Kong A-Level 3B287
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
The mass spectrum of pentan-2-one (CH3COCH2CH2CH3) is shown below:
What ions do the peaks at m/e 86, 71 and 43 represent? Explain your answer.
Answer
New Way Chemistry for Hong Kong A-Level 3B288
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
The relative molecular mass of pentan-2-one is 86. Therefore, the peak
at m/e 86 corresponds to the molecular ion of pentan-2-one. When the
C1 C2 bond is broken, the ion CH3CH2CH2CO+ (m/e = 71) is formed.
When the C2 C3 bond is broken, the ion CH3CO+ (m/e = 43) is formed.
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New Way Chemistry for Hong Kong A-Level 3B289
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
The mass spectrum of hydrocarbon X is shown below:
New Way Chemistry for Hong Kong A-Level 3B290
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
(a) What is the relative molecular mass of hydrocarbon X?
(b) Which peak is the base peak?
(c) How many mass units is the base peak less than the peak for the molecular ion?
(d) Deduce the structures of hydrocarbon X.
(e) Explain the peak at m/e 43.
(f) Propose the fragmentation pattern of the molecular ion which gives rise to the peaks at m/e 58, 43, 29 and 15. Answer
New Way Chemistry for Hong Kong A-Level 3B291
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
(a) The relative molecular mass of hydrocarbon X is 58.0.
(b) The base peak is at m/e 43.
(c) 15 mass units
(d) Since the compound is a hydrocarbon, the molecular formula of the
compound must be CxHy. From the relative molecular mass of the
compound (i.e. 58.0), we can deduce that the compound contains 4
carbon atoms only. (If the compound contains 5 carbon atoms, the
relative molecular mass would be more than 12.0 × 5 = 60.0). The
number of hydrogen atoms in the compound is (58.0 - 12.0 × 4 =
10) 10. Therefore, the hydrocarbon is butane.
New Way Chemistry for Hong Kong A-Level 3B292
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
(e) The peak at m/e 43 is 15 mass units less than the molecular ion. Thi
s suggests that a methyl group is lost during the fragmentation of th
e molecular ion. The peak at m/e 43 corresponds to CH3CH2CH2+.
(f)
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New Way Chemistry for Hong Kong A-Level 3B293
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120)
An organic compound is investigated. The structural formula of this compound is shown below:
New Way Chemistry for Hong Kong A-Level 3B294
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120)
The mass spectrum of the compound is shown below:
Interpret the peaks at m/e 134, 119, 91 and 43. Answer
New Way Chemistry for Hong Kong A-Level 3B295
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120)
The peak at m/e 134 corresponds to the molecular ion. The peak at m/e
119 corresponds to the ion that is 15 mass units less than the molecular
ion. This suggests that a methyl group is lost from the molecular ion. The
peak at m/e 91 is the base peak, which corresponds to the ion C6H5CH2+
. The peak at m/e 43 corresponds to the ion CH3CO+.
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New Way Chemistry for Hong Kong A-Level 3B296
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
Why would the molecular ion compound have two peaks, separated by two mass units, in the ratio 3 :
1?
AnswerChlorine has two isotopes, chlorine-35
and chlorine-37. Their relative
abundances are in the ratio of 3 : 1.
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New Way Chemistry for Hong Kong A-Level 3B297
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
Why would the molecular ion of a bromine-containing
compound have two peaks, separated by two mass units, having approximately equal intensities?
AnswerBromine has two isotopes, bromine-79
and bromine-81. Their relative
abundances are in the ratio of 1 : 1.
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New Way Chemistry for Hong Kong A-Level 3B298
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(a) What is base peak in a mass spectrum? Why is the m/e of the base peak not the molecular mass of the compound? Answer
(a) The base peak is the most intense peak in a mass spectrum. It
represents the most stable ion formed during fragmentation or the
ion that can be formed in various ways during fragmentation of
the molecular ion. As molecular ions are usually unstable and will
undergo fragmentation, they do not normally show up as base
peaks in mass spectra.
New Way Chemistry for Hong Kong A-Level 3B299
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(b) The following is the mass spectrum of bromomethylbenzene (benzyl bromide).
Interpret the peaks at m/e = 172, 170 and 91. Answer
New Way Chemistry for Hong Kong A-Level 3B300
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(b) The relative molecular mass of bromomethylbenzene (benzyl
bromide) is 171.0. However, as bromine contains equal
abundances of the 79Br and 81Br isotopes, the spectrum shows
two small peaks of equal intensity at m/e = 172 and 170. The
base peak at m/e = 91 is due to the formation of the ion
C6H5CH2+.
New Way Chemistry for Hong Kong A-Level 3B301
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(c) Study the following spectrum carefully and deduce what group of organic compound it is. The compound has a relative molecular mass of 114.
Answer
New Way Chemistry for Hong Kong A-Level 3B302
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(c) The base peak is at m/e = 57 which may be an oxonium ion or a
carbocation. This is a mass spectrum of a ketone, an aldehyde
or a hydrocarbon.
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