structural determination of organic compounds
DESCRIPTION
34. Structural Determination of Organic Compounds. 34.1 Introduction 34.2 Isolation and Purification of Organic Compounds 34.3 Tests for Purity 34.4 Qualitative Analysis of Elements in an Organic Compound - PowerPoint PPT PresentationTRANSCRIPT
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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
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The general steps to determine the structure of an organic compound
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Isolation and Isolation and Purification Purification of Organic of Organic
CompoundsCompounds
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Technique Aim
1. Filtration To separate an insoluble solid from a liquid (slow)
2. Centrifugation To separate an insoluble solid from a liquid (fast)
3. Recrystallization To separate a solid from other solids based on their different solubilities in suitable solvent(s)
4. Solvent extraction
To separate a component from a mixture with a suitable solvent
5. Distillation To separate a liquid from a solution containing non-volatile solutes
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Technique Aim
6. Fractional distillation
To separate miscible liquids with widely different boiling points
7. Steam distillation To separate liquids which are immiscible with water and decompose easily below their b.p.
8. Vacuum distillation ditto
9. Sublimation To separate a mixture of solids in which only one can sublime
10. Chromatography To separate a complex mixture of substances (large/small scale)
The mixture boils below 100C
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• 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
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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
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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
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34.2 Isolation and Purification of Organic Compounds (SB p.78)
FiltrationFiltration
The laboratory set-up of filtration
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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
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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
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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
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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
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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
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34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot 1. Crystallization by Cooling a Hot Concentrated SolutionConcentrated 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
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34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot 1. Crystallization by Cooling a Hot Concentrated SolutionConcentrated 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
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34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot 1. Crystallization by Cooling a Hot Concentrated SolutionConcentrated Solution
Crystallization by cooling a hot concentrated solution
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34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a 2. Crystallization by Evaporating a Cold Solution at Room TemperatureCold Solution 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
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34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a 2. Crystallization by Evaporating a Cold Solution at Room TemperatureCold Solution 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
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34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a 2. Crystallization by Evaporating a Cold Solution at Room TemperatureCold Solution at Room Temperature
Crystallization by slow evaporation of a solution (preferably saturated) at room
temperature
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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
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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
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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
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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
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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
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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
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34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
The laboratory set-up of distillation
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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
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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
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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
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34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional DistillationFractional Distillation
The laboratory set-up of fractional distillation
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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
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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
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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
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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
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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
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34.2 Isolation and Purification of Organic Compounds (SB p.83)
SublimatioSublimationn
A mixture of two compounds can be separated by sublimation
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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
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34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
The laboratory set-up of paper chromatography
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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
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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
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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
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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
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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
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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
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34.34.44 Qualitative Qualitative
Analysis of Analysis of Elements in an Elements in an
Organic Organic CompoundCompound
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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
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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
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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
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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
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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
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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
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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
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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
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34.34.55Determination of Determination of
Empirical Empirical Formula and Formula and
Molecular Molecular Formula from Formula from
Analytical DataAnalytical Data
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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
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34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
1. Carbon and 1. Carbon and
HydrogenHydrogen• 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
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34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
1. Carbon and 1. Carbon and
HydrogenHydrogen• The increases in mass in potassium hydroxide solution and calcium chloride represent
the masses of carbon dioxide and water vapour formed respectively
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34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
2. 2.
NitrogenNitrogen• 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
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34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
3. 3.
HalogensHalogens• 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
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34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
4. 4.
SulphurSulphur• The organic compound is heated with
fuming nitric(V) acid
• After cooling,
barium nitrate solution is added
the dry barium sulphate formed is weighed
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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
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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
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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
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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
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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
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34.34.66 Structural Structural
Information Information from Physical from Physical
PropertiesProperties
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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)
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34.6 Structural Information from Physical Properties (SB p.92)
Example 34-6Example 34-6 Check Point 34-6Check Point 34-6
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34.34.77 Structural Structural
Information Information from Chemical from Chemical
PropertiesProperties
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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
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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:
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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
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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
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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
91
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
92
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
93
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
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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
95
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
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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)
97
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)
98
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
99
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
100
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
101
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
102
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
103
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
104
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
105
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
106
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
107
The END
108
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
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109
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
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110
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
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111
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
112
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
7.61.06.7
33.316.053.3
13.333.33
23.336.7
13.333.33
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113
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
114
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
33.131.0
13.33 67.1
16.026.67
81.67
13.33 1
1.671.67
31.67
5
115
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.
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116
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
117
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
118
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
01.01.00.01
005.016.00.08
20.0050.01
10.0050.005
10.0050.005
119
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.
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120
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.
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121
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.
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122
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
123
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.
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124
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.
125
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.
126
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.
127
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°CAnswer(d)
1252,4-Dimethylpentan-3-one
144Heptan-4-one
155Heptanal
Boiling point (oC)Compound
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128
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.
129
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
130
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:
131
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.
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132
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
133
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
134
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
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135
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
136
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
137
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
138
34.7 Structural Information from Chemical Properties (SB p.98)
(b) As the compound contains a OH group, the hydrocarbon
skeleton of the compound becomes C2H5 after eliminating the
OH group from the molecular formula of C2H6O. The structural
formula of the compound is:
(c) The compound is optically inactive as both carbon atoms in the
compound are not asymmetric, i.e. both of them do not attach to
four different atoms or groups of atoms.
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139
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
140
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
38.20.138.2
19.10.14
67.16 38.2
0.1615.38
319.157.3
219.138.2
119.119.1
219.138.2
141
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)
142
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
143
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
144
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
145
(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)
146
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
5.140.15.14
1125.7125.7
2125.7
5.14
147
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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148
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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