Structural Determination of Organic Compounds

Structural Determination of Organic Compounds
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 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data 34.6 Structural Information from Physical Properties 34.7 Structural Information from Chemical Properties 34.8 Use of Infra-red Spectrocopy in the Identification of Functional Groups 34.9 Use of Mass Spectra to Obtain Structural Information

The general steps to determine the structure of an organic compound

Isolation and Purification of Organic Compounds

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

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 100C

Tests for Purity 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

Use of Infra-red Spectrocopy in the Identification of Functional Groups

Infra-red Spectroscopy
Arises from absorption of IR radiation by organic compounds Causes atoms and groups of atoms of organic compounds to vibrate with increased amplitude about the covalent bonds that connect them

Modes of vibrations Two basic modes : stretching and bending
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

Modes of vibrations Symmetrical stretching : -
Asymmetrical stretching : -

Modes of vibrations Bending (scissoring) : - two modes Behind paper
Out of paper

An out-of-plane bending
Modes of vibrations Rocking Wagging Symmetric stretching Asymmetric stretching An in-plane bending (scissoring) An out-of-plane bending (twisting)

For a diatomic molecule X – Y
The frequency() of a given stretching vibration of a covalent bond depends on Bond strength Reduced mass of the system For a diatomic molecule X – Y

Bond strength  k     C – C < C = C < C  C C – O < C = O Increasing frequency of vibration

Masses of bonding atoms      
C – H O - H N - H Chemical bonds containing H atoms have high frequency of vibration due to the small mass of H

E = h Infra-red Spectroscopy Molecular vibrations are quantized
the molecules absorb IR radiation of a particular amount of energy only. E = h Only IR radiation with the same frequency as the vibrational frequency can be absorbed by the molecules.

E = h Infra-red Spectroscopy
After the absorption of a quantum of energy (h), the amplitude of vibration  but the frequency of vibration () remains unchanged.

The electromagnetic spectrum
Wavelength (m)   Frequency (Hz / s1) UV / visible / Near IR  electronic transition E = hL = 400 – 1000 kJ mol1

Wavelength (m)   Frequency (Hz / s1) Mid IR  vibrational transition E = hL = 5 – 40 kJ mol1

Wavelength (m)   Frequency (Hz / s1) Far IR / microwave  rotational transition E = hL  0.02 kJ mol1

Wavelength (m)   Frequency (Hz / s1) translational transition E = hL  0 kJ mol1

Different functional groups (e.g. C=O, O-H) have different vibration frequencies  They have characteristic absorption frequencies Functional groups can be identified from their characteristic absorption frequencies

Range of wavenumber (cm-1)
Infra-red Spectroscopy The stretching vibrations of single bonds involving hydrogen (C  H, O  H and N  H) occur at relatively high frequencies 3350 – 3500 N  H 3230 – 3670 O  H 2840 – 3095 C  H Range of wavenumber (cm-1) Bond Characteristic absorption wavenumbers of some single bonds in infra-red spectra

Infra-red Spectroscopy Triple bonds are stronger and vibrate at higher frequencies than double bonds 1680 – 1750 C = O 1610 – 1680 C = C 2200 – 2280 C  N 2070 – 2250 C  C Range of wavenumber (cm-1) Bond Characteristic absorption wavenumbers of some double bonds and triple bonds in infra-red spectra

An IR spectrum is obtained by scanning the sample with IR radiations from 1.21013 Hz to 1.21014 Hz Or, Wavenumber : 400 cm1 to 4000 cm-1 % transmittance rather than absorbance is displayed

Dips(peaks) show absorptions by functional groups
C=C

Fingerprint region Band region

Infra-red spectrometer is used to  measure the amount of energy absorbed at each wavelength of the IR region An infra-red spectrometer

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

Use of IR Spectrum in the Identification of Functional Groups
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

In general, an IR spectrum can be split into four regions for interpretation purpose

Interpretation 400 – 1500 Often consists of many complicated bands (stretching and bending) 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 region between cm-1 and cm-1 is often used for  identification of functional groups from their characteristic absorption wavenumbers

Characteristic range of wavenumber (cm-1)
What is the characteristic range of wavenumber of C=N bond? Bond strength/wavenumber : C=C < C=N < C=O 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) 3230 – 3670 Primary amines N  H 3350 – 3500

Strategies for the Use of IR Spectra in the Identification of Functional Groups
Focus at the IR absorption peak at or above 1500 cm–1  Concentrate initially on the major absorption peaks

Strategies for the Use of IR Spectra in the Identification of Functional Groups
2. 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 groups or bonds in the molecule

Limitation of the Use of IR Spectroscopy in the Identification of Organic Compounds
1. Some IR absorption peaks have very close wavenumbers and the peaks always coalesce 2. Not all vibrations give rise to strong absorption peaks

Limitation of the Use of IR Spectroscopy in the Identification of Organic Compounds
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

C=C bonds in benzene are weaker  ~1610 cm1
Alkene Arene cm1 Y Y Absorption Slightly above 3000 cm1 C=C bonds in benzene are weaker  ~1610 cm1

Aldehyde : lower C-H absorption at 2720-2820 cm-1
Contain oxygen Y Carboxylic acid Y Y cm1 Broad cm1 cm1 N Aldehyde, Ketone, Ester N Aldehyde : lower C-H absorption at cm-1

Y Y N N Additional sharp peak ~3250 cm1 for termnal alkyne, -CC-H
Nitrile cm1 Y Contain nitrogen Y cm1 N cm1 Alkyne cm1 N Additional sharp peak ~3250 cm1 for termnal alkyne, -CC-H

Y Y N N Y Single peak for 2 amine; double peaks for 1 amine Amine
cm1 Y Single peak for 2 amine; double peaks for 1 amine Contain nitrogen Y cm1 Alkane cm1 N Y Alcohol, phenol cm1 N Contain oxygen

weak Several peaks ( cm1) due to 3300 different modes of Symmetrical & asymmetrical stretching 3100 medium Different modes of C-H bending (~1400 cm1)

Q.84 C3H5N IOU = /2 + ½ = 2 CC, CN, diene CC, CN

 C-H 1 amine CC, CN

Ozonolysis gives CO2  terminal C=CH2
Q.85 C6H10O IOU = /2 = 2 Ozonolysis gives CO2  terminal C=CH2

Q.85 C6H10O IOU = /2 = 2 +ve iodoform test 

C6H10O IOU = 6 + 1 -10/2 = 2 No broadband within 3200-3600  =C-H -C-H
Q.85 C6H10O IOU = /2 = 2 No broadband within  =C-H -C-H C=C C=O

No C-H stretching above 3000 cm1
No C=C-H  1 amine N-H bending

Broad band with peak at about 2900 cm1

1 amine

No aliphatic C-H stretching

No C-H stretching above 3000 cm1
No C=C-H  1 amine N-H bending

No aliphatic C-H stretching

2 amine

The IR spectrum of butanone

Interpretation of the IR spectrum of butanone
Wavenumber (cm-1) Intensity Indication 2983 Strong C  H stretching 2925 1720 Very strong C = O stretching 1416 Medium C  H bending (shifted as adjacent to C = O) Interpretation of the IR spectrum of butanone

5. Butan-1-ol  a broad band is observed
 the vibration of the O  H group is complicated by the hydrogen bonding formed between the molecules

Interpretation of the IR spectrum of butan-1-ol
Wavenumber (cm-1) Intensity Indication 3330 Broad band O  H stretching 2960 Medium C  H stretching 2935 2875 Interpretation of the IR spectrum of butan-1-ol

6. Butanoic Acid  a broad band is observed
 the vibration of the O  H group is complicated by the hydrogen bonding formed between the molecules

Interpretation of the IR spectrum of butanoic acid
Wavenumber (cm-1) Intensity Indication 3100 Broad band O  H stretching 1708 Strong C = O stretching Interpretation of the IR spectrum of butanoic acid

The IR spectrum of butylamine

Interpretation of the IR spectrum of butylamine
Wavenumber (cm-1) Intensity Indication 3371 Strong N  H stretching 3280 2960 – 2875 Weak C  H stretching 1610 Medium N  H bending 1475 C  H bending Interpretation of the IR spectrum of butylamine

The IR spectrum of butanenitrile

Interpretation of the IR spectrum of butanenitrile
Wavenumber (cm-1) Intensity Indication 2990 – 2895 Strong C  H stretching 2246 Very strong C  N stretching 1420 C  H bending 1480 Interpretation of the IR spectrum of butanenitrile

Example 34-8 Check Point 34-8

Use of Mass Spectra to Obtain Structural Information
34.9 Use of Mass Spectra to Obtain Structural Information

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Spectrometry 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

Mass Spectrometry In a mass spectrometric analysis, it involves:
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113) Mass Spectrometry In a mass spectrometric analysis, it involves: the conversion of molecules to ions 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

Mass Spectrometry 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

Components of a mass spectrometer
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry Components of a mass spectrometer

Mass Spectrometry In the vaporization chamber,
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry In the vaporization chamber, the sample is heated until it vaporizes  changes to the gaseous state

Mass Spectrometry 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

Mass Spectrometry Molecular ions are sometimes referred to as the parent ion

Mass Spectrometry  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

Mass Spectrometry e.g. if a molecule of methanol (CH3OH) is bombarded with a beam of fast-moving electrons  the following reaction will take place:

Mass Spectrometry 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

Mass Spectrometry The way that a molecular ion fragments
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry The way that a molecular ion fragments  give us highly useful information about the structure of a complex molecule

Mass Spectrometry 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

Mass Spectrometry 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

Mass Spectrometry In the ion detector,
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrometry 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

Mass Spectrometry The spectrometer shows the results by
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrometry 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

Mass Spectrum Generally published as bar graphs.
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrum Generally published as bar graphs. Mass spectrum of methanol

Interpretation of the mass spectrum of methanol
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrum Corresponding ion m/e ratio H3C+ 15 H  CO+ 29 H2C = OH+ 31 CH3OH 32 Interpretation of the mass spectrum of methanol

Formation of Fragments
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Formation of Fragments The molecular ions formed in the ionization chamber are energetically unstable  Some of them may break up into smaller fragments  Called the daughter ions

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Formation of Fragments 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

Mass spectrum of methanol
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments 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 Mass spectrum of methanol

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments The peak at m/e 31  corresponds to the ion H2C = OH+  formed by losing one hydrogen atom from the molecular ion

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments 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

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments The peak at m/e 29 corresponds to the ion HC  O+  formed by losing two hydrogen atoms from the ion H2C = OH+

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments The ion HC  O+ has two resonance structures:

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments The peak at m/e 15 corresponds to the ion H3C+  formed by the breaking of the C  O bond in the molecular ion

Mass spectrum of pentan-3-one
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments Mass spectrum of pentan-3-one

Interpretation of the mass spectrum of pentan-3-one
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments Corresponding ion m/e ratio CH3CH2+ 29 CH3CH2CO+ 57 CH3CH2COCH2CH3 86 Interpretation of the mass spectrum of pentan-3-one

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments The fragmentation pattern of pentan-3-one is summarized below:

Example 34-9A Example 34-9B Example 34-9C
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) Example 34-9A Example 34-9B Example 34-9C

Fragmentation Pattern
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) Fragmentation Pattern 1. Straight-chain Alkanes Simple alkanes tend to undergo fragmentation by  the initial loss of a • CH3 to give a peak at M – 15  This carbocation can then undergo stepwise cleavage down the alkyl chain

1. Straight-chain Alkanes
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 1. Straight-chain Alkanes Take hexane as an example:

2. Branched-chain Alkanes
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 2. Branched-chain Alkanes Tend to cleave at the “branch point”  more stable carbocations are formed

2. Branched-chain Alkanes
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 2. Branched-chain Alkanes e.g.

3. Alkyl-substituted Aromatic Hydrocarbons
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 3. Alkyl-substituted Aromatic Hydrocarbons Undergo loss of a hydrogen atom or alkyl group  yield the relatively stable tropylium ion Gives a prominent peak at m/e 91

3. Alkyl-substituted Aromatic Hydrocarbons
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 3. Alkyl-substituted Aromatic Hydrocarbons e.g.

4. Aldehydes and Ketones Frequently undergo fragmentation by losing one of the side chains  generate the substituted oxonium ion  often represents the base peak in the mass spectra

4. Aldehydes and Ketones

5. Esters, Carboxylic Acids and Amides
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 5. Esters, Carboxylic Acids and Amides 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)

5. Esters, Carboxylic Acids and Amides
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 5. Esters, Carboxylic Acids and Amides For carboxylic acids and unsubstituted amides,  characteristic peaks at m/e 45 and 44 are observed respectively

6. Alcohols 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

6. Alcohols For primary alcohols,
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 6. Alcohols For primary alcohols,  the peak at m/e 31, 45, 59 or 73 often appears  depends on what the R1 group is

7. Haloalkanes Haloalkanes simply break at the C  X bond
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 7. Haloalkanes Haloalkanes simply break at the C  X bond

7. Haloalkanes In the mass spectra of chloroalkanes,
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 7. Haloalkanes In the mass spectra of chloroalkanes,  two peaks, separated by two mass units, in the ratio 3 : 1 will be appeared

7. Haloalkanes In the mass spectra of bromoalkanes,
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 7. Haloalkanes In the mass spectra of bromoalkanes,  two peaks, separated by two mass units, having approximately equal intensities will be appeared

Check Point 34-9 Let's Think 4 Let's Think 5
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) Let's Think 4 Let's Think 5 Check Point 34-9

Isolation and Purification of Organic Compounds
34.2 Isolation and Purification of Organic Compounds (SB p.78) Isolation and Purification of Organic Compounds The selection of a proper technique  depends on the particular differences in physical properties of the substances present in the mixture

34.2 Isolation and Purification of Organic Compounds (SB p.78)
Filtration 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

The laboratory set-up of filtration
34.2 Isolation and Purification of Organic Compounds (SB p.78) Filtration The laboratory set-up of filtration

Filtration There are many small holes in the filter paper
34.2 Isolation and Purification of Organic Compounds (SB p.78) Filtration 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

Centrifugation When there is only a small amount of suspension, or when much faster separation is required  Centrifugation is often used instead of filtration

Centrifugation 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

Centrifugation 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

Crystallization Crystals are solids that have
34.2 Isolation and Purification of Organic Compounds (SB p.79) Crystallization Crystals are solids that have  a definite regular shape  smooth flat faces and straight edges Crystallization is the process of forming crystals

1. Crystallization by Cooling a Hot Concentrated Solution
34.2 Isolation and Purification of Organic Compounds (SB p.79) 1. Crystallization by Cooling a Hot Concentrated 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

1. Crystallization by Cooling a Hot Concentrated Solution
34.2 Isolation and Purification of Organic Compounds (SB p.79) 1. Crystallization by Cooling a Hot Concentrated 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

Crystallization by cooling a hot concentrated solution
34.2 Isolation and Purification of Organic Compounds (SB p.79) 1. Crystallization by Cooling a Hot Concentrated Solution Crystallization by cooling a hot concentrated solution

2. Crystallization by Evaporating a Cold Solution at Room Temperature
34.2 Isolation and Purification of Organic Compounds (SB p.80) 2. Crystallization by Evaporating a Cold 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

34.2 Isolation and Purification of Organic Compounds (SB p.80) 2. Crystallization by Evaporating a Cold 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

34.2 Isolation and Purification of Organic Compounds (SB p.80) 2. Crystallization by Evaporating a Cold Solution at Room Temperature Crystallization by slow evaporation of a solution (preferably saturated) at room temperature

Solvent Extraction 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

Solvent Extraction Often involves the use of a separating funnel
34.2 Isolation and Purification of Organic Compounds (SB p.80) Solvent Extraction 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

Solvent Extraction The organic product in an aqueous solution can be extracted by solvent extraction using diethyl ether

Solvent Extraction 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

Solvent Extraction 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

Distillation 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

Distillation The laboratory set-up of distillation
34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation The laboratory set-up of distillation

Distillation Before the solution is heated,
34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation 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

Distillation If bumping occurs during distillation,
34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation If bumping occurs during distillation,  some solution (not yet vaporized) may spurt out into the collecting vessel

Fractional Distillation
34.2 Isolation and Purification of Organic Compounds (SB p.81) Fractional Distillation A method used to separate a mixture of two or more miscible liquids

34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation The laboratory set-up of fractional distillation

34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional 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

34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation 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

34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation 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

Sublimation Sublimation is the direct change of
34.2 Isolation and Purification of Organic Compounds (SB p.82) Sublimation 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

Sublimation 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

A mixture of two compounds can be separated by sublimation
34.2 Isolation and Purification of Organic Compounds (SB p.83) Sublimation A mixture of two compounds can be separated by sublimation

Chromatography An effective method of separating a complex mixture of substances Paper chromatography is a common type of chromatography

The laboratory set-up of paper chromatography
34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography The laboratory set-up of paper chromatography

Chromatography A solution of the mixture is dropped at one end of the filter paper

Chromatography 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

Chromatography When the solvent moves across the sample spot of the mixture,  partition of the components between the stationary phase and the mobile phase occurs

Chromatography 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

A summary of different techniques of isolation and purification
34.2 Isolation and Purification of Organic Compounds (SB p.84) A summary of different techniques of isolation and purification 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
34.2 Isolation and Purification of Organic Compounds (SB p.84) A summary of different techniques of isolation and purification 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 Check Point 34-2

Qualitative Analysis of Elements in an Organic Compound
34.4 Qualitative Analysis of Elements in an Organic Compound

Qualitative Analysis of an Organic Compound
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Qualitative Analysis of an Organic Compound Qualitative analysis of an organic compound is  to determine what elements are present in the compound

34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Carbon and Hydrogen Tests for carbon and hydrogen in an organic compound are usually unnecessary  an organic compound must contain carbon and hydrogen

Carbon and Hydrogen 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

Halogens, Nitrogen and Sulphur
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Halogens, Nitrogen and Sulphur Halogens, nitrogen and sulphur in organic compounds can be detected  by performing the sodium fusion test

34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Halogens, Nitrogen and Sulphur 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

34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Halogens, Nitrogen and Sulphur 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

Results for halogens, nitrogen and sulphur in the sodium fusion test
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 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
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, as sulphide ion (S2-) Sodium pentacyanonitrosylferrate(II) solution A black precipitate is formed Check Point 34-4

34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data

Quantitative Analysis of an Organic Compound
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an Organic Compound 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

The organic compound is burnt in excess oxygen
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 1. Carbon and Hydrogen 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

 the masses of carbon dioxide and water vapour formed respectively
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 1. Carbon and Hydrogen The increases in mass in potassium hydroxide solution and calcium chloride represent  the masses of carbon dioxide and water vapour formed respectively

The organic compound is heated with excess copper(II) oxide
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 2. Nitrogen 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

The mixture is allowed to cool  then water is added
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 3. Halogens 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

The organic compound is heated with fuming nitric(V) acid
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 4. Sulphur The organic compound is heated with fuming nitric(V) acid After cooling,  barium nitrate solution is added  the dry barium sulphate formed is weighed

34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an Organic Compound After determining the percentage composition by mass of a compound,  the empirical formula of the compound can be calculated

34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an 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

34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an 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

34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88) Quantitative Analysis of an 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

Example 34-5A Example 34-5B Check Point 34-5
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88) Example 34-5A Example 34-5B Check Point 34-5

Structural Information from Physical Properties
34.6 Structural Information from Physical Properties

34.6 Structural Information from Physical Properties (SB p.89) Structural Information from Physical Properties 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

34.6 Structural Information from Physical Properties (SB p.89) Structural Information from Physical Properties From the physical properties of a compound,  obtain preliminary information about the structure of the compound

34.6 Structural Information from Physical Properties (SB p.89) Structural Information from Physical Properties e.g.  Hydrocarbons have low densities, often about 0.8 g cm–3  Compounds with functional groups have higher densities

34.6 Structural Information from Physical Properties (SB p.89) Structural Information from Physical 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

Physical properties of some common organic compounds
34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds 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

34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds 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

34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic compound Density at 20oC Melting point and boiling point Solubility In water or highly polar solvents In non-polar organic solvents Halo-alkanes • 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

34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds 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

34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic 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

34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic 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

34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic 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

34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic 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

34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic 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

34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic 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

34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic 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 (NH bond is less polar than O  H bond)

34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic 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)

Example 34-6 Check Point 34-6 Let's Think 1 Let's Think 2
34.6 Structural Information from Physical Properties (SB p.92) Let's Think 1 Let's Think 2 Check Point 34-6 Example 34-6

Structural Information from Chemical Properties
34.7 Structural Information from Chemical Properties

34.7 Structural Information from Chemical Properties (SB p.93) Structural Information from Chemical 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

34.7 Structural Information from Chemical Properties (SB p.93) Structural Information from Chemical Properties e.g. The molecular formula of C2H4O2 may represent a carboxylic acid or an ester:

34.7 Structural Information from Chemical Properties (SB p.93) Structural Information from Chemical Properties The next stage is  to find out the functional group(s) present  to deduce the actual arrangement of atoms in the molecule

Chemical tests for different groups of organic compounds
34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds Organic compound Test Observation Saturated hydrocarbons • Burn the saturated hydrocarbon in a non-luminous Bunsen flame • A blue or clear yellow flame is observed

34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Test 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

34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Test 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

34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Test Observation Alcohols (  OH) • Iodoform test for: Add iodine in sodium hydroxide solution • A yellow precipitate is formed

34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Test 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) Chemical tests for different groups of organic compounds Organic compound Test 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) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds Organic compound Test Observation Amides ( ) Boil with sodium hydroxide solution The colourless gas produced turns moist red litmus paper or pH paper blue

34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds 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

34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds 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

Example 34-7A Example 34-7B Example 34-7C Check Point 34-7
34.7 Structural Information from Chemical Properties (SB p.96) Example 34-7A Example 34-7B Example 34-7C Check Point 34-7

The END

34.1 Introduction (SB p.77) Check Point 34-1 What are the necessary information to determine the structure of an organic compound? Answer Molecular formula from analytical data, functional group present from physical and chemical properties, structural information from infra-red spectroscopy and mass spectrometry Back

Check Point 34-2 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 liquid Answer (a) Centrifugation (b) Chromatography (c) Fractional distillation (d) Sublimation (e) Filtration Back

Check Point 34-4 (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

34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
Example 34-5A 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

Example 34-5A Back 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

Example 34-5B 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. Element Carbon Hydrogen Oxygen Percentage by mass (%) 60.00 13.33 26.67 Answer

Example 34-5B (a) Let the mass of the compound be 100 g. Then, mass of carbon in the compound = g mass of hydrogen in the compound = g mass of oxygen in the compound = 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

Example 34-5B (b) The molecular formula of the compound is (C3H8O)n. Relative molecular mass of (C3H8O)n = 60.0 n × (12.0 × × ) = 60.0 n = 1 ∴ The molecular formula of compound Z is C3H8O. Back

Check Point 34-5 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

Check Point 34-5 Relative molecular mass of CO2 = × 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 × = 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

Check Point 34-5 ∴ 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

Check Point 34-5 Let the molecular formula of the compound be (CH2O)n. Relative molecular mass of (CH2O)n = 60.0 n × ( × ) = 60.0 n = 2 ∴ The molecular formula of the compound is C2H4O2. Back

34.6 Structural Information from Physical Properties (SB p.92)
Back Let's Think 1 Why do branched-chain hydrocarbons have lower boiling points but higher melting points than the corresponding straight-chain isomers? Answer Branched-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 Let's Think 2 Why does the solubility of amines in water decrease in the order: 1o amines > 2o amines > 3o amines? Answer The 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.

Example 34-6 Match the boiling points 65oC, –6oC and –88oC with the compounds CH3CH3, CH3NH2 and CH3OH. Explain your answer briefly. Answer

Example 34-6 Back 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.

Check Point 34-6 (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.

Ethanol, chloroethane, hexan-1-ol
34.6 Structural Information from Physical Properties (SB p.92) Check Point 34-6 (b) Arrange the following compounds in order of increasing solubility in water. Explain your answer. Ethanol, chloroethane, hexan-1-ol Answer (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.

Check Point 34-6 (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.

2,4-Dimethylpentan-3-one
34.6 Structural Information from Physical Properties (SB p.92) Check Point 34-6 Back (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) 125 2,4-Dimethylpentan-3-one 144 Heptan-4-one 155 Heptanal Boiling point (oC) Compound

34.7 Structural Information from Chemical Properties (SB p.96)
Example 34-7A The empirical formula of an organic compound is CH2O and its relative molecular mass is 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  (  ) = 60.0 n = 2 ∴ The molecular formula of the compound is C2H4O2.

Example 34-7A The empirical formula of an organic compound is CH2O and its relative molecular mass is It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky. (b) Deduce the structural formula of the compound. Answer

Example 34-7A 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:

Example 34-7A The empirical formula of an organic compound is CH2O and its relative molecular mass is 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

Example 34-7B 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

Example 34-7B (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 = ( ) cm3 = 45 cm3 Volume of carbon dioxide formed = ( ) cm3 = 30 cm3 CxHy + (x )O2  xCO H2O Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 15 : 30   x = 2

Example 34-7B 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: Back

Answer Example 34-7C 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. Calculate the molecular formula of the compound, assuming that all the volumes were measured under room temperature and pressure. The compound is found to contain a hydroxyl group ( OH) in its structure. Deduce its structural formula. Is the compound optically active? Explain your answer.

Example 34-7C (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 = ( ) cm3 = 60 cm3 Volume of carbon dioxide formed = ( ) cm3 = 40 cm3 Volume of water (in the form of steam) formed = ( ) cm3 = 40 cm3 CxHyOz + (x )O2  xCO H2O Volume of CxHyOz reacted : Volume of CO2 formed = 1 : x = 20 : 40   x = 2

Example 34-7C (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.

Back Example 34-7C 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.

Check Point 34-7 (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

Check Point 34-7 (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

Check Point 34-7 (a) (i) Let the molecular formula of the compound be (C3H2NO2)n. Molecular mass of (C3H2NO2)n = 168.0 n × (12.0 × × × 2) = 168.0 ∴ n = 2 ∴ The molecular formula of the compound is C6H4N2O4. (ii)

Check Point 34-7 (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

Check Point 34-7 (b) (i) Volume of hydrocarbon reacted = 30 cm3 Volume of unreacted oxygen = (95 – 60) cm3 = 35 cm3 Volume of oxygen reacted = ( ) cm3 = 105 cm3 Volume of carbon dioxide formed = 60 cm3 CxHy + (x )O2  xCO H2O Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 30 : 60   x = 2

Check Point 34-7 (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.

Check Point 34-7 (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

Check Point 34-7 (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

Check Point 34-7 Back (c) (i) Let the molecular formula of the hydrocarbon be (CH2)n. Molecular mass of (CH2)n = 56.0 n × ( × 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.

What is the relationship between frequency and
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102) Let's Think 3 What is the relationship between frequency and wavenumber? Answer The higher the frequency, the higher the wavenumber. Back

34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110)
Example 34-8 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.

Example 34-8 (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

Example 34-8 (a) Let the mass of the compound be 100 g. Then, mass of carbon in the compound = g mass of hydrogen in the compound = g mass of oxygen in the compound = 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

Example 34-8 Let the molecular formula of the compound be (C4H8O)n. Relative molecular mass of (C4H8O)n = 72.0 n × (12.0 × × ) = 72.0 ∴ n = 1 ∴ The molecular formula of the compound is C4H8O.

Example 34-8 From the IR spectrum, it can be observed that there are absorption peaks at cm–1 and cm–1. The absorption peak at cm–1 corresponds to the stretching vibration of the C  H bond, and the absorption peak at cm–1 corresponds to the stretching vibration 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 either an aldehyde or a ketone. If it is an aldehyde, its possible structure will be:

Example 34-8 If it is a ketone, its possible structure will be: Back

Check Point 34-8 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.

Check Point 34-8 Sketch the infra-red spectrum of a carboxylic acid based on the IR spectra of X and Y. Answer

Check Point 34-8 (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.

Check Point 34-8 (a) The infrared spectrum of a carboxylic acid is as follows:

Check Point 34-8 Answer (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.

Check Point 34-8 (b) Compound B could be an alcohol. From the two spectra given, compound B shows a broad band at 3300 cm–1 and several peaks at 2960 – 2875 cm–1. This broad band corresponds to the complication of the stretching vibration of the O — H bond by hydrogen bonding occurring among alcohol molecules.

Check Point 34-8 (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 – 1750 O  H 2500 – 3300 C  H 2840 – 3095 N  H 3350 – 3500 Answer

Check Point 34-8 Sketch the expected infra-red spectrum for an amino acid with the following structure:

Check Point 34-8 The infra-red spectrum of the amino acid is shown as follows:

Check Point 34-8 (d) A portion of the infra-red spectrum of an organic compound X is shown below. To which homologous series does it belong? Answer

Check Point 34-8 (d) In the IR spectrum of compound X, the wide absorption band at 3500 – 3000 cm–1 corresponds to the stretching vibration of the O — H bond. Besides, the absorption peak at 1760 – 1720 cm–1 corresponds to the stretching vibration of the C = O bond. Therefore, compound X is a carboxylic acid.

Check Point 34-8 (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

Check Point 34-8 (e) From the IR spectrum of compound Y, the two peaks in the 3300 – 3180 cm–1 region show that the compound contains the –NH2 group. Besides, the sharp peak at 1680 cm–1 implies that the compound also contains the C = O bond.

Check Point 34-8 (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:

Check Point 34-8 (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

Check Point 34-8 (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

Check Point 34-8 (i) Let the molecular formula of the compound be (C2H4O)n. Relative molecular mass of (C2H4O)n = 88.0 n × (12.0 × × ) = 88.0 n = 2 ∴ The molecular formula of the compound is C4H8O2. (ii)

Check Point 34-8 (iii) From the IR spectrum of compound Z, the absorption peak at – 2800 cm–1 corresponds to the stretching vibration of the C — H bond. Besides, the absorption peak at – 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. Back

Answer Example 34-9A 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.

Example 34-9A 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. Back

Example 34-9B The mass spectrum of hydrocarbon X is shown below:
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) Example 34-9B The mass spectrum of hydrocarbon X is shown below:

Example 34-9B (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

Example 34-9B (a) The relative molecular mass of hydrocarbon X is 58.0. (b) The base peak is at m/e 43. (c) 15 mass units 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 ( × 4 = 10) 10. Therefore, the hydrocarbon is butane.

Back Example 34-9B The peak at m/e 43 is 15 mass units less than the molecular ion. This suggests that a methyl group is lost during the fragmentation of the molecular ion. The peak at m/e 43 corresponds to CH3CH2CH2+. (f)

Example 34-9C An organic compound is investigated. The structural formula of this compound is shown below:

Example 34-9C Answer The mass spectrum of the compound is shown below:
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120) Example 34-9C The mass spectrum of the compound is shown below: Interpret the peaks at m/e 134, 119, 91 and 43. Answer

Back Example 34-9C 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+.

Let's Think 4 Why would the molecular ion compound have two peaks, separated by two mass units, in the ratio 3 : 1? Answer Chlorine has two isotopes, chlorine-35 and chlorine-37. Their relative abundances are in the ratio of 3 : 1. Back

Why would the molecular ion of a bromine-containing
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) Let's Think 5 Why would the molecular ion of a bromine-containing compound have two peaks, separated by two mass units, having approximately equal intensities? Answer Bromine has two isotopes, bromine-79 and bromine-81. Their relative abundances are in the ratio of 1 : 1. Back

Check Point 34-9 (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.

Check Point 34-9 (b) The following is the mass spectrum of bromomethylbenzene (benzyl bromide). Interpret the peaks at m/e = 172, 170 and 91. Answer

Check Point 34-9 The relative molecular mass of bromomethylbenzene (benzyl bromide) is 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+.

Check Point 34-9 Answer (c) Study the following spectrum carefully and deduce what group of organic compound it is. The compound has a relative molecular mass of 114.

Check Point 34-9 (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. Back

Structural Determination of Organic Compounds

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