structural determination of organic compounds

*
38.1 Introduction
38.3 Qualitative Analysis of Elements in an Organic Compound
38.4 Determination of Empirical and Molecular Formulae from Analytical Data
38.5 Chemical Tests for Functional Groups
38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups
Chapter 38
*
38.1 Introduction (SB p.137)
The determination of the structure of an organic compound involves four steps:
Isolation and purification of the compound
Qualitative analysis of the elements present in the compound
Determination of the molecular formula of the compound
Determination of the functional group present in the compound
New Way Chemistry for Hong Kong A-Level Book 3B
*
38.2 Isolation and Purification of Organic Compounds (SB p.138)
The solid/ liquid mixture to be filtered is poured onto the folded filter paper in the filter funnel
Liquids are allowed to pass through the filter paper (known as filtrate) while insoluble solids (known as residue) are retained on the filter paper
Filtration
*
The tubes containing undissolved solids are spun around very rapidly and are thrown outwards in the centrifuge
The denser solid collects as a lump at the bottom of the tube with the clear liquid above it
The liquid can be removed by decantation
Centrifugation
*
The solubility of most solids increase with a rise of temperature
As a hot concentrated solution cools, it cannot hold all of its dissolve solute. The excess solute separates out as crystals
Crystallization
*
When the solution becomes saturated after evaporation, further evaporation causes crystallization to occur
The crystallization process will be slow if it occurs at room temperature
Crystallization by Evaporating
*
To dissolve out a component from a mixture with a suitable solvent
An aqueous solution containing the organic product is usually shaken with diethyl ether in a separating funnel. The ether layer is then run off
Repeated extraction with fresh ether to extract any organic products remaining
The ether portions are combined and dried, and distilled away to obtain the organic products
Solvent Extraction
*
To separate a liquid from a solution of a liquid and a non-volatile solid or liquid
Only the liquid vapourizes to form vapours which condense to liquid on cold surface which collected as distillate
Anti-bumping granules are added to prevent bumping
Simple Distillation
*
To separate a liquid from a mixture of two or more miscible liquids which have large difference in boiling point
Fractionating column provides a large surface area for condensation and vapourization of the mixture to occur
Fractional Distillation
*
Many liquid or even solid compounds which are immiscible with water can be purified by distillation in a current of steam
Steam Distillation
*
Principle of steam distillation:
the total vapour pressure of an immiscible mixture is equal to the sum of the vapour pressures of individual components
the water and the compound will distil at a lower temperature than the boiling point of either one of them
*
Direct change of a solid to vapour on heating or vapour to solid on cooling
A mixture of two compounds is
heated in an evaporating dish, one
compound sublimes. The vapour
surface while the other compound
is not affected and remains in the
evaporating dish
*
To separate a complex mixture of substances
As the various components are being absorbed or partitioned at different rates, they move upwards to different extent
The ratio of the distance travelled by the substance to the distance travelled by the solvent is known as Rf value, which is the characteristic of the substance
Chromatography
*
Technique
Separation
Filtration Centrifugation Crystallization Solvent extraction Simple distillation Fractional distillation Steam distillation Sublimation Chromatography
An insoluble solid from a liquid (slow) An insoluble solid from a liquid (fast) A dissolved solute from its solution Dissolving out of a component from a mixture with a suitable solvent A liquid from a solution of non-volatile solutes Miscible liquids with widely different boiling points A liquid or a solid (that is immiscible with water and is volatile in steam) from impurities Two solids, only one of which can sublime A complex mixture of substances
*
Some of the dry solid is placed in a thin-walled glass melting point tube. The tube is attached to a thermometer
The temperature at which the solid melts is its melting point
Pure solid has a sharp melting point
Impure solid melts gradually over a wide temperature range
Tests for Purity
*
The boiling point of a liquid can be determined by using the simple distillation apparatus
The temperature at which the liquid boils steadily is its boiling point
The boiling point of a pure liquid is quite sharp, but varies with changes in external pressure
The presence of non-volatile solutes such as salts raises the boiling point of a liquid
Determination of Boiling Point
*
Carbon and hydrogen can be detected by heating a small amount of the substance with Cu2O
Carbon and hydrogen would be oxidized to CO2 and H2O respectively
CO2 turns lime water milky, and H2O turns anhydrous CoCl2 paper pink
Carbon and Hydrogen
38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.147)
*
Sodium fusion test is performed to test for the presence of halogens, nitrogen and sulphur
The compound under test is fused with a small piece of sodium metal in a small combustion tube and then heated strongly
The product of the test are extracted with water and analyzed
During sodium fusion, halogens, nitrogens and sulphur in the organic compound are converted to sodium halide, sodium cyanide and sodium sulphide respectively
Halogens, Nitrogen and Sulphur
*
Element
Test
Halogens, as Cl– Br– I–
Acidify and boil to remove any CN– and S2– present, then add AgNO3(aq) Ag+(aq) + X–(aq) AgX(s)
A white ppt., soluble in excess NH3(aq) A pale yellow ppt., sparingly soluble in excess NH3(aq) A creamy yellow ppt., insoluble in excess NH3(aq)
Nitrogen, as CN–
Add a mixture of iron(II) sulphate and iron(III) sulphate solutions
A blue-green colour is observed
Sulphur, as S2–
A black ppt. is observed
*
Quantitative analysis is to find out the percentage composition by mass of the compound
Methods to determine the mass of the elements:
38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.148)
1. Carbon and hydrogen
The organic compound is burnt in oxygen.
The carbon dioxide and water vapour formed are respectively absorbed by KOH solution and anhydrous CaCl2.
The increases in mass in KOH solution and CaCl2 represent the masses of carbon dioxide and water vapour formed respectively
*
The organic compound is heated with excess Cu2O
The NO and NO2 formed are passed over hot Cu and the volume of N2 formed is measured
*
3. Halogens
The organic compound is heated with fuming HNO3 and excess AgNO3 solution
The mixture is allowed to cool and then water is added
The dry AgX formed is weighed
*
After cooling, barium nitrate solution is added
The dry barium sulphate formed is weighed
*
The empirical formula of the compound can be calculated after the determination of percentage composition by mass of a compound
The molecular formula can be calculated after knowing relative molecular mass and the empirical formula
*
The empirical formula of a compound is the formula which shows the simplest whole number ratio of the atoms present in the compound
The molecular formula of a compound is the formula which shows the actual number of each kind of atoms present in a molecule of the compound
*
Example 38-1
On quantitative analysis, an organic compound was found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Calculate the empirical formula of the compound.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
*
Let the mass of the compound be 100 g. Then,
mass of C in the compound = 40.0 g
mass of H in the compound = 6.7 g
mass of O in the compound = 53.3 g
∴ The empirical formula of the organic compound is CH2O.
C
H
O
*
An organic compound Z has the following composition by mass:
(a) Calculate the empirical formula of compound Z.
(b) If compound Z is found to have a relative molecular mass of 60, determine the molecular formula of compound Z.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
Element
Carbon
Hydrogen
Oxygen
*
Let the mass of the compound be 100 g. Then,
mass of C in the compound = 60.0 g
mass of H in the compound = 13.33 g
mass of O in the compound = 26.67 g
∴ The empirical formula of compound Z is C3H8O.
C
H
O
*
Relative molecular mass of (C3H8O)n = 60
n (12.0 3 + 1.0 8 + 16.0) = 60
n = 1
∴ The molecular formula of compound Z is C3H8O.
*
Example 38-3
An organic compound is found to contain carbon, hydrogen and oxygen only. On complete combustion, 0.15 g of this compound gives 0.22 g of carbon dioxide and 0.09 g of water. If the relative molecular mass of this compound is found to be 60, determine the molecular formula of this compound.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
*
Mass of C in CxHyOz = Mass of C in CO2
Mass of H in CxHyOz = Mass of H in H2O
Mass of O in CxHyOz = Mass of CxHyOz – mass of C in CxHyOz
– mass of H in CxHyOz
Relative molecular mass of CO2 = 12.0 + 16.0 2
= 44.0
Mass of C in 0.22 g of CO2 = 0.22 g
= 0.06 g
Mass of H in 0.09 g of H2O = 0.09 g
= 0.01 g
*
Mass of O in the compound = (0.15 – 0.06 – 0.01) g
= 0.08 g
The simplest whole number ratio of x, y and z can be determined by following the steps in the table below.
∴ The empirical formula of the organic compound is CH2O.
C
H
O
*
Relative molecular mass of (CH2O)n = 60
n (12.0 + 1.0 2 +16.0) = 60
n = 2
∴ The molecular formula of the compound is C2H4O2.
*
The molecular formula does not give enough information on the structure of the compound
38.5 Chemical Tests for Functional Groups (SB p.152)
compounds having the same molecular formula may have widely different arrangement of atoms and even different functional groups
e.g.
*
∴ the following step is to find out the functional group(s) present so as to deduce the actual arrangement of atoms in the molecule
The presence of certain functional groups in an organic compound can be detected by simple chemical tests
*
Group
Test
Positive result
Saturated hydrocarbon
Combustion: burn a little saturated hydrocarbon in a non-luminous Bunsen flame
A blue or clear yellow flame is observed
Unsaturated hydrocarbon (C = C, C C)
Combustion: burn a little unsaturated hydrocarbon in a non-luminous Bunsen flame
A smoky flame is observed
Addition of Br2 in CH3CCl3 at room temp. and in the absence of light
Rapid decolourization of Br2
Addition of 1% (dilute) acidified KMnO4
Immediate decolourization of KMnO4 solution and/ or development of a brown colour (MnO2)
*
Group
Test
Haloalkanes (1°, 2° or 3°)
Hydrolysis followed by Ag+/HNO3(aq): boil with ethanolic KOH, then acidify with excess dilute HNO3 and add AgNO3 solution
Precipitation of AgX is observed (AgCl – white ppt.; AgBr – pale yellow ppt.; AgI – creamy yellow ppt. Remarks: Halobenzenes show no observable changes
Halobenzene
No precipitation of AgX is observed
*
Group
Test
Oxidation by K2Cr2O7/H+(aq)
1° and 2° alcohols: the clear orange solution becomes opaque and turns green almost immediately 3° alcohols: no observable changes
Test with sodium metal: add a small piece of Na metal to the alcohol
Evolution of H2 is observed
Esterification with ethanoyl chloride
A rise in temp. together with evolution of HCl is observed
Iodoform test for : Addition of I2 / NaOH(aq)
A yellow ppt. of CHI3 is produced
*
Group
Test
Positive result
Alcohol (–OH)
Lucas test: treat the alcohol with anhydrous ZnCl2 / conc. HCl (Lucas reagent)
1° alcohols: do not dissolve appreciably, the aqueous phase remains clear 2° alcohols: the clear solution becomes cloudy within 5 minutes 3° alcohols: a cloudy phase separates almost immediately
Ether (–O–)
—
*
Group
Test
Iodoform test for: Addition of I2 / NaOH(aq)
A yellow ppt. of CHI3 is produced
Carboxylic acid ( )
Ester formation: warm the carboxylic acid with an absolute alcohol in the presence of conc. H2SO4(aq), followed by adding Na2CO3(aq)
A sweet, fruity smell of an ester is detected
Addition of NaHCO3
*
Group
Test
Positive result
Amine (–NH2)
Test for 1° aliphatic amines: dissolve the amine in dilute HCl at 0 – 5°C, then add cold NaNO2(aq) slowly
Steady evolution of N2(g) is observed
Test for 1° aromatic amines: addition of naphthalen-2-ol in dilute NaOH
An orange or red ppt. is produced
Ester ( )
No specific test but can be distinguished by its characteristic smell
A sweet and fruity smell is detected
*
Group
Test
Positive result
Acyl halide ( )
Precipitation of AgX is observed (AgCl – white ppt.; AgBr – pale yellow ppt.; AgI – creamy yellow ppt. Remarks: Halobenzenes show no observable changes
Amide ( )
Phenol ( )
Addition of neutral FeCl3(aq)
Change in colour is observed (may be the transient appearance of a red, violet or blue colour)
*
Group
Test
A white ppt. of 2,4,6-tribromophenol is formed
Aromatic compound ( )
Combustion: burn a little aromatic compound in a non-luminous Bunsen flame
A smoky yellow flame with black soot is produced
Fuming H2SO4
*
Example 38-4
An organic compound is found to have an empirical formula of CH2O and a relative molecular mass of 60. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.
(a) Calculate the molecular formula of the compound.
Answer
Solution:
(a) Let the molecular formula of the compound be (CH2O)n.
Relative molecular mass of (CH2O)n = 60
n (12.0 + 1.0 2 + 16.0) = 60
n = 2
*
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
Solution:
(b) The compound reacts with sodium hydrogencarbonate to give carbon dioxide gas (which turns lime water milky), indicating that the compound contains a carboxyl group (–COOH). Eliminating the –COOH group from the molecular formula of C2H4O2, the atoms left are one carbon and three hydrogen atoms. This obviously shows that a methyl group (–CH3) is present. The structural formula of the compound is therefore CH3COOH.
(c) The IUPAC name for the compound is ethanoic acid.
*
Example 38-5
15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3 of oxygen which was in excess. The mixture was exploded and after cooling, the residual volume was 105 cm3. On adding concentrated potassium hydroxide solution, the volume decreased to 75 cm3.
(a) Calculate the molecular formula of the compound, assuming all volumes were measured under room temperature and pressure.
Answer
*
Solution:
(a) Let the molecular formula of the compound be CxHy.
Volume of hydrocarbon reacted = 15 cm3
Volume of unreacted oxygen = 75 cm3
Volume of oxygen reacted = (120 – 75) cm3 = 45 cm3
Volume of carbon dioxide formed = (105 – 75) cm3 = 30 cm3
Volume of CxHy reacted : volume of CO2 formed = 1 : x
= 15 : 30
*
Solution:
= 15 : 45
*
(c) Give the structural formula of the hydrocarbon.
Answer
Solution:
(c) The structural formula of the hydrocarbon is:
*
Example 38-6
20 cm3 of a gaseous organic compound containing only carbon, hydrogen and oxygen were mixed with 110 cm3 of oxygen which was in excess. The mixture was exploded at 105°C and the volume of the gaseous mixture was 150 cm3. After cooling to room temperature, the residual volume was reduced to 90 cm3. On adding concentrated aqueous potassium hydroxide, the volume further decreased to 50 cm3.
(a) Calculate the molecular formula of the compound, assuming all volumes were measured under room temperature and pressure.
Answer
*
Solution:
(a) Let the molecular formula of the compound be CxHyOz.
Volume of CxHyOz reacted = 20 cm3
Volume of unreacted oxygen = 50 cm3
Volume of carbon dioxide formed = (90 – 50) cm3 = 40 cm3
Volume of water (in the form of steam) formed = (150 – 90) cm3
= 60 cm3
Volume of CxHyOz reacted : volume of CO2 formed = 1 : x
= 20 : 40
*
Solution:
(a) Volume of CxHyOz reacted : volume of H2O formed = 1 :
= 20 : 60
= 20 : 60
*
Example 38-6 (cont’d)
(b) The compound is found to contain an –OH functional group in its structure. Deduce its structural formula.
(c) Is the compound optically active? Explain your answer.
Answer
Solution:
(b) As the compound contains a –OH group in its structure, the hydrocarbon skeleton of the compound becomes C2H5 after eliminating the –OH group from C2H6O. The structural formula of the compound is CH3CH2OH.
(c) The compound is optically inactive as both carbon atoms in the compound are not asymmetric, i.e. both of them do not attach to four different groups of atoms.
*
Check Point 38-1
(a) A substance contains 42.8% carbon, 2.38% hydrogen, 16.67% nitrogen by mass and the remainder consists of oxygen.
(i) Given the fact that the relative molecular mass of the substance is 168, deduce the molecular formula of the substance.
(R.a.m.: H = 1.0, C = 12.0, N = 14.0, O = 16.0)
Answer
*
(a) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the compound is C3H2NO2.
Let the molecular formula of the compound be (C3H2NO2)n.
Molecular mass of (C3H2NO2)n = 168
n (12.0 3 + 1.0 2 + 14.0 + 16.0 2) = 168
∴ n = 2
Carbon
Hydrogen
Nitrogen
Oxygen
*
Check Point 38-1 (cont’d)
(ii) The substance is proved to be an aromatic compound with only one type of functional group. Give the structural formulae for all isomers of the substance (the name of each isomer should also be included).
Answer
(a) (ii) 1,2-Dinitrobenzene 1,3-Dinitrobenzene
*
(b) 30cm3 of a gaseous hydrocarbon were mixed with 140 cm3 of oxygen which was in excess, and the mixture was then exploded. After cooling to room temperature, the residual gases occupied 95 cm3 by volume. Be adding potassium hydroxide solution, the volume was reduced by 60 cm3. The remaining gas was proved to be oxygen.
(i) Determine the molecular formula of the hydrocarbon.
Answer
*
Volume of unreacted oxygen = (95 – 60) cm3 = 35 cm3
Volume of carbon dioxide formed = 60 cm3
Volume of CxHy reacted : Volume of CO2 formed
= 1 : x
= 30 : 60
*
= 1 :
*
(ii) Is the hydrocarbon a saturated, unsaturated or aromatic hydrocarbon?
Answer
(b) (ii) From the molecular formula of the hydrocarbon, it can be deduced that the hydrocarbon is saturated because it fulfils the general formula of alkanes CnH2n+2.
*
Check Point 38-1
(c) A hydrocarbon having a relative molecular mass of 56 contains 85.5% carbon and 14.5% hydrogen by mass. Detailed analysis shows that it has two geometrical isomers.
(i) Deduce the molecular formula of the hydrocarbon.
(R.a.m.: H = 1.0, C = 12.0)
Answer
*
(c) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the hydrocarbon is CH2.
Let the molecular formula of the hydrocarbon be (CH2)n.
Molecular mass of (CH2)n = 56
n (12.0 + 1.0 2) = 56
n = 4
Carbon
Hydrogen
*
(ii) Name the two geometrical isomers of the hydrocarbon.
(iii) Explain the existence of geometrical isomerism.
Answer
(c) (ii)
(iii) Since the but-2-ene molecule is unsymmetrical and free rotation of the but-2-ene molecule is restricted by the presence of the carbon-carbon double bond, geometrical isomerism exists.
*
38.6 Uses of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.159)
Electromagnetic radiation has dual property: as wave or as photons
The relationship of the frequency () of an electromagnetic radiation, its wavelength () and velocity (c) is shown as follows:
The Electromagnetic Spectrum
The energy of a quantum of electromagnetic radiation is:
E = h where h is the Planck constant (i.e. 6.626 10–34 J s)
or
*
Visible light has wavelength between 400 nm and 800 nm
Infra-red radiation has wavelength between 800 nm and 300 m
*
Organic compounds absorb electromagnetic energy in the infra-red (IR) region of the spectrum
Infra-red Spectroscopy
IR radiation causes atoms and groups of atoms of organic compounds to vibrate with increased amplitude about the covalent bonds that connect them
The vibration is
*
Infra-red spectrometer:
amount of light absorbed at each
wavelength of the IR region
Infra-red spectrum:
shows the absorption of IR radiation by a sample at different frequencies
The IR radiation is specified by its wavenumber (unit: cm–1) which is the reciprocal of wavelength
*
Covalent bonds behave as if they were tiny springs connecting the atoms in the vibrations
The atoms only vibrate at certain frequencies, therefore covalently bonded atoms have only particular vibrational energy levels
The excitation of a molecule from one vibrational energy level to another occurs only when the compound absorbs IR radiation of the exact energy required
*
Molecules vibrate in a variety of ways:
Stretching vibration: two atoms joined by a covalent bond move back and forth as if they were joined by a spring
*
A variety of stretching and bending vibrations:
*
The frequency of a given stretching vibration of a covalent bond depends on:
1. The masses of the bonded atoms
2. The strength of the bond
Light atoms vibrate at higher frequencies than heavier ones
The stretching frequencies of groups involving hydrogen such as C – H, N – H and O – H occur at relatively high frequencies
Bond
C – H O – H N – H
2 840 – 3 095 3 230 – 3 670 3 350 – 3 500
*
Triple bonds are stronger (and vibrate at higher frequencies) than double bonds
The IR spectra of simple compounds contain many
absorption peaks
possibility of two different compounds having the same IR spectrum is very small
IR spectrum has been called the ‘fingerprint’ of a molecule
Bond
C C C N C = C C = O
2 070 – 2 250 2 200 – 2 280 1 610 – 1 680 1 680 – 1 750
*
A 100% transmittance in the spectrum implies no absorption of IR radiation
Absorption peak (or band): Due to the absorption of IR radiation, a decrease in percent transmittance is resulted and hence a dip in the spectrum
Use of IR Spectrum in the Identification of Functional Groups
*
The four regions of an IR spectrum
Range of wavenumber (cm–1)
Interpretation
400 – 1 500
This region often consists of many different complicated bands. This part of the spectrum is unique to each compound and is often called the fingerprint region. It is not used for identification of particular functional groups.
1 500 – 2 000
2 000 – 2 500
2 500 – 4 000
Absorption of single bonds formed by hydrogen, e.g. C – H, O – H and N – H
*
The region between 4 000 cm–1 and 1 500 cm–1 is often used for identification of functional groups from their characteristic absorption wavenumbers
Compound
Bond
Alkenes
*
Interpretation of IR Spectra
*
C = C stretching
*
The IR spectrum of hex-1-yne
*
Interpretation of the IR spectrum of hex-1-yne
Wavenumber (cm–1)
vs vs s
2 119
1 445
*
C – H bending
*
C – H stretching
*
C = O stretching
*
The absorption of the O – H group in alcohols and carboxylic acids usually appear as broad band instead of sharp peak
the vibration of the O – H group is complicated by the hydrogen bonding formed between the molecules
*
C – H stretching
*
C N stretching
*
Summary of strategies for the use of IR spectra in the identification of functional groups in an organic compound:
Focus at the IR absorption peak at or above 1500 cm–1. Concentrate initially on the major absorption peaks
For each absorption peak, try to list out all the possibilities using a correlation table or chart. Note that not all absorption peaks in the spectrum can be assigned
The absence and presence of absorption peaks at some characteristic ranges of wavenumbers are equally important. It is because the absence of particular absorption peaks can be used to identify certain functional groups or bonds do not exist in the molecule
*
Example 38-7
An organic compound with a relative molecular mass of 72 was found to contain 66.66% carbon, 22.23% oxygen and 11.11% hydrogen by mass. A portion of its infra-red spectrum is shown below.
(a) Determine the molecular formula of the compound.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
38.6 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.173)
*
Solution:
(a) Let the mass of the compound be 100 g. Then,
mass of carbon in the compound = 66.66 g
mass of hydrogen in the compound = 11.11 g
mass of oxygen in the compound = 22.23 g
∴ The empirical formula of the organic compound is C4H8O.
C
H
O
*
Relative molecular mass of (C4H8O)n = 72
n (12.0 4 + 1.0 8 + 16.0) = 72
∴ n = 1
∴ The molecular formula of the compound is C4H8O.
*
Example 38-7 (cont’d)
Deduce two possible structures of the compound, each of which belongs to a different homologous series.
Answer
*
Solution:
(b) From the IR spectrum, it can be observed that there are absorption peaks at 2950 cm–1 and 1700 cm–1. The absorption peak at 2950 cm–1 corresponds to the stretching vibration of the C – H bond, and the absorption peak at 1700 cm–1 corresponds to the stretching vibration of the C = O bond. As there is only one oxygen atom in the molecule of the compound, it is either an aldehyde or a ketone.
If it is an aldehyde, its possible structure will be:
*
Solution:
(b) If it is a ketone, its possible structure will be:
*
Check Point 38-2
(a) An organic compound X forms a silver mirror with ammoniacal silver nitrate solution. Another organic compound Y reacts with ethanoic acid to give a product with a fruity smell. The portions of infra-red spectra of X and Y are shown below.
*
(a)
Sketch the infra-red spectrum of a carboxylic acid based on the IR spectra of X and Y.
Answer
*
(a) From the information given, X would be an aldehyde and Y would be an alcohol.
Comparing the structures of an aldehyde and an alcohol with that of a carboxylic acid, some common features are found between the two.
In the IR spectrum of a carboxylic acid, it is expected that it contains the characteristic O – H (similar to the alcohol) and C = O (similar to the aldehyde) absorption peaks. Thus peak values at around 3300 cm–1 and 1720 cm–1 are predicted. A wide band at around 3300 cm–1 is observed due to the complication of the stretching vibration of the O – H group by hydrogen bonding and it overlaps with the absorption of the C – H bond in the 2950 – 2875 cm–1 region. The infra-red spectrum of a carboxylic acid is as follows:
*
(a)
*
(b) The infra-red spectra of two organic compounds A and B are shown below.
*
(b)
State which compound could be an alcohol. Explain your answer briefly.
Answer
(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.
*
(c) Given the following characteristic absorption wavenumbers of some covalent bonds in infra-red spectra:
Sketch the expected infra-red spectrum for an amino acid with the following structure:
Answer
Bond
C = O O – H C – H N – H
1680 – 1750 2500 – 3300 2840 – 3095 3350 – 3500
(c) The infra-red spectrum of the amino acid is shown as follows:
*
(d) Below is a portion of the infra-red spectrum of an organic compound X. To which homologous series does it belong?
Answer
(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.
*
(e) Below is a portion of the infra-red spectrum of an organic compound Y. Identify the functional groups that it contains.
Answer
(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.
*
(f) An organic compound Z with a relative molecular mass of 88 was found to contain 54.54% carbon, 36.36% oxygen and 9.10% hydrogen by mass. A portion of its infra-red spectrum is shown below:
(i) Determine the molecular formula of compound Z.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
*
(f) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the compound is C2H4O.
Let the molecular formula of the compound be (C2H4O)n.
Molecular mass of (C2H4O)n = 88
n (12.0 2 + 1.0 4 + 16.0) = 88
n = 2
∴ The molecular formula of the compound is C4H8O2.
Carbon
Oxygen
Hydrogen
*
(f) (ii) Based on the result from (i), draw two possible structures of the compound, each of which belongs to a different homologous series.
Answer
*
(f) (iii) Using the information from the IR spectrum, name the homologous series that compound Z belongs to .
Answer
(f) (iii) From the IR spectrum of compound Z, the absorption peak at 3200 – 2800 cm–1 corresponds to the stretching vibration of the C – H bond. Besides, the absorption peak at 1800 – 1600 cm–1 corresponds to the stretching vibration of the C = O bond. The absence of the characteristic peak of the O – H bond in the 3230 – 3670 cm–1 region indicates that compound Z is an ester.
*

structural determination of organic compounds

Documents