John Ng's Guide to Analytical Chemistry

Guide to Analytical Chem

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A.1.1 State the reasons for using analytical techniques. Uses should include structure

determination, analysis of composition of substances and to determine purity

determination of structure

determination of concentrations

identification of substances

Allow separation of substances.

analysis of (different) composition of substances/compounds/mixtures

determination of purity of substances

in medicine for body imaging

determining illegal drug use

forensic science for evidence in courts

monitoring of the environment

DNA testing

testing foods for levels of sugar / testing quality of food

Accept other specific examples.

A.2.1 Describe the electromagnetic spectrum. X-ray, ultraviolet (UV), visible, infrared (IR),

radioand microwave should be identified. Highlight the variation in wavelength, wave number,

frequencyand energy across the spectrum.

Ex.1 The figure below depicts the visible region of the electromagnetic spectrum and the two regions

nearest to it.

A B

vis

ible

Increasing wavelength

(a) Name the regions labelled A and B, identify the atomic or molecular processes associated

with each region and compare the energies of the photons involved in these processes.

(a) A is the ultraviolet / UV;

Radiowave


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electronic transitions;

B is the infrared / IR;

molecular vibrations;

A is higher energy than B / OWTTE

(b) State, giving a reason, which region (A or B) could be used to

(i) test for metal ions.

(ii) obtain information about the strengths of bonds.

(b) (i) A (because) electron transitions occur;

(ii) B from vibration frequencies;

confuse A and B perhaps seen the electromagnetic spectrum drawn the opposite way but the

question did clearly indicate how the wavelength increased

Ex.2 1H NMR and IR spectroscopy both involve the absorption of electromagnetic radiation.

(a) (i) Identify the region of the electromagnetic spectrum used in 1H NMR

spectroscopy.

(ii) Identify which of these two techniques involves higher energy radiation.

(b) Identify which of the following molecules absorbs IR radiation and explain your

choice.

H2 O2 HCl

(a) (i) radiowaves;

(ii) IR / infrared;

(b) HCl;

vibration/stretching of bond/molecule produces a change in dipole

moment/polarity;

Do not accept contains a polar bond.

Ignore reference to bending. although the correct selection of HCl was usually made in (b), did not refer to the change of

dipole during the absorption of IR radiation.

A.2.2 Distinguish between absorption and emission spectra and how each is produced.

absorption spectrum:

energy required to move/excite (electrons) from lower/ground state to higher energy

level/excited state;

emission spectrum:

radiation emitted by electrons from higher/excited state to lower/ground energy level;

OR

absorption spectrum: continuous spectrum with missing regions/lines corresponding to

energies absorbed;

emission spectrum: regions/lines corresponding to energies given out/emitted;


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Similarity:

lines / only certain frequencies/wavelengths/energies / discrete / not continuous /

absorptions and emissions occur at the same frequencies/wavelengths;

Difference:

emission spectra have only a few frequencies/bright lines while absorption spectra show all

frequencies/have a coloured background except for a few missing/black lines /

Accept emission spectra have fewer frequencies included than absorption spectra.

Accept a suitable diagram to illustrate one or both points.

A.2.3 Describe the atomic and molecular processes in which absorption of energy takes place.

The description should include vibrations, rotation and electronic transitions.

Gamma-rays: nuclear transitions

X-rays: electronic transitions in atoms and molecules

ultraviolet / UV: electronic transitions;

visible electron transition from lower energy level to higher energy level;

Accept electron excited/promoted/absorbs energy.

Accept electron moves from bonding/to antibonding/* orbital.

infrared / IR: molecular vibrations;

Microwave: molecular rotation;

radiowave: change in the orientation of a nucleus relative to a magnetic field,

e.g. used in 1H NMR

IR involves vibrations of bonds / IR involves shorter wavelength/ more energy than 1H

NMR;

whereas 1H NMR involves transitions between different energy states in the nucleus

which are lower in energy / 1H NMR occurs in the radio region therefore energy is

lower;

Radio waves: lowest frequency

Microwaves/IR/infra red cause molecules to rotate faster

Ex.3 (a) Identify the atomic or molecular processes associated with microwave and ultraviolet

radiation.

(a) Microwave:

(molecular) rotation;

Do not allow mark if incorrect rotations (i.e. not molecular) are stated.

Ultraviolet:

electronic transition;

(b) State which region of the electromagnetic spectrum can be used to identify the functional

groups present in a molecule..

(b) infrared/IR;

(c) Explain why the absorptions in infrared (IR) spectroscopy occur at much higher

frequency than those in 1H NMR spectroscopy.

(c) IR involves vibrations of bonds / IR involves shorter wavelength/ more energy than


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1H NMR;

whereas 1H NMR involves transitions between different energy states in the nucleus which are lower in energy / 1H NMR occurs in the radio region therefore

energy is lower;

Ex.4 Explain why the absorptions obtained from 1H NMR spectroscopy occur at much lower

fequencies than those obtained from IR spectroscopy. 1H NMR absorptions are due to transitions between different energy states in the

nucleus (when an external magnetic field is applied);

IR absorptions are due to bond vibrations;

nuclear transitions are at a much lower energy than bond vibrations;

Ex.5 Compare infrared radiation and visible light in terms of the processes that occur in atoms and

molecules upon absorption.

infrared: vibrations (stretching and bending) of bonds increase;

visible light: electronic transitions to higher energy levels / electrons excited to higher

energy levels;


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A.3.1 State the uses of atomic absorption (AA) spectroscopy. Include uses such as the

identification of metals in water, blood, soils and foods.

identification/detection/concentrations of metal/metal ions;

identification of (named) metal in blood/soil/food;/ measure low concentration of

metals;

e.g. Mg/Al/Cu/Na/K in blood (serum)

Hg/Cu in alloys

Cu in water from Cu pipes

Ba/Cd/Cr/Mn/Pb/Zn/Hg/named heavy metal in (sea)

water

Al/Fe in plants / soil

Pb in paint

Mg (for hardness) in water [1 max]

A.3.2 Describe the principles of atomic absorption.

A.3.3 Describe the use of each of the following components of the AA spectrophotometer: fuel,

atomizer, monochromatic light source, monochromatic detector and read-out.

Monochromatic

light source

Monochromatic

detector

Atomizer


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The fuel to form a combustion mixture / to produce heat;

Atomizer;

converts liquid sample into small droplets / converts metal ions into atoms;

turns all the ions in the sample into atoms / ions are converted/dissociated into atoms;

(sample) dehydrated/vaporized/atomized / solvent/water evaporated;

e.g. Cu2+

ions converted to Cu atoms / Cu2+

(g) + 2e– → Cu(g) / Cu atoms are produced;

monochromatic light source;

produces radiation/light of the same frequency/wavelength as is absorbed by the species

(being detected);

Hollow cathode lamp specific to the element to be analysed;

light source emits wavelength of light that will be absorbed by the element/ Cu atoms / must

be a Cu lamp / hollow cathode Cu lamp;

light source emits wavelength of light that is absorbed by the element/Al atoms

absorbs/re-emits radiation / (electromagnetic radiation) excites electrons to higher energy

levels/states;

monochromator

only allows the particular/specific/required wavelength/frequency/colour to pass through

monochromatic detector;

detects radiation/light of the same frequency/wavelength absorbed /converts photons into electric

current/signal;

provides light that is absorbed by the lead/metal to be detected / provides light from excited atoms

of lead/metal being detected;

Atomizer


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Explain how the student can determine the concentration of Cu2+

(aq) ions in the water sample using an

atomic absorption spectrophotometer and a solution of 0.10 mol dm–3

CuSO4.

make up different solutions of known concentrations (from the 0.10 mol dm–3

CuSO4);

measure the absorbance for each concentration;

plot a calibration/absorbance against concentration curve;

read the value of unknown concentration from its absorbance /

compare the absorption of the unknown with the standard solutions;

After measuring the absorbance due to the one metallic ion, e.g. Cu,to measeure another

metallic ion, e.g. Hg, need to change the (monochromatic) light source/wavelength/frequency;

from Cu (atoms/ions) to mercury (atoms/ions)

A.3.4 Determine the concentration of a solution from a calibration curve.

e.g. According to recommendations from the World Health Organization (WHO), the maximum

allowed concentration of lead(II) cations, Pb2+(aq), in drinking water is 0.001 mg dm–3.

The tap water taken from a building was analysed using atomic absorption (AA)spectroscopy

to determine the concentration of Pb2+(aq). An AA spectrophotometer was calibrated and the

following results were obtained.

[Pb2+(aq)] / mg dm–

3 Absorbance

0.25 0.110

0.50 0.220

0.75 0.340

1.00 0.450

1.25 0.560

Sample 0.170

Draw the calibration curve and determine whether or not the water is within the WHO

recommended maximum allowed concentration of lead(II) cations.

graph of absorbance versus concentration showing all five points plotted

and connected;

Do not penalize if graph is not extended to origin.

Points should be plotted closest to within one-half of a small sqaure


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determination of concentration (~ 0.38 mg dm

–3) corresponding to absorbance of 0.170;

Allow a range of 0.35 to 0.42.

concentration not within/more than WHO limit / OWTTE;

Ex.6 A calibration curve was plotted using water with known concentrations of lead ions.

100 dm3 of the contaminated drinking water was reduced by boiling, to 7.50 dm

3.

It was found that when the reduced volume was tested it had an absorbance of 0.55.

Calculate the concentration of lead ions (in mg dm–3

) in the original contaminated drinking

water.

concentration of Pb2+

from graph = 1.15 (mg dm–3

);

Allow between 1.13 and 1.17 mg dm–3

.

original concentration = 1.15 × 100

5.7 = 8.63 × 10

–2 (mg dm

–3);


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Ex.7 Ore samples may be analysed for iron using AAS. An ore sample was prepared in acid and

diluted to 1 part in 10. The diluted solution gave an absorbance reading of 0.80. Determine the

concentration of iron in the sample in mg cm–3

.

50 100 150 2000

0.60

0.80

1.00

0.40

0.20

1.20

0.00

Concentration / g cm–3

Abso

rpti

on

(2)

absorbance reading of 0.80 = 170 g cm–3

;

sample diluted by 10, therefore concentration of iron =10×170 g cm–3

= 1700 g cm–3

=) 1.7 mg cm–3

;

watch video on UV-Vis) Spectrsocopy

A.4.3 State that organic molecules containing a double bond absorb UV radiation. Refer to

conjugated and delocalized systems,including arenes, alkenes and chlorophyll.

colour due to electrons absorbing energy when they are promoted to a higher energy level /

colour observed/transmitted is complementary colour to colour of light absorbed;


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pigments all contain much conjugation/delocalization/alternate C–C single and C=C double

bonds/ which absorbs light of lower energy/frequency in visible region;

preservatives show little conjugation so absorb light of higher energy/frequency in

ultraviolet region;

A.4.4 Describe the effect of the conjugation of double bonds in organic molecules on the

wavelength of the absorbed light. Examples should include retinol and phenolphthalein. The

application of this in sun creams could be discussed.

Ex.8 Explain why phenolphthalein is colourless in acidic solutions but coloured in alkaline solutions.

O

O

O

O

O

OH

OH

C

C

C

C

HO

–

structure in acidic solution structure in alkaline solution

phenolphthalein in alkaline solution is more conjugated than it is in acidic solution; the

more conjugation / delocalization of electrons the less energy is required to excite the

electrons;

Most realised that phenolphthalein is more conjugated in alkaline solution, but some failed to

state that the consequence of this is that less energy is required to excite the electrons so the

absorption occurs in the visible region of the spectrum.

e.g. The structures of retinol and cholesterol are given in Table 21 of the Data Booklet.

Both are slightly soluble in a colourless non-polar solvent. Explain why retinol forms a

coloured solution whereas the solution of cholesterol remains colourless.

retinol (absorbs visible light as it) contains conjugated double bonds/delocalized electrons;

cholesterol absorbs in the UV region as it contains one double C=C bond / does not absorb

in the visible region as it has no conjugated double bonds;

did not refer to the presence of double bonds in retinol, idea of conjugation or cholesterol.

A.4.5 Predict whether or not a particular molecule will absorb UV or visible radiation.

e.g. β-carotene is involved in the formation of vitamin A. Its sources include carrots,

broccoli and dark, leafy vegetables. Its structure is shown below.


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Explain whether β-carotene absorbs ultraviolet or visible radiation.

extensive conjugation of (C=C) double bonds / alternate single and

double (carbon–carbon) bonds / involving delocalization of π electrons;

less energy is required (to excite the electrons);

absorption occurs in the visible region;

Few candidates could explain that less energy was required to excite the electrons due to the

conjugation.

A.4.6 Determine the concentration of a solution from a calibration curve using the Beer-Lambert

law.

e.g. A student wanted to determine a more accurate value for the concentration of a solution of

Mn2+

(aq) which was known to be between 0.10 and 0.010 mol dm–3

. She was provided with a

solution of 1.00 mol dm−3

manganese(II) sulfate, MnSO4. Describe how she could determine the

unknown concentration using a visible spectrometer and explain the importance of the Beer-

Lambert law in the method used.

determine λmax ;

make up different solutions of known concentrations from the standard;

measure the absorbance for each concentration at λmax;

plot a calibration curve and read off value of unknown concentration

from its absorbance; at a fixed wavelength the absorption is directly

proportional to the concentration provided the same path length is

used / Beer-Lambert law only works for dilute solutions;

A.5.1 State the reasons for using chromatography. The qualitative and quantitative aspects of

chromatography should be outlined.

Qualitative:

identification of an unknown substance /

identify presence/verify purity of an individual substance /

determination of the qualitative composition of a mixture /

Quantitative:

measurement of the concentration/amount/level of a substance in a solution/ mixture/

biological material

determination of the ratio of components/ percentage composition of a mixture /

Accept other general or specific uses.

A.5.2 Explain that all chromatographic techniques involve adsorption on a stationary phase and

partition between a stationary phase and a mobile phase. Components in a mixture have

different tendencies to adsorb onto a surface or dissolve in a solvent. This provides a means

of separating the components of a mixture.

Retention factor

The ratio between the distance moved by the spot and the distance moved by the solvent

front

Accept this expressed as a correct equation.


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e.g.1 Explain why the value of the retention factor for the same component can be very different if

different solvents (eluents) are used for the mobile phase.

Rf value depends on the intermolecular forces that the component has with the mobile phase

compared to the stationary phase / relative attraction of the component to mobile phase

compared to the stationary phase / partition of the component between the mobile phase and

the stationary phase

if polarity of the solvents is different the intermolecular forces/attraction to mobile phase/

partition will be different

Accept “Components have different solubilities in different solvents”

e.g.2 If the components of the mixture are coloured then they can be seen with the naked eye.

Describe two different ways in which a chromatogram can be developed if the components

are colourless.

(viewing under) ultraviolet/UV light;

(staining with) a dye/ninhydrin/potassium manganate(VII);

(exposing to) iodine (vapour);

Accept “staining with a developing (re)agent”.

Do not accept just staining.

components are adsorbed onto (solid) stationary phase/alumina/Al2O3;

components dissolve in mobile phase/propanone / components partition between stationary

phase and mobile phase/propanone;

separation depends on different abilities of the components to adsorb and dissolve/partition

Ex.1 (a) All chromatographic techniques involve the phenomena of adsorption or partition. They

all use a stationary phase and a mobile phase, but these phases can include solids, liquids

or gases. Complete the following table to show which states of matter are used in the two

phenomena.

Stationary phase Mobile phase

Adsorption

Partition

(b) Explain the term Rf value used in some chromatographic techniques.

(a) Stationary phase Mobile phase

Adsorption solid; liquid;

Partition liquid; liquid / gas;

(b) ratio of distances moved by solute and solvent / OWTTE;

A.5.3 Outline the use of paper chromatography, thin-layer chromatography (TLC) and column

chromatography. An outline of the operation for each technique will be assessed. This should

include an understanding and calculation of Rf values where relevant. Students should be

aware that, in some instances, paper chromatograms may need to be developed, for example,

in the separation of sugars.


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Paper Chromatography

Ex.2 Paper chromatography may be used to separate a mixture of sugars.

(a) State the stationary phase and an example of a mobile phase used in paper

chromatography.

Stationary phase:

Mobile phase:

(b) The identity of two sugars in a mixture can be determined by measuring their Rf values,

after staining.

(i) Describe how an Rf value can be calculated.

(ii) Calculate the Rf value for sugar 2 in the chromatogram below.

(c) Explain how the Rf value of sugar 2 could be used to identify it.

(a) Stationary phase: water in the paper;

Mobile phase: water / any other non corrosive solvent or solvent mixture;

(b) (i) Rf = frontlvent solvent/soby moved distance

substanceby moved distance;

(ii) Rf = 7.5

0.3 = 0.53;

Accept answers between 0.5 and 0.55 (± 1mm on measurements).

(c) compare Rf of unknown to known values;

(conducted) under the same conditions (of stationary/mobile phase);

Allow second mark for specifying a particular condition to keep

the same (e.g. solvent).

must be obtained under identical conditions.

Ex.3 A student used the technique of ascending paper chromatography in an experiment to

investigate some permitted food dyes(labelled P1 - P5). The result is shown below.


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6

4

2

0

Distance / cm

P1 P2 P3 P4 P5

(a) By reference to the diagram above, describe how the experiment would be carried out and

explain the meaning of the terms stationary phase, mobile phase, partition, solvent front

and Rf value.

(b) (i) Calculate the Rf value of P1.

(ii) State, giving a reason, whether P4 is a single substance or a mixture.

(a) (stationary phase) – water in the fibres of the paper

(do not accept just paper as the stationary phase);

(mobile phase) – the solvent;

(partition) – distribution between the two phases;

(solvent front) – how far the solvent moves up the paper;

(Rf value) – the distance travelled by one component divided by the distance traveled

by the solvent;

Above five points essential.

dyes spotted near bottom of paper;

bottom of paper placed in solvent;

solvent front below base line at beginning;

use of container with lid;

left until solvent near top of paper;

(b) (i) Rf value) )1.0(

)1.0(

9.48.0

;

= 0.16 (accept answer in range 0.14 – 0.20);

Allow [1] for 9.58.1 = 0.3

(ii) mixture as more than one spot;

Column Chromatography

Explain how the components of the mixture are separated

components dissolve in solvent/mobile phase;

components adsorb onto stationary phase/silica(gel)/silicon

dioxide/SiO2;

components have different affinities for stationary phase /

different solubility in mobile phase;

distribution/partition between a stationary phase and a mobile

phase;


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components move only when they are in the mobile phase / components don't move when

they are in/on the stationary phase;

better soluble/less adsorbed components elute earlier / less soluble/better adsorbed

component elute later

each component takes a different amount of time to emerge / components move through

column at different rates

e.g.3 Outline how the technique of column chromatography could be used to separate a mixture of two

coloured substances in solution

tube / column with alumina / silica (gel);

saturated with solvent;

mixture / solution added at top;

tap opened (at bottom);

more solvent added;

substances collected in separate containers;

e.g.4 Outline how a mixture containing two different coloured components could be separated

quantitatively into its two pure components using column chromatography. Assume the

column is packed with alumina and the mobile phase is propanone.

mixture dissolved in propanone/eluent/solvent and added to top of column (packed with

alumina / Al2O3) / mixture placed on top of column;

propanone/eluent/solvent (continually) added to top of column;

components collected separately as they leave column;

components move at different rates/have different retention times;

propanone/eluent/solvent is removed (by evaporation);

components are weighed / use UV absorption/fluorescence to measure the concentration of

components

Thin-Layer Chromatograpghy (TLC)


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e.g.5 Outline the principle of thin-layer chromatography. Refer in your answer to the nature of

the mobile and stationary phases and the reason why a mixture of amino acids can be

separated using this method.

stationary phase is thin layer of /silica/silicon dioxide/Al2O3/alumina/aluminium oxide (on

support) (and mobile phase a solvent);

distance travelled/separation (of components) depends on adsorption/bonding to stationary

phase / polarity of component / relative solubility between two phases;

e.g.6 Describe how you would show that a sugar solution was a mixture of two substances using

the technique of thin-layer chromatography (TLC).

place a spot of the liquid on a thin-layer chromatography/TLC plate;

place plate in a suitable solvent/eluent;

wait until solvent/eluent has almost reached the top of the plate;

spray/immerse plate in concentrated sulfuric acid/H2SO4/reagent to show spots /

Do not accept ninhydrin.

observe the plate to see how many spots are there on it;

e.g.7 Given solutions of glucose and fructose, explain how you would show that these were the

two sugars present in the mixture.

put spots of the pure sugars on the same plate as the mixture / produce chromatograms of the

pure sugars under the same conditions as the mixture;

see if the spots in the mixture have moved the same distance/have same Rf as the pure

sugars;

e.g.8 State one advantage of thin-layer chromatography over paper chromatography.

separation is better/more effective/faster/efficient / separated component can be

more easily recovered / withstands strong solvents / develops better;

Ex.4 A sample of food colouring was analysed using thin-layer chromatography to check whether it

contained a banned substance. The Rf value of the banned substance is 0.25 under the same

conditions.


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(a) State the number of components used to produce the food colouring.

(b) Identify a stationary phase commonly used in thin-layer chromatography.

(c) Identify the component in this chromatogram that has the greatest attraction for the

stationary phase.

(d) Explain what is meant by the term Rf value.

(e) Predict where you would expect the banned dye to appear on the chromatogram and mark

this spot with a circle on the diagram above.

(a) two;

(b) silica/silicon dioxide/SiO2;

alumina/aluminium oxide/Al2O3;

Accept either of the above.

(c) (component) A;

(d) the ratio between the distance moved by the spot and the distance

moved by the solvent front / OWTTE;

Accept this expressed as a correct equation.

(e) a spot drawn with its centre 1.4 cm from the start line;

Accept in the range 1.3 to 1.5 cm from the start line.

did not mark their spot with sufficient precision (±1 mm).

Ex.5 The following diagram represents a thin-layer chromatogram of an amino acid.

(iii) Calculate the Rf of the amino acid.

Rf = 46(mm)

mm)(40 = 0.87;

Allow 0.86 to 0.88.

No mark if Rf has units.

did not measure the distance from the centre of the spot, leading to the wrong answer.

Ex.6 Chromatography is used to test for the presence of illegal drugs in sport.

(a) A chromatogram of a concentrated urine sample from an athlete shows the

presence of a banned substance known to have an Rf value of 0.75.

(i) Define the term Rf value.

(ii) Calculate the Rf values for A and B, and so deduce which of the spots

corresponds to the banned substance.


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Rf of A:

Rf of B:

Banned substance:

(iii) Suggest how the results could change if the experiment was repeated with a

different solvent.

(b) Paper and column chromatography both have stationary and mobile phases.

Identify the stationary phases in the different techniques.

Paper chromatography:

Column chromatography:

(a) (i) solventby moved distance

soluteby moved distance / ratio of distances moved by solute and

solvent / OWTTE;

Accept sample/substance/compound/component for solute. 1

(ii) Rf of A:

0.8

7.75.70.93 – 0.96;

Rf of B:

0.8

0.60.75 / 0.74 and banned substance = B;

Penalize if units are given for Rf of A and B only once.

(iii) different number of spots/Rf values / OWTTE;

Do not accept solvent moves different distances.

Accept spots move different (relative) distances / OWTTE.


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(b) Paper chromatography:

water (in the fibres of the paper);

Do not accept just paper.

Column chromatography:

alumina/Al2O3/aluminium oxide / silica (gel)/SiO2/silicon dioxide;

. failed to indicate that the pattern of spots would be different. commonest error was to state that

the Rf values would be the same but that the solvent would move a different distance.

A.5.4 Describe the techniques of gas-liquid chromatography (GLC) and high-performance liquid

chromatography (HPLC).An outline of the operation for each technique will be assessed. This

should include an understanding of Rt value and its dependence on other factors where

relevant.

can detect very small quantities;

substance not volatile;

substance decomposes at high temperature/does not decompose at low temperature;

Gas-Liquid Chromatography (GLC)

sample vaporized/heated;

assumes sample does not decompose at high temperature;

different components carried through apparatus in stream of carrier gas;

mobile phase inert gas/nitrogen/helium/argon;

stationary phase liquid/long chain alkane/hydrocarbon/grease adsorbed on solid support;

each component has a different attraction for the hydrocarbons / attraction of components

depends on molecular mass and polarity;

less volatile components have longer retention time / components pass through at different

rates ;

components detected by flame ionizer;

converted to electrical current /connected to chart recorder /connected to mass spectrometer

/ spectrophotometer;

stationary phase is high boiling point liquid/non-volatile liquid/long-chain alkane/high boiling point

alkane/hydrocarbon/grease/wax adsorbed/coated on solid/silica/silicon dioxide/ SiO2/ alumina /

aluminium (III) oxide / Al2O3 (support);


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Allow high boiling point oil.

Allow oxide instead of silica or alumina.

mobile phase is inert gas/nitrogen/helium/argon;

Do not accept just gas.

liquids vaporized in oven/at high temperature;

liquids/components have different retention times/move through tube at different speeds;

Accept area under peak is proportional to quantity/amount as another marking point.

sample is vaporized/heated;

mobile phase is inert/nitrogen/helium gas;

stationary phase is a liquid/long chain alkane/grease on a solid support;

less volatile components have longer retention time;

components detected by flame ionizer;

converted to electrical current / connected to chart recorder;

HPLC

(non-polar or polar) mobile/liquid phase and (short, polar or non-polar) solid/stationary phase;

mobile phase forced (through the column) under pressure (at constant temperature);

sample placed in liquid stream / components separate as they pass through tube;

detected by absorption of UV/fluoresence/conductivity;

components reach detector at different times/Rt / compare retention times to standard samples

(under identical conditions);

the relative attraction of the stationary and mobile phases are different of the various components of

the mixture;

long tube packed with stationary phase;

liquid forced through this under high pressure;

sample injected at start of column;

different components emerge at different times;


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Ex.7 A mixture of two alcohols was analysed using high-performance liquid chromatography, HPLC, and

produced chromatogram 1 below. In the space provided in chromatogram 2, sketch the chromatogram

if the column of the chromatograph was less tightly packed and all other variables were kept constant.

(Total 2 marks)

Chromatogram 2 should be:

two broader peaks as chromatogram 1 for each peak;

two peaks of same area as chromatogram 1 for each peak;

two peaks of the same area as chromatogram 1 are closer to each other;

eluted earlier;

A.5.5 Deduce which chromatographic technique is most appropriate for separating the components

in a particular mixture. HPLC can identify compounds that are temperature-sensitive. Uses

include: analysis of oil; alcoholic beverages; antioxidants, sugars and vitamins in foods;

pharmaceuticals; polymers; biochemical and biotechnology research; and quality control of

insecticides and herbicides. GLC can identify compounds that can vaporize without

decomposing. Uses include: analysis of urine samples from athletes for drugs, underground

mine gases and blood alcohol levels.

GLC cannot be used as sugars decompose/are not stable at high temperatures / cannot be used

as sugars react at the high temperature used in GLC;

e.g.9 Explain why gas-liquid chromatography (GLC) is suitable for the determination of blood

alcohol levels but not for the separation of sugar samples.

GLC involves vaporizing a sample at high temperature;

alcohols can (easily) vaporize without decomposing / sugar molecules are temperature

sensitive/ decompose at high temperature/non-volatile;

GLC involves vaporizing a sample at high temperature;

alcohols can (easily) vaporize without decomposing / sugar molecules are temperature

sensitive/ decompose at high temperature/ non-volatile;


22

Ex.8. (a) Identify a possible mobile phase and stationary phase for HPLC and gas-liquid

chromatography, GLC.

Chromatographic

technique Stationary phase Mobile phase

HPLC

GLC

(a)

Chromatographic

technique Stationary phase Mobile phase

HPLC

Silica/silicon

(di)oxide/SiO2 and/or long

chain hydrocarbon

(on column);

hexane / water / alcohol /

solvent / liquid;

GLC Alkane / long chain

hydrocarbon (on column); N2(g) / He(g) / Ar(g);

Accept names of gases nitrogen, helium, argon.

Accept air or H2(g) as the mobile phase of GLC.

Do not accept inert gas for GLC.

(b) Deduce which technique, HPLC or GLC, can be used to analyse the urine sample of an

athlete for the anabolic steroid, tetrahydrogestrinone, THG.

GLC;

(c) Outline how the technique selected in part (b) would be carried out to confirm the

presence of the steroid THG in the urine sample..

measure retention times for THG and sample;

if THG present in sample then (retention) times will be the same

(under the same conditions);

OR

measure peak sizes for pure THG and mixture of THG and sample;

if THG present in sample then peak size greater for mixture

(under same conditions); Details of the stationary phase and mobile phase of the HPLC and GLC were known by many

candidates. Candidates confused TLC with GLC in part (c). The syllabus is clear in this area but

candidates just did not know the detail required.

Ex.9 An analgesic tablet contains 400 mg of aspirin and 80 mg of caffeine. The molecular

formula of aspirin is C9H8O4 and that of caffeine is C9H10N4O2.

(a) State and explain which method, gas-liquid chromatography (GLC) or high

performance liquid chromatography (HPLC), would be best for the separation

and mass determination of aspirin and caffeine in the tablet.

(b) State and explain which of the two components would have the shorter retention

time.

(c) Sketch a chromatograph for the separation of the aspirin and caffeine in the

analgesic tablet.


23

time (a) HPLC;

component would decompose (because of the higher temperature) in GLC;

(b) aspirin;

lower molecular mass;

(c)

timeaspirin caffeine relative position of peaks;

relative size of peaks;


24

Nuclear Magnetic Resonance (NMR) Spectroscopy A.8.1 Deduce the structure of a compound given information from its 1H nuclear magnetic

resonance (NMR) spectrum. Students will only be assessed on their ability to deduce the

number of different hydrogen (proton) environments and the relative numbers of hydrogen

atoms in each environment. They should be familiar both with a word description of a

spectrum and with a diagram of a spectrum,including an integration trace. The

interpretation of splitting patterns will not be assessed. Data banks could be used here.

number of peaks: the number of different hydrogen/proton environments

chemical shift: the environment of proton / neighbouring group

ratio of peak areas: the ratio of the numbers of protons in each environment;

splitting pattern: the number of (identical) protons on the neighbouring carbon atom(s);

e.g.1 Distinguish between the 1H NMR spectra of 1-bromopropane and 2-bromopropane

(splitting patterns are not required).

2-bromopropane will show two separate absorptions/peaks;in the ratio 6:1;

1-bromopropane will show three separate absorptions/peaks;in the ratio 3:2:2;

A.8.2 Outline how NMR is used in body scanners. Protons in water molecules within human cells

can be detected by magnetic resonance imaging (MRI), giving a three-dimensional view of

organs in the human body.

Magnetic resonance imaging (MRI) is a diagnostic technique in which spin/ magnetic

moment of protons, in water and other molecules inside a patient, interact with a magnetic

field.

MRI is (usually) a proton NMR/1H NMR;

(strong) magnetic field and radio waves/frequency are used;

human body consists of 70 %/mostly/a lot of water;

(the states of) protons in water molecules/lipids/carbohydrates in human cells

can be detected by MRI / cells have different water to lipid ratios / protons in water have

different chemical environments/give different signals;

different organs have different water concentration;

(by focusing the scanner on different parts of the body) three-dimensional/3-D images of

(organs in) the body are produced ;

Advantages: reducing health risks, can detect soft tissues/body organs/different properties

of (soft) tissue (and bone);

Accept “contrast can be a problem in X-ray”

Accept three-dimensional image

uses no ionizing radiation / uses low-energy radio waves / radio waves safer than x-rays;

Accept “does not damage body tissue”

e.g.1 Outline the principle behind magnetic resonance imaging (MRI) used to diagnose and

monitor conditions such as cancer in humans.

protons/hydrogens in water/lipids/carbohydrates (within cells) can be detected by (1H)

NMR/MRI;

to give a (three-dimensional) view/image of organs/body;

protons/hydrogens in cancerous cells are in a different chemical environment so give

different signals;


25

A.8.3 Explain the use of tetramethylsilane (TMS) as the reference standard.

A.8.4 Analyse 1H NMR spectra. Students should be able to interpret the following from 1H NMR

spectra: number of peaks, area under each peak, chemical shift and splitting patterns.

Treatment of spin–spin coupling constants will not be assessed, but students should be

familiar with singlets, doublets, triplets and quartets.

Ex.1 The 1HNMR spectrum of a compound with the formula C4H8O2 with chemical shifts, areas and

splitting patterns given below exhibits three major peaks.

chemical shift / ppm peak area splitting pattern

0.9 3 triplet

2.0 2 quartet

4.1 3 singlet

Using information from Table 19 in the Data Booklet, determine the types of proton present in

the molecule.

0.9 ppm H on C attached to a second C / alkyl group / R—CH3

2.0 ppm H on C attached to carboxyl C / C of an ester / CH3—CO—OR

4.1 ppm H on C attached to O of carboxyl group / ester group / R—CO—OCH2R

Deduce a structure consistent with the information indicated in (b). Explain your answer.

Structure

HH

H

HH

H

H

H

CC CC

O

O

Explanation [3 max]

two CH3 groups in different environments;

one CH3 group is not next to a C / has no neighbouring protons;

because it is a singlet (at 4.1ppm);

CH2 next to CH3 / C2H5;

because of triplet and quartet;

The 1H NMR information asked for was generally given correctly. The most common mistake

made by candidates was to interpret the splitting patterns and the chemical shifts incorrectly and

to give the wrong isomer (ethyl ethanoate) as the structure in G3(c) rather than the correct

structure (methyl propanoate).

Ex.2 A feature of some 1H NMR spectra is the electron-withdrawing effect of electronegative

atoms. These atoms cause nearby protons to produce peaks at higher chemical shift

values, often in the range 2.5 to 4.5 ppm.


26

Consider the 1H NMR spectrum of an unknown compound, D, which has a molecular

formula C4H8O2 and is known to have an absorption in its IR spectrum corresponding to

a C=O absorption.

Use this information and the values in Table 18 of the Data Booklet to deduce the

structure of D.

D could be CH3CH2COOCH3 or CH3COOCH2CH3;

this is because there are 3 peaks / 3:2:3 ratio;

explanation of splitting into a singlet, a triplet and a quartet;

methyl propanoate/CH3CH2COOCH3 is correct isomer because of higher chemical

shift value of singlet (3.6 instead of 2.0–2.5);

From this question it is clear that proton NMR is poorly understood in some schools. While

many candidates could identify the correct structure, few could write a description of how they

had obtained it from the information provided. Very few candidates explicitly mentioned that

there were 3 peaks or explained how the structure had been determined, especially neglecting to

explain the splitting patterns.


27

Infrared (IR) Spectroscopy

A.6.1 Describe the operating principles of a double-beam IR spectrometer.A schematic diagram of

a simple double beam spectrometer is sufficient.

Monochromator

ensures that only light with a single/particular/narrow range of wavelength/ frequency is

passed through;

Accept “light of only one colour”.

allows only a narrow band/one frequency/wavelength/wavenumber (of IR radiation) to pass

through;

Rotating mirror

ensures that the beam of (monochromatic) radiation is (alternatively) passed through the

sample and the reference;

Accept “splits/deflects the beam into two beams”.

Splitter:


28

splits the (infrared) light into two beams (with the same wavelength);

Reference:

absorbance/transmittance (of the reference) compared with/subtracted from absorbance/

transmittance of sample / (the reference is) used to set the baseline /compare with

sample/current / compensation for solvent;

Photmultiplier (photodiode)

converts light/radiation into an electrical current/signal;

Accept “detects the radiation”

Operating principles

(using a rotating mirror) beam of monochromatic radiation / radiation of one frequency/

wavelength /wavenumber;

splitter splits (IR) light into two beams (of same wavelength);

which passes through sample and reference;

photomultiplier converts radiation/photons into current/signal/voltage (output) /

photomultiplier/photodiode used as detector;

absorbance/transmittance of reference compared with/subtracted from

absorbance/transmittance of sample / reference used to set baseline;

IR spectrum generated by varying wavelength/frequency/wavenumber;

Description should include:

monochromator (to create single wavelength);

(rotating) mirrors/beam splitting;

sample and reference (to compare);

photomultiplier detector;

Allow diagram to show this.

infrared light is split into two (separate) beams;

one passes through the sample;

one through a reference;

a detector compares the two signals;

Accept a suitable diagram.

A.6.3 Explain what occurs at a molecular level during the absorption of IR radiation by molecules.

H2O, –CH2–, SO2 and CO2 are suitable examples. Stress the change in bond polarity as the

vibrations (stretching and bending) occur.

change in bond length / bond stretching / asymmetric stretch;

change in bond angle / bending (of molecule);

induces molecular polarity/dipole moment;

e.g.1 Explain the following observation: Hydrogen iodide is infrared active whereas iodine is

infrared inactive.

the bond in both molecules vibrates/stretches;

only the stretching in H—I causes a change in dipole moment;


29

e.g 2. Explain why the nitrogen molecule, N2, does not absorb infrared radiation.

no change in dipole moment/bond polarity;

as vibration/stretching occurs;

e.g.3 Explain at the molecular level why oxygen molecules do not absorb infrared radiation.

O2 is non-polar/has no dipole moment/symmetrical;

the stretching of the O=O bond would not change the polarity/dipole moment;

e.g.4 Describe two vibrations in the water molecule that absorb infrared radiation.

symmetrical stretching;

asymmetrical stretching;

bending/change in bond angle;

Accept diagrams of the water molecules which illustrate the bending and

stretching.

e.g.5 Outline the reasons why compounds containing C=C bonds absorb infrared radiation.

(C=C) bond vibrates;

Accept stretch/bend.

(vibration) must involve a change in dipole moment/polarity

e.g.6 Explain why different compounds containing C=C bonds absorb infrared radiation at slightly

different wavenumbers.

(energy/frequency of the C=C bond vibration) would depend on groups

attached/rest of the molecule;

e.g.7 Explain what occurs at a molecular level during the absorption of infrared (IR)

radiation by the sulfur dioxide molecule, SO2.

(O–S–O) bond angle changes;

(S–O) bond (length) stretches;

polarity of SO2 molecule changes;

A.6.4 Analyse IR spectra of organic compounds. Students will be assessed using examples

containing up to three functional groups. The Chemistry data booklet contains a table of IR

absorptions for some bonds in organic molecules.Students should realize that IR absorption

data can be used to identify the bonds present, but not always the functional groups present.

A.6.2 Describe how information from an IR spectrum can be used to identify bonds.

e.g.8 The infrared spectrum of C2H4O2.shows the following absorptions: 2920, 2765 and 1710cm−1

. Use

the information in Table 18 of the Data Booklet to assign each absorption to a particular vibration.

2920 cm−1

.....................................................................................................................

2765 cm−1

.....................................................................................................................

1710 cm−1

......................................................................................................................

Deduce the structures which would demonstrate the infrared absorptions above. Explain your

answer.

2


30

2920 cm−1

C—H;

2765 cm−1

O—H (in hydrogen bonded acids);

1710 cm−1

C==O;

H

H

H

HH

C C

O

O

;

(because of) the position of O—H vibration;

Ex.1 Consider the IR spectra of the following three compounds.

A = CH3(CH2)3COOH B = CH3COOC(CH3)3 C = (CH3CH2)3COH

Determine which IR spectrum corresponds to each compound A, B and C. Explain your

reasoning. IR data can be found in Table 17 of the Data Booklet.

Compound Spectrum Reason

A

B

C

A is Spectrum I and B is Spectrum III and C is Spectrum II;

A Spectrum I:

only spectrum with a (broad) peak in the range 2500–3300 (cm–1

) corresponding

to the carboxylic acid functional group / –OH in carboxylic acid / H-bonding in

carboxylic acid (so must be a carboxylic acid);

B Spectrum III:

peak in the range 1700–1750 (cm–1

) corresponding to the carbonyl/C=O group;

but no peak for O–H/no peak at 2500–3300 (cm–1

) or 3200–3600 (cm–1

);

C Spectrum II:

peak in the range 3200–3600 (cm–1

) corresponding to the alcohol functional

group/OH / the only one without a peak at 1700–1750 (cm–1

) corresponding

to an alcohol;


31

Identification of Unknown Organic Compounds Using Various Analytical Techniques

A.1.2 State that the structure of a compound can be determined by using information from a

variety of analytical techniques singularly or in combination.

e.g.1 Identify which analytical technique is regularly used for

(i) separation of a mixture of sugars.

(i) paper chromatography / thin layer chromatography / column chromatography /

high-performance liquid chromatography/HPLC;

(ii) 14C isotopic dating.

(ii) MS/mass spectrometry/spectroscopy;

(iii) scanning of the human body to detect diseases such as cancer and multiple sclerosis.

(iii) (1H/proton) NMR/nuclear magnetic resonance / MRI/magnetic resonance imaging;

e.g.2 Identify the most suitable spectroscopic technique to

(i) distinguish between butan-1-ol and butan-2-ol.

(i) (1H/proton) NMR/nuclear magnetic resonance;

(ii) determine the concentration of cadmium ions in polluted water. (ii) AA(S)/atomic absorption;

1. Organic compounds are often identified by using more than one analytical technique. Some of

these techniques were used to identify the compounds in the following reactions.

C3H7Br→ C3H8O→ C3H6O

A B C

(a) Using H2O as an example, describe what happens, at a molecular level, during the

absorption of infrared radiation.

(H—O—H) bond angle changes / bending;

(H—O) bond length changes / stretching;

polarity (of bond or molecule) changes;

(b) The infrared spectrum of B showed a broad absorption at 3350 cm–1

. The infrared

spectrum of C did not show this absorption, but instead showed an absorption at

1720 cm–1

. Explain what these results indicate about the structures of B and C.

B has O—H group / is an alcohol;

C has C==Ois a carbonyl compound / aldehyde or ketone

(c) The mass spectrum of A showed two lines of approximately equal height, one of which

was at m/z = 122. State the m/z value of the other line and explain these observations.

124 (ECF);

bromine has isotopes;

exists as 79

Br and 81

Br / peaks at M and M+2;

(d) The evidence in (b) and (c) indicates that each compound (A, B and C) could have two

possible structures. Draw the two possible structures of C.

CH3CH2CHO;

CH3COCH3;


32

(e) Fragmentation of C in a mass spectrometer produced lines with m/z values of 15 and 28,

but none at values of 14 or 29. Identify C and explain how you used this information to

do so.

C is CH3COCH

no 14 or 29 means no CH2 or C2H5 / 15 and 28 indicates CH3 and CO;

(f) State the number of lines in the 1HNMR spectrum of each of the structures in (d).

CH3COCH would have one line;

CH3CH2CHO would have three lines / accept splitting pattern;

2. Describe the IR and 1H NMR spectra of propanal and propanone. Your answer should include

both the similarities and the differences between the spectra of both compounds.

IR:

both have (sharp) C==O absorption at about 1680 – 1750 cm-1

1610 −1680 cm−1

;

and C—H absorptions at about 2840 − 3095 cm−1

;

only obvious difference will be in the “fingerprint region”;

1H NMR :

propanal will show three separate absorptions;

a triplet at 0.9 ppm, a quartet at 1.3 ppm (accept 1.1 – 1.6 ppm)

and a singlet at 9.7 ppm;

in the ratio of 3:2:1;

propanone will show one absorption;

a singlet at 2.1 ppm;

3. There are four isomeric alcohols with the molecular formula C4H10O. They can be distinguished

using a variety of analytical techniques.

(a) The structures of two of the alcohols (A and B) are shown below. Draw a structure for

each of the other two alcohols (C and D).

CH CHCH CHCH CHCH CH

OH OH

2 32 2 23 3

alcohol A alcohol B

alcohol C alcohol D

(C/D) − (CH3)3COH;

(C/D) − (CH3)2CHCH2OH;

C and D can be either way round.

(b) Explain why the four compounds could not easily be distinguished by looking at their

infrared spectra.

they have same functional groups / they all have an absorption in the range 2840 – 3095 /

1000 – 1300 / 3230 – 3550 cm–1

;

(c) The 1H NMR spectra of A and B both show the same number of peaks, but with a

different ratio of areas under the peaks.

(i) State what can be deduced from the number of peaks in an 1H NMR spectrum.

the number of different chemical environments of the hydrogen atoms / protons


33

(ii) Deduce the number of peaks in the 1H NMR spectra of A and B.

5;

Accept 6 (if TMS has been included).

(iii) Determine the ratio of areas under the peaks for A and B.

A

3 : 2 : 2 : 2 : 1;Order not important

B

3 : 3 : 2 : 1 : 1;Order not important

(d) One of the alcohols, C or D, has a high resolution 1H NMR spectrum that shows only

peaks that are singlets. Identify which alcohol it is, and explain why there is no splitting

of the peaks.

C/D;

Depending on how (CH3)3COH is labelled in (a);

no adjacent carbon atoms with hydrogen atoms

(e) Explain the following features of the mass spectra of A and B.

(i) Both spectra show a peak at m/z = 74.

(this is due to) the molecular ion / C4H10O+ / C4H9OH

+;

(ii) One spectrum shows a prominent peak at m/z = 45 but the other shows a prominent

peak at m/z = 31.

peak at 45 due to CH3CHOH+ / loss of C2H5;

peak at 31 due to CH2OH+ / loss of C3H7;

(f) Another compound, E, with molecular formula C4H10O, has an 1HNMR spectrum as

follows:

5 4 3 2 1 0

Chemical shift / ppm

TMS

3

2

(i) Explain the splitting patterns in the spectrum.

the triplet means next C has 2 H atoms / is CH2 group;

the quartet means next C has 3 H atoms / is CH3 group;

so presence of ethyl group / C2H5 / CH3CH2;

(ii) Compound E has an absorption in its infrared spectrum close to 1150 cm–1

.

Deduce which bond in E is responsible for this and use the information from

both spectra to deduce the structure of E.

Bond in E ........................................................................................................

Structure of E


34

bond in E

C—O;

structure of E

CH3CH2OCH2CH3;

4. There are two alcohols, A and B, with molecular formula C3H8O. The following information was

obtained from a mass spectrum of each alcohol.

A: peaks at m/z = 29, 31, 60

B: peaks at m/z = 45, 60

(a) Write the formula for the species responsible for the peak at m/z = 60.

C3H8O+;

(b) Deduce the formula of the species with m/z = 31.

CH3O+

/ CH2OH+;

(c) Deduce the structures of the two alcohols.

Structure of A

Structure of B

(A) CH3CH2CH2OH;

(B) CH3CH(OH)CH3;

(d) The peak at m/z = 45 is more prominent than that at m/z = 29. Suggest a reason for this.

(for m/z = 45) two possible bonds to break / two CH3 groups could be lost / two ways

to make this;

OR

(for m/z = 29) only one bond could break to produce this / one way to make this;

(e) The 1HNMR spectrum of one of the alcohols shows four peaks with areas in the ratio

3:2:2:1.

(i) State what can be deduced from this information.

Four peaks

Areas in ratio 3:2:2:1

(four) different environments for hydrogen atoms/protons;

the number of hydrogen atoms/protons in each environment

(are in the ratio 3:2:2:1)

(ii) Predict the number of peaks, and the ratio of their areas, in the 1HNMR spectrum

of the other alcohol.

3 peaks;

6:1:1;

order not important

5. This question is about the three organic compounds involved in the following reaction.

C2H4O2 + C2H6O → C4H8O2 + H2O

W X Y (a) The infrared spectra of all three compounds showed several absorptions. Describe what

happens on a molecular level when molecules absorb infrared radiation.

bond angles change / bonds bend;

bond lengths change / bonds stretch;

bond polarity changes;


35

(b) Use the following information about their infrared spectra to deduce which bonds are

present in the three compounds.

All three compounds showed an absorption close to 1200cm–1

.

There were broad absorptions in both W and X. The one in W was centred around 3000

cm–1

, and in X around 3400 cm–1

.

Compounds W and Y showed absorptions close to 1700 cm–1

.

bonds in W ...........................................................................................................................

bonds in X ............................................................................................................................

bonds in Y ............................................................................................................................

W: C––O, C==O, O––H;

X: C––O, O––H;

Y: C––O, C ==O;

(c) The 1H NMR spectra of the three compounds were available. State what can be deduced

from each of the following.

(i) The presence of two peaks in the spectrum of W.

two different chemical environments for hydrogens / two different ‘types’ of proton

(ii) The presence of a triplet and a quartet, with areas in the ratio 3:2, respectively, in

the spectra of both X and Y.

CH3 and CH next to each other / CH3CH2/ ethyl group;

(d) Use your answers to (b) and (c) to deduce the structures of the three compounds.

W ...........................................................................................................................

X ..........................................................................................................................

Y ..........................................................................................................................

W: CH3COOH

X: CH3CH2OH;

Y: CH3COOCH2CH3;

(e) The infrared spectrum of compound Z, an isomer of Y, is shown below. 100

80

60

40

20

Transmittance / %

3 000 2 000 1 000

Wavenumber / cm–1 (i) Estimate the wavenumber values of the three most prominent absorptions in this

spectrum and suggest which bonds are responsible for them.

absorption 1 ............................................................................................


36

absorption 2 ............................................................................................

absorption 3 ............................................................................................

3000 – 2900 (cm–1

) C––H;

1750 – 1700 (cm–1

) C==O;

1150 – 1200 (cm–1

) C––O;

(ii) Deduce the structure of Z.

CH3CH2COOCH3;

Accept C2H5COOCH3 and HCOOCH(CH3) and CH3COCH2OCH3 .

6. The 1H NMR spectrum of X with molecular formula C3H6O is shown below.

(a) Deduce which of the following compounds is X and explain your answer.

CH3–CO–CH3 CH3–CH2–CHO CH2=CH–CH2OH

Compound:

Explanation:

Compound:CH3–CH2–CHO;

Explanation:

only this compound would give 3 peaks;

only this compound has H–atoms in 3 different chemical environments ;

only this compound has protons in ratio 3:2:1 in each environment;

only this compound would give a peak in the 9.4–10 ppm region ;

(b) Deduce which one of the peaks in the 1H NMR spectrum of X would also occur in the

spectrum of one of the other isomers, giving your reasoning.

2.5 ppm peak;

CH3–CO–CH3 also has hydrogen atoms on a carbon next to the >C=O group;

(c) The infrared and mass spectra for X were also recorded.

(i) Apart from absorptions due to C–C and C–H bonds, suggest one absorption, in

wavenumbers, that would be present in the infrared spectrum.

1700–1750 cm–1

(>C=O);


37

(ii) Apart from absorptions due to C–C and C–H bonds, suggest one absorption, in

wavenumbers, absent in this infrared spectrum but present in one of the other

compounds shown in part (a).

1610–1680 cm–1

(>C=C<) / 3200–3600 cm–1

(–O–H);

(c) Suggest the formulas and m/z values of two species that would be detected in the mass

spectrum.

Species: .......................................................................................................................

m/z: .............................................................................................................................

Species: .......................................................................................................................

m/z: .............................................................................................................................

C3H6O+ and m/z = 58;

C2H5+ and m/z = 29;

CHO+ and m/z = 29;

CH3+ and m/z = 15;

Penalize missing + sign once only.

7. The IR spectrum, mass spectrum and 1H NMR spectrum of an unknown compound, X,

of molecular formula C3H6O2 are as follows.


38

Source: SDBSWeb: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and

Technology)] (i) Identify the bonds responsible for the peaks A, B and C in the IR spectrum of X.

A: O–H

B: C=O

C: C–O

(ii) In the mass spectrum of X, deduce which ions the m/z values at 74, 45 and 29

correspond to.

m/z = 74 :C2H5COOH+ / C3H6O2

+;

m/z = 45 :COOH+;

m/z = 29 :C2H5+;

(iii) Identify the peak at 11.73 ppm in the 1H NMR spectrum. –COOH;

(iv) Deduce the structure of X.

. . CH3CH2COOH / CH3CH2CO2H;


39

8. The infrared spectrum of a substance, X, with empirical formula C3H6O is given below.

[Source: NIST http://webbook.nist.gov/chemistry]

(i) Explain why the structural formula of X cannot be:

absence of peak between 3200–3600 cm–1/above 3000 cm–1/peak for OH;

presence of peak between 1700–1750 cm–1/peak for C=O;

absence of peak between 1610–1680 cm–1 /peak for C=C;

(ii) The 1H NMR spectrum of X consists of three peaks. Deduce the structural

formula of X and the relative areas under each peak.

Accept CH3CH2CHO.

3:2:1;

9. (a) The mass spectrum of an unknown compound, X, of empirical formula C2H4O is

shown below.


40

[Source: Cleapss Guides: L202 Spectra (Cleapss School Science Service), Sept 2000.]

(i) Determine the relative molecular mass of X from the mass spectrum and

deduce the formula of the molecular ion. 88;

C4H8O2+;

(ii) Identify a fragment which gives rise to the peak at m/z = 29.

CH3CH2+ /C2H5

+/ CHO

+;

(iii) Comment on the absence of a peak at m/z = 59. C2H3O2 produced has no charge / fragment produced after loss of C2H5 from

molecular ion has no charge;

(b) The IR spectrum of X is shown below.

[Source: http://modbo1.ibase.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng]

(i) Use Table 17 of the Data Booklet to identify the bonds which correspond to

the absorptions A and B.

A: C=O and B: C–O;


41

(ii) Deduce the name of the functional group present in X.

ester;

(c) (i) The 1H NMR spectrum of X shows three peaks. State the information that

can be obtained from the number of peaks.

the number of different hydrogen/proton environments

(ii) The 1H NMR spectrum of X includes peaks at 2.0 and 4.1 ppm. Use Table

18 of the Data Booklet to suggest the chemical shift of the third peak and

state its relative peak area. Show your answers in the table below.

Peak Chemical shift / ppm Relative peak area

First 2.0 3

Second 4.1 2

Third

Peak Chemical shift / ppm Relative peak

First 20 3

Second 4.1 2

Third 0.9 –1.0; 3;

(iii) Deduce a possible structure for X that is consistent with the mass, IR and 1H

NMR spectra.

10. Butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol are four structural isomers with

the molecular formula C4H10O.

(a) Details of the 1H NMR spectra of two of these alcohols are given below.

Spectrum 1

Two peaks: One at 1.3 ppm (relative to the TMS reference) with an integration trace

of nine units, and the other at 2.0 ppm with an integration trace of one unit.

Spectrum 2

Four peaks: The first at 0.9 ppm with an integration trace of six units.

The second at 1.7 ppm with an integration trace of one unit.

The third at 2.1 ppm with an integration trace of one unit.

The fourth at 3.4 ppm with an integration trace of two units.

Consider the proton environments present in each of the alcohol molecules when

answering the following questions.

(i) Identify which alcohol gives spectrum 1 and explain your answer by stating which

hydrogen atoms in the molecule are responsible for each of the two peaks.

(2-)methylpropan-2-ol;

the (H atoms in the three) –CH3 groups are responsible for the

peak at 1.3 ppm;

the –OH hydrogen atom is responsible for the peak at 2.0 ppm;

Accept explanations with suitable diagram. (ii) Deduce which alcohol gives spectrum 2. Explain which particular hydrogen atoms


42

in the molecule are responsible for the peaks at 0.9 ppm and 3.4 ppm.

(2-)methylpropan-1-ol;

the first peak (at 0.9 ppm) is due to the (H atoms in the)

two –CH3 groups (bonded to the second carbon atom) /

(CH3)2CHCH2OH;

the peak at 3.4 ppm is due to the (H atoms in the) –CH2– group /

(CH3)2CHCH2OH;

(b) The mass spectrum of one of the alcohols shows peaks at m/z values of 74, 59 and 45.

(i) Deduce which two of the alcohols could produce this spectrum and identify the

species responsible for the three peaks.

butan-1-ol and butan-2-ol;

74: M+ / C4H10O

+ / CH3CH2CH2CH2OH

+ and CH3CH2CH(OH)CH3

+;

59: C3H7O+ / (M – CH3)+ / CH2CH2CH2OH

+ and

CH2CH(OH)CH3+/CH3CH2CH(OH)

+;

45: C2H5O+ / (M – C2H5)

+ / CH2CH2OH

+ and CH(OH)CH3

+;

(ii) The spectrum also shows a significant peak at m/z = 31. Suggest which alcohol is

responsible for this spectrum and deduce the species responsible for the peak at

m/z = 31.

butan-1-ol;

CH2OH+ / (M – C3H7)

+;

(c) Explain why the infrared spectra of all four alcohols are very similar.

they all contain O–H;

they all contain C–H;

they all contain C–O;


43

A.4.1 Describe the effect of different ligands on the splitting of the d orbitals in transition metal

complexes. (HL only) The ligands should include NH3, H2O and Cl–

increase in oxidation state causes greater splitting;

change from H2O to NH3 causes greater splitting;

the greater the splitting, the higher the frequency (of absorbed light);

(complexes of) Cr(III) absorb higher-frequency light than (complexes of) Cr(II) /

(complexes with) NH3 absorb higher-frequency light than (complexes with)

H2O ; .

A.4.2 Describe the factors that affect the colour of transition metal complexes. Include the identity

of the metal ion (for example,Mn2+ or Fe2+ ), the oxidation number of the metal (for example, for

Fe, +2 or +3) and the identity of the ligand (for example, NH3 or H2O). These factors will be

assessed only for octahedral complexes in aqueous solution.

oxidation state of transition element/number of d electrons/charge on ion;

type/identity/charge density of ligands;

stereochemistry/shape of complex/number of ligands;

e.g.1 (a) The colour of transition metal complexes depends on several factors.

(i) Use [Mn(H2O)6]2+

and [Fe(H2O)6]2+

as examples to outline why the colour depends on

the identity of the transition metal itself.

(H2O)6]2+

is pink / colourless and [Fe(H O)6]2+

is green /the colours they show are

complementary to the colours they absorb;

colour is caused by transitions between the d orbitals;

different metals cause the d orbitals to split differently

(due mainly to the different number of protons in the nucleus);

(ii) Outline why the colour depends on the oxidation state of the transition metal.

the oxidation state affects the size of the d orbital splitting due to the different number

of electrons present

(iii) Outline why the colour depends on the identity of the ligand.

the more electron-dense the ligand the greater the splitting of the d orbitals

e.g.2 Aqueous solutions containing complexes of transition metals are usually coloured. This is due

to the absorption of part of the spectrum of white light passing through the solution.

(a) Three factors help to determine the colour absorbed.

For each of the following pairs, state the difference between the two complexes that is

responsible for the difference in colour.

[Co(NH3)6]2+

and [Ni(NH3)6]2+

[Fe(H2O)6]2+

and [Fe(H2O)6]2+

[Cu(NH3)4(H2O)2]2+

and [Cu(H2O)6]2+

(difference in) metal (ion);

(difference in) oxidation number (not charge);

(difference in) ligand;

(b) The wavelength of colour absorbed by the complex can be explained in terms of the

splitting of the d orbitals in the metal ion.

The arrangement of electrons in the d orbitals of the Cu2+

ion is shown in the following

diagram.


44

Draw a diagram to show how the electrons are arranged in Cu

2+ when it is present in the

[Cu (H2O)6]2+

ion.

(c) Predict whether the splitting of the d orbitals in [Cu(NH3)4(H2O)2]2+

and [CuCl4]2−

would

be less than or greater than the splitting in [Cu(H2O)6]2+

.

splitting in [Cu(NH3)4(H2O)2]2+

..........................................................................

splitting in [CuCl4]2−

..........................................................................

[Cu(NH3)4(H2O)2]2+

greater / [CuCl4]2−

less

e.g.3 Copper(II) sulfate forms a pale blue aqueous solution. When aqueous ammonia is added to this,

initially a pale blue precipitate forms; this precipitate then dissolves in excess ammonia to form

a deep blue solution. Explain why these solutions are coloured and the colour with excess

ammonia is a deeper blue.

electron transitions between (split, partially filled) d orbitals;

absorption depends on energy difference between the split d orbitals; waters replaced by

ammonias;

ammonia (ligands) increase the splitting between the d orbitals/larger energy difference;

absorption moves to shorter wavelength/higher frequency/towards blue end of spectrum;

e.g.4 When excess ammonia solution is added to a solution of copper(II) sulfate the oxidation number

of the copper ion does not change but there is a noticeable colour change.

Outline the reasons for this change in colour.

the H2O ligand is exchanged for NH3;

Accept suitable equation with co-ordination numbers of 4 and 6.

colour is due to electron transitions between the split d orbitals;

NH3 causes a greater/different splitting than H2O;

e.g.5 Although both lead, Pb, and chromium, Cr, are metals, only chromium is classified as a

transition metal and forms transition metal complexes, such as [Cr(H2O)6]3+.

(i) The energy level diagram showing the electrons in the five 3d orbitals of a

chromium atom is represented below. Draw the completed diagram showing the d

orbitals in [Cr(H2O)6]3+ after splitting.

(ii) State and explain what happens to the splitting of the d orbitals if the ligand is


45

changed from H2O to NH3.

(ii) increases / greater;

Award [1] for one of the following:

NH3 has greater electron/charge density;

NH3 higher in spectrochemical series;

NH3 stronger base;

e.g.6 The wavelength of colour absorbed by a transition metal complex can be explained in

terms of the splitting of the d orbitals in the metal ion.

(i) For the complex, [Ni(NH3)6]Cl2, draw a diagram showing the splitting of the d

orbitals.

(i) drawing showing splitting of levels into a low-energy triply degenerate set;

high-energy doubly degenerate set;

(ii) Outline why colour depends on the oxidation state of the transition metal.

(ii) the oxidation state affects the size of the d orbital splitting (due to the different

number of electrons present);

(iii) The complex [Ni(NH3)6]Cl2 is purple. Predict the colour of the [Ni(H2O)6]Cl2

complex and explain your answer. (iii) colour will be yellow/green/red/pink/orange;

NH3 ligand causes more splitting of d orbitals than H2O;

e.g.7 The complex ion [Ni(H2O)6]2+

is green and [Ni(NH3)6]2+

is blue. Explain why the [Ni(H2O)6]2+

complex ion is coloured and outline why changing the identity of the ligand changes the colour

of the ion.

d orbitals splits (into two levels);

due to repulsion between d electrons and non-bonding electrons on ligand / due to interaction

with electric field of ligands;

difference in energy between levels corresponds to visible light;

(visible light absorbed as) electrons move from lower to higher energy d orbitals;

colour observed complementary to absorbed;

Why changing ligand changes colour:

more electron-dense ligand greater splitting of d orbitals;

NH3 ligand has greater (crystal field/ligand) splitting energy / NH3 ligand at higher energy in

spectrochemical series;

Documents

John Ng's Guide to Analytical Chemistry