Chapter 29: Analytical chemistry
Type of proton Chemical shift, δ / ppm
R CH3 0.7–1.6
N H R OH 1.0–5.5(a)
R CH2 R 1.2–1.4
R3CH 1.6–2.0
O O O 2.0–2.9
H3C C RCH2 C R2CH C
CH3 CH2R CHR2 2.3–2.7
2.3–2.9
N CH3 N CH2R N CHR2 3.3–4.3
O CH3 O CH2R O CHR2 3.0–4.2
Br or Cl CH3 Br or Cl CH2R Br or Cl CHR2
4.5–10.0(a)
OH
45.–6.0
CH CH O
5.0–12.0(a)
O C
C NH
NH2 441
H 6.5–8.0
O 9.0–10
C 11.0–12.0(a)
H
O
C
H
Table 29.1 1H NMR chemical shifts relative to TMS. Chemical shifts are typical values that can vary slightly depending on the
solvent, concentration and substituents. (a)OH and NH chemical shifts are very variable (sometimes outside these limits and are
often broad. Signals are not usually seen as split peaks).
question High-resolution NMR
5 P redict the number of peaks and the relative areas As you can see in Table 29.1, the chemical shifts are given over
under each peak, where appropriate, on the low- ranges, and the ranges for different types of hydrogen atoms
resolution proton NMR spectrum of: do overlap. In some molecules where there is heavy shielding
of the hydrogen nuclei by lots of electrons in surrounding
a methanol, CH3OH d propan-1-ol atoms, peaks are shifted beyond their usual range. In such
b benzene, C6H6 e propan-2-ol cases high-resolution NMR is useful. High-resolution NMR
c chloroethane, C2H5Cl f propanone. gives us more information to interpret. Peaks that appear
as one ‘signal’ on a low-resolution NMR spectrum are often
revealed to be made up of a cluster of closely grouped peaks.
This is because the magnetic fields generated by spinning
nuclei interfere slightly with those of neighbouring nuclei.
Cambridge International A Level Chemistry
This interference is called spin–spin coupling. The exact Table 29.2 shows the relative intensities and distribution of the
splitting pattern of a peak depends on the number of splitting patterns you are likely to meet.
hydrogen atoms on the adjacent carbon atom or atoms.
Figure 29.19 shows another high-resolution NMR
The number of signals a peak splits into equals n + 1 spectrum. You should try to interpret it by following
where n is the number of 1H atoms on the adjacent these steps:
carbon atom.
Step 1 Use δ values to identify the environment of the
The high-resolution NMR spectrum of ethanol illustrates this equivalent protons (1H atoms) present at each
n + 1 rule used to interpret splitting patterns (Figure 29.18). peak (remembering the peak at zero is the TMS
standard reference peak).
■■ The CH3 peak is split into three because there are two 1H
atoms on the adjacent CH2 group. n + 1 = 3 (as n = 2); this is Step 2 Look at the relative areas under each peak
called a triplet. to determine how many of each type of non-
The CH2 equivalent protons (1H atoms) are present.
■■ 1H atoms on peak is split into four because there are three
the adjacent CH3 group. n + 1 = 4 (as n = 3); Step 3 Apply the n + 1 rule to the splitting patterns to see
this is called a quartet. which protons (1H atoms) are on adjacent carbon
■■ The OH peak is not usually split as its 1H atom is atoms in the unknown molecule.
constantly being exchanged with the 1H atoms of other
Step 4 Put all this information together to identify the
unknown molecule.
ethanol molecules and any water present. This results in
one average peak being produced.
3H
1H
442 Absorption of energy
2H
6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0
δ / ppm
Figure 29.18 The high-resolution NMR spectrum of ethanol, showing the splitting pattern in two of the peaks. The area under
each series of peaks still represents the number of equivalent 1H atoms in the molecule, as in low-resolution NMR.
Number of Using the n + 1 rule, the Relative intensities in the Observed on the NMR spectrum as …
adjacent 1H atoms peak will be split into … splitting pattern
0 1 peak, called a singlet 1
1 2 peaks, called a doublet 1:1
2 3 peaks, called a triplet 1:2:1
3 4 peaks, called a quartet 1:3:3:1
Table 29.2 Splitting patterns in high-resolution NMR spectra.
Chapter 29: Analytical chemistry
worked example question
1 An ester is used as a solvent in a glue. A chemist was 6 A pathologist was given a sample of a white tablet
given a sample of the ester to analyse. The NMR found at the scene of a suicide. In order to complete
spectrum of the ester is shown in Figure 29.19. her report the pathologist received an NMR spectrum
of the sample (Figure 29.20a) and information from the
3H police that the tablets involved were either aspirin or
paracetamol. The displayed formulae of both drugs
3HAbsorption of energy are also shown in Figure 29.20b.
2H
Absorption of energy a 3H
6543210 1H 2H 2H
Chemical shift, δ / ppm 1H
Figure 29.19 The high-resolution NMR spectrum of
an unknown ester in a glue.
Step 1 Identify possibilities for the three major
peaks that appear at chemical shifts of 1.3,
1.9 and 4.1 ppm. Using Table 29.1, these 11 10 9 8 7 6 5 4 3 2 1 0
could be: Chemical shift, δ / ppm
1.3 ppm R CH3, R CH2 R b HO
1.9 ppm RRC3CHH2C(pOo ss iobrlyRH2C3HC
CO H C H H H
CO O H
, H H 443
) O N
4.1 ppm O CH3, O CH2R, O CHR2 HO
H CCH
HC H
Step 2 Use the relative numbers of each type of O C HO H
proton (1H atom) labelling each major peak H H
to narrow down possibilities.
aspirin paracetamol
1.3 ppm labelled 3H, so could be R CH3
1.9 ppm labelled 3H, so could be H3C CO Figure 29.20 a NMR analysis of the unknown drug
4.1 ppm labelled 2H, so could be O CH2R sample. b Aspirin and paracetamol.
Step 3 By applying the n + 1 rule to the splitting a Using this information, which drug was in the white
patterns we can see which protons tablet? Explain your answer.
(1H atoms) are on adjacent carbon atoms.
1.3 ppm labelled 3H and split into triplet, so R CH3 b Sketch the NMR spectrum you would expect to
would be next to a C atom bonded to two see if the other drug had undergone NMR analysis.
Label each peak with its relative area and the type
of proton that caused it.
1H atoms (2 + 1 = 3, triplet). CO
1H atoms
1.9 ppm labelled 3H and a singlet, so H3C
would be next to a C atom with no
attached (0 + 1 = 1, singlet). It could well be Identifying the OH or NH signal in an
next to the C O, with the carbonyl carbon NMR spectrum
also bonded to an O atom, as in an ester, i.e. The OH signal in the high-resolution NMR spectrum
H3C COOR. of ethanol appears as a single peak. As we have seen on
4.1 ppm labelled 2H and split into quartet, O CH2R page 442, the peak is not split by the 1H atoms (protons)
would be next to a C atom bonded to three on the neighbouring e xCcHha2 nggesrovuepry. Trhapeirdelayswonithfoprrtohtiosniss
1H atoms (3 + 1 = 4, quartet). that the OH proton
Step 4 Putting this information together we get the in any traces of water (or acid) present, as follows:
ester ethyl ethanoate, CH3COOCH2CH3.
CH3CH2OH + H2O CH3CH2OH + HOH
Cambridge International A Level Chemistry
The hydrogen atoms involved in this reversible proton question
exchange have been coloured red and blue. The exchange takes
place so rapidly that the signal for the OH protons becomes 7 a Look back to Figure 29.18 on page 442. The high-
a single peak. This exchange also happens in amines and resolution NMR spectrum shown is from a sample
amides which contain the NH group. of ethanol containing traces of water. How would
the NMR spectrum differ if D2O had been added to
Table 29.3 shows the chemical shift ranges for the the sample of ethanol?
different OH and NH signals.
b Look back to Question 6. How would repeating
the NMR paantahlyosloisguisstindgisatinsogluviesnhtboeftDw2eOebneaaspbilreinto
Different OH and NH protons Range of chemical help the
in alcohols, R OH protons shift (δ) / ppm and paracetamol?
1.0–5.5
in phenols, arene OH protons 6.5–7.0 Carbon-13 NMR spectroscopy
in carboxylic acids, R COOH protons 11.0–11.7
in amines, NH2 / NH 1.0–5.5 In addition to proton NMR. carbon-13 NMR is another
analytical tool used frequently by organic chemists. The
in aryl amines, arene NH2 vast majority of carbon atoms in any organic compound
in amides, CONH2 , CONH 3.0–6.0 will be the carbon-12 isotope. This isotope has an even
5.0–12.0 mass number (12). Therefore it has no signal on an NMR
Table 29.3 Chemical shift ranges for OH and NH spectrum, as NMR only works with atoms with an odd
protons in different molecular environments. mass number (such as 1H, as we have already seen).
However, about 1% of the carbon atoms in any sample of
an organic compound will be the carbon-13 isotope. Its
As you can see from Table 29.1 (page 441), these ranges nuclei will interact with the magnetic field applied in NMR
444 overlap with the chemical shifts of other types of analysis so can produce a NMR spectrum.
proton. The signals can also appear outside the quoted Carbon-13 NMR produces a spectrum with different
ranges under certain conditions, e.g. choice of solvent chemical shifts for non-equivalent carbon atoms in a
or concentration. This makes NMR spectra difficult to molecule. Typical carbon-13 NMR shifts are shown in
interpret. However, there is a technique for positively Table 29.4. As in proton NMR, the chemical shifts are
identifying OH or NH groups in a molecule. measured with reference to the TMS peak at 0 ppm on the
Their peaks ‘disappear’ from the spectra if you add a spectrum (see page 439).
Tsemxhceahldalenaugmteeorruieunvmetrosafitbdolemyuwstei(rt2hiHutm)hienopxDrido2Oeto,,nDcsa2iOlnle,dttho‘ehteh aev OsyaHwmaoptrleer.’,
Analysis of carbon-13 NMR spectra is similar to that
NH groups. For example: of proton NMR, looking to match different chemical shifts
to characteristic molecular environments. However, the
OH + D2O OD + HOD signals produced in carbon-13 NMR appear as discrete
vertical lines on the spectra (without the complication of
NH CO + D2O ND CO + HOD the splitting patterns caused by the protons in 1H atoms
within the molecules). Take care in interpreting the
The deuterium atoms do not absorb in the same region carbon-13 NMR spectra because the heights of the lines
of the electromagnetic spectrum as protons (1H atoms).
This makes the OH or NH signal disappear from the NMR are not usually proportional to the number of equivalent
spectrum. By checking against the peaks in the original NMR 13C atoms present.
The solvent used to prepare samples for 13C NMR
spectrum, without Din2Oth,ewseamcapnlete. lTlhifeth1He at oOmHionrt he N H O H or 8an0 aplpymsistihsaCt DcaCnl3b.eTihginsoarcecdowunhtesnfoinrttehreprsemtianlgl saigsnpaelcntreuamr ,
groups are present as it is caused by the atoms of 13C in the solvent molecules.
NH group is referred to as a ‘labile’ proton.
Chapter 29: Analytical chemistry
Hybridisation of Environment of carbon atom Example structures Chemical / ppm
carbon atom shift range (δ)
alkyl CH3 , CH2 , CH
sp3 next to alkene/arene CH2 C C, CH2 0–50
sp3 10–40
sp3 next to carbonyl/carboxyl CH2 COR, CH2 CO2R 25–50
sp3 next to nitrogen CH2 NH2, CH2 NR2, CH2 NHCO 30–65
sp3 next to chlorine ( CH2 Br and CH2 Cl 30–60
CH2 I are in the same range
as alkyl)
sp3 next to oxygen CH2 OH, CH2 O CO 50–70
sp2 alkene or arene C C C 110–160
C C C
C C ,
sp2 carboxyl R CO2H, R CO2R 160–185
sp2 carbonyl R CHO, R CO R 190–220
sp2 alkyne R C C 65–85
sp2 nitrile R C N 100–125
Table 29.4 Typical 13C chemical shift values (δ) relative to TMS = 0. Note that chemical shifts are typical values and can vary 445
slightly depending on the solvent, the concentration and substituents present.
Figure 29.21 shows the 13C NMR spectrum for propanone, Figure 29.22 shows another example of a carbon-13 NMR
(CH3)2CO. spectrum, that of ethylbenzene, C6H5CH2CH3.
Note that there are only two peaks: one for the carbon
atom in the carbonyl group, C O, and the other for the The carbon atoms in the benzene ring are almost
carbon atoms in the methyl ginropurposp,ano CnHe,3.thAelythaorue gbhoth equivalent but will be affected to slightly different extents
there are two CH3 groups by the presence of the ethyl group in the molecule. Hence
equivalent and so appear as only a single peak (just as the series of lines clustered near 125 ppm.
equivalent H atoms do in proton NMR).
C6H5CH2CH3
H3C O
C
H3C
δ 200 100 0 δ 200 100 0
Figure 29.21 The carbon-13 NMR spectrum of propanone. Figure 29.22 The carbon-13 NMR spectrum of ethylbenzene.
Cambridge International A Level Chemistry
question The peak at the highest mass-to-charge ratio is caused by
the molecular ion (M+). This ion is formed by the sample
8 a Look at the series of lines clustered at about molecule with one electron knocked out. It gives us the relative
125 ppm in Figure 29.22. molecular mass of the sample. We can assume the
ions detected carry a single positive charge, so the reading
i One line is separated slightly from the main on the horizontal axis gives us the mass. In the case of
cluster – explain which carbon atom in mopfrao(3spsa×onf1o52n8.e0.,0)C.+TH(h13iCs×cO1oC6r.rH0e)s3p+, ot(hn6ed×ms t1oo.0leC)c.Hul3aCrOioCnHha3+s,awrietlhataivme ass
ethylbenzene is most likely to have produced
that signal. We also get large peaks at 15 and 43 on the mass
spectrum. These peaks are due to fragments that are
ii Predict how many lines make up the tightly produced when propanone molecules are broken apart
clustered grouping of the tallest line on the 13C by the electron bombardment. Knowing the structure
spectrum and explain your reasoning. of propanone we should be able to identify the fragment
responsible for each peak (Figure 29.24).
b P redict the number and location of signal lines in
the carbon-13 NMR spectrum of benzene, C6H6.
c Explain the number of signal peaks you would
expect to see in the carbon-13 NMR spectrum of:
i propan-1-ol
ii propan-2-ol.
Mass spectrometry
You have already seen how a mass spectrometer
works (see page X). The mass spectrum of an element
can be used to measure relative isotopic masses and
446 their relative abundances. This information is used to Figure 29.24 TahnedfCraHg3mC+eOnctaatuiosensotfhperpoepaaknoatn4e3: +.CH3 causes
the peak at 15
calculate relative atomic masses. However, the main
use of mass spectrometry is in the identification of
organic compounds. As in other forms of spectroscopy, The electron bombardment has caused the C C single
a substance can be identified by matching its spectrum bonds in the propanone molecules to break. This has
against the spectra of known substances stored in a resulted in the fragments at m/e 15 and 43 that are
database. This technique is known as ‘fingerprinting’. observed in Figure 29.22. The breaking of single bonds,
In a mass spectrometer the sample is first vaporised. such as C C, C O or C N, is the most common cause
When vapour from the sample enters the machine it of fragmentation.
is bombarded by high-energy electrons. This knocks
electrons from the molecules and breaks covalent bonds,
fragmenting the molecule. Figure 29.23 shows the mass question
spectrum produced by propanone.
9 L ook at Figure 29.25 on page 447, which shows the
Relative abundance (%) 100 43 mass spectrum of ethanol, C2H5OH. A structural
80 isomer of ethanol is methoxymethane, an ether with
the formula CH3OCH3.
60
58 a Predict the mass-to-charge ratio of a fragment
that would appear on the mass spectrum of
40 methoxymethane but does not appear on
ethanol’s mass spectrum.
20 15
b Give the formula of the ion responsible for the
peak in your answer to part a.
0 100
0 20 40 60 80
Mass-to-charge ratio (m/e)
Figure 29.23 The mass spectrum of propanone.
Chapter 29: Analytical chemistry
question (continued) 100 [CH2OH]+
c Look at the mass spectrum of ethanoic acid:
Relative abundance (%) 80
Relative abundance (%)
100 43 60
80 45
60 15 40 [C2H5]+ [C2H5O]+
40 60 20 [C2H3]+ [C2H5OH]+
20
[M + 1]
0
10 20 30 40 50
Mass-to-charge ratio, m/e
20 40 60 80 100 Figure 29.25 The mass spectrum of ethanol, showing the
Mass-to-charge ratio (m/e) [M + 1] peak.
Identify the fragments with mass-to-charge In any organic compound there will be 1.10% carbon-13.
ratios of: We can use this fact to work out the number of carbon
atoms (n) in a molecule. We apply the equation:
i 15 iii 45
ii 43 iv 60. n = _1 10_.10_ × _ a _b_ua _nb_du_an_nd_ ca_en_oc_ef_[o_ Mf_M_ +_+ _1i_]o_+n_i o_n_
High-resolution mass spectra worked example 447
High-resolution mass spectrometers can distinguish 2 An unknown compound has a molecular ion peak, M+,
between ions that appear to have the same mass on a low- with a relative abundance of 54.5% and has an [M + 1]+
resolution mass spectrum. Table 29.5 shows the accurate peak with a relative abundance of 3.6%. How many
relative isotopic masses of the most common atoms found carbon atoms does the unknown compound contain?
in organic molecules.
Substituting the values of relative abundance into
Isotope Relative isotopic mass the equation:
1H 1.007 824 6 n = _1 1_0._10_ × _ a _b_a u_bn_ud_na_nd_ca_en__co e_f_ [oM_f_M +_ +_1_i]o_+n_io _n_
12C 12.000 000 0 (by definition)
14N 14.003 073 8 we get:
16O 15.994 914 1 n = _1 1_0._10_ × _5 3_4._6.5_ = 6.0
Table 29.5 Accurate masses of isotopes. There are 6 carbon atoms in each molecule.
These accurate isotopic masses enable us to measure question
the mass of the molecular ion so accurately that it can
only correspond to one possible molecular formula. For 10 A hydrocarbon has a molecular ion peak at a mass-to-
example, a molecular ion peak at 45 could be caused by charge ratio of 84 (relative abundance of 62.0%) and
sCthp2eeHcCt7rNHu3moNrOwCo+Hup3leNdaOskh.aoHtw4o5twh.0ee2v1Ce r42,H6a27h.NiWg+hep-ecraeaskno,altutht4ei5or.en0f5om7re a8,s4sb6easnudre an [M + 1] peak with a relative abundance of 4.1%.
which molecule is being analysed.
a How many carbon atoms are in the hydrocarbon?
Using the [M + 1] peak b What is its molecular formula?
c The hydrocarbon does not decolourise bromine
There will always be a very small peak just beyond the
molecular ion peak at a mass of [M + 1]. This is caused water. Name the hydrocarbon.
by molecules in which one of the carbon atoms is the 13C
isotope. This is shown in the mass spectrum of ethanol in
Figure 29.25.
Cambridge International A Level Chemistry
Using [M + 2] and [M + 4] peaks 100 112
[C6H535Cl]+
If the sample compound contains chlorine or bromine
atoms, we also get peaks beyond the molecular ion peak Relative abundance (%) 80
because of isotopes of chlorine and bromine. Chlorine
has two isotopes, 35Cl and 37Cl, as does bromine, 79Br and 77
81Br. Table 29.6 shows the approximate percentage of each 60 [C6H5]+
isotope in naturally occurring samples.
40
Isotopes Approximate % 51 114
35Cl 75 [C6H537Cl]+
37Cl 25 20 [C4H2]+ [C4H3]+
79Br 50 110
81Br 50 0 35 56
30
50 70 90
Mass-to-charge ratio, m/e
Table 29.6 Naturally occurring isotopes of chlorine Figure 29.26 The mass spectrum of chlorobenzene, showing
and bromine. the [M + 2] peak. (Note that there are also tiny [M + 1] and
[M + 3] peaks corresponding to 13C in the molecule.)
One Cl or Br atom per molecule The M, [M + 2] and [M + 4] peaks also occur in
dibromomethane but the relative heights of peaks are
Im(2m5oa%leg)ci.nuTelehaseosmafmoClpHecle3u3o5laCfrclih(o7lno5r%wo)milalenbtdheamCnHoel,3eC3c5uHClle3+Cs, aol.nfWdCeHtww3o3i7lCul hnl aitvse easier to work out. Because the ratio 79Br : 81Br is 1 : 1, the
beyond that on the mass spectrum will be the peak for M : [M + 2] : [M + 4] height ratio is 1 : 2 : 1.
448 CheHig3h37tColf+.thTehempoelaeckufloarr CioHn3.3T7Chils+ will be one-third the question
is the [M + 2] peak.
In the mass spectrum of bromomethane, CH3Br, we 11 a List the ions responsible for the M, [M + 2] and
will have two molecular ion peaks of approximately the [M + 4] peaks in a mass spectrum of dibromomethane.
same height – one for CH379Br+ and the other for CH381Br+
(the [M + 2] peak). b What would be the mass-to-charge ratio and
You should look out for the relative heights mentioned relative abundances of the major peaks with the
here when interpreting mass spectra. highest charge-to-mass ratios in the mass spectrum
of chloroethane?
■■ if the [M + 2] peak is one-third the height of the M peak, this
suggests the presence of one chlorine atom per molecule c How many peaks would you see beyond the
molecular ion peak in 1,1-dibromoethane? What
■■ if the [M + 2] peak is the same as the height of the would be their mass-to-charge ratios and
M peak, this suggests the presence of one bromine abundances relative to the molecular ion? (Ignore
atom per molecule. peaks due to 13C.)
An example of the [M + 2] peak is shown on the mass
spectrum of chlorobenzene (Figure 29.26).
Two Cl or Br atoms per molecule Applications of the mass spectrometer
The situation is a little more complex with two chlorine
atoms in a molecule, as there are three possibilities. To identify the components in a mixture, we can link
Considering dichloromethane, CH2Cl2, we have: gas–liquid chromatography (GLC) or high-performance
liquid chromatography (HPLC) apparatus directly to a
33337755CCCCllllCCCCHHHH222233337575CCCCllll++++ tttthhhheeee M peak peak mass spectrometer.
[M + 2] peak
[M + 2] peak This combined technique is very sensitive, and any two
[M + 4] solutes that can be separated with a time gap of 1 second
on a GLC column can be identified almost instantly by
The relative heights of the peaks must take into account the mass spectrometer without the need to be collected.
the natural abundances: it works out as 9 : 6 : 1 for Identification is by comparing the mass spectrum of
molecules with two Cl atoms. each solute with the mass spectra of known compounds,
using a computer’s spectral database. The data generated
Chapter 29: Analytical chemistry
is complex. There can be many components in a mixture, question
each with a peak at its particular retention time on the 12 Look at Figure 29.27.
chromatogram, and each peak will generate its own a What is the retention time of the
characteristic series of lines in the mass spectrometer.
We can combine the chromatogram and the mass spectra compound shown?
to display the data on a three-dimensional (3-D) graph b What is the approximate relative molecular mass
(Figure 29.27).
of the compound shown?
GLC linked to a mass spectrometer (GLC–MS) is c How would the compound be identified?
used for analysing complex mixtures, for example the
identification of the hydrocarbons in a sample of crude oil. As with electrophoresis and NMR spectroscopy, mass
The combined technique is fast and gives reliable results spectrometry is also helping in medical research – to both
that can identify trace quantities of pollutants, drugs, identify and research the amino acid sequences in proteins.
biochemical molecules and toxins. This means it is used in: It can be used to analyse the whole protein molecule or the
peptides left after breaking down the protein with specific
■■ forensics enzymes. Figure 29.28 shows the mass spectrum of a
■■ environmental monitoring of pollutants pentapeptide that is made into a charged compound by the
■■ drug testing in sport addition of a proton, hence the MH+ peak.
■■ geological and archaeological dating
■■ airport security. In research laboratories synthesising new compounds,
a combination of instrumental techniques will need to be
Mass spectrometry has even been used on space probes to used to confirm the structure of a previously undiscovered
analyse rocks on Mars and in 2005 a mass spectrometer molecule (as no spectral records exist in databases, so
was used to analyse the frozen hydrocarbon surface of identification by ‘fingerprinting’ is not an option).
Titan, one of Saturn’s moons. The technique is also used to
analyse the isotopes in the solar wind on board the Solar 449
and Heliospheric Observatory (SOHO) satellite.
100 200
80 180
160
Relative abundance (%)
m/e
60 140
120
40 100
20 80
60
0 40
8.8 8.9 9.0 9.1
Time / min
Figure 29.27 The x-axis shows retention time, the y-axis the amounts and the z-axis is the charge/mass ratio of the mass spectra.
These 3-D data show the peaks on a mass spectrum for one component in a gas–liquid chromatogram.
Cambridge International A Level Chemistry
Abundance (%)100 question
MH+
13 L ook at Figure 29.28.
50 a Calculate the relative molecular mass of
381.1 leucine enkephalin (C28H37N5O7) using relative
0 atomic masses.
(Ar values C = 12.0, H = 1.0, N = 14.0, O = 16.0)
150 200 250 300 350 400 450 500 550 600 650 b i How is the peptide ionised before detection in
m/e
the mass spectrometer?
Figure 29.28 The mass spectrum of leucine enkephalin ii Why is this known as ‘soft ionisation’?
(C28H37N5O7), a peptide made up of five amino acids. It has c Why is there a peak at [MH + 1]?
been charged by adding a proton instead of by ionisation by d An unexpected peak occurs at charge-to-mass
high-energy electrons (which would fragment the molecule).
This is known as a ‘soft ionisation’ method. ratio 578.1. This is caused by ionisation of the
pentapeptide by a metal ion instead of an H+ ion.
Which metal ion is responsible for this ionisation?
Summary
■ Chromatography separates mixtures of substances ■ Protons in different chemical environments
for identification. In chromatography, the mobile produce signals at different chemical shifts. The
phase moves the components of a mixture through chemical shift provides information about the
450 or over the stationary phase. Separation occurs by proton’s environment.
the transfer of the components to the stationary ■ Protons on neighbouring carbon atoms cause signals
phase either by: to be split. The splitting pattern establishes which
– partition between two liquids (due to the groups of protons are on adjacent carbon atoms. The
different solubility of solutes in the mobile phase n + 1 rule predicts the splitting pattern.
and stationary phase) ■ Protons on OH and NH can be identified
– partition between a gas and a liquid by the addition of D2O to the NMR sample, which
– adsorption on a solid surface. collapses the peak due to an OH or an NH
■ The stationary phase may be solid or liquid; the proton.
mobile phase may be liquid or gas. ■ Carbon-13 NMR can also help to determine the
■ In paper and thin-layer chromatography (TLC) the structure of organic molecules.
components of a mixture are identified by their ■ The mass spectrum of a compound enables
Rf values.
the relative molecular mass of the compound
■ In gas–liquid chromatography (GLC) and high- to be determined using the molecular ion peak.
performance liquid chromatography (HPLC), the The molecular ion peak, M, is the peak produced
components of a mixture are identified by their by the loss of one electron from a molecule of
retention times; the amount of each component is the compound.
found by measuring the area of each peak (estimates ■ We can deduce the number of carbon atoms in a
can be made from peak heights). compound using the [M + 1] peak and the presence
■ The proton NMR spectrum of a compound provides of a single bromine or chlorine atom using the
detailed information about the structure of the [M + 2] peak (and two Cl or Br atoms by the [M + 4]
compound. In particular, the spectrum for the peak as well).
protons, 1H, in a compound can provide a complete ■ We can also use mass spectroscopy to identify
determination of the compound’s structure. unknown organic compounds by ‘fingerprinting’
Chapter 29: Analytical chemistry
(matching the spectrum to other known spectra). times but can be ‘fingerprinted’ by their unique
The fragmentation peaks give us clues as to the mass spectra). It is used in airport security checks,
structure of the original molecule. food industries and in forensic, environmental and
medical testing.
■ Gas–liquid chromatography/mass spectrometry
(GLC–MS) provides a more powerful tool for ■ A combination of techniques (such as infra-red, NMR
identifying the components in a mixture than GLC and mass spectroscopy) must be used to confirm the
alone (compounds can have similar retention structure of newly discovered compounds.
End-of-chapter questions
1 a Identify the fragments that would cause peaks in the mass spectrum of HOCH2COCH3 with the following [1] 451
m/e values: [1]
[1]
i m/e = 15 [1]
ii m/e = 17 [1]
iii m/e = 31 [1]
iv m/e = 43 [1]
v m/e = 57
vi m/e = 59
b At what value for m/e would you find the molecular ion peak?
Total = 7
2 The gas–liquid chromatogram for a mixture of organic compounds is shown below.
90 pentane
80
70 octane
60 pentan-1-ol
50
A 40
30
20
10 B
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 [3]
C [6]
[3]
a Give the correct labels for a, B and C. [1]
b What percentage of the mixture is pentan-1-ol? [2]
c Give an explanation for the different retention times. [2]
d i How would the chromatogram change if the liquid in the stationary phase was much more polar? [2]
[2]
ii Explain your answer.
e Why is gas–liquid chromatography useful in testing for anabolic steroids in the blood of athletes?
f Explain why the use of gas–liquid chromatography linked to a mass spectrometer is so useful.
g Why is it difficult to separate dyes using gas–liquid chromatography?
Total = 21
Cambridge international a level Chemistry
3 Paper chromatography was used to separate a mixture of amino acids. The mixture was run in two dimensions
using two different solvents. The chromatogram obtained is shown below.
solvent front
for solvent 1
C DE
B
solvent 1
F
AG solvent front
for solvent 2
sample solvent 2
spot
a Explain briefly how the chromatography was carried out. [4]
b Which amino acids were inseparable using solvent 1? (Give just the corresponding letters.) [2]
c How could the amino acids be located? [1]
d Give the Rf value for amino acid C in each solvent. [2]
e Which amino acids would have been inseparable by solvent 2 alone? [1]
f In another experiment, a mixture of radioactively labelled amino acids was separated using paper
chromatography. The results obtained are shown below.
Radioactive counts per second452 80
70 X
60 Z
50 Y solvent
40 front
30
20
10
0
starting point
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [3]
Distance travelled / cm [4]
i Find the Rf values for acids X, Y and Z. Show your working.
ii Explain how paper chromatography is used to separate the components of a mixture.
Total = 17
Chapter 29: Analytical chemistry
When answering questions 4–7, you will need to use the values in Table 29.1 (page 441).Relative abundance
4 Compound B has the composition 62.1% carbon, 10.3% hydrogen and 27.6% oxygen. Its mass and 1H NMR
spectra are shown below.
a 100
80
60
40
20
0
10 15 20 25 30 35 40 45 50 55 60 65 70 75
Mass-to-charge ratio (m/e)
b
Absorption of energy triplet (1H) triplet (3H)
(2H)
10 9 8 7 6 5 4 3 2 1 0
Chemical shift, δ / ppm
a Calculate the empirical formula of B. [2]
b From the mass spectrum, find the molecular mass of B and hence its molecular formula. [2]
c i Draw displayed formulae for the possible isomers of B which contain a carbonyl group. [2]
ii Use the 1H NMR spectrum of B to decide which isomer is B. [1] 453
iii Explain your reasoning. [3]
d Explain what caused the peak at δ = 1.1ppm and why it is split into a triplet in the 1H NMR spectrum of B. [2]
e Predict the number of signal lines present on the carbon-13 NMR spectrum of:
i compound B, stating the origin of each line [2]
ii the isomer of compound B identified in your answer to c part i, stating the origin of each line. [2]
Total = 16
Relative abundance5 Arene C has the composition 90.6% carbon and 9.4% hydrogen. Its mass and 1H NMR spectra are shown below.
a
100
80
60
40
20
0
10 20 30 40 50 60 70 80 90 100 110
Mass-to-charge ratio (m/e)
b
Absorption of energy
10 9 8 7 6 5 4 3 2 1 0 [2]
Chemical shift, δ / ppm [2]
a Calculate the empirical formula of C.
b From the mass spectrum, find the molecular mass of C and hence its molecular formula.
Cambridge international a level Chemistry
c i Draw displayed formulae for the possible aromatic isomers of C. [4]
ii When C is treated with chlorine in the presence of AlCl3 it undergoes electrophilic aromatic [5]
substitution. In this reaction one of the hydrogen atoms bonded directly to the benzene ring is [4]
replaced with one chlorine atom, and one single aromatic product is formed. Use this evidence
and the NMR spectrum of C to decide which isomer is C.
d Explain the main features of the 1H NMR spectrum of C.
Total = 17
6 Compound D has the composition 77.8% carbon, 7.41% hydrogen and 14.8% oxygen. It mass and 1H NMR spectraRelative abundance
are shown below.
Absorption of energy
a 100
80
60
40
20
0
10 20 30 40 50 60 70 80 90 100 110
Mass-to-charge ratio (m/e)
b
5H
2H
1H
454 10 9 8 7 6 5 4 3 2 1 0
Chemical shift, δ / ppm
a Calculate the empirical formula of D. [2]
b From the mass spectrum, find the molecular mass of D (ignoring the 13C peak) and hence its
[2]
molecular formula. [5]
[1]
c i Draw displayed formulae for five possible isomers of D that contain a benzene ring. [3]
ii Use the 1H NMR spectrum of D to decide which isomer is D. [4]
iii Explain your reasoning.
d Explain the main features of the NMR spectrum of D.
Total = 17
Chapter 29: Analytical chemistry
7 Compound E has the composition 69.8% carbon, 11.6% hydrogen and 18.6% oxygen. Its mass and 1H NMRRelative abundance
spectra are shown below.
a 100
80
60
40
20
0
10 20 30 40 50 60 70 80
Mass-to-charge ratio (m/e)
b singlet (9H)
Absorption of energy singlet (1H)
12 11 10 9 8 7 6 5 4 3 2 1 0 –1
Chemical shift, δ / ppm
a Calculate the empirical formula of E. [2]
b From the mass spectrum, find the molecular mass of E (ignoring the 13C peak) and hence its
molecular formula. [2]
455
c Compound E reacts with 2,4-dinitrophenylhydrazine to give a yellow-orange precipitate. Draw displayed
formulae for the seven possible isomers of E. [7]
d Compound E gives a silver mirror with Tollens’ reagent. Identify the functional group in E. [3]
e Use the 1H NMR spectrum to identify E. Explain your reasoning. [4]
Total = 18
456
Chapter 30:
organic synthesis
Learning outcomes ■■ for an organic molecule containing several
functional groups:
You should be able to:
– identify organic functional groups using key
■■ state that most chiral drugs extracted from reactions
natural sources often contain only a single
optical isomer – predict properties and reactions
■■ state reasons why the synthetic preparation of drug ■■ devise multi-stage synthetic routes for
molecules often requires the production of a single preparing organic molecules using key reactions.
optical isomer, e.g. better therapeutic activity,
fewer side effects
Chapter 30: Organic synthesis
Introduction
Chemists play a vital role in developing new materials
to improve our lives. In this chapter you can find out
about their work in developing new medicinal drugs.
Figure 30.1 Chemists around the
world work in large teams to model,
develop and test new medicines.
Designing new medicinal drugs medicine involved far more trial and error. Chemists had 457
to prepare many more possible medicines for testing.
How do we go about designing new molecules to fight With molecular modelling, only those molecules that
diseases? One way is to identify the structural features the are definite possibilities are made and tested. Molecular
new drug will need to stop particular bacteria or viruses modelling on a computer is now a powerful tool, used
working. The structural features may be associated with when designing medicines and many other compounds
the active site on a particular enzyme needed for an (e.g. pesticides and polymers).
essential function of the pathogen. Once these structural
features have been identified we can then predict the shape This type of research was used in the fight against
of a molecule that would fit into, and hence block, the AIDS in the late 1980s and 1990s. Scientists using
active site. X-ray crystallography (a method in which a sample is
irradiated with X-rays and the pattern is analysed by
The functional groups present would also be crucial to computer) worked out the shape of HIV protease in
ensure the drug could bind into the active site effectively. 1988 (Figure 30.2). This enzyme plays an important role
The intermolecular bonds formed between the drug and when the virus becomes infectious. Researchers realised
its target molecule could involve: that, if a molecule could be discovered that could block
its active site, this might be one step on the route to
■■ hydrogen bonding finding a cure for AIDS. Knowing the molecule that the
■■ ionic attraction enzyme worked on (its substrate), researchers were able
■■ dipole–dipole forces construct similar molecules on the computer screen to fit
■■ instantaneous dipole–induced dipole forces (van der Waals’ the active site.
forces, see Chapter 4, page 62). The first attempts fitted perfectly, but were not water
soluble. This meant the drug could not be delivered to its
Computers are now used to judge the fit between a target, the HIV protease. Eventually a soluble molecule
potential drug molecule and a receptor site on its that would interfere with the enzyme was found. In less
target molecule. Such molecular modelling has greatly than 8 years pharmaceutical companies had developed
speeded up the process of designing new medicines. three new anti-viral drugs for people with HIV/AIDS.
The interactions and fit of a potential medicine with This would have taken about twice as long if the structure
a biological receptor molecule can be studied before of HIV protease had not been determined. Traditional
the medicine is ever made in the lab. Before molecular
modelling became available, the synthesis of a new
Cambridge International A Level Chemistry
active site
Figure 30.2 A symmetrical HIV protease molecule, with its
active site in the centre of the molecule. Knowing its structure
made the search for a drug to fight AIDS much quicker and
cheaper than traditional trial-and-error methods.
trial-and-error methods involve the testing of many Figure 30.4 A protein made up of 153 amino acid residues.
thousands of possible drugs. This NMR analysis takes place in solution, so it is
The death rate from AIDS dropped significantly. particularly useful for medical research. Many human
proteins exist in solution in the body so we can mimic the
However, the virus developed resistance to the new drugs interactions that take place in cells or in the bloodstream.
as it mutated. So scientists now have to model the new
drug-resistant strains of the infection and are developing
new drugs to inhibit the mutant versions of HIV protease.
These inhibitors are one part of a cocktail of drugs that question
458 can be used to treat the disease now. 1 a Which method was used to determine the
Identifying macromolecules structure of HIV protease in 1988?
b Which method could be used to show the shape of
NMR spectroscopy is also used extensively in finding
out the structures of biological macromolecules such the enzyme in solution?
as proteins and nucleic acids. As well as identifying the c What type of natural polymer is an enzyme such as
different types of 1H atoms present, more sophisticated
data can yield, for example, the distance between atoms in HIV protease?
d How did the new anti-viral drugs work?
macromolecules. Large amounts of data are collected and
analysed by computer programs to reveal the shape of the
molecules under investigation.
Figures 30.3 and 30.4 show images obtained from Chirality in pharmaceutical synthesis
NMR analysis of two protein molecules made up from The pharmaceutical industry is constantly searching for
over 100 amino acids. These are called ribbon diagrams. new drugs. Their research chemists have discovered that
most of these drugs contain at least one chiral centre (see
page 195). Remember that a molecule containing a carbon
atom bonded to four different atoms or groups of atoms
can exist as two non-superimposable mirror images.
These two isomers are called enantiomers and they will
be optically active. They differ in their ability to rotate the
plane of polarised light to the left or to the right.
Using conventional organic reactions to make the
desired product will yield a 50 : 50 mixture of the two
enantiomers. We call this a racemic mixture. Although the
physical properties of the enantiomers will be identical,
each differs in its ‘pharmaceutical activity’, i.e. the effect
Figure 30.3 A protein made up of the drug has on the body. For example, naproxen is a drug
106 amino acid residues.
Chapter 30: Organic synthesis
used to treat the pain caused by arthritis (Figure 30.5). non-toxic, is easily removed by reducing the pressure and
One enantiomer will ease the pain but the other can cause then recycling it to use in the process again.
liver damage.
We can also use high-performance liquid
As another example, one enantiomer of a drug used chromatography (HPLC, see page 436) to separate a
to treat tuberculosis (TB) is effective, whereas the racemic mixture, as long as the stationary medium (e.g.
other can cause blindness. Therefore, chemists ideally the solid that packs the column) is itself optically active.
need a single pure enantiomer to put in their drug
product. Note that about 80% of new drugs patented are Using optically active starting materials
single enantiomers.
This technique uses starting materials that are themselves
Using pure enantiomers will be beneficial as it: optically active and in the same orientation as the desired
product. These are often naturally occurring compounds
■■ reduces the patient’s dosage by half as the pure enantiomer such as carbohydrates or L-amino acids. The biochemist
is more potent, i.e. has better therapeutic activity (thereby will choose from this ‘chiral pool’. The synthetic route is
cutting costs of production) designed to keep any intermediates and the final product
formed in the same enantiomeric form. As a result, there is
■■ minimises the risk of side effects (thereby protecting no need to carry out the costly separation process needed
patients from further problems and drugs companies when a racemic mixture is produced.
from possible legal action for damages if serious side effects
do occur). Chiral catalysts
There are three ways to prepare pure enantiomers: Chemists are also developing new chiral catalysts that
ensure only one specific enantiomer is formed in a
■■ optical resolution reaction. The benefits of these catalysts are that only small
■■ using optically active starting materials quantities are needed and they can be used over and
■■ using a chiral catalyst. over again, although the catalyst itself can be expensive.
A ruthenium (Ru) organometallic catalyst is used in the
Optical resolution production of naproxen (see Figure 30.5).
This method involves the chemists following a traditional
Often a combination of optical resolution and 459
chiral synthesis is needed in the production of a
synthetic route to make the compound, resulting in a pharmaceutically active, pure enantiomer.
racemic mixture. Then they separate the two enantiomers
in a process called optical resolution. This involves using
a pure enantiomer of another optically active compound
(called a chiral auxiliary) that will react with one of the
isomers in the mixture. The new product formed will
now have different properties and so can be separated HH
by physical means. For example, the solubility in a C
given solvent will differ so the unwanted enantiomer
C
COOH
and the new product can be separated by fractional CH3O + H2
crystallisation. The new product is then converted back
to the desired enantiomer in a simple reaction (e.g. by organometallic
adding dilute alkali). ⎯⎯→ruthenium catalyst
The crystallisation is repeated many times to ensure (a chiral catalyst)
purity. This method is difficult, time-consuming, uses CH3 H
extra reagents and involves the disposal of half the original C
racemic mixture. * COOH
Large volumes of organic solvents (often harmful
to the environment) are used in the process. However, CH3O
chemists are now using supercritical carbon dioxide as a naproxen
solvent, which is much safer. At 31 °C and 73 atmospheres
dprreusgsudreer,ivCaOti2veiss a suitable non-polar solvent for many (drug for arthritis)
in the racemic resolution process. The
solubility of the derivatives can be changed, simply by Figure 30.5 The chiral catalyst (an organometallic ruthenium
varying the density of the solvent. The solvent, which is compound) ensures only the desired enantiomer is formed –
in this case, naproxen for the treatment of arthritis. The chiral
centre in naproxen is marked with an asterisk. The marked
carbon atom is known as an ‘asymmetric carbon’.
Cambridge International A Level Chemistry
The pharmaceutical industry can also use enzymes to question (continued)
promote stereoselectivity and produce single-enantiomer b Why are modern enzyme-based processes
products. The specific shape and the nature of the
molecular interactions at the active site of an enzyme for manufacturing pure enantiomers more
ensure only one enantiomer will be formed (as in living sustainable (environmentally friendly) than
things). The enzymes are often immobilised (fixed in traditional synthetic routes used by the
place) on inert supports. This enables the reactants to be pharmaceutical industry?
passed over them without the need to separate the product c Find out why the drug thalidomide resulted in legal
from the enzymes after the reaction. action against its manufacturer.
However, it can be expensive isolating enzymes from living Synthetic routes
things. Using whole organisms, such as bacteria, can reduce
this cost. Nowadays, synthetic enzymes can also be made, When research chemists want to make a new compound,
designed for a particular synthesis. Therefore a search for a they usually work backwards from the desired compound
suitable enzyme from the limited pool available from natural to create a series of reactions, starting with a compound
sources is not always necessary. extracted from a commonly available raw material. In
industry, common starting materials are hydrocarbons
Overall, using an enzyme process might take longer from crude oil and its refining, and compounds extracted
to develop than a conventional synthetic route, but from plants, such as esters from fats and vegetable oils.
in the long run the benefits generally outweigh the
disadvantages. There are fewer steps needed in the You will need some of the skills of a research chemist
synthesis route, resulting in a ‘greener’ process. when tackling questions that involve:
question ■■ predicting the reactions of complex molecules you
have never seen before, containing more than one
460 2 a Why are pure enantiomers rather than functional group
racemic mixtures the better option for use as
pharmaceutical drugs from the point of view of: ■■ suggesting a series of reactions to make a given compound
from a given starting compound.
i a patient
ii a pharmaceutical company? In order to be successful in answering these questions,
you will need to be familiar with all the reactions and
conditions of each homologous series mentioned in the
syllabus. The flow chart in Figure 30.6 is a summary of
alkanes
H2(g), heat ultraviolet
under pressure
light with amines
with nickel
catalyst halogen, room heat under pressure heat with
temperature with ammonia in alcohol alcohol
and pressure and acid
catalyst
alkenes treat with HX(aq), halogenoalkanes carboxylic acids esters
room temperature distil from mixture warm with heat with water heat with
and pressure of alcohol, conc. water and
NaBH4 in and acid catalyst acid catalyst
heat with dilute H2SO4 and KX(s) water reflux with excess
aqueous alkali
aldehydes acidified potassium
pass alkene and hot pumice warm with NaBH4 dichromate(VI)
steam over heated and AI2O3 in water
phosphoric acid catalyst alcohols distil on adding to
under pressure acidified potassium dichromate(VI)
Figure 30.6 Some of the important organic reactions you have to remember.
Chapter 30: Organic synthesis
some of the most useful reactions you need to know, The RCN molecule can be either hydrolysed to make
but you could be asked about others, so it is a good idea a carboxylic acid (by refluxing with dilute hydrochloric
to make your own summary spider charts for all the acid) or reduced to make an amine (by adding LiAlH4 in
reactions in Chapters 15 to 28. For example, for Chapter 16 dry ether).
write the word ‘Halogenoalkane’ in a box and draw arrows
radiating out from the box to the products made in their We can add an alkyl or acyl side-chain to a benzene
reactions, labelling the arrows with other reactants and the ring by carrying out a Friedel–Crafts reaction, which is
reaction conditions. Displaying these on a wall and using another useful reaction when planning synthetic routes.
different colours will help you remember them. For example:
Adding carbon atoms Cbe6nHzen6e + CH3CH2Cl AlCl3 catalyst C6Heth5yClbHen2zeCneH3 + HCl
Sometimes the starting compound from a raw material
does not have enough carbon atoms in its molecules to
make the desired product. An extra carbon atom can be
added by adding the nitrile functional group, C N.
Remember that these can be made from halogenoalkanes:
reflux with ethanolic KCN
RBr + HCN RCN + HBr
where R is an alkyl group, so RCN has an extra
carbon atom.
question
3 a Name the functional groups in the molecules of: 4 a Name the organic products A to D in the synthetic 461
i aspirin route below:
ii paracetamol.
boil with NaOH(aq) add excess HCl(aq)
O
C2H5COOC3H7 add PCl5 C AB
C H H H add conc. ammonia
O O N
D
H b i Devise a three-stage synthetic route to convert
benzene into benzenediazonium chloride.
O CCH
O CH OH ii How would you convert the benzenediazonium
C chloride into an orange dye?
H H
aspirin paracetamol
b i Both the molecules in part a are broken down
when refluxed with dilute hydrochloric acid. Write
an equation each for the reaction of aspirin and
paracetamol with H2O.
ii What do we call the type of reaction in b i ?
Cambridge International A Level Chemistry
Summary ■ Molecular design of a new medicinal drug is
made possible with a sound understanding of the
■ Both natural biochemicals and modern medicinal structural features that produce beneficial effects.
drugs contain chiral molecules. Generally, only one The computerised study of the interactions between
of the enantiomers of a drug is beneficial to living molecules and biological receptors has become a
organisms and the other isomer may have undesirable powerful tool in the search for new medicines.
effects. The beneficial isomer has the appropriate
shape and pattern of intermolecular forces to interact ■ Recognising the functional groups in a given organic
with a receptor molecule in a living organism. molecule enables us to predict its reactions.
■ Chemists are now producing drugs containing ■ Knowing the reactions of the different functional
single enantiomers rather than a racemic mixture groups in organic reactions enables us to devise
of isomers. This enables the dose to be halved, synthetic routes to make given compounds.
improves pharmacological activity (behaviour of
molecule in an organism), reduces side effects and
minimises litigation against manufacturers.
End-of-chapter questions
1 A sample of lactic acid (CH3CH(OH)COOH) was extracted from a natural source and found to be optically active. It
462 was then subjected to two reactions, as shown below.
step A step B
CH3CH(OH)COOH CH3COCOOH CH3CH(OH)COOH
sample 1 sample 2
Sample 1 was optically active but sample 2 was not optically active. [1]
a i Give the systematic name for lactic acid.
ii The structure of one optical isomer of the lactic acid is:
OH
C H
H3C
COOH
Draw the other optical isomer. [1]
[1]
iii Explain why lactic acid can form optical isomers. [2]
[2]
b i Give the reagents and conditions necessary for step A. [2]
ii Give the balanced equation for the reaction. [2]
c i Give the reagents and conditions necessary for step B. [5]
ii Give the balanced equation for the reaction. [3]
[2]
d i Give the mechanism for step B. The first step involves nucleophilic attack on the carbon of the [1]
ketone group by an H– ion from NaBH4. [1]
ii Explain why sample 2 does not show any optical activity – it does not rotate plane-polarised light.
e i Explain why lactic acid can be polymerised.
ii State the type of polymerisation reaction that this an example of.
iii Draw the repeat unit of poly(lactic acid).
Total = 23
Chapter 30: Organic synthesis
2 Explain how 2-aminopropanoic acid can be prepared from lactic acid in two steps. You should give [9]
the reagents and conditions necessary plus balanced symbol equations for the reactions taking place.
Total = 9
3 Write short notes on the following methods of synthesising chiral molecules: [3]
a using a chiral auxiliary [3]
b use of a chiral pool.
Total = 6
4 The structure of the compound known as thalidomide can be shown as:
O
N N O
OO H
a Copy the molecule and mark the chiral centre on your drawing. [1]
b Name two functional groups found in a molecule of thalidomide. [2]
c Suggest the type of reaction that might change the molecule into an alcohol and any reagents needed. [2]
d Explain why the chirality of drugs has been such an important issue in the pharmaceutical industry,
[3]
giving one benefit to patients and one benefit to pharmaceutical companies of using pure enantiomers.
Total = 8
5 The flow charts below show how poly(ethene) can be obtained by two different routes. 463
process 1 ethene process 2 poly
Route I decane (ethene)
substance
A
Route II substance process 3 ethene process 2 poly
B (ethene)
H2O [3]
[1]
a i Identify substance A and give the equation for the reaction taking place in process 1. [3]
ii What term is used to describe process 1? [1]
b i Name substance B and give the equation for the reaction taking place in process 3. Total = 8
ii What term is used to describe process 3?
464
Chapter P2:
Practical skills 2
Practical skills 2
Introduction
The ability to plan, analyse and evaluate practical
work requires higher-order thinking skills. You will
need plenty of practice in carrying out tasks requiring
these skills as you progress through your course.
Figure P2.1 Many chemists are employed to plan, carry out
and evaluate tests on substances found in the environment.
Written examination of Methods 465
practical skills
■■ describe the method to be used to vary the independent
Paper 5 in the Cambridge International Examinations variable, and the means that you propose to ensure that you
International A Level Chemistry syllabus is the written have measured its values accurately
examination of practical skills. It focuses on the higher-order
experimental skills of planning, analysis and evaluation. ■■ describe how you will measure the dependent variable
Some questions on this paper may be set in areas of chemistry ■■ describe how you will control each of the other
that are difficult to investigate experimentally and may be
based on novel situations. However, no question will require key variables
knowledge of theory or equipment that is beyond the syllabus. ■■ explain how you will use any control experiments to verify
Before going through the following section, it would that it is the independent variable that is affecting the
be a good idea to read Chapter P1, Practical skills 1, dependent variable and not some other factor
pages 246–256. This will remind you of the structure of ■■ describe the arrangement of apparatus and the steps in the
investigations – which is essential to understand before procedure to be followed
you can apply the higher-order practical skills tested in ■■ suggest appropriate volumes and concentrations
Paper 5. of reagents
■■ assess the risks of your proposed methods
Planning ■■ describe precautions that should be taken to keep risks to a
minimum
Expectations ■■ draw up tables for data that you might wish to record
■■ describe how the data might be used in order to reach a
You should be able to: conclusion.
Defining the problem As the expectations above show, all plans contain two
distinct sections:
■■ identify the independent variable in the experiment
or investigation ■■ defining the problem – this is worth 4 marks
■■ the actual method – this is worth a further 8 marks.
■■ identify the dependent variable in the experiment
or investigation Defining the problem
■■ formulate the aim in terms of a prediction or hypothesis, The question generally begins with a stem, which will
and express this in words or in the form of a predicted graph contain information relevant to the plan and the aim of
the investigation. The context of the task to be covered may
■■ identify the variables that are to be controlled. or may not be familiar to you, but the stem will contain all
of the detail required.
Cambridge International A Level Chemistry
Almost invariably, a prediction is required together apparatus in a simple and understandable format. It
with an explanation of the basis of the prediction, forming is essential that each piece of equipment can be easily
a hypothesis to test. For example, in a question about rate recognised in the diagram. Practice drawing sessions are a
of reaction, you might make a quantitative prediction, very good idea.
such as the direct proportionality between the rate Relevant calculations are an almost routine
and concentration. Here, the supporting explanation requirement, and these must be logically presented, step
would need to be given in terms of the doubling of the by step.
frequency of particle collisions as a result of doubling the The preparation of a standard solution of a solid acid,
concentration. The planning question might also ask for a such as ethanedioic acid, requires the calculation of the
supporting graph to be given. mass to be dissolved in a chosen volumetric flask in order
You will have to specify the independent and the to produce the required concentration. The described
dependent variables. You might also have to consider the procedure should indicate that the solid needs to be
need to control other variables. dissolved completely in distilled water before adding the
In a ‘rate’ exercise, the independent variable might be solution to the volumetric flask (see Figure P2.2). You
the concentration of a reactant or the temperature. The should make clear how you can ensure that all of this
measured rate of the reaction at different values of the solution is transferred successfully to the volumetric
independent variable is the dependent variable. Alternative flask – for example, by rinsing any solution stuck on the
ways of measuring the rate of reaction, such as the time inside of the original container into the volumetric flask
to collect a set volume of gas (giving average rates) or the using distilled water. Once in the volumetric flask, the
monitoring of gas released over time (then calculating the solution is made up to the required mark, and must then
initial gradient of lines on a graph to measure initial rates), be thoroughly mixed.
are acceptable. You are expected to calculate the amounts of substance
used to suit the volume or size of the apparatus that you
466 Methods specify. For example, if the plan involves collecting a gas
from the decomposition of a carbonate in a 100 cm3 gas
When creating a method, you should produce one that can
be followed by another student without the need to ask any syringe, the calculated mass of the solid heated should be
extra questions. This means that fine detail is required and such as to produce slightly less than 100 cm3 of the gas.
nothing should be omitted on the basis that it is ‘obvious’.
The question might provide further information
that builds on the stem, but you are expected to have
experience of basic laboratory apparatus. The best way
to achieve this is to follow a programme of experimental
work throughout the course. This also has the advantage of
promoting the understanding of concepts underlying the
practical work.
Basic techniques, such as titrations, standard solution
production and rate measurement, should be thoroughly
understood in detail. You also need to be aware of the
various methods of measuring volume and mass available
in a laboratory. For example, an experiment involving the
production and collection of a gas, such as the effect of
heat on nitrates, would require you to be aware that gas
syringes usually have a maximum capacity of 100 cm3 and
that the alternative of collecting the gas over water in a
burette would allow for the collection of only 50 cm3. This
example shows how it is important to perform meaningful
practical work in order to prepare for Paper 5. There is no
substitute for familiarity in handling real apparatus.
On occasion, a diagram of the apparatus to be used
Figure P2.2 Volumetric flasks are used to make up standard
is requested and you need to be able to represent the solutions of known concentration accurately.
Practical skills 2
An integral part of any plan is the collection and paper before writing a bullet pointed list or flow chart that 467
presentation of results. Any processing of the results explains exactly how you would carry out the investigation.
must be specified, and the figures obtained should also be These could include diagrams of assembled apparatus that
presented as the ‘derived results’. The derived results may can save complicated written explanations of the set-up.
then be processed in order to confirm or deny (falsify) ■■ When describing how to vary your independent variable,
the original prediction or hypothesis. This processing ensure you consider how to measure its values accurately.
may involve calculating means of repeat readings and the For example, in a rate of precipitation investigation, when
graphical presentation of results. varying the concentration of a solution you might choose
to measure volumes of water and solution using
The recorded results are those collected from the volumetric pipettes and fillers rather than just using a
various steps taken in carrying out the procedure. For measuring cylinder (see Figure P2.3). You should also be
example, in an exercise to find the enthalpies of solution of familiar with the volumes and masses of reagents that are
Group 1 hydroxides, the necessary results columns to be reasonable to use in normal everyday laboratory work. For
recorded are: example, a volume of 15 cm3 of solution would be suitable
for an experiment in a boiling tube but not in a 250 cm3
■■ the name or nature of the hydroxide conical flask.
■■ the mass of a weighing bottle ■■ When deciding how to measure the dependent variable, you
■■ the mass of the weighing bottle and a sample of the hydroxide should also consider accuracy – how you intend to measure
■■ the initial temperature of the water its ‘true’ value. In a rate of precipitation investigation you
■■ the final temperature of the solution. might choose to carry out the reaction in a conical flask on
top of a piece of paper with a pencil mark drawn on it. You
If you plan to repeat any tests to get replicate results, extra could then time how long it takes for the pencil mark to be
columns must be incorporated into the results table. Each no longer visible when viewed from above the flask. This
column should have the correct units – represented as will involve a subjective judgement by the observer. If you
either ‘/ g’ or ‘(g)’ for any mass, for example. have done an experiment like this before, you will know
just how difficult it is to decide the exact moment when
The next stage is to process the results: you can no longer see through the solution. So including a
1 to calculate the mass of hydroxide used Figure P2.3 Volumetric and graduated pipettes are used to
2 to calculate the number of moles of hydroxide used transfer accurate volumes of liquid. Volumetric pipettes give
3 to calculate the number of joules of energy released greater accuracy than graduated pipettes, due to the single
line to measure to on the thin section above the bulge.
into the solution
4 to use the results to parts 2 and 3 to deduce the
standard enthalpy change per mole of hydroxide.
All plans have to be assessed for risk. Examination
questions can ask for:
■■ an assessment of the risks involved in a particular
experiment
■■ how to deal with the risks
■■ both of the above.
In an exercise to determine the enthalpy of neutralisation
of hydrochloric acid (the question will tell you that this is
‘harmful’) and sodium hydroxide (which is ‘corrosive’),
you could be asked about a suitable precaution.
Alternatively, an experiment on solubility may have hot
water as a possible hazard or if a toxic gas is formed, as in
the decomposition of Group 2 nitrates, the investigation
should be carried out in a fume hood.
Points to remember
■■ When planning an investigation, your method must be
ordered in a logical sequence, so you might prefer to
make some rough notes at the back of the examination
Cambridge International A Level Chemistry
very accurate timing device, reading to one-thousandth of a ■■ within familiar contexts, suggest possible explanations for
second, would not be appropriate for this method. anomalous readings
■■ You might also be required to sketch the axes you would use
to graph the data collected and describe how you would use ■■ identify the extent to which provided readings have been
the data to draw a conclusion. adequately replicated, and describe the adequacy of the
range of data provided
question
■■ use provided information to assess the extent to which
selected variables have been effectively controlled.
1 In an experiment to determine the formula of an oxide Conclusions
of copper, the copper oxide was reduced by hydrogen
gas. The hydrogen was passed over the oxide of ■■ draw conclusions from an investigation, providing a
copper in a boiling tube with a hole near its end. The detailed description of the key features of the data and
excess hydrogen was burnt off at this hole. The copper analyses, and considering whether experimental data
oxide was placed inside the boiling tube, which was supports a given hypothesis
clamped in a horizontal position, in a porcelain boat
and was heated strongly. ■■ make detailed scientific explanations of the data, analyses
and conclusions that you have described
a Name the products of the reaction.
b Draw a labelled diagram of the apparatus. ■■ make further predictions, ask informed and relevant
c Make a list of the measurements you would make questions and suggest improvements.
to find the formula of the oxide of copper. The three remaining skills are Dealing with data,
d Using the measurements made, how would you Evaluation and Conclusions cover these areas. You
should also revise the equivalent section in Chapter P1
work out the mass contained in the oxide of on pages 251–253 before reading the following advice.
copper of:
i copper Dealing with data
468 ii oxygen
e The experiment was carried out by ten groups In a written practical paper, typically you will be presented
of students, each with a different initial mass with some data, usually in a table, derived from an
of the oxide of copper. Sketch a graph with the experiment. The details of the experiment are given to
line you would expect to enable you to work out you, and then you will be asked to process the data and
the formula. produce a system that incorporates all the results and
allows not only correct patterns to become apparent but
Analysis, conclusions and also detects anomalous results. The tabulated results will
evaluation be the measured quantities of the processes carried out in
a laboratory.
For example, in an experiment to confirm the formula
of zinc iodide by reacting excess zinc with iodine, the
measurements would be:
Expectations ■■ mass of an empty test tube
■■ mass of the test tube and zinc powder
You should be able to: ■■ mass of the test tube, zinc powder and iodine
■■ mass of the test tube and excess of zinc.
Dealing with data
■■ identify the calculations and means of presentation of data From these you would calculate:
that are necessary to be able to draw conclusions from ■■ the mass of zinc used
provided data ■■ the mass of iodine that reacted with that mass of zinc.
■■ use calculations to enable simplification or explanation of These results would be tabulated, possibly alongside
data the original data, with each column showing the
appropriate units and an expression to show how the
■■ use tables and graphs to draw attention to the key points in values were calculated.
quantitative data, including the variability of data.
All calculations should be correct and recorded to an
Evaluation appropriate number of significant figures and/or decimal
places. The question will on occasion tell you the degree
■■ identify anomalous values in provided data and suggest
appropriate means of dealing with such anomalies
Practical skills 2
of accuracy required, but often you will need to decide as a curve or straight line. This line may not pass through
this yourself. all, or indeed any, of the plotted points. As a rough guide,
Many balances weigh to the nearest 0.01 g, and an equal number of points should lie on the left of the line
hence masses should be recorded to 2 decimal places, and on the right of the line, producing an ‘average’ line.
remembering that whole number masses should be
recorded to the same accuracy, e.g. 4.00 g. In other Evaluation
circumstances, derived data should be recorded to the
same number of significant figures as the least accurate The evaluation of a set of results can be approached in a
item of supplied data. Where you are in doubt, 3 number of ways. The aim of any experiment is to draw an
significant figures are appropriate for most calculations. appropriate conclusion, either to verify a relationship or
Having processed the raw data, the next step involves to establish a new relationship. This is accomplished by
either a series of calculations to be averaged or some form looking at the nature of the results and the quality of the
of graphical plot. experiment itself.
In the zinc iodide experiment already discussed, the
ratio of the number of moles of iodine to the number of When anomalous results are identified, they should
moles of zinc can be used to produce a series of ratios. be clearly labelled as such. The source of the anomaly will
These ratios can either then be averaged, excluding any often need to be identified and will usually fall into one
anomalies to produce an appropriate result, or a graph of two categories: either a positive or a negative deviation
could be plotted of the two molar values. In this case the from the general trend.
appropriate gradient of the line would indicate the formula
of zinc iodide. For example, if you plot the number of In a reaction involving the reduction of a metal oxide,
moles of zinc on the x-axis, and the moles of iodine on the the anomaly might arise because the reduction was
y-axis, the gradient of the line on the graph should be 2. incomplete, leading to a larger than expected mass of
This shows that the formula of zinc iisodanidaepipsrZonpIr2ia. te metal. The excess mass is the unreacted oxide.
The first step in plotting a graph choice 469
If the reaction involves the thermal decomposition of
of scales for the two axes, with the independent variable hydrated iron(II) sulfate two, opposite, errors are possible:
along the x-axis (the horizontal axis). When choosing ■■ insufficient heating – this will cause the residual mass to be
a suitable scale it is useful to remember that the plotted
points should be spread over at least half of each axis. too great
You must decide whether or not the origin (0,0) is a ■■ overheating – this will cause the residual mass to be
point on your graph. Whether or not to include the origin
will usually be clear from the nature of the results. For too small.
example, if the concentration of a reactant is shown to
affect the rate of a reaction, it is reasonable to assume that The former result corresponds to the incomplete removal
if the reactant were absent the reaction would not occur, of the water of crystallisation, whereas the latter relates to
so the rate would be zero when the concentration was the decomposition of the iron(II) sulfate into iron oxide.
zero. The origin can then provide a further useful point
to be plotted. Chosen scales should be easily readable, You may be called upon to consider whether the
but errors are counted if unhelpful scales are used. For actual experiment data under consideration is of high
example, the division of each of the two-centimetre enough quality to produce reliable results, and appropriate
squares on the graph grid into three makes accurate improvements may be requested. If the results provide poor
plotting very difficult. support for a conclusion, it may be that further repeats are
Plotted points are best shown as small crosses, made needed, or that the range of results needs to be extended.
with a hard, sharp, pencil.
Most plotted points will show a clear trend, indicating Apparatus may need to considered from two viewpoints:
whether the graph should be a straight line or a curve; a ■■ is it appropriate?
straight line is the more usual. The plotted points must ■■ is it accurate?
first be assessed to identify any anomalous points. If any
anomalous points are identified, they should be clearly In the measurement of volume, burettes and pipettes are
labelled as such. The line of best fit is then drawn, either intrinsically more accurate than measuring cylinders,
but if the volume to be measured is large, for example, a
measuring cylinder could be perfectly adequate.
An experiment measuring a small temperature change
in an enthalpy exercise often involves volumes of the order
of 25 cm3 to 50 cm3 and a typical temperature change of
the order of 5 °C to 10 °C, using a thermometer accurate to
the nearest degree. In this case, the volumes could safely
be measured with a measuring cylinder, as the percentage
Cambridge International A Level Chemistry
error of the temperature change will be greater than that to [H+] in the rate equation. This conclusion would then
of the measuring cylinder. be considered in the light of an original hypothesis. This,
The volumes measured during titrations are about in turn, could lead to further predictions and experiments
as accurate as simple exercises can be. If these are (see page 253 in Chapter P1).
involved in an experiment the source of any error is likely If the experiment is considered to be too approximate,
to be elsewhere. suggested improvements to the exercise might be requested.
Measuring small volumes or masses generally
produces high percentage errors, whichever item of Points to remember
simple laboratory apparatus is used. As an example, an ■■ Calculations involving data collected may ask for the mean,
experiment to investigate the rate of reaction between
median, mode, percentage loss or percentage gain.
hydrochloric acid and magnesium by measuring the ■■ To calculate the mean, add up the individual values in a set
and divide by the number of values. Take care not to quote
volume of hydrogen produced requires less than 0.10 g the mean to an unrealistic number of significant figures –
of magnesium if the gas produced is to be collected in a the mean should reflect the precision of the measurements
100 cm3 syringe. This is a very small mass of magnesium. from which it was calculated.
Using a typical balance accurate to 0.01 g would give a 10% ■■ The median is the middle result when all the results are put
error and consequently the accuracy of the syringe is of in order of magnitude.
negligible significance. The error in measuring the mass of ■■ The mode is the most common value.
the magnesium will be the greatest percentage error. ■■ Percentage loss or gain is calculated by dividing the actual
Conclusions loss or gain by the original value, then multiplying by 100.
The conclusion of an exercise draws upon the key features ■■ The interval of the independent variable can be consistent
when planning an investigation, but additional values can
be selected to look more closely where a pattern seems
of data collected and the subsequent analyses. Usually the to change (e.g. the gradient of a line changes on a graph
data given will support a given hypothesis. However, it of your results) or to extend the limits of your original
470 must be clearly understood that data that do not support range below its minimum value or above its maximum
an initial suggestion might also have to be considered. value. For example, you might decide to test a much lower
dilution of 0.01 mol dm–3 when investigating the effect of
In the magnesium/acid reaction, processing could concentration on the rate of reaction. This would enable
involve plotting a graph of rate against the relative
concentration of the acid. Inspection of the graph would
allow a deduction to be made about the order with respect you to test if the trend seen within your original range of
data continues as expected.
question c Nine out of the ten groups who tackled the experiment
obtained results consistent with those of the group
2 a In the experiment described in Question 1 on page described in part b. The other group’s measurements
468, a student weighed a mass of 11.35 g of the oxide resulted in a ratio that suggested the formula of the
of copper on a balance reading to the nearest 0.05 g. oxide was Cu2O.
What was the percentage error in this measurement?
i What do we call their result when plotted on a class
b Another group of students carrying out the same graph of moles of oxygen against moles of copper?
experiment found that their sample of the oxide of
copper contained 13.24 g of copper and 3.26 g of ii How would you deal with this result when drawing
oxygen. What is the most likely formula of the oxide of the line of best fit?
copper? (Relative atomic mass of
Cu = 63.5; O = 16.0.) ii i Give a possible explanation for the Cu2O
result obtained.
Practical skills 2
Summary
■ The following table summarises the breakdown of skills and the marks allocated to each skill area as it is assessed
in the Cambridge International A Level Chemistry Planning, Analysis and Evaluation examination (Paper 5).
Skill 12 marks Breakdown of marks(a) 4 marks
Planning 12 marks Defining the problem 8 marks
Methods 6 marks
Analysis, conclusions and evaluation Dealing with data 4 marks
Evaluation 2 marks
Conclusion
(a)The remaining 6 marks will be allocated across the skills in this grid and their allocation may vary from paper to paper.
End-of-chapter question
1 Diffusion in a gas is the random motion of particles involved in the net movement of a substance from an area 471
of high concentration to an area of low concentration. The process of diffusion also takes place in solution.
Medical scientists are interested in the rate of diffusion of pharmaceutical compounds through tumours. They
can model the factors that affect the rate of diffusion of these drugs by conducting investigations using coloured
compounds (to model the drugs) and gelatin, a jelly-like substance (to model the tumours).
The kinetic energy of particles depends on their mass and the speed they travel at. So at a given temperature,
large particles will travel slower on average than smaller particles.
Imagine that you are a member of a research team trying to find out how the rate of diffusion through gelatin
depends on the relative molecular mass (Mr) of a drug.
a i Predict how you think that the rate of diffusion will be affected as the relative molecular mass of a drug
increases. Explain your reasoning.
ii Display your prediction on a sketch graph, including any relevant units. [4]
b In the experiment you are about to plan, identify the:
i the independent variable
ii the dependent variable. [2]
Cambridge international A Level Chemistry
c The research team can make thin discs of gelatin to cover the central area of Petri dishes, leaving space
around the edge of each disc to place a solution of the coloured dyes under investigation:
TOP VIEW
Petri dish
gelatin disc
solution of dye
placed in gap between
the gelatin disc and
the edge of the Petri dish
CROSS-SECTIONAL SIDE VIEW gelatin disc of
Petri dish 3 mm height
You have been given five coloured powders of dyes, labelled A to E, to test. These have relative molecular
472 masses of 486, 534, 686, 792 and 886, respectively. You are also provided with a stopclock/watch and a ruler
with a millimetre scale. You can also use any other common laboratory apparatus needed to complete the
investigation. The diffusion is a slow process and the team carry out some trial runs to get a rough idea how
quickly the dyes diffuse through the gelatin. They decide to monitor the experiment for 72 hours.
Describe how you would carry out the experiment, making sure that you include how to:
■ ensure the same number of dye molecules is used in each test
■ ensure that the Petri dish is kept under the same conditions throughout all of the experiments
■ measure the rate of diffusion
■ produce reliable results. [6]
d Two of the dyes are classified as ‘harmful’ and are hazardous if absorbed through the skin or are inhaled.
State any precautions you would take to minimise the risk. [2]
e Draw a table with headings that show clearly the data you would record when carrying out your
experiments and any values you would need to calculate in order to construct a graph to check your
prediction in part a. The headings must include the appropriate units. Ensure that the table covers [2]
all the detail relating to the five dyes listed in part c.
Total = 16
Appendix 1
The Periodic Table of the Elements
473
Cambridge International A Level Chemistry
Appendix 2
Selected standard electrode
potentials
Electrode reaction E —O/V Electrode reaction E —O/V
– 1.18
Ag+ + e– Ag + 0.80 Mn2+ + 2e– Mn + 1.52
– 0.25
Br2 + 2e– 2Br– + 1.07 MnO4– + 8H+ + 5e– Mn2+ + 4H2O + 0.81
Ca2+ + 2e– Ca – 2.87 Ni2+ + 2e– Ni + 0.87
+ 1.23
Cl2 + 2e– 2Cl– + 1.36 NO3– + 2H+ + e– NO2 + H2O + 0.40
ClO– + H2O + 2e– + 0.89 – 0.13
Cr2+ + 2e– Cr Cl– + 2OH– – 0.91 NO3– + 10H+ + 8e– NH4+ + 3H2O + 1.47
– 0.74 – 0.14
+ 1.33 O2 + 4H+ + 4e– 2H2O + 0.15
+ 0.52 + 0.17
Cr3+ + 3e– Cr + 0.15 O2 + 2H2O + 4e– 4OH– + 2.01
+ 0.09
Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O Pb2+ + 2e– Pb – 1.20
– 0.26
Cu+ + e– Cu PbO2 + 4H+ + 2e– Pb2+ + 2H2O + 0.34
Sn2+ + 2e– Sn + 1.00
474 + 1.00
– 0.76
Cu2+ + e– Cu+
Cu2+ + 2e– Cu + 0.34 Sn4+ + 2e– Sn2+
F2 + 2e– 2F– + 2.87 SO42– + 4H+ + 2e– SO2 + 2H2O
Fe2+ + 2e– Fe – 0.44
+ 0.77 S2O82– + 2e– 2SO42–
– 0.04
Fe3+ + e– Fe2+ S4O62– + 2e– 2S2O32–
Fe3+ + 3e– Fe V2+ + 2e– V
2H+ + 2e– H2 0.00 V3+ + e– V2+
2H2O + 2e– H2 + 2OH– – 0.83 VO2+ + 2H+ + e– V3+ + H2O
H2O2 + 2H+ + 2e– 2H2O + 1.77 VO2+ + 2H+ + e– VO2+ + H2O
I2 + 2e– 2I– + 0.54 VO3– + 4H+ + e– VO2+ + 2H2O
K+ + e– K – 2.92 Zn2+ + 2e– Zn
Mg2+ + 2e– Mg – 2.38
Appendix 3
Qualitative analysis notes
1 Reactions of aqueous cations
Cation Reaction with NH3(aq) 475
NaOH(aq) white precipitate
aluminium, Al3+(aq) white precipitate insoluble in excess
ammonium, NH4+(aq) soluble in excess –
barium, Ba2+(aq) no precipitate
calcium, Ca2+(aq) NH3 produced on heating no precipitate
chromium(III), Cr3+(aq) faint white precipitate is nearly always
copper(II), Cu2+(aq) observed unless reagents are pure no precipitate
iron(II), Fe2+(aq) white precipitate with high [Ca2+(aq)] grey-green precipitate
grey-green precipitate insoluble in excess
iron(III), Fe3+(aq) soluble in excess pale blue precipitate
magnesium, Mg2+(aq) pale blue precipitate soluble in excess giving dark blue solution
manganese(II), Mn2+(aq) insoluble in excess green precipitate turning brown on contact with air
green precipitate turning brown on insoluble in excess
zinc, Zn2+(aq) contact with air
insoluble in excess red-brown precipitate
red-brown precipitate insoluble in excess
insoluble in excess white precipitate
white precipitate insoluble in excess
insoluble in excess off-white precipitate rapidly turning brown on contact
off-white precipitate rapidly turning with air
brown on contact with air insoluble in excess
insoluble in excess white precipitate
white precipitate soluble in excess
soluble in excess
2 Reactions of anions
Ion Reaction
carbonate, CO32–(aq)
chloride, Cl–(aq) CO2 liberated by dilute acids
bromide, Br –(aq) gives white precipitate with Ag+(aq) (soluble in NH3(aq))
iodide, I–(aq) gives cream precipitate with Ag+(aq) (partially soluble in NH3(aq))
nitrate, NO3–(aq) gives yellow precipitate with Ag+(aq) (insoluble in NH3(aq))
nitrite, NO2–(aq) NH3 liberated on heating with OH–(aq) and Al foil
NH3 liberated on heating with OH–(aq) and Al foil; NO liberated by dilute acids (colourless
sulfate, SO42–(aq) NO → (pale) brown NO2 in air)
sulfite, SO32–(aq) gives white precipitate with Ba2+(aq) (insoluble in excess dilute strong acids)
SexOc2elsibsedrialutetedsotnrownagramciindgs)with dilute acids; gives white precipitate with Ba2+(aq) (soluble in
3 Tests for gases Test and test result
turns damp red litmus paper blue
Gas gives a white precipitate with limewater (precipitate dissolves with excess CO2)
ammonia, NH3 bleaches damp litmus paper
carbon dioxide, CO2 ‘pops’ with a lighted splint
476 chlorine, Cl2 relights a glowing splint
hydrogen, H2 turns acidified aqueous potassium manganate(VII) from purple to colourless
oxygen, O2
sulfur dioxide, SO2
Glossary
acid a proton (hydrogen ion) donor. atomic orbitals regions of space outside the nucleus
acid–base indicator a substance that changes colour over that can be occupied by one or, at most, two
electrons. Orbitals are named s, p, d and f. They
a narrow range of pH values. have different shapes.
constant, Ka the equilibrium constant
acid dissociation Ka = _ [H_[_H+_]A[_A]_ –_] average bond energy a general bond energy value used for
for a weak acid: a particular bond, e.g. a C H, when the exact
bond energy is not required. Average bond energies
activation energy the minimum energy that colliding are often used because the strength of a bond between
particles must possess for a successful collision that two particular types of atom is slightly different in
results in a reaction to take place. different compounds.
active site (of an enzyme) the ‘pocket’ on an enzyme
surface where the substrate binds and undergoes Avogadro constant the number of atoms (or ions,
catalytic reaction. molecules or electrons) in a mole of atoms (or ions,
acyl chloride a reactive organic compound related to molecules or electrons): its numerical value
a carboxylic acid, with the OH group in the acid is 6.02 × 1023.
replaced by a Cl atom, for example ethanoyl chloride,
CH3COCl. azo dyes coloured compounds formed on the
addition polymerisation the reaction in which monomers addition of phenol (or another aryl compound) to
a solution containing a diazonium ion. They contain
containing carbon-to-carbon double bonds react the N N group.
together to form long-chain molecules called polymers.
addition reaction an organic reaction in which two base a proton (hydrogen ion) acceptor.
reactant molecules combine to give a single bidentate ligands that can form two co-ordinate bonds 477
product molecule. from each ion or molecule to the central transition
adsorption (in catalysis) the first stage in heterogeneous metal ion.
catalysis – molecules of reactants (usually gases) form biofuels renewable fuels, sourced from plant or
bonds with atoms on the surface of the catalyst. animal materials.
boiling point the temperature at which the vapour
alkali a base that is soluble in water. pressure is equal to the atmospheric pressure.
alkaline earth metals the elements in Group 2 of the Boltzmann distribution a graph showing the
distribution of energies of the particles in a sample at a
Periodic Table. given temperature.
alkanes saturated hydrocarbons with the general formula bond energy/bond enthalpy the energy needed to
alkeCnneHs 2nu+n2.saturated hydrocarbons with a carbon–carbon break 1 mole of a particular bond in 1 mole of
double bond. Their general oforrmmoulleaciuslaCrnfHor2nm. s gaseous molecules.
allotrope different crystalline of the Born–Haber cycle a type of enthalpy cycle used to
same element. Graphite and diamond are allotropes calculate lattice energy.
of carbon. Brønsted–Lowry theory of acids acids are proton donors
alloy a mixture of two or more metals or a metal with a and bases are proton acceptors.
non-metal. buffer solution a solution that minimises changes in
amino acid residue an amino acid unit within a pH when moderate amounts of acid or base are
polypeptide chain. added. Common forms of buffer consist of either a
amphoteric able to behave as both an acid and a base. weak acid and its conjugate base or a weak base and
Aluminium oxide is amphoteric. its conjugate acid.
anion a negatively charged ion. carbocation an alkyl group carrying a single positive
anode the positive electrode. charge on one of its carbon atoms, e.g. +CH2CH3
arenes hydrocarbons containing one or more
benzene rings.
Cambridge International A Level Chemistry
catalyst a substance that increases the rate of a reaction cracking the process in which large, less useful
but remains chemically unchanged itself at the end of
the reaction. hydrocarbon molecules are broken down into smaller,
cathode the negative electrode. more useful molecules.
cation a positively charged ion. dative covalent bond another name for a co-ordinate bond.
chiral centre a carbon atom with four different groups degenerate orbitals atomic orbitals at the same
attached, creating the possibility of optical isomers. energy level.
closed system a system in which matter or energy is not dehydration a reaction in which a water molecule is
lost or gained, e.g. gases in a closed vessel. removed from a molecule, e.g. in the dehydration of an
cofactor a small molecule that is not a substrate but that is alcohol to give an alkene.
delocalised electrons electrons that are not associated
essential for an enzyme-catalysed reaction. with a particular atom – they can move between three
common ion effect the reduction in the solubility of a or more adjacent atoms.
denaturation the process by which the three-
dissolved salt by adding a compound that has an ion in dimensional structure of a protein or other biological
common with the dissolved salt. This often results in macromolecule is changed, often irreversibly. Relatively
precipitation of the salt. high temperatures, extremes of pH and organic solvents
competitive inhibition enzyme inhibition by molecules often cause denaturation.
that bind to the active site, preventing the normal desorption the last stage in heterogeneous catalysis.
substrate from reacting. They have a structure similar to The bonds holding the molecule(s) of product(s) to
the substrate molecule. The inhibition is reversible.
complementary base pairing In nucleic acids, bases are the surface of the catalyst are broken and the product
said to be complementary to each other if they form
specific hydrogen-bonded pairs. In DNA adenine (A)
always pairs with thymine (T) and cytosine (C) always molecules diffuse away from the surface of the catalyst.
478 pairs with guanine (G).
diazotisation the reaction between phenylamine
complex a central transition metal ion surrounded and nitrous acid (nitric(III) acid), HNO2, to give a
by ligands. diazonium salt in the first step in making an azo dye.
compound a substance made up of two or more elements
bonded (chemically joined) together. dipeptide the product formed when two amino acids
condensation the change in state when a vapour changes react together.
to a liquid.
dipole a separation of charge in a molecule. One end of
condensation reaction a reaction in which two organic the molecule is permanently positively charged and the
molecules join together and in the process eliminate a other is negatively charged.
small molecule, such as water or hydrogen chloride.
discharge the conversion of ions to atoms or molecules at
conjugate pair (acid/base) an acid and base on each side electrodes during electrolysis, for example, during
of an equilibrium equation that are related to each the electrolysis of concentrated sodium chloride
other by the difference of a proton; e.g. the acid in the solution, chlorine is discharged at the anode by the
forward reaction and the base in the reverse reaction conversion of Cl– ions to Cl atoms, which then combine
or the base in the forward reaction and the acid in the to form Cl2 molecules.
reverse reaction.
displayed formula a drawing of a molecule that shows all
co-ordinate bond a covalent bond in which both electrons the atoms and bonds within the molecule.
in the bond come from the same atom.
disproportionation the simultaneous reduction and
co-ordination number the number of co-ordinate (dative) oxidation of the same species in a chemical reaction.
bonds formed by ligands to the central transition metal
ion in a complex. dissociation the break-up of a molecule into ions, for
example, when HCl molecules dissolve in aqueous
coupling reaction when a diazonium ion reacts with an solution, they dissociate completely into H+ and Cl– ions.
alkaline solution of phenol (or similar compound) to
disulfide bridge an S S bond formed when the SH
make an azo dye. groups on the side-chain of two cysteine residues in a
covalent bond a bond formed by the sharing of pairs of protein combine. Disulfide bridges help maintain the
electrons between two atoms. tertiary structure of some proteins.
Glossary
DNA (deoxyribonucleic acid) a polymer with a double electrophile a species that can act as an acceptor of a pair
helical structure containing two sugar–phosphate chains of electrons in an organic mechanism.
with nitrogenous bases attached to them. The sequence electrophoresis the separation of charged particles by
of bases forms a code, which is used to form more DNA their different rates of movement in an electric field.
by replication or to encode mRNA (transcription). electrovalent bond another name for an ionic bond.
element a substance made of only one type of atom.
dot-and-cross diagram a diagram showing the elimination a reaction in which a small molecule, such as
arrangement of the outer-shell electrons in an ionic empHir2OicaolrfHorCml,uilsar etmheofvoerdmfurolamthaant organic molecule.
or covalent element or compound. The electrons are tells us the simplest
shown as dots or crosses to show their origin. ratio of the different atoms present in a molecule.
endothermic term used to describe a reaction in which
double covalent bond two shared pairs of electrons energy is absorbed from the surroundings: the enthalpy
bonding two atoms together. change is positive.
energy levels (of electrons) the regions at various
dynamic (equilibrium) in an equilibrium mixture, distances from the nucleus in which electrons have a
molecules of reactants are being converted to particular amount of energy. Electrons further from the
products at the same rate as products are being
converted to reactants.
electrochemical cell two half-cells in separate nucleus have more energy. See principal quantum shells.
enhanced global warming the increase in average
compartments joined by a salt bridge. When the poles temperatures around the world as a consequence of
of the half-cells are joined by a wire, electrons travel
in the external circuit from the half-cell with the more the huge increase in the amounts of CO2 and other
negative E —O value to the half-cell with the more greenhouse gases produced by human activity.
enthalpy change of atomisation ΔH —aOt ; the enthalpy
positive E —O value. change when 1 mole of gaseous atoms is formed from
electrode a rod of metal or carbon (graphite) which
conducts electricity to or from an electrolyte. its element under standard conditions.
enthalpy change of hydration ΔHh—Oyd; the enthalpy change 479
electrode potential the voltage measured for a half-cell when 1 mole of a specified gaseous ion dissolves in
compared with another half-cell.
electrolysis the decomposition of a compound into its sufficient water to form a very dilute solution.
ΔH —O ; the energy absorbed
elements by an electric current. enthalpy change of solution
sol
or released when 1 mole of an ionic solid dissolves in
electrolyte a molten ionic compound or an aqueous sufficient water to form a very dilute solution.
solution of ions that is decomposed during
electrolysis. enthalpy change the energy transferred in a chemical
reaction (symbol ΔH).
electron tiny subatomic particles found in orbitals around enthalpy cycle a diagram showing alternative routes
the nucleus. They have a negative charge but have
negligible mass. between reactants and products that allows the
determination of one enthalpy change from other
electron affinity (first electron affinity) ΔH —O ; the known enthalpy changes by using Hess’s law.
ea1
enthalpy change when 1 mole of electrons is added to
1 mole of gaseous atoms to form 1 mole of gaseous enthalpy profile diagram a diagram showing the
enthalpy change from reactants to products along the
1– ions under standard conditions. reaction pathway.
ΔH —O ; the
electron affinity (second electron affinity) entropy a measure of the dispersal of energy or disorder of
ea2
enthalpy change when 1 mole of electrons is added to
1 mole of gaseous 1– ions to form 1 mole of gaseous a system. The system becomes energetically more stable
when disordered.
2– ions under standard conditions. enzyme a protein molecule that is a biological catalyst.
electronegativity the ability of an atom to attract the
bonding electrons in a covalent bond. Most act on a specific substrate.
enzyme activity a measure of the rate at which substrate is
electronic configuration a way of representing the converted to product in an enzyme-catalysed reaction.
arrangement of the electrons in atoms showing the
principal quantum shells, the subshells and the number equilibrium constant a constant calculated from the
equilibrium expression for a reaction.
of electrons present, e.g. 1s2 2s2 2p3. The electrons may equilibrium expression a simple relationship that links
also be shown in boxes. Kc to the equilibrium concentrations of reactants and
electropherogram the physical results of electrophoresis. products and the stoichiometric equation.
Cambridge International A Level Chemistry
equilibrium reaction a reaction that does not go to functional group an atom or group of atoms in an organic
completion and in which reactants and products are molecule that determine the characteristic reactions of a
present in fixed concentration ratios. homologous series.
esterification the reaction between an alcohol and a
carboxylic acid (or acyl chloride) to produce an ester gene a length of DNA that carries a code for making a
particular protein.
and water. general formula a formula that represents a homologous
eutrophication an environmental problem caused by series of compounds using letters and numbers; e.g.
fertilisers leached from fields into rivers and lakes. The sthuebsgteitnuetrinalgfoarnmuumlabeforrfothr enailnkathneesgiesnCernaHl 2fno+r2m. Buyla you
fertiliser then promotes the growth of algae on the get the molecular formula of a particular compound in
surface of water. When the algae die, bacteria thrive that homologous series.
and use up the dissolved oxygen in the water, killing general gas equation an equation relating the volume of a
aquatic life. gas to the temperature, pressure and number of moles
exothermic the term used to describe a reaction in which of gas. Also called the ideal gas equation.
energy is released to the surroundings: the enthalpy
change is negative. pV = nRT
Faraday constant the charge carried by 1 mole of electrons genetic code a code made up of sets of three consecutive
(or 1 mole of singly charged ions). It has a value of nitrogenous bases that provides the information to
make specific proteins.
96 500 coulombs per mol (C mol–1). genetic engineering the deliberate alteration of one or
Faraday’s laws first law: the mass of a substance produced more bases in the DNA of an organism, leading to an
at an electrode during electrolysis is proportional to the altered protein with improved properties. Scientists
quantity of electricity passed in coulombs. Second law:
the number of Faradays needed to discharge 1 mole of hope to be able to use genetic engineering to eliminate
genetic diseases which are caused by mutations in DNA.
480 an ion at an electrode equals the number of charges on genetic fingerprinting a technique based on matching the
the ion.
feasibility (of reaction) the likelihood or not of a reaction minisatellite regions of a person’s DNA to a database of
occurring when reactants are mixed. We can use E —O reference samples.
values to assess the feasibility of a reaction. giant molecular structure/giant covalent
Fehling’s solution an alkaline solution containing structure structures having a three-dimensional
network of covalent bonds throughout the
copper(II) ions used to distinguish between aldehydes whole structure.
and ketones. A positive test is one in which the clear
blue solution gives a red/orange precipitate when Gibbs free energy the energy change that takes into
warmed with aldehydes, but no change is observed account both the entropy change of a reaction and
with ketones. enthalpy change. Reactions are likely to be feasible if
fragmentation the breaking up of a molecule into smaller the value of the Gibbs free energy change of reaction
parts by the breaking of covalent bonds in a mass is negative. The Gibbs free energy change of reaction is
spectrometer. given by the relationship
free energy see Gibbs free energy. ΔG —O = ΔH —O – TΔS —O
free radical very reactive atom or molecule that has a
single unpaired electron. GLC gas–liquid chromatography.
GLC/MS a technique in which a mass spectrometer is
free-radical substitution the reaction in which halogen
atoms substitute for hydrogen atoms in alkanes. The connected directly to a gas–liquid chromatograph to
identify the components in a mixture.
mechanism involves steps in which reactive free radicals
are produced (initiation), regenerated (propagation)
and consumed (termination). haemoglobin the iron-containing protein found in red
Friedel–Crafts reaction the electrophilic substitution of blood cells that transports oxygen around the body.
half-cell half of an electrochemical cell. The half-cell with
an alkyl or acyl group into a benzene ring. the more negative E —O value supplies electrons. The half-
fuel cell a source of electrical energy that comes directly cell with the more positive E —O value receives electrons.
from the energy stored in the chemicals in the cell, one half-equation in a redox reaction, an equation showing
of which is oxygen (which may come from the air). either an oxidation or a reduction.
Glossary
half-life the time taken for the amount (or concentration) ionic product of wofatweart,eKr.w the equilibrium constant for
of the limiting reactant in a reaction to decrease to half the ionisation
its value.
Kw = [H+][OH–]
halogens Group 17 elements.
Hess’s law the total enthalpy change for a chemical ion1ismatioolenoefneelregcytr,oΔnHs fi rothme energy needed to remove
1 mole of atoms of an element
reaction is independent of the route by which the in the gaseous state to form 1 mole of gaseous ions.
reaction takes place. isotopes atoms of an element with the same number of
heterogeneous catalysis the type of catalysis in which the protons but different numbers of neutrons.
catalyst is in a different phase from the reactants. For
example, iron in the Haber process. kinetic theory the theory that particles in gases and
homogeneous catalysis the type of catalysis in which liquids are in constant movement. The kinetic theory
the catalyst and reactants are in the same phase. For can be used to explain the effect of temperature and
example, sulfuric acid catalysing the formation of an pressure on the volume of a gas as well as rates of
ester from an alcohol and carboxylic acid. chemical reactions.
HPLC high-performance liquid chromatography.
hybridisation of atomic orbitals the process of mixing lattice a regularly repeating arrangement of ions, atoms or 481
atomic orbitals so that each has some character of each molecules in three dimensions.
of the orbitals mixed.
hydrocarbon a compound made up of carbon and lattice energy the enthalpy change when 1 mole of an
hydrogen only. ionic compound is formed from its gaseous ions under
hydrogen bond the strongest type of intermolecular force standard conditions.
– it is formed between molecules having a hydrogen
atom bonded to one of the most electronegative le Chatelier’s principle when any of the conditions
elements (F, O or N). affecting the position of equilibrium are changed,
hydrolysis the breakdown of a compound by water, which the position of that equilibrium shifts to minimise
is often speeded up by reacting with acid or alkali. the change.
hydrophobic the non-polar part of a molecule that has no
attraction for water molecules (‘water hating’). ligand a molecule or ion with one or more lone pairs of
hydroxynitrile an organic compound containing both an electrons available to donate to a transition metal ion.
OH and a CN group, e.g. 2-hydroxypropanenitrile, lock-and-key mechanism a model used to explain why
CH3CH(OH)CN. enzymes are so specific in their activity. It is suggested
ideal gas a gas whose volume varies in proportion to the that the active site of the enzyme has a shape into which
temperature and in inverse proportion to the pressure. the substrate fits exactly – rather like a particular key
Noble gases such as helium and neon approach ideal fits a particular lock.
behaviour because of their low intermolecular forces.
infra-red spectroscopy a technique for identifying lone pairs (of electrons) pairs of electrons in the outer
compounds based on the change in vibrations of shell of an atom that are not bonded.
particular atoms when infra-red radiation of specific
frequencies is absorbed mass number see nucleon number.
initiation step the first step in the mechanism of free- mass spectrometer an instrument for finding the relevant
radical substitution of alkanes by halogens. It involves
the breaking of the halogen–halogen bond by UV light isotopic abundance of elements and to help identify
from the Sun. unknown organic compounds.
intermolecular forces the weak forces between molecules. metabolism the series of linked chemical reactions taking
ion polarisation the distortion of the electron cloud on place in living organisms.
an anion by a neighbouring cation. The distortion is metalloid elements that have a low electrical conductivity
greatest when the cation is small and highly charged. at room temperature but whose conductivity increases
ionic bond the electrostatic attraction between oppositely with increasing temperature. Metalloids are found in a
charged ions. diagonal band running from the top left to nearly the
bottom right of the p-block in the Periodic Table.
mobile phase the solvent in the chromatography process,
which moves through the column or over the paper or
thin layer.
molar mass the mass of a mole of substance in grams.
Cambridge International A Level Chemistry
mole the unit of amount of substance. It is the amount of open system a system in which matter is lost or gained,
substance that has the same number of particles (atoms, e.g. a mixture of solids and gases in an open beaker.
ions, molecules or electrons) as there are atoms in optical isomers stereoisomers that exist as two non-
exactly 12 g of the carbon-12 isotope.
molecular formula the formula that tells us the actual superimposable mirror images.
numbers of each type of atom in a molecule.
optical resolution the separation of optically active
molecular ion the ion that is formed by the loss of an isomers (enantiomers) from a mixture.
electron from the original complete molecule during
mass spectrometry and that gives us the relative order of reaction the power to which the concentration
molecular mass of an unknown compound. of a reactant is raised in the rate equation. If the
concentration does not affect the rate, the reaction is
monodendate ligands, such as water and ammonia, that zero order. If the rate is directly proportional to the
can form only one co-ordinate bond from each ion or reactant concentration, the reaction is first order. If the
molecule to the central transition metal ion. rate is directly proportional to the square of the reactant
concentration, the reaction is second order.
monomer a small, reactive molecule that reacts to make
long-chain molecules called polymers. oxidation the addition of oxygen, removal of electrons or
increase in oxidation number of a substance; in organic
chemistry refers to a reaction in which oxygen atoms
nanotechnology the design and production of machines are added to a molecule and/or hydrogen atoms are
that are so small we measure them in nanometres (nm), removed from a molecule.
where 1 nm = 1 × 10–9 m.
Nernst equation an equation used to predict oxidation number (oxidation state) a number given to
quantitatively how the value of an electrode potential an atom in a compound that describes how oxidised or
varies with the concentration of the aqueous ion. reduced it is.
neutron a subatomic particle found in the nucleus of an oxidising agent a reactant that increases the
atom. It has no charge and has the same mass as oxidation number of (or removes electrons from)
another reactant.
482 a proton.
nitrogenous bases nitrogen-containing bases found in
DNA and RNA. In DNA they are adenine (A), guanine
(G), thymine (T) and cytosine (C). In RNA uracil (U) partial pressure the pressure that an individual gas
replaces thymine. contributes to the overall pressure in a mixture of gases.
NMR nuclear magnetic resonance spectroscopy. partition coefficient the ratio of the concentrations of
non-degenerate orbitals atomic orbitals that have been
split to occupy slightly different energy levels. a solute in two different immiscible solvents when an
non-polar (molecule) a molecule with no separation equilibrium has been established.
of charge; it will not be attracted to a positive or
negative charge. peptide bond the link between the amino acid residues
nucleon number the total number of protons and in a polypeptide or protein chain. The link is formed
by a condensation reaction b eCtOweOeHn tghreo up NoHf a2ngortohuepr
neutrons in the nucleus of an atom. of one amino acid and the
nucleophile species that can act as a donor of a pair amino acid.
of electrons.
periodicity the repeating patterns in the physical and
nucleophilic addition the mechanism of the reaction chemical properties of the elements across the periods
in which a nucleophile attacks the carbon atom in a of the Periodic Table.
carbonyl group and adds across the C O bond, e.g.
aldehydes or ketones reacting with hydrogen cyanide. permanent dipole–dipole forces a type of intermolecular
force between molecules that have permanent dipoles.
nucleotide a compound consisting of a nitrogenous base,
a sugar (ribose or deoxyribose) and a phosphate group. pH the hydrogen ion concentration expressed as a
logarithm to base 10.
Nucleotides form the basic structural units of DNA pH = –log10[H+]
and RNA.
nucleus the small dense core at the centre of every pi (π) bonds multiple covalent bonds involving the
atom, containing protons (positively charged) and sideways overlap of p atomic orbitals.
neutrons (no charge). Nuclei are therefore always
pKa values of Ka expressed as a logarithm to base 10.
positively charged. pKa = –log10Ka
Glossary
polar (covalent bond) a covalent bond in which the two rate-determining step the slowest step in a
bonding electrons are not shared equally by the atoms reaction mechanism.
in the bond. The atom with the greater share of the real gases gases that do not obey the ideal gas law,
especially at low temperatures and high pressures.
electrons has a partial negative charge, δ–, and the other redox reaction a reaction in which oxidation and
has a partial positive charge, δ+.
polarising power (of a cation) the ability of a cation to reduction take place at the same time.
reducing agent a reactant that decreases the oxidation
attract electrons and distort an anion. number of (or adds electrons to) another reactant.
polyamides polymers whose monomers are bonded to
each other via the amide link, CONH . reduction the removal of oxygen, addition of electrons
or decrease in oxidation number of a substance; in
polyesters polymers whose monomers are bonded to each organic chemistry it is the removal of oxygen atoms
other via the ester link, COO .
polymer a long-chain molecule made up of many from a molecule and/or the addition of hydrogen atoms
to a molecule.
repeating units. relative atomic mass the weighted average mass of
polypeptides natural polymers whose monomers are
bonded to each other via the amide link, CONH , the atoms of an element, taking into account the
proportions of naturally occurring isotopes, measured
and whose monomers are amino acids. on a scale on which an atom of the carbon-12 isotope
primary alcohol an alcohol in which the carbon atom
bonded to the OH group is attached to one other has a mass of exactly 12 units.
relative formula mass the mass of one formula unit of a
carbon atom (or alkyl group). compound measured on a scale on which an atom of
primary structure (of proteins) the sequence of amino
acids in a polypeptide chain. the carbon-12 isotope has a mass of exactly 12 units.
relative isotopic mass the mass of a particular isotope
principal quantum shells, n regions at various distances of an element on a scale in which an atom of the
from the nucleus that may contain up to a certain
number of electrons. The first quantum shell contains carbon-12 isotope has a mass of exactly 12 units.
relative molecular mass the mass of a molecule measured 483
up to 2 electrons, the second up to 8 and the third up
to 18. on a scale in which an atom of the carbon-12 isotope
has a mass of exactly 12 units.
propagation step a step in a free-radical mechanism in residue see amino acid residue.
which the radicals formed can then attack reactant
molecules generating more free-radicals, and so on. retention time the time taken for a component in a
mixture to travel through the column in GLC or HPLC.
protein condensation polymer formed from amino reversible reaction a reaction in which products can be
acids and joined together by peptide bonds. Proteins
can be structural (e.g. cartilage), catalysts (enzymes), changed back to reactants by reversing the conditions.
value the ratio of the distance a component has
hormones (e.g. insulin) or antibodies. Rf travelled compared with the distance travelled by the
proton a positively charged subatomic particle in solvent front during paper chromatography or TLC.
the nucleus.
rate constant the proportionality constant in the rate salt bridge a piece of filter paper soaked in potassium
equation (see rate equation). nitrate solution used to make electrical contact between
the half-cells in an electrochemical cell.
rate equation an equation showing the relationship
between the rate constant and the concentrations of saturated hydrocarbons compounds of hydrogen and
those reactants that affect the rate of reaction. The carbon only in which the carbon–carbon bonds are
general form of the rate equation is: all single covalent bonds, resulting in the maximum
rate = k[A]m[B]n number of hydrogen atoms in their molecules.
where k is the rate constant, [A] and [B] are the secondary alcohol an alcohol in which the carbon atom
concentrations of those reactants that affect the rate of bonded to the OH group is attached to two other
reaction, m is the order of the reaction with respect to A carbon atoms (or alkyl groups).
and n is the order of reaction with respect to B.
secondary structure (of proteins) the second level of
rate of reaction a measure of the rate at which reactants protein structure. The folding of a polypeptide chain
are used up or the rate at which products are formed. into specific structures (e.g. α-helix and β-pleated
The units of rate are mol dm–3 s–1. sheet), which are stabilised by hydrogen bonds formed
between CO and NH groups in peptide bonds.
Cambridge International A Level Chemistry
shielding the ability of inner shells of electrons to standard conditions conditions of temperature and
reduce the effective nuclear charge on electrons in the pressure that must be the same in order to compare
outer shell. moles of gases or enthalpy changes accurately. Standard
conditions are a pressure of 105 pascals (100 kPa) and a
sigma (σ) bonds single covalent bonds, formed by the temperature of 298 K (25 °C).
‘end-on’ overlap of atomic orbitals. standard electrode potential the electrode potential of
single covalent bond a shared pair of electrons bonding a half-cell when measured with a standard hydrogen
two atoms together. electrode as the other half-cell.
skeletal formula a simplified version of the displayed
formula that has all the symbols for carbon and standard enthalpy change an enthalpy change that takes
hydrogen atoms removed, as well as the carbon to place under the standard conditions of pressure (105 Pa)
hydrogen bonds. The carbon to carbon bonds are left in and temperature (298 K).
place as are the bonds to other atoms.
SN1remaectcihoanniinsmw htihche steps in a nucleophilic substitution standard hydrogen electrode a half-cell in which
the rate of the reaction (which is hydrogen gas at a pressure of 1 atmosphere (101 kPa)
determined by the slow step in the mechanism) involves bubbles into a solution of 1.00 mol dm–3 H+ ions.
only the organic reactant, e.g. in the hydrolysis of a This electrode is given a standard electrode potential
tertiary halogenoalkane. of 0.00 V. All other standard electrode potentials are
SN2remaectcihoanniinsmw htihche steps in a nucleophilic substitution measured relative to this value.
the rate of the reaction (which is
determined by the slow step in the mechanism) involves state symbol a symbol used in a chemical equation that
describes the state of each reactant and product: (s) for
solid, (l) for liquid, (g) for gas and (aq) for substances in
aqueous solution.
two reacting species, e.g. in the hydrolysis of a primary stationary phase the immobile phase in chromatography
that the mobile phase passes over or through.
halogenoalkane. Examples are the surface of the thin-layer particles in
solushboilwitiynpgrtohdeupcrto, dKuscp t the equilibrium expression TLC or the involatile liquid adsorbed onto the column
484 of the concentrations of each ion
in a saturated solution of a sparingly soluble salt at in GLC or HPLC.
298 K, raised to the power of the relative concentrations: stereoisomers compounds whose molecules have the
Ksp = [C y+(aq)]a[Ax–(aq)]b same atoms bonded to each other but with different
arrangements of the atoms in space.
stoichiometry the mole ratio of the reactants and products
where a is the number of C y+ ions in one formula unit in the balanced equation for a reaction.
of the compound and b is the number of Ax– ions in one
formula unit of the compound. strong acid/base an acid or base that is (almost)
completely ionised in water.
solute a substance that is dissolved in a solution.
specific most enzymes are described as specific because structural formula the formula that tells us about the
atoms bonded to each carbon atom in an organic
they will only catalyse one reaction involving one
particular molecule or pair of molecules. strumctoulercaul lies,oem.ge.rCs Hc3oCmHp ou CnHds2.with the same molecular
spectator ions ions present in a reaction mixture that do formula but different structural formulae.
not take part in the reaction.
spin-pair repulsion electrons repel each other as they subshells regions within the principal quantum shells
have the same charge. Electrons arrange themselves where electrons have more or less energy depending on
their distance from the nucleus. Subshells are given the
so that they first singly occupy different orbitals in the letters s, p, d and f.
same sublevel. After that they pair up with their spins
opposed to each other. substitution a reaction that involves the replacement of
one atom, or group of atoms, by another.
splitting pattern the pattern of peaks that main signals are substrate a molecule that fits into the active site of an
divided into in high-resolution NMR. enzyme and reacts.
stabfoilrimtyactoionnstoafntth,eKcsotamb tphleexeqiouniliibnraiusmolvceonntstfaronmt foitrsthe
sucrceeqssuiivreediotnoirseamtioonveetnheergfiyrs tΔ, tHhie1n, ΔthHei2s,eectocn.:dth, teheennetrhgey
constituent ions or molecules. third electrons and so on from a gaseous atom or ion,
standard cell potential the difference in standard producing an ion with one more positive charge each
electrode potential between two half-cells. time. Measured in kJ per mole of ions produced.
Glossary
surroundings in enthalpy changes, anything other than two-way chromatography a technique used in paper 485
the chemical reactants and products, for example the or thin-layer chromatography in which one spot of a
solvent, the test tube in which the reaction takes place, mixture is placed at the corner of a square sheet and
the air around the test tube. is developed in the first solvent as usual. The sheet is
then turned through 90° and developed in the second
termination step the final step in a free-radical solvent, giving a better separation of components
mechanism in which two free radicals react together to having similar Rf values.
form a molecule.
unsaturated hydrocarbons compounds of hydrogen and
tertiary alcohol an alcohol in which the carbon atom carbon only whose molecules contain carbon-to-carbon
bonded to the OH group is attached to three other double bonds (or triple bonds).
carbon atoms (or alkyl groups).
van der Waals’ forces the weak forces of attraction
tertiary structure (of proteins) the third level of protein between molecules caused by the formation of
structure. It involves further folding of the polypeptide temporary dipoles.
chain, which is stabilised by interactions between the
amino acid side-chains (ionic interactions, hydrogen vaporisation the change in state when a liquid changes
bonding, van der Waals’ forces and disulfide bonds). to vapour.
titre in a titration, the final burette reading minus the vapour pressure the pressure exerted by a vapour in
initial burette reading. equilibrium with a liquid.
TLC thin-layer chromatography. weak acid/base an acid or base that is only slightly ionised
TMS tetramethylsilane. An inert, volatile liquid used as a in water.
reference in NMR, given a chemical shift of zero. X-ray crystallography an analytical technique that uses
Tollens’ reagent an aqueous solution of silver nitrate the diffraction pattern of X-rays passed through a solid
sample to elucidate its structure.
in excess ammonia solution, sometimes called
ammoniacal silver nitrate solution. It is used to
distinguish between aldehydes and ketones. It gives
a positive ‘silver mirror’ test when warmed with
aldehydes, but no change is observed with ketones.
triple covalent bond three shared pairs of electrons
bonding two atoms together.
Index
acid dissociation constant (Ka), 307–9 arenes, 382
acid rain, 184–5, 205, 341–2 arenes, reactions of, 384–7
acid–base equilibria, 130–4 aromatic hydrocarbons, 203, see also arenes
acid–base indicators, 309–10 aryl compounds, 193, 382–3
acids, definitions of, 130–4 atomic mass, 2–5
activation energy, 100, 141–5, 182, 340 atomic orbitals, 37–8, 57–8
active site, 457–8 atomic structure, 25–9
acyl chlorides, 396–8, 406, 410, 416 atoms, 2, 5–9, 25–6, 28
addition polymerisation, 211–3, 426–7 average bond energies, 100
addition reactions, 198, 208–12, 237, 383, 397, 412, 422, 424 Avogadro number or Avogadro constant, 5–6, 277–8
addition reactions of alkenes, 198, 208–12 Avogadro’s hypothesis, 18–19
adsorption, 342–3 azo dyes, 403–4
adsorption chromatography, 436
alanine, 404, 415 base units, SI, 5
alcohols, formation of, 219 bases, definitions of, 130–4
alcohols, reactions of, 226–31 benzene, 193, 382–3
aldehydes, formation of, 210, 236 benzene, reactions of, 384–7
aldehydes, reactions of, 225, 237–40 benzene, structure of, 382–3
alkalis, definition of, 130 benzene, nitration of, 385
alkaline earth metals, 164–167 benzene, π bonding in, 383
486 alkanes, reactions of, 204–7 bidentate ligands, 372–3, 376
alkanes, sources of, 202–3 biochemical analysis, 407
alkenes, production of, 207–8 biochemical polymers, 414–8
alkenes, reactions of, 208–12 biochemical reactions, 145
allotropes, 80, 82 biofuels, 227
alloys, 79–80 bleach, 178
aluminium, 53–4, 79–80, 84, 153 boiling points, trends in, 62–65, 66
aluminium, reactions of, 154–6 Boltzmann distribution, 143–4
aluminium–air batteries, 293 bonds, breaking and making of, 99–100, 196
amides, formation of, 398 bond energy, 54, 100–1
amides, reactions of, 406 bond length, 54–6, 382
amines, basicity of, 401–2 Born–Haber cycle, 259–62
amines, classes of, 401 boron trifluoride, structure and shape of, 51–2, 56
amines, formation of, 402–3 bromoethane, 209, 218, 220, 227, 373
amines, reactions of, 403–4 Brønsted–Lowry theory of acids and bases, 131–2
amino acid residues, 416–7, 435 buckminsterfullerene, 82–3
amino acids, 404–5, 407, 414–6, 434 buffer solutions, 313–5, 405, 407
ammonia, basicity of, 130, 134
ammonia, hydrogen bonding and, 64 calcium chloride, 50
ammonia, production of, 122, 128–9, 182 calorimetry, 94
ammonia, reactions of, 101, 401–2, 406 carbocations, 197, 209–10, 221–2, 386
ammonia, shape of, 52, 53, 55–6 carbon dioxide, structure and shape of, 52–3, 56
ammonium compounds, 182–4 carbon-neutral fuel, 227
amphoteric materials, 131, 156, 405, 415 carbonyl group, testing for, 238–9
anions, 151, 263, 266, 295–6 carboxylic acids, acidity of, 394–5
anodes, 275, 295–7 catalysis, 142, 144–5, 340–3
Index
catalysts, definition of, 142 coupling reaction, 404 487
catalytic converters, 184, 205, 343 covalent bonding, 51–5
catalytic crackers, 208 covalent bonds, dative, 53–4
cathode ray tube, 26 covalent bonds, multiple, 52–3
cathodes, 275, 295 covalent compounds, 11, 66–7, 109, 263
cations, 151, 295–6, 316, 384 crude oil, 185, 202–3, 207, 227, 449, 460–1
cells and batteries, 293–5 cycloalkanes, 202
ceramics, 158 dative covalent bonds, 53–4
CFCs, see chlorofluorocarbons, 222–3 degenerate orbitals, 376–7
chain isomerism, 195 dehydration reactions, 230
chemical bonding, types of, 49– 55 delocalised electrons, 59–60, 383–5, 398, 402–4
chemical equations, balancing of, 12, 97–8, 108, 112, 117–8 desorption, 342–3
chemical formulae, 10–12 diamond, 81–2, 352
chiral catalysts, 459 diazonium dyes, 403–4
chiral centres, 195–6 diazotisation reactions, 403–4
chiral drugs, 458–9 diffusion, 342–3, 351
chlorides, of Period 3 elements, 158–60 dipeptides, 405
chlorine, disproportionation of, 177 displacement reactions, 174
chlorobenzene, 382, 384, 398, 448 displayed formula of compounds, 53, 190, 194
chlorofluorocarbons (CFCs), 222–3 disproportionation, 177–8
chromatography, 319–20, 434–8 DNA (deoxyribonucleic acid), 418–20
cis–trans isomerism, 195, 373 dot-and-cross diagrams, 50–4
closed system, 119 drugs, delivery of, 83
collision theory, 141–2 drugs, designing new, 457–9
combustion, enthalpy change of, 92–3 drugs, testing for, 449
combustion, in determination of empirical formulae, 9 dynamic equilibrium, 117–8, 319
combustion, of alcohols, 226 E—O values, defining standard electrode potentials, 279, 281
combustion, of alkanes, 204–5 E—O values, for oxidising and reducing agents, 284
combustion, of fuels, 90–1 E—O values, impact of ion concentration on, 280–1
common ion effect, 318–9 E—O values, measuring standard electrode potentials, 282
complementary base pairs, 419 E—O values, table of, 474
complex ions, 176, 369, 371–8 E—O values, use of, 284–8
complexes, colour of, 376–8 electrical conductivity, improving through doping, 426
compounds, properties of covalent, 66–7 electrical conductivity, patterns of, 67, 83, 151-2, 326, 369
compounds, properties of ionic, 66–7 electrochemical cells, 280–1, 284–8, 293–4
compounds, naming of, 10–11 electrode, pH, 306, 311
concentration, effects on equilibrium, 120, 126 electrode, standard hydrogen, 279
concentration of reactant against time graphs, 332–4 electrode potentials, 278–88, 296
condensation polymerisation, 412–4, 419, 424 electrode reactions, 275, 474
condensation polymers, 412, 415, 421, 426–8 electrodes, 275–6, 407
condensation reactions, 238–9, 397, 405, 412–3, 415 electrolysis, 84, 275
conjugate acids, 132–3, 314, electrolysis, of aqueous solutions, 295–6
conjugate bases, 132–3 electrolysis, of brine, 296–7
conjugate pairs, 132–3 electrolysis, of molten electrolytes, 295
Contact process, 129, 185 electrolysis, quantitative, 276–8
co-ordinate bonds, 53–4, 371–2, 376–7 electrolytes, 275–6, 293–6
co-ordination number, 372 electrolytic cells, 275
copper, colour of complex ions of, 368, 374
copper, in electrolysis, 122
copper, uses of, 367
Cambridge International A Level Chemistry
electrolytic purification of copper, 84 equilibrium and sulfuric acid production, 129–30
electron affinity, definition of, 258–9 equilibrium constant (Kc), 123–6
electronegativity, 60–61, 64, 204, 209, 220, 238, 394, 397 equilibrium constant (Kp), 127–9
electronic configurations of atoms and ions, 11, 33–4, equilibrium expressions, 123
38–40, 164–5, 172–3, 193 equilibrium reactions, 117
electronic configurations of transition elements, esterification reactions, 229–30
367–8, 377 esters, 229–30
electronic structure, 33–4, 38–40 ethane, 58, 193
electron-pair repulsion theory, 55–6 ethanoic acid, 229, 231–2
electrons, 26–9 ethene, 53, 58, 193–4
electropherograms, 407 ethylamine, 220, 241, 401–3
electrophiles, 197, 204–9, 384–6, 388, 403–4 eutrophication, 183–4
electrophilic addition, 209–10 exothermic reactions, 90–1, 99, 122, 126, 141–2, 258, 265,
electrophilic substitution, 384–6 266, 354, 357–8
electrophoresis, 407–8
electrovalent bond, 50 faraday, unit (F), 276
elements, 2–4 Faraday constant, 277–8
elimination reactions, 222, 230, 397, 412 Fehling’s solution, 239, 395
empirical formulae of compounds, 9–10, 190 forensic science, use of chromatography in, 436, 449
endothermic reactions, 90–1, 99, 122, 141–2, 265, 354, 358 free-radical substitution reactions, 207, 385
energy changes, 90 free radicals, 196, 206–7, 223
energy levels (quantum shells), 33 fullerenes, 82–3
enthalpy change, calculation using bond energies, 99–101 functional group isomerism, 195
488 enthalpy change, definition of, 90 functional groups, 192, 412, 426
enthalpy change, Hess’s law of, 97
enthalpy change, measurement of, 94–6 gas reactions, equilibria in, 127–8
enthalpy change, of atomisation, 93 gas volumes and stoichiometry, 19
enthalpy change, of combustion, 92, 95, 98 gaseous state of matter, 73–7
enthalpy change, of formation, 92, 98 gas–liquid chromatography (GLC), 437–8, 448–9
enthalpy change, of hydration, 93–4, 99 gasohol, 227
enthalpy change, of neutralisation, 93, 94 gel electrophoresis, principles of, 407–8
enthalpy change, of reaction, 92 general formulae of compounds, 192
enthalpy change, of solution, 93, 95 general gas equation, 75–6
enthalpy change, of vaporisation and boiling point, 65, 77 genetic code, 419
enthalpy change, standard conditions for, 92–4 genetic information, 421
enthalpy change, types of, 92–5 giant molecular structures, 80–2
enthalpy cycles, 97–100, 259–260, 266 Gibbs free energy, 357–60
enthalpy profile diagrams, 91 Gibbs free energy, calculating changes in, 360–2
entropy, 350–3 global warming, 84, 205–6, 212
entropy, and free energy, 357–8 graphene, 83
entropy, and temperature, 357 graphite, 80–1
entropy, calculating changes in, 354–7 greenhouse gases, 205
enzymes, 145 Group 2 carbonates and nitrates, thermal decomposition
enzymes, digestive, 417 of, 168
enzymes, in DNA replication, 420 Group 2 compounds, uses of, 169
equilibria, acid–base, 130–4 Group 2 elements, physical properties of, 164–5
equilibrium, 117–135 Group 2 elements, reactions of, 165–8
equilibrium, dynamic, 118 Group 17 elements, physical properties of, 172
equilibrium, position of, 119–23 Group 17 elements, reactions of, 173–5
equilibrium and ammonia production, 129 Group 17 elements, uses of, 178
Index
Haber process, 101, 129, 182, 183, 342–3 isotopes, 2–4, 28–9 489
haemoglobin, 204–5, 315 isotopes, and mass spectrometry, 3–4, 446–8
half-cells, 279–85, 290 isotopic mass, 2–3
half-equations, 108–9, 177, 274, 279–85, 369–70 ketones, 235
half-life, 335–6 ketones, reactions of, 237–8
halide ions, reactions of, 174–6, 290–1 ketones, preparation of, 210, 230, 236–7
halogenoalkanes, elimination reactions, 222 ketones, testing for, 238–40
halogenoalkanes, nucleophilic substitution kinetic theory of gases, 73–5
lattice energy, definition of, 258
reactions, 220–1 lattice energy, and Born–Haber cycles, 259–63
halogenoalkanes, uses of, 222–3 lattice enthalpy, see lattice energy
halogens, 172–8, 206–7, 209 lattice structures, 78–80
halogen compounds, uses of, 178, 222–3 Le Chatelier’s principle, 119–23
Hess’s law, 97–9, 259–62, 266 lead–acid cells, 293
heterogeneous catalysis, 144–5, 340, 342–3 ligands, 371–8
heterolytic fission, 196–7 liquid state of matter, 77–8
high-performance liquid chromatography (HPLC), 436–7, liquids, behaviour of, 77–8
lone pairs, and bonding, 51, 53, 64, 384, 397–8, 402,
448–9, 459 lone pairs, and ligands, 376–8
homogeneous catalysis, 145–6, 340–2 lone pairs, and molecular shapes, 56, 182, 388
homolytic fission, 196, 206, macromolecules, 424, 458
hydrocarbon fuels, 204 mass, 2
hydrocarbons, 19, 189–190, 192, 202, 213, 460 mass, molar, 5–7
hydrocarbons, saturated, 202 mass, percentage composition by, 8
hydrocarbons, unsaturated, 207 mass, relative atomic, 2–5
hydrofluorocarbons (HFCs), 223 mass, relative formula, 3
hydrogen bonding, 64–7, 226, 242, 387, 413–21 mass, relative isotopic, 2–3
hydrogen electrode, standard, 279 mass, relative molecular, 3, 76–7
hydrogen–oxygen fuel cells, 294–5 mass spectrometry, 3–4, 446–50
hydrolysis, 67, 198 medicinal drugs, design of, 457–61
hydrolysis, of acyl chlorides, 397 metallic bonding, 58–60
hydrolysis, of amides, 406, 425 metallic lattices, 79
hydrolysis, of esters, 229 metals, electrical conductivity of, 67
hydrolysis, of halogenoalkanes, 219 metals, properties of, 59–60, 66–7
hydrolysis, of nitriles, 231 metals, solubility of, 67
hydrolysis, of phosphorus(V) chloride, 159 molar gas volume, 18–19
hydrolysis, of proteins, 421 molar mass, 5–7
hydroxynitrile, 238 molecular formulae, of compounds, 9–10, 190
ice, structure of, 66 molecular ions, 446
ideal gas laws, limitations of, 75 molecular lattices, 80
ideal gases, 74–6 molecular modelling, 457
intermolecular forces, 49, 60–6, 74–8, 152, 405, 422 molecules, shapes of, 55–6
iodine–peroxodisulfate reaction, 341 moles, 5–6
ion polarisation, 263–4 moles, calculations of, 6–10
ionic bonding, 49–50, 60, 77, 158, 258 monodendate ligands, 372
ionic compounds, 13, 49–51, 66–7, 77–8, 295, 316 monomers, 211–13, 412–14, 421–8
ionic equations, balancing of, 13–14
ionic lattices, 50, 78–9, 258, 263–4
ionic product of water, 304
ionic radii, periodic patterns of, 151
ionisation energies, 34–6, 41–2, 153, 260, 369
Cambridge International A Level Chemistry
nanotechnology, 25 Period 3 oxides, 156–8
neutrons, 2, 26–9 Periodic Table, 10–11, 28, 149, 150, 473
nickel–cadmium cells, 293 Periodic Table, patterns in, 40–2, 149–60
nitrate fertilisers, 183–4 periodicity, 149–60
nitrogen, 181–4 periodicity, of chemical properties, 154–60
nitrogen oxides, 182–5, 205, 343 periodicity, of physical properties, 149–53
NMR (nuclear magnetic resonance) spectroscopy, 439–45 permanent dipole–dipole forces, 60, 63–4
non-degenerate orbitals, 376–8 pH, definition of, 305
non-polar alkanes, 204 pH calculations, 305–6
nuclear magnetic resonance (NMR) spectroscopy, 439–45 pH change, monitoring of, 309–13
nucleic acids, 418–9, 458 pH of buffer solution, 313
nucleon number, 2, 28 pH of strong acids, 306
nucleons, 25–6, 28 pH of strong bases, 306
nucleophile, definition of, 197 pH scale, 305
nucleophilic addition reactions, 237–8 pharmaceuticals, synthesis of, 457–61
nucleophilic substitution reactions, 218–22, 227 phenol, 387–8
nucleus, atomic, 25–6 phenol, reactions of, 388–9
nucleus, cell, 420 phenylamine, 401–4
nucleus, in NMR, 439–40 pi (π) bonds, 54, 57–8, 193–4
nylons, 413 pickling, 185
ppKolaavriasliunegs,p3o0w7e–r8o, f31c5at,i3o8n7s–, 8263
open system, 119
optical isomerism, 195–6 polarity and chemical reactivity, 61
490 optical resolution, 459 polarity in molecules, 60–1
orbitals, atomic, 37–8 pollution, 204–6, 314
order of reaction, 330–1, 333–7 poly(alkene) plastics, disposal of, 212
order of reaction, calculation of, 334–7 polyamides, 413–4, 422, 425–6
order of reaction, prediction of, 340 polyesters, 421, 425–6
organic compounds, techniques for identification of, polymer deductions, 426–8
241–2, 439–50 polymerisation, addition, 211–3, 426–7
organic compounds, naming of, 192–3 polymerisation, condensation, 412–4, 419, 424
organic molecules, bonding in, 193–4 polymers, 211–13, 412, 414–5, 422–4
organic molecules, representation of, 189–91 polymers, degradable, 425–6
organic molecules, shapes of, 57–8, 189–91 position isomerism, 194
organic reactions, mechanisms for, 196–7 primary alcohols, 226
organic reactions, types of, 198 primary alcohols, oxidation of, 230–1
Ostwald, Friedrich, 307 principal quantum shells, 33–4, 37
oxidation, 107–12, 184, 198, 205, 210–11, 230–1, 236–7, proteins, 182, 404–5, 407–8, 413–8
242–3, 275, 387, 395 proteins, hydrolysis of, 418
oxidation numbers, 110–111, 156, 158, 177, 340–1, 368 proteins, structure of, 415–8
oxidation states, 113, 368 proteins, synthesis of, 414–5
oxides, of Period 3 elements, 156–8 proton number, 28, 35
ozone layer, 206, 223 protons, 2, 25–9
paper chromatography, 320, 434–6 quantitative electrolysis, 276–8
partial pressure, 127–8 quantum shells, principal, 33–4, 37
partition coefficients, 319–20, 435 quantum sub-shells, 37–8
pepsin, 417 rate constant (K) 330
peptides, 405, 412–3 rate constant, calculations involving, 334–5
Period 3 chlorides, 158–60
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Cambridge International AS and A Level Chemistry Coursebook 2nd Edition
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Cambridge International AS and A Level Chemistry Coursebook 2nd Edition
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