Worked solutions

Chapter 11
Exercises

 1 The scale on the measuring cylinder is marked  7 D It is harder to read the true value from the
in 1 cm3 divisions, so we should be able to top of the meniscus so both the accuracy
estimate the value to ±0.5 cm3. and the precision will be affected.

 2 9.92 ± 0.05 g gives a range of 9.87–9.97 g.  8 A Wet burettes and pipettes would affect the
10.1 ± 0.2 g gives a range of 9.9–10.3 g. As the concentration of the solution delivered from
two ranges overlap, they could be measuring the them (making the solution more dilute);
same mass. however, a wet conical flask would not lead
to a systematic error as the total number of
 3 (a) 4 × 10–2 g
moles of reactants in the conical flask would
(b) 2.22 × 102 cm3 remain unchanged.
(c) 3.0 × 10–2 g. Note how the number of
 9 A Random errors only can cancel when
significant figures is retained, and compare
differences in quantities are calculated. This
with answer (a).
wouldn’t have any effect on systematic
(d) 3 × 10 °C or 3.0 × 10 °C as we do not know errors.
if the 0 is significant or not. (In practice,
‘× 10’ and ‘× 10–1’ are very rarely used). 10 D All three of the statements are true.

 4 (a) Four 11 C The volumes are not accurate as they fall

outside ±0.1 cm3 of 38.5 cm3. However, they
(b) Two or three. Is the zero significant? Or is it
are precise as they all fall within ±0.1 cm3 of
just there to show where the decimal point
each other.
is? This question shows the value of using
standard notation. 12 The average value = 49.0 s. The uncertainty in
(c) Three. The zeroes just show where the the measurements is given as ±0.1 s but the
significant figures are relative to the decimal results show that there is additional uncertainty,
point. suggesting that the value could be anywhere
(d) Four between 48.8 s and 49.2 s. The value should be
quoted as 49.0 s ± 0.2 s.
 5 A Random errors will eventually average out if
multiple measurements are made. However, 13 Note that the correct solution to this question is
systematic errors will always be present, no not one of the possible answers listed.
matter how many measurements are made. The thermometer has a smallest division of 0.05
°C. The uncertainty of an analogue scale is half
 6 A The absolute uncertainty =
the smallest division so the readings will have an
2%
× 57.536 kJ mol–1 = 1.15072. uncertainty of ±0.025 °C. (As the uncertainty has
100%
As the % uncertainty lies in the range of 2% 3 d.p. the thermometer can be read to 3 d.p.)
or greater, the absolute uncertainty should Initial temperature = −0.650 ± 0.025 °C
only have 1 s.f.; absolute uncertainty = 1 Final temperature = 1.400 ± 0.025 °C
(see page 538). Temperature change = final temperature − initial
The value must have the same number of temperature = 1.400 °C − (−0.650 °C) = 2.050
decimal places as the absolute uncertainty so °C = 2.050 K
the value can only be expressed as 59 kJ mol–1.

1
 Uncertainty = uncertainty in final temperature + Final answer is (1.00 ± 0.06) × 10–2 mol
uncertainty in initial temperature = 0.025 °C +
21 (a) ΔT = 43.2 − 21.2 °C = 22.0 °C
0.025 °C = 0.050 °C = 0.050 K
absolute uncertainty = (±)0.2 °C
The temperature change expressed to the
0.2
appropriate precision is 2.050 ± 0.050 K. (b) % uncertainty =  × 100% ≈ 1%
22.0
14 D Since the precision is ±0.5 °C −4.18 × 22.0
(c) ΔH = = −184 kJ mol−1
the percentage uncertainty is 0.500
0.5 + 0.5 (d) 1%
 × 100% 1
temperature difference
(e) absolute uncertainty =  × 184 =
1 100
i.e.  × 100% = 5% ±2 kJ mol−1
20
(f) experimental value for ΔH =
15 B ‘Heat exchange’ includes heat loss in
−184 ± 2 kJ mol−1
exothermic reactions and heat gain in
endothermic reactions. The literature value is outside this range.
The random errors involved in reading
16 B The overall number of significant figures the thermometer do not account for this
depends on the measurement with the difference.
smallest number of significant figures in the
There are systematic errors. The
calculation.
assumptions on which the calculation is
17 C Because this is a subtraction of two values based are not strictly valid. Some of the heat
with four decimal places the answer should of reaction passes into the surroundings and
also have four decimal places. 11.6235 g – the other uncertainties in the measurements
10.5805 g = 1.0430 g. This answer has five cannot be ignored. It should also be noted
significant figures. that the standard value for ∆H refers to
standard conditions of 298 K and 100 kPa.
18 A Mass lost = 0.266g – 0.186 g = 0.080 g
0.080 g 22 B If PV = nRT is rearranged to obtain a general
% mass lost = × 100% = 30% (as
0.266 g equation of a straight line (y = mx) then P =
0.080 g has 2 s.f. the answer can only have 1
2 s.f.) constant ×
V
19 C The answer has to be quoted to 2 significant 23 The scale of the graph does not allow for an
figures and the error associated is 0.5% + accurate determination of the two coordinates
0.5% = 1% = ±0.001. for two points on the line.
Estimates for two points on the line are (16, 50)
20 number of moles = volume × concentration
and (73, 200).
= 1.00 mol dm–3 × 10.0 × 10–3 dm3
Density of object = mass/volume, which
= 10.0 × 10–3 mol = 1.00 × 10–2 mol corresponds to the slope of the line.
% uncertainty in concentration is 0.05 mol Slope = (y2 − y1)(x2 − x1) = (200 − 50) g/(73 − 16)
dm–3/1.00 mol dm–3 × 100% = 5% cm3 = 150 g/57 cm3 = 2.6 g cm−3
% uncertainty in volume is 0.1 cm3/10.0 cm3 × The density of 2.6 g cm−3 that we obtain from
100% = 1% the slope indicates that the objects are made of
aluminium as it is the only metal listed that has a
The overall uncertainty in the number of moles =
similar density, 2.70 g cm−3.
5% + 1% = 6%
The fact that the best-fit line passes through all
6% of 1.00 × 10–2 mol is 0.06 × 10–2 mol of the points, and their associated error bars,

2
 indicates that there were no significant random peak at 88. The peaks at 29 (C2H5+) and
errors in the experiment. 73 (C2H5CO2+) suggest that the spectrum
The scale of the graph does not allow us to corresponds to C2H5CO2CH3.
determine a reliable uncertainty for the slope. 29 A (the spectrum on the left) corresponds to
Because of this we are unable to determine if CH3CH2CHO
the experimental value obtained agrees within
B (the spectrum on the right) corresponds to
uncertainty with the literature values and this
CH3COCH3
prevents us from concluding if significant
systematic errors were present. Similarities

Answers A and B are therefore both acceptable Both have a molecular ion corresponding to 58.
without the extra detail required to determine a Differences
reliable uncertainty for the density obtained from A has peaks corresponding to 29 (CH3CH2+), 28
the slope of the line. (loss of CH3CH2) and 57 (CH3CH2CO+).
24 B The results are inaccurate as distilled B has a peak corresponding to 43 (loss of CH3).
water (at 25 °C) should have a pH of 7.00;
however, they are precise as they are all 30 (a) Mass / charge Fragment

within 0.02 of a pH unit of each other. 15 CH3+

25 Concentration of chromium = 3.34 μg dm–3. 29 C2H5+

Plot a graph of absorbance against chromium 43 C3H7+
concentration using the values given. Then use
58 C4H10+
the graph to find the chromium concentration of
the sample. (b) CH3CH2CH2CH3 – this can be deduced from
0.4 the peaks or from the molecular ion peak
0.35 corresponding to the molecular mass of
0.3 58. The empirical formula is C2H5 and the
0.25 molecular formula is C4H10.
Absorbance

0.215
0.2
31 Corresponding saturated
0.15 Molecule IHD
non-cyclic molecule
0.1
C6H6 C6H14 4
0.05

0
3.34
CH3COCH3 C3H8O 2
0.00 1.00 2.00 3.00 4.00 5.00 6.00
Chromium concentration / μg dm–3 C7H6O2 C7H16O 5

26 B Taking logs of both sides of the equation ln k C2H3Cl C2H5Cl 1

E C4H9N C4H9NH2 1
= ln A – a
RT 1 C6H12O6 C6H14O6 1
A graph of ln k versus will give a straight
T
line.
32 B With single bonds between large atoms, the
27 C C2H3O will produce a peak at 43, C3H5O at
+ + molecules will vibrate the lowest frequency
57 and C4H8O+ at 72. Ion peaks are present and so will have the longest wavelength
at 57 and 72. absorption.

28 C The formula masses of the compounds 33 D The absorptions given in the question
are A = 44, B = 88, C = 88, D = 88. A correspond to carbon–carbon triple bonds
cannot be correct as the spectrum has a (2100 cm–1), carbonyl groups (1700 cm–1)

3
 and carbon–oxygen bonds (1200 cm–1) (see 40 C–H bond (see table on page 557).
table on page 557).
41 The structure is CH3OCH3 (methoxymethane /
34 B Both aldehydes and esters contain carbonyl dimethyl ether). The IHD is 0, so there are no
groups, which absorb at 1720 cm–1 (see double bonds (neither C=O nor C=C). The
table on page 557). absorption at 1000–1300 cm–1 is due to a C–O
single bond so the molecule must be an ether or
35 (a) C H O an alcohol. It cannot be an alcohol as there are
% composition 40.0 6.7 53.3 no absorbances above 3000 cm–1.
Moles 40.0 6.7 53.3 42 C The peak area corresponds to the relative
12.01 1.01 16.00
numbers of hydrogen atoms in that
Ratio 3.33 6.6 3.33 particular environment. The unknown
Simplest ratio 1 2 1 compound has two different CH3 groups and
a CH2 group. The only molecule that satisfies
Empirical formula CH2O. This has a relative this is molecule C. A would have two peaks,
formula mass of 30. The relative molecular in a 2 : 6 ratio, B would have three peaks,
mass is 60 so the molecular formula is in a 3 : 2 : 1 ratio (and the CHO group would
C2H4O2. have chemical shift of 9.4–10), while D
(b) IHD = 1 would have three peaks, but in a 2 : 4 : 6
(c) The absorptions correspond to a ratio.
carbonyl group and hydrogen-bonded
43 A This is the only molecule with two different
hydroxyl group. Molecular structure is
hydrogen environments. Molecules B and D
O
have one environment and molecule C has
CH3 C OH, ethanoic acid. three different environments.

36 44 (a) 2, the two CH3 groups are in different

environments
(b) 1, the molecule is symmetrical so the CH3
symmetric stretch asymmetric stretch symmetric bend
IR active IR active IR active groups are in the same environment
(c) 1, the molecule is symmetrical so the CH3
37 B C–O and C=O bonds absorb strongly groups are in the same environment
whereas C=C only has a medium-weak
(d) 2, the two CH3 groups are in a different
absorption (see table on page 557).
environment to the H atom bonded to the
38 IR radiation can only be absorbed if it results central carbon atom
in a change in the dipole moment of a bond
45 The H atoms are in three different environments
or molecule therefore whenever it is absorbed
so there are three peaks in the 1H NMR
by a bond or molecule the polarity (of bond or
spectrum, in the ratio of peak area of 3 : 2 : 1
molecule) changes due to the change in dipole
corresponding to the CH3, CH2, and H atom.
moment as the bonds are bent or stretched.
Chemical shifts will be around 1.0 for the –CH3,
39 Hex-1-ene shows an absorption in the range around 3.5 for –CH2–O, and between 1.0 and
1610–1680 cm−1 due to the presence of the 6.0 for the O–H (see Section 27 of the IB data
C=C bond. Cyclohexane won’t show an booklet and the table on pages 561–562).
absorption in this range as it only has C–C
bonds.

4
46 (a) CH3COCH2CH3 49 X-ray crystallography
H O H H 50 Monochromatic means all the X-rays have
the same wavelength. The angle of diffraction
H C C C C H
depends on the wavelength. If the X-rays
H H H have different wavelengths, different diffraction
angles/pattern would be obtained. It would
(b) be impossible to match the angles with the

No. of H
Type of Chemical

atoms
Splitting wavelengths.
hydrogen shift /
pattern
atom ppm
51 Hydrogen atoms have a low electron density.
CH3CO 2.2–2.7 3 1
52 The atoms must have a regular arrangement if
COCH2CH3 2.2–2.7 2 4, in 1 : 3 :3 : 1 an ordered diffraction pattern is to be produced.
pattern
Crystals are used if possible.
CH2CH3 0.9–1.0 3 3, in 1 : 2 : 1
pattern 53 (a) C6H5CH3 H3C

47
of peaks

No. of H

Chemical
Number

atoms

Splitting
Compound shift / (b) Hydrogen atoms have a low electron density
pattern
ppm
and so do not appear.
CH3CHO 2 2.2–2.7 3 2, in 1 : 1 (c) The saturated non-cyclic compound is
pattern C7H16. IHD = 12(16 − 8) = 4 (the IHD of a
9.4–10.0 1 4, in benzene ring = 4 due to the presence of
1 : 3 :3 : 1 three double bonds and one ring)
pattern

CH3COCH3 1 2.1 6 1

48 IHD is 2, so either one double bond (either Practice questions

C=C or C=O) or a ring structure (unlikely as
the largest ring would be a four-membered  1 If the burette reading is recorded as 27.70 ±
ring, which would be very strained). Possible 0.05 cm3 the actual value could be any value
structures (esters and carboxylic acid) are between 27.65 cm3 and 27.75 cm3.
CH3CH2COOH, CH3COOCH3, HCOOCH2CH3. Correct answer is B.
Could also be H2C=CHCH2OH.
 2 The mass is 10.044 g, which is precise to five
●● The peak at 8.0 ppm corresponds to R–
significant figures.
COOH. There is no splitting as there are no
The volume is 3.70 cm3, which is precise to
hydrogen atoms bonded to neighbouring
three significant figures.
carbon atoms.
Because the final value is calculated through a
●● The peak at 1.3 ppm corresponds to a –CH3
division the answer can only be presented to the
group. The peak is split into a triplet because
same precision as the least precise piece of data.
there is a neighbouring –CH2– group.
mass 10.044 g
●● The peak at 4.3 ppm corresponds to the density = = = 2.71 g cm–3
volume 3.70 cm3
R–CH2–COO group. The peak is split into
Correct answer is C.
a quartet as there is a neighbouring –CH3
group.  3 The procedures in A, B and D are all conducted
Molecular structure: CH3CH2COOH with specialized and highly precise pieces of

5
 equipment: burette, pipette and gas syringe. Comparing X to compound C: The
The piece of equipment used in procedure C is infrared spectrum of compound
a lab beaker, which is not precise for measuring C, CH2=CH–CH2OH, would have
volume or a suitable reaction vessel for absorptions at 1620–1680 cm–1 due
measuring temperature changes that occur with to the C=C bond and 3200–3600 cm–1
neutralization reactions. due to the O–H bond. These two peaks
Correct answer is C. would be absent in the spectrum of X.
The mass spectrum of compound X will
(d)
 4 Random uncertainties can only be decreased
show peaks at m/z values of:
through the conduction of repeat measurements.
* 58 due to the molecular ion CH3CH2CHO+
The methods suggested in B, C and D are all
associated with the reduction of systematic errors. * 29, which could be due to the fragment
CHO+ or the fragment C2H5+
Correct answer is A.
* 15 due to the fragment CH3+.
 5 (a) The low-resolution H NMR spectrum shows
1

three peaks so the compound contains three  6 (a) To be IR active the absorption of energy
different chemical environments. by the vibration must involve a change in
the bond dipole. As H–Br is a polar bond
A CH3–CO–CH3 has one chemical
the absorption of IR radiation by the bond
environment (the CH3 hydrogens are all
stretch will result in a longer bond and a
equivalent).
change in the bond dipole, so it is IR active.
B CH3–CH2–CHO has three chemical As Br–Br is a non-polar bond there will be
environments. (These environments will no dipole change by the bond stretch if it
give an integration of 3:2:1, consistent absorbs IR radiation, so it is IR inactive.
with the spectrum observed. The
(b) (i) I is a strong, very broad absorption at
spectrum also gives a peak at 9.8 ppm
3000 cm–1. The bond responsible is
due to one hydrogen that is consistent
O–H (2500–3000 cm–1 for acids).
with an aldehyde.)
II is a strong absorption at 2750 cm–1.
C CH2=CH–CH2OH has four chemical
The bond responsible is C–H (2850–
environments.
3090 cm–1). Although this absorption
X is compound B, CH3–CH2–CHO. is outside the range in the data
The peak at 2.5 ppm is due to hydrogen
(b) booklet this bond is most likely to be
atoms that are on a carbon adjacent to a responsible.
carbonyl group (–C=O). This would also III is a strong absorption at 1700 cm–1.
occur for compound A, where the CH3 The bond responsible is C=O (1700–
hydrogens are adjacent to a C=O group. 1750 cm–1).
(c) (i) The infrared spectrum of compound The peak at m/z = 102 is due to the
(ii)
X, CH3–CH2–CHO, would have an molecular ion C5H10O2+.
absorption at 1700–1750 cm–1 due to
The peak at m/z = 57 is due to fragment
the C=O bond.
C4H9+.
Comparing X to compound A: The
(ii)
The peak at m/z = 45 is due to fragment
infrared spectrum of compound A, CH3–
COOH+.
CO–CH3, would have an absorption
at 1700–1750 cm–1 due to the C=O The peak in the 1H NMR spectrum at
(iii)
bond. This peak would be absent in the 11.5 ppm is due to the hydrogen in the
spectrum of X. acid group, –COOH. (Acid hydrogens
occur in the region 89.0–13.0 ppm.)
6
 The peak at 1.2 ppm has an integrated
(iv) The peak at m/z = 29 is due to fragment
area of 9. This tells us that nine C2H5+.
hydrogen atoms share this chemical The peak in the 1H NMR spectrum at 11.73
(c)
environment, which is consistent with a ppm is due to the hydrogen in the acid
–C(CH3)3 group. group, –COOH. (Acid hydrogens occur in
From (ii) and (iii) we have deduced that X
(v) the region 89.0–13.0 ppm.)
contains a –C(CH3)3 group and an acid IR peaks at 2900 cm–1, 1700 cm–1 and 1200
(d)
group, –COOH. The structure must be: cm–1 along with a mass spectrometer peak
CH3 O at m/z = 45 and an 1H NMR peak at 11.73
ppm are all consistent with an acid group
H3C C C OH being present.
A mass spectrometer peak at m/z = 29
CH3
along with 1H NMR peaks at 1.2 ppm
CH3COOCH2CH2CH3 has four chemical
(vi) (3H triplet) and 2.4 ppm (2H quartet) is
environments so its 1H NMR spectrum consistent with CH3–CH2– being present.
will have four peaks with integration Structure is:
traces 3:2:2:3. CH3COOCH2CH2CH3
is an ester so it will not have a peak in H H O
the 9.0–13.0 ppm region due to an acid
hydrogen. H C C C OH

 7 The absorption of IR radiation by a molecule H H

must result in a change in the molecular dipole.
For CO2 this can only happen with absorption  9 (a) The structure given contains an alcohol
of the asymmetric stretch and the bending group. The spectrum is not of this structure
vibrations. The symmetric stretch is IR inactive as a strong, very broad peak would be
as it results in no dipole change. observed in the 3200–3600 cm–1 region if an
O–H bond was present.
The absorption of IR energy due to the
asymmetric stretch of CO2 will result in a change The structure given contains an alkene. The
in the C=O bond lengths. spectrum is not of this structure as a strong
peak would be observed in the 1620–1680
The absorption of IR energy due to the bends of
cm–1 region if a C=C bond was present.
CO2 will result in a change in the O=C=O bond
angles. The spectrum has a strong peak at 1750
cm–1, which indicates that a C=O bond is
 8 (a) A is a strong, very broad absorption at 2900 present in the compound. As the structure
cm–1. The bond responsible is O–H (2500– provided does not contain a C=O bond it is
3000 cm–1 for acids). not consistent with the IR spectrum.
B is a strong absorption at 1700 cm–1. The The molecular formula of C3H6O indicates
(b)
bond responsible is C=O (1700–1750 cm–1). that the compound contains one double
C is a strong absorption at 1200 cm–1. The bond. As the C=C peak is not observed in
bond responsible is C–O (1050–1410 cm–1). the IR spectrum the compound must be a
The peak at m/z = 74 is due to the molecular
(b) ketone or an aldehyde. Possible structures
ion C3H6O2+. are therefore:
The peak at m/z = 45 is due to fragment
CO2H+.

7
 CH3 C CH2 CH3

H
butan-2-ol

H O H OH

H C C C H CH3 C CH3

H H CH3

I 2-methylpropan-1-ol

H H O (a) (i) Spectrum 1 has two peaks in the

1
H NMR spectrum so the alcohol
H C C C H responsible must contain two chemical
environments. From the structures
H H we can see that the only alcohol
II with two chemical environments is
Structure I is not the correct structure as it 2-methylpropan-2-ol.
would give an 1H NMR spectrum with only (Butan-1-ol and butan-2-ol each
one peak as all the CH3 hydrogens have the have five chemical environments and
same chemical environment. 2-methylpropan-1-ol has four chemical
Structure II is the correct structure as this environments.)
would give an 1H NMR spectrum with The peak at 1.3 ppm, with an
three peaks as there are three chemical integration trace of nine units, is due
environments: CH3–, –CH2– and –C(O)–H. to the nine hydrogens on the three
The integrated areas of these peaks would chemically equivalent –CH3 groups.
be 3:2:1. The peak at 2.0 ppm, with an
integration trace of one unit, is due to
10 The structures of the four alcohols are given
the alcohol hydrogen, –OH.
below.
OH Spectrum 2 has four peaks in the
(ii)
1
H NMR spectrum so the alcohol
H C CH2 CH2 CH3 responsible must contain four chemical
environments. From the structures
H we can see that the only alcohol
butan-1-ol with four chemical environments is
2-methylpropan-1-ol.

OH H The peak at 0.9 ppm, with an

integration trace of six units, is due to
H C C CH3 the six hydrogens on the two chemically
equivalent –CH3 groups.
H CH3 The peak at 3.4 ppm, with an
2-methylpropan-1-ol integration trace of two units, is due
to the two hydrogens on the carbon
attached to the alcohol, –CH2OH.
OH
(b) (i) The peak at m/z = 74 is due to the
CH3 C CH2 CH3 molecular ion C4H10O+.
The peak at m/z = 59 is due to fragment
H
C3H7O+ (loss of CH3).
butan-2-ol

8
OH
 The peak at m/z = 45 is due to fragment reasonably distant from the ester group, R–O–
C2H5O+ (loss of CH2CH3). CO–CH2–CH3. (From the data booklet, –CH3
From the structures we can see that the occurs at 0.9–1.0 ppm.)
loss of both CH3 and CH2CH3 to give The structure is:
C3H7O+ and C2H5O+ can occur for two H O H H
of the alcohols, butan-1-ol and butan-
2-ol. H C O C C C H
The peak at m/z = 31 is due to fragment
(ii)
H H H
CH3O+ (loss of CH2CH2CH3).
From the structures we can see that 12 (a) (i) The peak with highest m/z occurs at 88.
loss of CH2CH2CH3 to give CH3O+ This will be due to the molecular ion so
(CH2OH+) can only occur for butan-1-ol. the relative molecular mass of X is 88.
The IR spectra will be similar for the four
(c) The empirical formula given is C2H4O,
alcohols as they contain the same bonds: which has a relative formula mass of 44.
C–H, C–C, C–O and O–H. As m/z is double the formula mass
11 With the molecular formula C4H8O2 the the molecular formula is 2 × C2H4O =
compound could be an acid or an ester. There C4H8O2.
is no 1H NMR peak in the acid region, 9.0–13.0 The molecular ion is therefore C4H8O2+.
ppm, so the compound is an ester. The peak at m/z = 29 is due to
(ii)
The peak at 3.7 ppm is a singlet with an fragments C2H5+ or CHO+.
integrated area of three units, which indicates a A peak at m/z = 59 would be due to
(iii)
methyl group, –CH3. Because it is a singlet the the fragment C3H7O+ that would result
n + 1 rule tells us there are no hydrogens on from the loss of CH2CH3. As CH2CH3+ is
the neighbouring atoms. Because it is at a high observed at m/z = 29 this indicates that
chemical shift this indicates it is adjacent to the the C3H7O fragment formed with loss
oxygen atom of the ester group, CH3–O–CO–. of CH2CH3 is not charged. (If it is not
(From the data booklet, R–CO–O–CH2– occurs charged it cannot be deflected by the
at 3.7–4.8 ppm.) magnetic field and be observed at the
The peak at 2.3 ppm is a quartet with an detector.)
integrated area of two units, which indicates a (b) (i) A is a strong absorption at 1750 cm–1.
methylene group, –CH2–. Because it is a quartet The bond responsible is C=O (1700–
the n + 1 rule tells us there are three hydrogens 1750 cm–1).
on a neighbouring carbon(s), –CH2–CH3. B is a strong absorption at 1250 cm–1.
Because it is at a moderately high chemical shift The bond responsible is C–O (1050–
this indicates it is adjacent to the carbon atom of 1410 cm–1).
the ester group, R–O–CO–CH2–CH3. (From the
The IR spectrum shows that C=O
(ii)
data booklet, RO–C(O)–CH2– occurs at 2.0–2.5
and C–O are both present, which
ppm.)
indicates that X contains an acid or
The peak at 1.0 ppm is a triplet with an an ester functional group. The strong,
integrated area of three units, which indicates a broad O–H absorption observed for
methyl group, –CH3. Because it is a triplet the acid O–H bonds at 2500–3000 cm–1
n + 1 rule tells us there are two hydrogens on is absent from the IR spectrum so we
a neighbouring carbon(s), –CH2–CH3. Because can conclude that X contains the ester
it is at a low chemical shift this indicates it is functional group.
9
 (c) (i) A possible structure is: table will be due to the hydrogens on
H O H H the methyl group at the end of the alkyl
chain, –CH2–CH3.
H C C O C C H The n + 1 rule tells us this peak will be
a triplet and it will have a relative peak
H H H
area of 3 as three hydrogens share this
Reasoning: chemical environment. From the data
From the IR spectrum it was determined booklet we see that –CH3 occurs at
that the compound was an ester. 0.9–1.0 ppm.
From the 1H NMR data the peak at The peak at 4.1 ppm is due to the
(iii)
2.0 ppm is a singlet with an integrated methylene hydrogens, CH3–CO–O–
area of three units, which indicates a CH2CH3. This peak is a quartet as
methyl group, –CH3. Because it is a there are three hydrogens on the
singlet the n + 1 rule tells us there are no neighbouring atoms.
hydrogens on the neighbouring atoms.
Because it is at a moderately high
chemical shift this indicates it is adjacent Challenge yourself
to the carbon atom of the ester group,
CH3–CO–O–. (From the data booklet, 1 Y1ave = Y2ave = 3
RO–CO–CH2– occurs at 2.0–2.5 ppm.)
(–2×(–2)) + (–1×(–1)) + 0 + (1×1) + (2×2)
The peak at 4.1 ppm is a quartet with R= =1
(–2)2 + (–1)2 + 02 + 12 + 22
an integrated area of two units, which
indicates a methylene group, –CH2–.
2 Y1ave = Y2ave = 3
Because it is a quartet the n + 1 rule
tells us there are three hydrogens on (–2×2) + (–1×1) + 0 + (1×(–1)) + (2×(–2))
R= = –1
the neighbouring atoms. Because it is (–2)2 + (–1)2 + 02 + 12 + 22
at a high chemical shift this indicates it
is adjacent to the oxygen atom of the 3 Y1ave = Y2ave = 3
ester group, –CH2–O–CO–. (From the (–2×(–2)) + (–1×2) + 0 + (1×1) + (2×(–1))
R= = 0.10
data booklet, R–CO–O–CH2– occurs at 22 + 12 + 02 + 12 + 22
3.7–4.8 ppm.)
Because the peak at 4.1 ppm is a 4 The index of hydrogen deficiency (IHD) can
quartet and has three hydrogens on the be determined for a molecule with the formula
neighbouring atoms we can deduce CnHpOqNrXs by seeing how it changes from
that the –CH2– is adjacent to a methyl that of an alkane, CnH2n+2, where IHD = 0, by
group, –CH3. successively adding functional groups containing
each element of interest and seeing how it
(ii) Peak Chemical Relative Splitting affects the IHD.
shift/ppm peak pattern
area For CnHp:
first 2.0 3 singlet H atoms needed = number required for
saturated hydrocarbon – number present
second 4.1 2 quartet
= 2n + 2 – p
third 0.9–1.0 3 triplet
H2 molecules needed = ½(2n + 2 – p)
From the structure determined in (ii) we
can see that the peak absent from the
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For CnHpOq: Consider chloroethane, CH3CH2Cl, which is
O is typically present with two single bonds saturated:
(ether or alcohol) or a double bond (ketone, If we use 2n + 2 – p + r then:
aldehydes, esters) H atoms needed = 2n + 2 – p + r
Consider ethanol, CH3CH2OH, which is = 2(2) + 2 – 5 = 1
saturated (IHD = 0): This answer is incorrect so the equation
H atoms needed =
 2n + 2 – p = 2(2) + 2 – 6 needs to be modified to 2n + 2 – p – s + r.
=0 H atoms needed = 2n + 2 – p – s + r
H2 molecules needed = 0 = 2(2) + 2 – 5 – 1 = 0
Consider propanal, CH3CH2CHO, which is H2 molecules needed = ½(2n + 2 – p – s
unsaturated due to the C=O bond (IHD = 1): + r) = 0
H atoms needed =
 2n + 2 – p = 2(3) + 2 – 6 Therefore we find that the IHD for a general
=2 formula CnHpOqNrXs can be calculated using:
H2 molecules needed = ½(2n + 2 – p) = 1 IHD = ½(2n + 2 – p – s + r)
In both cases, CH2CH2OH and CH3CH2CHO, we 5 As we are given the frequency, ν, of the photon
find that the correct IHD is obtained even though of light we can calculate the energy of a single
the equation does not include the number of photon using the equation E = hν:
oxygen atoms, q.
E = hν
(This answers the second part of the question.)
= 6.63 × 10−34 J s × 3.0 × 1014 s−1
For CnHpOqNr: = 2.0 × 10−19 J
N is typically present with three single bonds We can calculate the energy in kJ mol–1 by
(amine) or a triple bond (nitrile). multiplying the energy of one photon by
Consider ethylamine, CH3CH2NH2, which is Avogadro’s number:
saturated (IHD = 0): E = 6.02 × 1023 mol−1 × 2.0 × 10−19 J
If we use 2n + 2 – p then: = 1.2 × 105 J mol−1= 120 kJ mol−1
H atoms needed = 2n + 2 – p = 2(2) 6 The IR absorption is given as 2100 cm–1, which
+ 2 – 7 = –1 is in wavenumber units.
This answer is incorrect so the equation needs Wavenumber = 1/λ, where λ is the
to be modified to 2n + 2 – p + r. wavelength of the light absorbed.
H atoms needed = 2n + 2 – p + r 1 1
λ = =
= 2(2) + 2 – 6 + 1 = 0 wavenumber 2100 cm–1
H2 molecules needed = ½(2n + 2 – p + r) = 0 = 4.762 × 10–4 cm = 4.762 × 10–6 m
Consider propanenitrile, CH3CH2CN, which is Frequency is related to wavelength by the
unsaturated due to the C≡N bond (IHD = 2): equation ν = c/λ
H atoms needed =
 2n + 2 – p + r 3.00 × 108 m s–1
ν= = 6.30 × 1013 s–1
= 2(3) + 2 – 5 + 1 = 4 4.762 × 10–6 m
H2 molecules needed = ½(2n + 2 – p + r) = 2 7 For constructive interference to occur light
For CnHpOqNrXs: waves must be travelling parallel to each other
X represents a halogen atom, which is typically and be in phase with each other.
present with one single bond. In the diagram below the blue lines represent
two light waves travelling through a solid sample.
The extra distance travelled by the lower light
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 wave, the path difference, is represented by The distance a can be calculated based on d,
the green lines. The distance between the recognizing that they form two sides of a right-
layers of atoms, d, is represented by the red angled triangle, therefore:
line. If the lower light wave is to exit the sample a
sinθ =
in phase with the upper light wave then the d
path difference (2a) must be a multiple of the a = dsinθ
wavelength of the light involved: The extra distance travelled by the lower wave is
2a = nλ (i) 2a, therefore:
2a = 2dsinθ  (ii)
θ Combining (i) and (ii) by substituting for 2a gives
the Bragg equation:
nλ = 2dsinθ
θd

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