ANSWERS TO EXAM-STYLE QUESTIONS A 1CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011

CH

H

H

C

O

H

propan-2-ol

C

H

H

H

HCH

H

H

C

H

H

propan-1-ol

C

H

H

O

H

Answers to exam-style questionsb The three peaks in the NMR spectrum of

propan-2-ol arise because there are three di� erent environments for the protons (H atoms) in propanone.

c The integration trace gives us the relative number of protons (H atoms) in each environment. This is obtained by � nding the ratio between the heights of the vertical steps in the integration trace.

d i

Option A1 a atomic absorption spectroscopy (AAS)

b chromatography c mass spectrometry would be easiest, as they have

di� erent relative molecular masses

2 a CO and HCl are polar molecules, and there will therefore be a change in dipole moment as they vibrate. As there is a change in dipole moment, they will be able to absorb IR radiation.

b SO2 is bent but CO2 is linear. This means that SO2 is a polar molecule, but CO2 is non-polar, as the dipoles cancel. Both molecules can have the following vibrational modes: symmetric stretch; asymmetric stretch; bend. In SO2 all three vibrational modes involve a change in dipole moment and are IR active. In CO2 the symmetric stretch does not involve a change in dipole moment and does not absorb IR radiation.

bond lengths decrease bond lengths increase

dipoles cancel

dipole moment remains zero

δ+δ – δ –O C O

δ+δ – δ –O C O

δ+δ – δ –O C O

CH

H

H

C

O

H

C

H

H

H

H

CH

H

H

C

O

H

C

H

H

H

C

H

H

H

c There is an absorption band in the range 1700–1750 cm−1, which suggests the presence of a C=O group; therefore molecule I can be eliminated (also no O–H absorption). There is no very broad band in the range 2400–3400 cm−1, which is characteristic of carboxylic acids; therefore molecule III can be eliminated. There is no absorption band in the range 1000–1300 cm−1, indicating no C–O bond present; thus IV (and III and I) can be eliminated. Therefore this must be the spectrum of molecule II.

d Propanal and propanone both contain the same bonds and would have very similar absorption bands in the region around 1500 cm−1. Propanal, however, has three peaks in the NMR spectrum but propanone has only one, so they may be easily distinguished using NMR.

3 a

ii The molecule is non-symmetrical and there are therefore � ve di� erent environments for the protons and thus � ve peaks in the NMR spectrum. The ratios of the areas under the peaks is the same as the ratios of the number of H atoms in each environment, which is 3 : 1 : 1 : 2 : 3.

4 a The relative molecular mass is obtained from the highest-mass peak in the mass spectrum, i.e. 60 in this case.

b Z contains C, H and O, and the only two possible combinations of these that adds up to a relative molecular mass of 60 are C3H8O and C2H4O2.

c 31 could be CH3O+ or CH2OH+; 29 could be C2H5

+ or CHO+.d The absorption band at about 3350 cm−1 suggests

the presence of an O–H bond. No absorptions in the range 1600–1800 cm−1 suggests that there is no C=C or C=O present.

e Z cannot have the formula C2H4O2, as this formula requires the presence of a double bond (assuming it is not an epoxide). Z therefore has the formula C3H8O and possible structural formulas containing an OH group are:

f There are four di� erent chemical environments for H in propan-1-ol but only three in propan-2-ol, as it is symmetrical. There are four peaks in the NMR spectrum, indicating four di� erent chemical environments for H. We can, therefore, deduce that Z is propan-1-ol.

Note: the full structural formula must show the O–H bond.

2 CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011ANSWERS TO EXAM-STYLE QUESTIONS A

7 a [V(H2O)6]3+ will have greater splitting of the d orbitals. The ligands are pulled in more strongly by the higher charge on the central transition metal ion. There is, therefore, greater repulsion between the ligand electrons and the transition metal electrons, and therefore greater splitting of the d orbitals.

b NH3 causes greater splitting of the d orbitals than H2O (it is a stronger � eld ligand), so there is greater splitting of the d orbitals in [Ni(NH3)6]2+.

8 a Energy is absorbed to promote an electron from a lower-energy molecular orbital to a higher-energy molecular orbital.

b For a molecule to absorb ultraviolet light it must contain a double bond (or a benzene ring). I and III contain double bonds: C=O in A and C=C in C. Therefore A and C will absorb ultraviolet radiation. II does not contain any double bonds and will not absorb ultraviolet radiation.

c i

We could also use the integration trace and, by measuring the vertical heights of the steps, we get a ratio of the areas under the peaks of 1 : 2 : 2 : 3, which corresponds to the numbers of H atoms in each environment in propan-1-ol.

5 a A series of solutions of known concentration is made up. The absorbance of each solution is measured in an atomic absorption spectrophotometer. Absorbance is plotted against concentration to obtain the calibration curve.

b The absorbance is 0.320. A horizontal line is drawn across from 0.32 to the line of best � t and the concentration of nickel is read as 0.64 μg dm−3. The total volume of the solution of the sample that was made up was 100.0 cm3, i.e. 0.1000 dm3. If the concentration of Ni in this solution is 0.64 μg dm−3, then in 0.1000 dm3 there must be 0.1 × 0.64, i.e. 6.4 μg.

The mass of shell� sh that was originally used was 0.200 g, so we can conclude that this mass of shell� sh contains 6.4 μg of nickel. In 1.0 g of shell� sh there is therefore: 1

0.200 × 6.4 μg, i.e. 32 μg, of nickel.

6 a TLC stands for thin-layer chromatography.b The stationary phase is the solid particles of silica/

silica gel/alumina coating the plate. The mobile phase is the solvent.

c Amino acids are colourless and they are sprayed with ninhydrin (a locating agent) so that they can be seen – they show up as pink/purple spots.

d distance solute moves

Rf = distance solvent front moves

The distances are shown on the diagram:

solvent front

Z

10.7 cm

8.1 cm

4.8 cm

3.1 cm

Y

X

C

H

H

C

H

H

C C

Hpenta-1,3-dieneH

HC

H

C

H

H

C

H

C

H

H

C

penta-1,4-dieneH

HC

H

ii Penta-1,3-diene will absorb the longer wavelength of UV radiation as the double bonds are conjugated. The double bonds in penta-1,4-diene are separated by two single bonds, so they are not conjugated and delocalisation is not possible. Penta-1,3-diene has a longer conjugated system, and so a smaller energy gap between lower and upper molecular orbitals. This means that lower energy radiation (longer wavelength) must be absorbed to promote an electron to a higher molecular orbital in penta-1,3-diene.

d X is coloured as it has a longer conjugated/delocalised system. The conjugated system in X extends over the whole molecule, but in Y there are three single bonds in a row in the middle of the molecule and the conjugated system is thus broken. The conjugated system in X is long enough that it absorbs visible light and thus appears coloured. Y absorbs only UV radiation and is therefore colourless. Thus, for X we can calculate:

3.1 = 0.29 10.7

The other Rf values are: Y = 0.45; Z = 0.76.e Comparing the Rf values with those in the table

we get: X = glycine; Y = arginine; Z = tyrosine.

ANSWERS TO EXAM-STYLE QUESTIONS A 3CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011

With ten protons there must, therefore, be two ethyl groups and the isomer is structure I above.

c The isomers are:

9 a The signal due to the proton in bold is not split by the protons (H atoms) on the same C as they are chemically equivalent. There is one H on the adjacent C atom, so the signal due to the H in bold is split into a 1 : 1 doublet (if we add 1 to the number of H on the adjacent C we get 2, i.e. a doublet).

b This problem can be approached in two ways. First, draw out the isomers of C4H10O:

I

C C

O

C C C

O

C C C

C

O C C

C

C

C

O C

C

CO

C C

C

H

H

H

H

H

H

HH

H

HH H

H

H

H

H

H H

H

H

H

HH H

H

C C O C C

II

III IV

V

VII

VI

C

H

H

H

H

H

H

H

H

C

H

H

H

H

H

H

C

H

H

H

H

H

H

H

H

H

H

H

H

O C

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

There are two sets of peaks in the NMR spectrum and only two of these isomers (I and III) have two chemical environments for H. These may be distinguished by using the integration trace. The ratio between the heights of the steps is 1 : 1.5, i.e. 2 : 3, so the isomer must be I. This has protons in two di� erent chemical environments and the number of protons in each environment is in the ratio 4 : 6 i.e. 2 : 3.

Second, consider the splitting pattern and the integration trace. The ratio between the heights of the steps in the integration trace is 1 : 1.5, i.e. 2 : 3. There are 10H in this molecule, so, if they are divided in the ratio 2 : 3, there are four protons in one environment and six in the other. The splitting pattern is a quartet and a triplet. The quartet indicates 3 Hs on the adjacent C and the triplet indicates 2 Hs on the adjacent C. These two sets of peaks together are characteristic of an ethyl group.

C

H

H

I O

HCC

H

H

H O

C

H

H

II O

CC

H

H

H O H

Isomer I has two di� erent chemical environments for H, and therefore there will be two sets of peaks in the NMR spectrum. The ratio between the areas under these peaks will be 1 : 1. There are no H atoms on adjacent C atoms, so the spectrum will consist of two singlets.

Isomer II has three di� erent chemical environments for H, and therefore there will be three sets of peaks in the NMR spectrum. The ratio between the areas under these peaks will be 3 : 2 : 1. There are 3 Hs with 2 Hs next to them, so the peak with area 3 will be a triplet and the peak with area 2 will be a quartet. The third peak will be a singlet. The quartet will have a chemical shift di� erent to any of the protons in I as the protons are in a type of environment not found in I.

d The isomers are:

C

H

H

II

IV

O

C C

H

H

HC

H

H

H OC

H

H

I O

CC

H

H

H O C

H

H

H

C

H

H

III O

CC

H

H

C

H

H

H O H O

CO H

C

H C

C

H

H

H

H H

H

The NMR spectrum shows three sets of peaks and therefore three di� erent environments for H. This means that isomer III can be eliminated, as it has four di� erent environments. The ratios of the heights of the steps in the integration trace are 3 : 2 : 3. There is a total of eight H atoms in the molecule, so 3 : 2 : 3 indicates the number of H atoms in each environment. Thus isomer IV can be eliminated, as the ratio of the number of H atoms in each environment is 6 : 1 : 1. The splitting pattern is singlet, quartet, triplet.

Putting this together with the information from the integration trace, we get: 3 Hs with 0 Hs on

4 CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011ANSWERS TO EXAM-STYLE QUESTIONS A

10 a The mobile phase and stationary phase in HPLC are both liquids. A mixture is separated, as the di� erent components have di� erent solubilities in the stationary and mobile phase solvents. Partition refers to the principle that a solute will distribute itself between two immiscible solvents according to its solubility in each.

b Propane will have the shorter retention time, as it has a lower boiling point (lower relative molecular mass and, therefore, weaker van der Waals’ forces). Propane will, therefore, spend more time in the gas phase and be carried through the column more quickly.

c Because the carbamates have high melting points, they will not be su� ciently volatile to be separated by GLC, so HPLC must be used. In order to separate a mixture using GLC it must be possible to vaporise the components relatively easily.

the adjacent C; 2Hs with 3Hs on the adjacent C; 3Hs with 2 Hs on the adjacent C. The latter two signals indicate the presence of an ethyl group. Both I and II have the splitting pattern shown in the spectrum and must, therefore, be distinguished using the chemical shift values.

If we consider the singlets in the spectrum of I and II, we can predict the following chemical shifts using Table A6:

The singlet in the spectrum occurs at δ = 3.7 ppm, so we can conclude that the compound responsible for the spectrum is II, i.e. methyl propanoate.

H

H

H

C

H

H

C O

O

C CC

H

H

H H

H

H

C

H

H

C C

O

O C

H

H

H

O

CO

CR

H O

CCR O

H

δ = 2.0–2.5 ppmδ = 3.7–4.8 ppm