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NATIONAL QUALIFICATIONS CURRICULUM SUPPORT

Chemistry

Answers to Additional Questions

[ADVANCED HIGHER]

AcknowledgementsThis document is produced by Learning and Teaching Scotland as part of the National Qualifications support programme for Chemistry. The work of John Briggs and Douglas Buchanan in particular is acknowledged with thanks.

First published 2003Electronic version 2003

© Learning and Teaching Scotland 2003

This publication may be reproduced in whole or in part for educational purposes by educational establishments in Scotland provided that no profit accrues at any stage.

CONTENTS

Unit 1: Electronic Structure and the Periodic Table 1

Unit 2: Principles of Chemical Reactions 27

Unit 3: Organic Chemistry 45

Extra Questions 63

ANSWERS TO ADDITIONAL QUESTIONS (AH CHEMISTRY) III

© Learn ing and Teaching Scot land 2003

IV ANSWERS TO ADDITIONAL QUESTIONS (AH CHEMISTRY)


ELECTRONIC STRUCTURE AND THE PERIODIC TABLE

UNIT 1

1. Br–Br bond enthalpy = 194 kJ mol –1 (1)E = Lh for one mole of bonds (See units above.) (1) = E/Lh = 194000/6.63 10–34 6.02 1023 J/Js

(Showing these units helps to get the units right for the next line.) (1)

= 4.86 1014 s–1 (or Hz) (1)

(4)

2. (a) An excited electron returns to ground state, emitting energy difference as visible light of a specific wavelength. 1

(b) The energy gaps between energy levels decrease with increasing energy, i.e. the higher energy levels get closer and closer together. 1

(c) (i) E = h = 6.63 10–34 1.26 1015 J = 8.35 10–19 J (1)

But IE = L E= 6.02 1023 8.35 10–19 J mol–1

= 50.27 104 J mol–1

= 502.7 kJ mol–1 (1)2

(ii) The first ionisation energy of the element.

(Check page 10 of the Data Booklet to confirm that this is a likely answer.) 1

(5)

3. (a) Line A has a longer wavelength than all the others shown. (1) (Since it is the Balmer series, ground state is n = 2.) This represents the smallest energy jump, i.e. from n = 3 to n = 2. (1)

2

ANSWERS TO ADDITIONAL QUESTIONS (AH CHEMISTRY) 1



(b) E = Lhc/ (1)(Note: The question tells you to include ‘L’ in this equation.)

= (6.02 1023 6.63 10–34 3 108 )/656 10–9 (1)= 0.1825 106 kJ mol–1

= 182.5 kJ mol–1 (1)3

(5)

4. (a) The ultra-violet region 1(See page 14 of the Data Booklet: ultra-violet includes = 310 nm; visible runs from ~700 nm to ~400 nm.)

(b) E = Lhc/ (1)= 6.02 1023 6.63 10–34 3 108/284 10–9 (1)= 0.422 10 (23 – 34 + 8 + 9) J mol–1 = 422 kJ mol–1 (1)

3

(c) Energy from the spark excites some electrons to a higher energy level. (1)When these electrons return to ground state a specific amount of energy is released and this shows up as a line of measurable wavelength (or frequency) in the spectrum. (1)

2

(d) To gain a wider range of properties, e.g. harder, resistant to corrosion, etc. 1

(7)

5. (a) The Balmer series is, for n1= 2:

1/ = Rh (1/22 – 1/42)

= 1.097 107 (1/4 – 1/16)= 1.097 107 3/16= 2.06 106 (1)

= 1/(2.06 106) = 0.485 10–6

= 485 nm (1)

This line will be blue-green (see page 14 of the Data Booklet). (1)

3

2 ANSWERS TO ADDITIONAL QUESTIONS (AH CHEMISTRY)



(b) n1 = 1 (1)

All the jumps must be of shorter wavelength, i.e. of higher energy than those for n

1= 2. (1)

2

(5)

6. (a) An electron 1

(b) Each letter represents an orbital orientated along the x-, y- or z-axis. 1

(c) (i) s orbitals are spherical and symmetrical around the nucleus. (1)p orbitals are dumb-bell shaped and are symmetrical around each axis. (1)

2

(ii) The p orbitals are arranged mutually at right angles. 1

(d) Electrons are placed singly in degenerate orbitals before pairing occurs in one orbital. 1

(e) Error 1: The ‘4p’ orbitals should be labelled ‘3p’. (1)Error 2: Hund’s rule states that electrons will occupy

degenerate orbitals singly before any one is doubly filled. (1)

The correct configuration is:

1s 2s 2px 2py 2pz 3s 3px 3py 3pz

2(8)

7. (a) (i) 1 1(ii) 5 1

(b) (i) 6 1(ii) 2 1(iii) 18 1

(c) (i) 1s 1(ii) 4s 4p 4d 4f 1




(7)




8. (a) (ii) is wrong because the 2s electrons have identical ‘addresses’, i.e. all four quantum numbers the same for both.or Electrons cannot have parallel spin in the same orbital. (1)This violates Pauli’s exclusion principle. (1)

2(b) The two 2p electrons should occupy two degenerate 2p orbitals

(with parallel spin). (1)Hund’s rule (1)

2

(c) (vi) has violated the Aufbau principle since the s sublevel (1)is not yet full but the 2p sublevel is filling. (1)

2

(6)

9. (a)

2

(b) A region where one or (at most) two electrons are likely to be found. 1

(c) It signifies the second energy level. 1

(d) Of equal energy 1



zzz

x

yz

2s

x

y

2p x

yy

x x

2p z2p y


(e)

(–1 for any error) 2

(7)

10.

only one area correctly labelled (1) (2)

11. (a) Manganese 1(b) 1s2 2s2 2p6 3s2 3p6 4s2 3d5 or [Ar] 4s2 3d5 1(c) [Ar] 4s0 3d4 (the 4s term may be omitted) 1

(3)

12. (a) 380 kJ mol–1 (20 kJ mol–1 ) 1

(b) There is a huge energy requirement to break the noble gases into a stable octet.or It is very difficult to remove an electron from a full energy level. 1

(c) (i) The Group 1 metal has the largest radius in that period and has the smallest nuclear charge in that period. Both facts lead to a lesser attraction for the outermost electron. 1

(ii) Each new energy level means a larger radius (less attraction




for the outermost electron) and provides a greater shielding effect (again reduced attraction by the nucleus). 1

(d) Two factors apply: the steady increase in nuclear charge and the slight decrease in atomic radius from Li to Ne makes the attraction of the nucleus for outer electrons greater. 1

(e) (i) Be 1s2 2s2

B 1s2 2s2 2p1, i.e. B has started a new subshell so its outermost electron is relatively easier to remove than that of Be, where a complete subshell has to be broken into. 1

(ii) Half-full shells are relatively stable so N (with a half-filled p subshell) has a higher IE than O, which has one electron more. 1

(7)

13. (a) Electrons are excited by electric discharge to a higher level. (1)These electrons emit energy as they return to a lower energy level. (1)The quantity of energy emitted depends on the energy values of the two energy levels involved. (Many lines may be produced and each will represent a specific electronic jump.) (1)

3

(b) Pass the light through a prism and examine the spectrum produced on a screen. (1) Spectral lines characteristic of sodium and of neon would be seen. (1)

2

(5)

14. (a) Since E = hc/ or E1/, shorter wavelengths correspond to higher energy. (1)or It can be assumed that these lines represent a part of a series of lines which converge (at a continuum, not shown) at the higher energy end, i.e. to the left on the diagram shown in the question. Both arguments lead to the conclusion than 393 nm represents the highest energy value shown. (1)




2




(b) Instead of coloured lines on a black background, there would be black lines on a coloured background (of the visible spectrum). 1

(c) = 1/= 1/620 10–7 (1)= 16129 cm–1 (Remember the units.) (1)

2

(d) The orange-red of the 620 lines would probably swamp the less intense blue-green lines. 1

(6)

15. (a) By spraying as a solution into a Bunsen flame or by electric discharge through a gaseous sample or by electric sparking between graphite electrodes 1

(b) Valence electrons are excited and promoted to higher energy levels. 1

(c) (i) Electrons return to a lower energy level, including the ground state, emitting energy equal to the energy difference between the two levels involved as light. 1

(ii) A spectral line on the spectrum for each ‘jump’ or a series of characteristic spectral linesor the intensity of light of one spectral line. 1

(d) By using an appropriate filter. 1

(e) The intensity of the light emitted. 1

(f) Make up standard solutions of Ca 2+(aq). (1)Use the solutions to make a graph of intensity of radiation vs concentration of solution. (1)

2

(g) Measure the intensity of radiation of the water sample (being tested). (1)The concentration is read from the graph. (1)

2




(h) (i) Each lamp gives out radiation characteristic of a specific metal. Particles of the metal in the sample absorb a measurable amount of this light in proportion to their concentration as electrons are promoted. 1

(ii) The quantity of energy absorbed, i.e. the difference inintensity, between the incident light and the transmitted light. 1

(12)

16. (a)

1

(b) (i)

2

(ii)

1

(iii) In (ii) two of the bonding electrons are delocalised over the two oxygen atoms and the carbonyl carbon:

so the negative charge is considered to be shared equally between all three atoms and not to be residing on either oxygen.

(In (i) the negative charge is attached completely, i.e. localised, on only one of the oxygens.) 1




(5)




17. (a)

1

(b) (i)

1

(ii)

1(3)

18. (a) Trigonal planar (trigonal is acceptable). 1

(b) In CF4 all four outer electrons in carbon are used in

bonding pairs with fluorines. All four bonds are identical. The bond angle will be the tetrahedral angle of 109.5°. (1)

In NF3

two of the electrons in nitrogen form a lone pair. The three others each form bonding pairs. The lone pair exerts a greater repulsion ‘downwards’ so the bond angle is slightly less than 109.5°. (1)

2

(3)




19. (a)

(1) (1)

(Do not rely on your skill as an artist – label the shapes also.) 2

(b) Nitrogen has an extra pair of electrons (a lone pair). These exert a strong repulsive force downwards on the bonding pairs hence they are pushed down ‘below’ the tetrahedral angle, creating a pyramid: (1)

Boron has only three outermost electrons, so BCl3 has

only three pairs of bonding electrons. They spread themselves symmetrically (or as far from each other as possible), i.e. pointing to the corners of an equilateral triangle. (1) 2

(4)




20. (a) There are five pairs of electrons around the central chlorine atom (since an atom of Cl has seven outermost electrons and each F atom contributes one electron to the total 10 electrons 5 pairs). 1

(b) Five pairs lead to a trigonal bipyramid. 1

(c)

(Any two of these three shapes are acceptable.) 2

(4)




21. (a) Sodium chloride has a face-centered cubic arrangement.or Each ion is surrounded by six near neighbours of the oppositely charged ions. (1)

(Any one of these equivalent sketches is acceptable.) (1)2

(b) The different sizes (radius ratio) of the two ions involved cause CsCl to have a co-ordination of 8:8 (and not 6:6 as in NaCl). 1

(c) The closer match of iron(II) oxide to the radius ratio of NaCl suggests a sodium chloride structure.

Na+/Cl– = 95/181 = 0.52 Cs+/Cl– = 174/181 = 0.96 Fe2+/O2– = 61/136 = 0.45 2

(5)




22. (a) Each Na+ ion

has six Cl– ions as near neighbours and each

Cl– ion has six Na+ ions similarly arranged. Each ion therefore has six near neighbours 6:6 co-ordination. 1

(b) Since Na and Cl have a valency of 1 they cannot have six bonds. or In the sketch each Na+ has 6 Cl– near neighbours and none of these can be considered to be the partner for the Na+ any more than any other therefore the lines represent the direction along which the forces of ionic bonding apply (and do not represent shared pairs of electrons as in covalent bonds of covalent molecules.) 1

(c) Molecules of sodium chloride do not exist. Only covalent substances can be correctly labeled as molecular, but no single Na+

has a single identifiable special

neighbour among the six Cl – ions. 1

(3)

23. (a) H2 + SiHCl

3 Si + 3HCl 2

(–1 for each error)

(b) (i) Boron 1(ii) Phosphorus 1

(c) (i) Boron p-type semiconductor. (ii) Phosphorus n-type semiconductor. 1

(both correct for 1)

(5)

24. (a) C, E (both correct for 1) 1

(b) F, G (both correct for 1) 1

(c) B 1

(d) A, D (both correct for 1) 1

(Note: while graphite is undoubtedly a non-metal it conducts electricity by a mechanism that we associate with metals, called metallic conductivity.)

(4)




25. (a) Doping 1

(b)

1

(c) (i) An n-type semiconductor. (1) P has five outer electrons (one more than Si) and these extra electrons are the charge carriers. (1)

2

(ii) Boron 1

(d) n-type semiconductors have a surplus of electrons (from the dopant) in some areas and these fill the lowest unfilled conduction band, making the semiconductor a better conductor. (1)

p-type semiconductors have positive holes (introduced by the dopant) and these are the charge carriers. (1)

2

(e) (i) A layer of an n-type semiconductor, e.g. Si containing P impurities, is attached to a layer of a p-type semiconductor, e.g. Si containing B impurities. 1

(ii) When a p–n junction is irradiated with light, electron hole pairs are formed and electrons migrate. (1)

The electrons migrate n-type semiconductor and the holes p-type. The n-type is therefore now negative with respect to the p-type, i.e. a potential difference has been created. (1)

2

(10)




26. (a) Tc is the temperature at which the electrical resistance falls

to zero. 1

(b) Superconductors are materials that present no resistance at all to the flow of an electric current. 1

(c) Liquid helium has the lowest boiling point and this was needed to achieve the very low temperatures that were needed initially. 1

(d) Tc for some new materials was above the boiling point of

nitrogen – a much cheaper coolant. 1

(e) (i) This would bring superconductivity within reach at ambient temperatures, with no need for expensive coolants.

1

(ii) Distribution of electricity, electric trains/trams, electric motors (big and small), etc. 1

(6)

27. (a) B 1(b) A 1(c) C 1(d) A 1(e) D 1

(5)

28. (a) Graph 1 1(b) Graph 3 1(c) Graph 2 1

(3)




29. (a) C (diamond or graphite) has a covalent network structure so covalent bonds have to be broken on melting, i.e. the melting point is very high. (1)

N2 has a covalent molecular structure so only van der

Waals’ forces need to be broken to melt solid nitrogen, i.e. the melting point is very low. (1)

2

(b) (i) FCl 1

(ii) LiCl 1

(iii) BeCl2

1

(iv) Polarity falls from Li to N since the difference in electronegativity decreases to zero. (1)

Polarity rises from N to F since the difference in electronegativity increases again (with the polarity reversed). (1)

2

(7)

30. The H+ ion would consist of only a nucleus and as such cannot have a separate existence. (In fact it exists in aqueous solution attached to a water molecule, i.e. H

3O+.)

(1)

31. (a) AlCl3 = 27 + (3 35.5) = 133.5 but the relative molecular

mass = 267, i.e. twice the value for the empirical formula. The structure must therefore be a dimer of AlCl

3,

i.e. Al2Cl

6. 1

(b) Covalent molecular structure with dipole–dipole interactions. 1

(c) Aluminium hydroxide. (1)

The chloride is hydrolysed by the water and HCl(g) is evolved. (1)

2




(4)




32. (a) Li+H– or

Mg2+(H–)

2 (1)

Each reacts vigorously (at least) with water, releasing H

2 gas, leaving the metal hydroxide in solution. (1)

2

(b) Al2O

3+ 6HCl 2AlCl

3+ 3H

2O (1)

base + acid salt + water (½)(Note: there is no need to show the dimeric nature of the salt since it will be in solution as separate ions.)

Al

2O

3 + 3H

2O + 2NaOH 2NaAl(OH)

4(1)

acid + base salt (½) sodium aluminate

or sodium tetrahydroxoaluminate(III) 3

(c) SiCl4, PCl

3, SCl

2 are all hydrolysed by water. (1)

Fumes of HCl(aq) are seen and an oily liquid falls to the bottom of the tube. (1)(BCl

3 and CCl

4 are not hydrolysed; the chlorides of As,

Se, Br, Sb and Te are all covalent chlorides and simply dissolve in non-polar solvents without reaction, but are hydrolysed by water, since it is polar, giving fumes of HCl(aq), as well as an oxyacid or an oxychloride – both seen as an oily droplet.) 2

(d) The hydrogen bond is the force of attraction between an H atom of one molecule (which is directly attached to an element of high electronegative value) (1)and an atom of high electronegative value on another molecule. (1)For water (or HF or HCl) (1)the boiling points are higher than predictions based only on molecular size. (1)(This is also true of melting point, of viscosity, of surface tension and of specific heat capacity. Hydrogen bonding also causes water to have its maximum value of density at 4oC, (ice floats), by holding neighbouring molecules a little further apart than liquid molecules.) 4




(e) Na2O MgO Al

2O

3SiO

2P

2O

3SO

2Cl

2O

strong weak amphoteric acidic acidic acidic acidicbase base

or strong or alkali amphoteric strongbase acid

1

(12)

33. (a)

(1)

(1)2

(b) Fe3+ is the more stable because it has a d sub-shell that is exactly half-filled. This is a more stable arrangement than that with one more electron. 1

(3)

34. (a) +1 1

(b) 10 1

(c)

When the ion is irradiated with white light, energy is absorbed and promotes one electron from a lower 3d energy level to the higher 4s energy level. 1

(3)




35. (a) Water molecules and hydroxide ions. (1) They use a lone pair of electrons to form a dative covalent bond with the central metal ion, Fe 2+. (1)

2(b)

(1) octahedral (1)

2

(c) When a water molecule acts as a ligand its O–H bonds will become more polarised than before so H + ions will tend to form. (1)

However, the hydroxide ion donates a complete negative charge to the central metal ion, effectively reducing the positive charge on it and so reducing the polarising effect on the O–H bonds. H+ ions are therefore less likely to form, i.e. the solution is less acidic. (1)

2

(6)

36. (a) The 3d orbitals are each half-filled before one is doubly filled, i.e. maximum unpairing before pairing, etc. 1

(b)

(Note: the 4s box may precede the 3d or it may be omitted altogether since it is empty.) 1

(c) Only one 1

(3)




37. (a) 1. [Cu(NH3)

4]2+ (1)

2. [Cu(NH3)

3(Cl)]+ (1)

3. [Cu(NH3)

2(Cl)

2]0 (You may omit the zero in

this formula.) (1)

Ligands are held to a central metal ion by dative covalent bonds, not by ionic bonds. The ionic chloride ions are free to form a precipitate with Ag+ ions whereas the covalently bonded chlorides are not. (1)

4

(b) The degree of d–d splitting is controlled by the ligands involved. (1)NH

3 on its own will therefore be different from NH

3

with varying proportions of Cl. (1)2

(6)

38. (a) CrO4

2– Cr = ? oxidation state

= ? + (4 –2) = –2? = –2 + 8

= + 6 (1)

i.e. Cr atom has lost six electrons. 1s2 2s22p6 3s2 3p63d0 4s0 (You may omit the 3d0 4s0.) (1)

2

(b) Cr in CrO4

2– is in oxidation state +6 (see above).In Cr

2O

72– , Cr = ?

(2 ?) + (7 –2) = –2? = +12/2 = +6So Cr in Cr

2O

72– is also in oxidation state +6. (1)

This is therefore not a redox reaction. (1)2

(c) d–d splitting can only apply when there is at least one electron in a d orbital (this theory similarly does not apply to acompletely filled d sub-shell). 1

(5)




39. (a) Colour intensity is at a maximum when the concentration of the complex is at a maximum. (1)Ratio of [Ni2+]/[ligand] = 25/75 = 1/3. (1)i.e. x = 1, y = 3. (1)

3

(b) The N atoms in diaminoethane each have one lone pair. (1)Each lone pair forms a dative covalent bond with the Ni2+ ion. (1)

2

(c) The complex is purple, i.e. it absorbs in the green region. (1)The filter transmits green to get the best absorption from the complex. (1) 2

(7)

40. The H2O ligands (1)

split the degenerate d orbitals. (1)Energy from the red end of the visible spectrum is absorbed as electrons are promoted across the small energy gap, , now existing in the d orbitals. Hence the green colour is seen. (1)

3




41. (a) Green 1

(b)

Labels/units on axes (½ + ½)Subtracting 3 from each colorimeter reading (1)Concentration of unknown = 0.0068 0.0001 g l–1 (1)(Remember to subtract 3 from the colorimeter reading for the unknown.) 3

(c) There is some iron in the water.or Impurities with similar absorption patterns are present. 1

(d) Duplicates should be prepared or Large volumes should be used (to reduce error). or A blank should be carried out on the buffer and solvent without any sample present. 1

(e) d–d splitting or d–d transitions 1

(7)

42. (a) Five degenerate orbitals are split into two orbitals of higher energy and three of lower energy. Electrons can be promoted across this gap by absorbing energy from the visible spectrum. (1)

The peak around 410 nm represents the wavelengths absorbed (and equals the energy value of the d–d split). (1)

2




(b) Since the chloro complex leads to a lower value of d–d splitting (1)less energy is needed to make the jump (1)so absorption moves to the lower or red end of the spectrum. (1)

3

(5)

43. (a) [VO2]+ V + (2 –2) = +5

V = +5 (or V) (1)[V(H

2O)

6]2+ V = +2 (or II) (1)

2

(c) Hexaaquavanadium(III) (1)

(Make sure you show that the O atom (1)is donating the lone pair, not the H atom.) 2

(c) (i) 1s2 2s22p6 3s23p6 4s0 3d0 (or omit the last two terms) 1(ii) It has no d electrons to be promoted. 1

(d) Absorption should cover the red/yellow and possibly the green regions.

1




(7)




44. (a) Ni2+ [Ar] 4s2 3d8 or 1s2 2s2 2p6 3s2 3p6 4s2 3d8. 1

(b) Cl– splits the d orbitals and energy (a selection of frequencies) from the visible spectrum promotes electrons across the gap. (1)Remaining frequencies are transmitted and these produce the colour. (1) 2

(c) The size of the d–d split is altered by the presence of different ligands, so frequencies absorbed are different and the colour transmitted is altered. 1

(4)

45. (a) [Cu(Cl )4]2– ((1) for formula and (1) for charge) 2

(b) (i) Hexaamminecopper(II) 1

(ii)

1

(4)

46. (a) E = Lhc/ (1) = Lhc/E = 6.02 1023 6.63 10–34 3 108/239 103(1)

= 119.74 10 –6/239 = 5.01 10–7m = 501 nm (Note: this is in the visible spectrum

so it is a sensible answer.) (1)3

(b) Green is absorbed (1)so purple or magenta seen. (1)

2




(5)




47. (a) Green (since red is absorbed). 1

(b) E = Lhc/ (1)= 6.02 1023 6.63 10–34 3 108/540 10–9 J mol–1(1)= 0.222 106 J mol–1

= 222 kJ mol–1 (This is a likely order of magnitude – see Data Booklet.) (1)

3

(4)

48. (a) It has a full d sub-shell. 1

(b) It contributes no colour to compounds. It has only one valency. Neither the element nor its compounds show significant catalytic properties.(any two) 2

(c) Fe Haber process Pt Ostwald process/oxidation of ammonia. Pt catalytic converters in car exhausts.Ni hardening of vegetable oils fats.(accept others) 3

(d) Vacant d orbitals are readily available for (reversible) bonding, i.e. show variable oxidation states. 1

(7)

49. (a) They are compounds of the transition metals. 1

(b) One reactant may be adsorbed (or held by covalent bonds, often with d orbitals of the catalyst) (1)in a suitable orientation for a more probably successful collision. or in such a way that bonds within the reactant are weakened (and there is a new arrangement of bonds, i.e. a reaction occurs). (1)

2

(3)



PRINCIPLES OF CHEMICAL REACTIONS

UNIT 2

1. (a) Iodide ions 1

(b) Starch indicator 1

(c) n = v c = 24.8 10–3 0.1 = 2.48 10–3 1

(d) Moles of Cu2+ = 2.48 250/25 = 2.48 10–2 (1)Mass of Cu = 2.48 10–2 63.5 = 1.575 g (1)Percentage of Cu = 1.575/2.63 100 = 59.89% (1)

3(6)

2. Fe(NH4)

2(SO

4)

2.6H

2O = 392

Moles of Fe2+ = 1.8/392 = 4.6 10–3 (1)From redox equation Fe2+: MnO

4– = 5:1 (1)

Moles of MnO4

– = 4.6 10–3/5 = 9.2 10–4 (1)Concentration = 9.2 10–4/0.04 = 0.023 mol l–1 (1)

(4)

3. (a) (i) AgCl =143.4 Mass of Ag in precipitate = 107.9/143.4 0.6

= 0.451 g (1)Mass of Ag in coin = 1000/100 0.451 = 4.51 g (1)Percentage of Ag in coin = 4.51/10 100 = 45.1% (1)

3

(ii) Add some AgNO3 to filtrate. There should be no

more precipitate. 1

(b) CuCNS = 121.6 Mass of Cu = 63.5/121.6 0.31 = 0.1618 (1)Mass of Cu in coin = 1000/100 0.1618 = 1.62 g (1)Percentage of Cu in coin = 16.2% (1)

3(7)




4. (a) Add more AgNO3 – no further precipitate. 1

(b) 2AgNO3 + MgCl

2 2AgCl + Mg(NO

3)

2

Moles of AgCl = 2.01/143.4 = 0.014 (1)Moles of MgCl

2= 0.014 1/2 500/100 = 0.035 (1)

Mass of MgCl2

= 0.035 95.3 = 3.336 g (1)%MgCl

2= 3.336/4.5 100 = 74.1% (1)

4

(c) Moles AgNO3 = 0.014/4 = 0.0035 (1)

Volume = moles/conc. = 0.0035/0.1 = 0.035 litres (35 cm3) (1)

2 (7)

5. (a) Moles of HCl = 15 10–3 1.0 = 1.5 10–2 (1)Moles of Na

2CO

3= 1/2 1.5 10–2 = 7.5 10–3 (1)

Mass of Na2CO

3= 7.95 10–3 106 250/25 (1)= 7.95 g (1)

4

(b) Mass of water = 16 – 7.95 = 8.05 gMoles of water = 8.05/18 = 0.447Moles of Na

2CO

3= 7.95/106 = 0.075 (1)

Ratio of moles = 0.447/0.075 = 5.96 = 6 (i.e. 6H2O) (1)

2(6)

6. (a) Moles of NaOH = 18.2/1000 0.1 = 1.82 10–3 (1)Moles of CH

2(COOH)

2= 9.1 10–4 (1)

2

(b) Mass of acid = 9.1 10–4 250/25 104= 0.946 g (1)Mass of water = 1.28 – 0.946 = 0.334 g (1)

2

(c) Moles of water = 0.334/18 = 0.0186Moles of acid = 0.946/104 = 0.0091 (1)

n = 0.0186/0.0091 = 2 (1)2

(6)




7. (a) Mass of BaSO3

= 1.09 g (1) Moles of BaSO

3= 1.09/217 = 0.005 (1)

2

(b) Moles of Na2SO

3= 0.005 250/50 = 0.025 (1)

Mass of Na2SO

3= 0.025 126 = 3.15 g (1)

%Na2SO

3= 3.15/5.02 100 = 62.75% (1)

3(c) It is a good oxidising agent.

or It can oxidise SO3

2– but not SO4

2–. (1)It acts as its own indicator. (1)

2(7)

8. (a) (i) Yield = 50% 1(ii) side reactions or impurities 1

(b) The suggestion is wrong. (1)A catalyst only brings the reaction to the same equilibrium more quickly. (1)

2

(c) K is a constant at a fixed temperature and altering the alcohol concentration will not change the value of K. (1)It will, however, increase the yield of the ester as the forward reaction will be increased. (1)

2(6)

9. (a) (i) K = [SO3]/[SO

2]2[O

2] 1

(ii) No units 1

(b) (i) Le Chatelier’s principle states that if a system is subjected to any change, the system readjusts itself to counteract the applied change. 2

(ii) Increased temperature favours H +ve. Thus thebackward reaction is favoured and the equilibrium position moves to the left. 1

(iii) K will decrease. 1 (6)




10. (a) A 1; B 3; C 1; D 2; E 2Reactions A and C are going from a larger volume to a smaller volume therefore an increase in pressure favours the products.

Reaction B goes from a small volume (no gas) to a large volume therefore an increase in pressure favours the reactants.

Reaction D has two volumes of gas on both sides therefore pressure does not affect the result.

Reaction E has no gas therefore pressure change has no effect.

(5 (1) for correct answer with correct explanation) 5

(b) It will have the same general shape but it will be less steep. 1

(6)

11. (a) K = [NO2]2/[NO]2 [O

2] 2

(b) K > 1 therefore equilibrium favours the products. 1

(c) Increasing temperature favours H +ve therefore the products are favoured (1)therefore K decreases. (1)

2(d) 15 = [NO

2]2/[0.1]2 [0.1]

[NO2] = 0.015 (1)

= 0.12 mol l–1 (1)2

(7)

12. (a) [Ba2+] = 1 10–10 = 1 10–5 mol l–1 1

(b) BaSO4 = 233.4 g mol–1 (1)

Mass dissolving = 233.4 1 10–5 (1)= 2.334 10–3g (1)

3




(c) (i) Ksp

is a constant therefore stays the same. 1

(ii) The equilibrium moves to the left therefore [Ba 2+] decreases. 1

(6)

13. (a) (i) Ether 1(ii) Ether is less dense than water. 1

(b) K = [Xether

]/[Xwater

] = 12 (let V grams be extracted)12 = (V/100)/(1.1 – V)/100 = V/(1.1 – V) (1)V = 1.0154 g (1)

2

(c) 12 = V/50/(1.1 – V)/100V = 0.943 gMass now dissolved in the water = 1.1 – 0.943 = 0.157 g (1)12 = V/50/(0.157 – V)/100 V = 0.106 gTotal extracted = 0.943 + 0.106 = 1.049 g (1)

2(d) Many organic compounds are non–polar and dissolve

in non-polar solvents such as ether (water is polar). 1(7)

14. (a) Moles of acid originally = 2.36/118 = 2 10–2 1

(b) (CH2COOH)

2 + 2NaOH (CH

2COONa)

2 + 2H

2O

Moles of NaOH = 34.8/1000 1 = 3.48 10–2 (1)Moles of acid = 1.74 10–2 (1)Moles in ether = (2 10–2) – (1.74 10–2) = 2.6 10–3 (1)

3

(c) K = 2.6 10–3/0.1/1.74 10–2/0.1 = 0.149 1

(d) Use two 50 cm3 portions of ether (or 4 25 cm3),

i.e. do two or four extractions from the aqueous layer. 1(6)




15. (a) (i) The solvent mixture 1(ii) The chromatography paper 1(iii) 3 1(iv) Glycine – if the solvent were allowed to move

further up the paper then the component may not travel the same distance as glycine. 2

(v) Solvent front 1

(vi) Base line

1

(b) (i) To prevent the solvent evaporating away or to ensure the atmosphere in the tank is saturated with the solvent. 1

(ii) Rf values are used to identify substances. The same

substance has the same Rf value for the same

conditions. 1

(iii) The ink may dissolve in the solvent. 1

(iv) The component is held very strongly by the paper on the baseline. (Do not answer in terms of solubility.) 1

(v) A mixture of solvents gives better separation; more components are likely to be soluble. 1

(vi) Yes (mobile and stationary phases) 1

(vii) The components are colourless and so must be reacted with a developing agent to make them visible. 1

(14)

16. (a) (i) Nitrogen or argon 1(ii) It will not react with the alkanes. 1




(b) The time it takes for the compound to move through the column from the point it was injected at to the point where it is detected. 1

(c) (i) A – methane, E – 2,2-dimethylpropane,F – 2-methylbutane 3

(ii) Peak F is at 9.3, which is the value for 2-methylbutane. Peak E is at about 8.7 and therefore the alkane must be an another isomer of pentane. Peak A is at about 1.7 and therefore the molecule must be smaller than ethane. 3

(9)

17. (a) H2O 1

(b) OH– 1(c) NH

3 1

(3)

18. (a) (i) The HSO3

–(aq) provides an H+(aq) ion. 1(ii) The SO

32–(aq) is the conjugate base. 1

(b) HSO3

–(aq) + H+(aq) H2SO

3(aq) 1

(3)

19. (a) pH = ½pKa – ½ logc (1)

= ½ 4.77 – ½ log0.01 (1)= 2.385 – (–1) = 3.4 (1)

or Ka

= [H+]2/[CH3COOH] (1)

Therefore [H+] = (1.75 10–5 0.01) (1)= 1.7 10–7 = 4.12 10–4

pH = – log H+ = –log 4.12 10–4 = 3.4 (1)3

(b) CH3COOH(aq) CH

3COO–(aq) + H+(aq) acid

CH3COO–Na+(s) CH

3COO–(aq) + Na+(aq) salt

(i) Add HCl – the H+ (aq) ions added react with the excess CH

3COO–(aq) from the salt. (1 + 1) 2

(ii) Add NaOH – the OH–(aq) added reacts with the H+(aq) from the acid and more acid ionises to replace the H+(aq) ions removed. (1 + 1) 2




(c) pH = pKa – log[acid]/[salt] = 4.77 – log 0.25/0.15 (1)

= 4.55 (1)2

(9)

20. (a) The product of [H+(aq)] [OH–(aq)] 1

(b) [H+]2 = 51.3 10–14, therefore [H+] = 7.16 10–7 and pH = 6.15 (1+1+1)

3

(c) An increase in temperature favours H +ve. As the temperature increases so does the value of K

w and therefore more ions form.

Thus the ionisation is endothermic. 1(5)

21. (a) NaH2PO

4; Na

2HPO

4;

Na

3PO

4 (any two, 1+1)

2

(b) Ka = [H+][ H

2PO

4–]/[H

3PO

4] (1)

7.08 10–3 = [H+]2/0.1 [H+]2 = 7.08 10–4 [H+] = 2.66 10–2 (1) pH = 1.6 (1)

3(You could use pH = ½pK

a – ½log[C].) (5)

22. Kw at 288K = 0.45 10–14 (1)

[H+] = 6.71 10–8 (1)pH = –log 6.71 x 10–8 = 7.17 (1)

(3)

23. (a) [Ca(OH)2] = 0.126/74 moles per 100 cm3

= 0.017 mol l–1 (1) [OH–] = 2 0.017 = 0.034 mol l –1 (1)

2

(b) [H+] = 10–14/0.034 = 2.94 10–13 (1) pH = – log 2.94 10–13 = 12.53 (1)

2(4)




24. (a) Methyl yellow pH = 3.3; bromothymol blue pH = 7; thymol blue pH = 8.9 (3 correct = 2; 1 or 2 correct = 1) 2

(b) Thymol blue. (1)The titration involves a weak acid and strong alkalitherefore the salt will have a pH above 7. (1)

2(4)

25. (a) 25 0.1/20 = 0.125 mol l–1 1

(b) Weak (1)Acid neutralised at pH 9 (1)

2

(c) Indicator must change colour at a pH corresponding to the neutralisation point. 1

(d)

2

(e) No suitable indicator 1(7)



Small point of inflect ion about pH7.


26. (a) 1.6/64.1 = 0.025 mol in 250 cm3 (1)= 0.1 mol l–1 (1)

2

(b) pH = ½ pKa – ½logc (1)

= (½ 1.8) – (½ log 0.1) (1)= 1.4 (1)

or Ka

= [H+][HSO3

–]/[H2SO

4] (1)

H+ = Ka [H

2SO

4] = (1.5 10–2) (0.1) (1)

= 3.87 10–2 pH = – log [H+] = 1.4 (1)(If the wrong equation is used then zero marks.) 3

(c) It is volatile, i.e. SO2 is given off from solution. or It is

unstable. 1

(d) Accept range within 7.5 – 10.5 but must have at least 1.5 of a difference between values. 1

(7)

27. Ka

= [H+][A–]/[HA] = [H+]2/[HA] (1)[H+] = K

a [HA] = 1.3 10–5 0.1 (1)

= 1.14 10–3

pH = 2.9 (1)or pH = ½pK

a – ½logc (1)

= ½ 4.9 – ½log 0.1 (1)= 2.95 (1)

(If the wrong equation is used then zero marks. ) (3)

28. (a) Find MgCl2 (s) Mg2+(aq) + 2Cl–(aq)

Given (1) Mg2+(g) Mg2+(aq) H –1923(2) Cl–(g) Cl–(aq) H –338(3) Mg2+(g) + 2Cl–(g) Mg2+(Cl–)

2 (s) H –2326

H = – H3 + H

1 + 2H

2 (1)

= –(–2326) + (–1923) + 2(–338) (1)= –273 kJ mol–1 (1)

3

(b) MgCl2(s) + 6H

2O(l) MgCl

2.6H

2O(s)

H = Hp – H

r(1)

= –2500 – (–642) + 6(–286) (1)= –142 kJ mol–1 (1)

3(6)




29. (a) Ionic bonding becomes weaker going down the series. or Halide ion increases in size going down the series (must mention ionic/ion). 1

(b) Enthalpy hydration = enthalpy solution – lattice enthalpy (breaking) = 17.2 – 701 = –683.8 kJ mol –1 1

(c) The hydration enthalpy for Ca2+ will be more negative. (1)Ca2+ has a larger charge than K+ and therefore there is stronger attraction between the Ca 2+ and the water. (1)

2 (4)

30. (a) C(s) + H–O–H(g) C~O(g) + H–H(g) H = 130 kJ mol–1 Bond breaking C(s) C(g) = 715; two O–H bonds = 2 458 = 916; total = 1631 kJ (1)Bond making C~O; H–H = –432; total = –432 + C~O; (1)C~O = 130 –1631 + 432 = –1069 kJ mol –1 Energy given out in forming one mole of C~O bonds = –1069 kJ (1)

3

(b) C–O = 358; C=O = 798; therefore the bond in CO is C–O + C=O, i.e. CO 1

(c) CO poisonous; H2 could be explosive; CH

3OH is flammable

(any two correct = 1+1) 2(6)

31. (a) (i) 3C(s) + 4H2(g) C

3H

8(g) –140 (1)

(ii) C(s) + O2(g) CO

2(g) –394 (1)

(iii) H2(g) + ½O

2(g) H

2O(g) –286 (1)

3

(b) Find C3H

8(g) + 5O

2 (g) 3CO

2(g)

H = H(i) + 4 H(ii) + 3H(iii) = – (–104) + (4 –286) + (3 –394) (1)= –2222 kJ mol–1 (1)

2 (5)




32. (a) (i) Sublimation enthalpy 109 kJ mol –1

(ii) Bond dissociation enthalpy 248.5 kJ (iii) 1st ionisation energy 502 kJ mol –1 (iv) 1st + 2nd electron affinity 703 kJ mol –1

(v) Lattice formation enthalpy –2481 kJ mol –1 (5 1)5

(b) 2Na(s) + ½O2(g) (Na+)

2O2–(s) 1

(c) (i)

1

(ii) H(b) = H(i) + H(ii) + H(iii) + H(iv) + H(v)

= 2(109) + 2(502) + ½(497) + 703 + (–2481)= –307.5 kJ mol–1 1

(8)

33. (a) (i) S = Sp – S

r = 208 – (201 + 2(187))

= – 367 J K–1 mol–1 1(ii) H =H

p – H

r = (–166 – (227 + 2(–92.3)))

= – 209.4 kJ mol–1 1

(b) T = H/S = 209.4/0.367 (1)= 570.6K (1)

2(4)

34. (a) Large entropy change or gas to solid 1

(b) Spontaneous G –ve (1)H +ve or endothermic, therefore drop in temperature (1)

2




(c) G = H – TS S = (H – G)/T = (–92 – (–95))/298 (1)= 10 J K–1 mol–1 (1)

2(5)

35. (a) C8H

18 C

8H

10 + 4H

2(1)

S = Sp – S

r = (352 + 4(131)) – 463 (1)

= 413 J K–1 mol–1 (1)3

(b) At this temperature G = 0T = H/S (1)

= 227/0.413 (1)= 550K (1)

3

(c) A small quantity of 2,3-dimethylhexane will form even if K << 1. Removal of 2,3-dimethylhexane in reaction B disturbs the equilibrium in reaction A, promoting the forward reaction. 1

(7)

36. (a) S = Sp – S

r = (136 + 188.7 + 2213.8) – 2(102.1) (1)

= 334.1 J K–1 mol–1 (1)2

(b) T = H/S (1)= 129/0.3341 (1)= 386.1K (1)

3

(c)

Volume of gas/cm3

290 390 490 Temperature/K

labels (1) shape at correct point (1) 2 (7)




37. (a)Zn(s) + O

2(g)

ZnO

2

G/kJ mol–1

2C(s) + O

2(g) 2CO(g)

Temperature/K

scale (1) labels (1) accurate points (1) 3

(b) (i) Minimum temperature (where graphs cross) 1160K 1

(ii) G for A +300; G for B –490 (1)G for reaction = –190 kJ mol–1 (1)

2

(c) Advantage – carbon is cheap and readily available. (1)Disadvantage – it is a solid therefore the reaction isslower. (1)

2(8)

38. (a) At temperatures above 2250K 1

(b) ZnO + H2 Zn + H

2O 1

(c) 1400K (above) 1

(d) Below A solid + gas solidAbove A liquid + gas solid (1)therefore there is a greater decrease in entropy, i.e. S –ve. (1)As gradient is –S then an increase in gradient means that S is more –ve. 2

(e) Although the reaction is feasible at room temperature it does not occur in practice because the activation energyfor the reaction is too high. or The rate of the reactionat room temperature is too slow for the reaction to occurat a measurable rate. 1

(6)




39. (a) G = (275 to 280) – 555 (1)= –270 to –280 kJ

or –135 to –140 kJ mol–1 (1)2

(b) Accept between 950 and 1020K 1

(c) Gas/solid reaction better than solid/solid reaction. 1

(d) To remove impurities (silicates) or forming slag or to produce more CO

2 (any one) 1

(e) CO could be burned to (pre)heat the air. 1(6)

40. (a) Ecell

= EAg

– EM E

M = –1.03 + 0.8 = – 0.23 V 1

(b) G = –nFE = –2 96500 1.03 (1)= –198.79 kJ mol–1 (1)

2(If student shows clearly that Ag+ is used and has n = 1 then accept 99.39 kJ.)

(c) Maximum work would not be obtained from cell or thermodynamically reversible conditions do not apply or current would be drawn from cell (any one). 1

(d) A Group 1 metal (or NH4

+) nitrate. 1(5)

41. (a) –0.28 V 1

(b) G = –nFE (1)= –2 96500 0.48 (1)= 92.6 kJ mol–1 (1)

(–1 if n = 1; –1 if wrong voltage; –1 if G = +nFE ) 3 (4)




42. (a) Combustion reaction or oxidation of fuel. 1

(b) To prevent the CO2 from reacting with the potassium hydroxide

(OH–). 1

(c) H2 + ½O

2 H

2O 1

(d) Cheaper than pure platinum or increase the surface area or platinum is unreactive (any one). 1

(e) Fuel cells do not produce CO2, SO

2 or radioactive waste;

fuels cells are not corrosive/toxic; fuels cells are silent (any one). 1

(5)

43. (a) Rate = k[I2][H

2] 1

(b) Rate = 6.43 102 0.5 0.5 (1) = 1.61 10–2 mol l–1 (1)

2

(c)T

1

Number (fraction) T2

of molecules with given energy

(1)Energy E

a (1)

label axis (1)(If drawn on two separate graphs, then 1 only.) 3

(6)




44. (a) (i) Rate doubles as [H2] doubles (exp. 1 and 2)

therefore first order. 1

(ii) Rate increases four-fold as [NO] doubles (exp. 4 and 5) therefore second order. 1

(b) Rate = k[H2][NO]2 1

(c) k = 0.012/(0.005)(0.03)2 = 2666.67 = 2.67 103 mol–2 l2 s–1

value (1) units (1) 2

(d)Initial rate of N

2 /mol l–1s–1

0.0 0.01 0.02 0.03Initial [NO]/mol l–1 1

(6)

45. (a)Experiment 1 2 3Av. rate/mol l–1 min–1 0.1 0.2 0.4

Av. rate mol l–1 s–1 1.67 10–3 3.33 10–3 6.67 10–3

method (1) arithmetic (1) 2

(b) A – second order (1)B – first order (1)Rate = k[A]2[B] (1)

3

(c) k = 0.1/(0.1)2(0.1) (1)= 100 mol–2 l2 min–1 (1.67 mol–2 l2s–1) (1)

2(7)

46. (a) NO3 + CO NO

2 + CO

2 (1)

1

(b) x = 4 (1)y = 1 (1)

2(3)




47. (a) Rate does not depend on A. (1)Rate is proportional to [B]2. (1)

(2)

48. (a) Step 1 (1)Both propanone and OH– are in the rate-determining step as both are in the rate expression. (1)

2

(b) (i) A blue/black colour (1)The Br

2 displaces the I

2 from the KI. The I

2 reacts

with the starch to give a blue/black colour. (1)2

(ii) At the end-point it turns colourless. 1 (5)



ORGANIC CHEMISTRY

UNIT 3

1. (a) T – Br*; U – HBr; V – C2H

5*; W – Br*; X – Br

2; Y – C

4H

10;

Z – Br* (any 6 ½) 3

(b) Endothermic (1)Energy required to break bonds (1) 2

(c) (i) Propagation (1)(ii) Termination (1)

2

(d) Once the Br* radical is produced, the propagation step produces more of these radicals to keep the reaction going. 1

(8)

2. (a) 2 1

(b) The 2s and 2p orbitals combine to form four hybridised sp3 orbitals. (1)Each occupied by one electron. (1) 2

(c) The C=C bond consists of a bond and a bond (1)but the C–C is just a bond. (1) 2

(d)

1(6)

3. (a) Nuclophilic substitution 1

(b) They have a polar C–X bond. 1



ORGANIC CHEMISTRY

(c) React the alcohol with an alkali metal (Na). 1

(d) (i) Butanoic acid (1)(ii) Propanoic acid (1)

2(5)

4. (a) CH3CHBrCH

3 or full structural formula 1

(b) Nucleophile 1(c) An ether (alkoxyalkane) 1(d) Any hexanol isomer, e.g. CH

3CH

2CH

2CH

2CH

2CH

2OH 1

(e)

1

(f) Use infra-red to detect the presence of C=O in R or its absence in Q; ether has lower boiling point than ester; ester can be hydrolysed with NaOH but ether cannot be 1(any one).

(g) Propanone 1(7)

5. (a) Ethoxide ion 1

(b) (i) Sodium/alkali metal (1)(ii) PCl

5/PCl

3/AlCl

3(1)

2

(c)

1(4)



ORGANIC CHEMISTRY

6.

(i) (ii)

(1) Substitution HBr/PBr3/PBr

5

(2) Elimination Alcoholic KOH

(3) Substitution KCN/NaCN/CN–

(4) Hydrolysis H+(aq)

(5) Condensation/

esterification

CH3OH

(concentrated H2SO

4)

(6) Neutralisation NH3

(7) Chlorination PCl5/PCl

3 /SOCl

2

(7 1) (7 1) (14)

7. (a)

(1 + 1)2

(b) Electrophilic (aromatic) attack (substitution) or nitration 1(3)

8. (a) Phenol or hydroxy benzene 1

(b) (concentrated) H2SO

4 + HNO

3 1

(Note: phenols will react with HNO3 alone.)

(c) Reduction 1

(d) C6H

4NH

2OH C

6H

4NHCOCH

3OH

111 g 153 g (1)43.6 g 60.1 g (1)for 80% yield get 60.1 80/100 = 48.1 g (1) 3



ORGANIC CHEMISTRY

(e)

(1 + 1)2

(8)

9. (a) A reaction in which the carbon atoms of a hydrocarbon are rearranged to produce a different carbon skeleton by the action of heat and a catalyst. 1

(b) (Friedel–Crafts) alkylation (1)using an alkyl halide (1)in the presence of aluminium chloride. 2

(c) (catalytic) hydrogenation 1

(d) Detergent 1

(e) (i)

2

(ii) Condensation polymerisation 1(8)

10. (a) (i) OH(ii) RCHO(iii) RCOR

(Note: if B were another primary alcohol then it maybe oxidised to a carboxylic acid and D = RCO

2H.)

(3 1) 3

(b) Reagent acidified dichromate or Tollens’ or Benedict’s Change orange to blue-green or silver mirror formed

or blue to orange Name of a correct reagent (1)correct change described for C (1)no change for D (1)



ORGANIC CHEMISTRY

or make a derivative (1)measure melting point (1)check with literature (1)

3

(c) 3500 to 2500 cm–1 and 1725 to 1700 cm–1 (1+1)2

(8)

11. (a) Ratio of molesMoles C = 69.77/12 = 5.81 5Moles H = 11.63/1 = 11.63 10 (1)Moles O = 18.60/16 = 1.16 1

C5H

10O (1)

2

(b) OH/hydroxyl/alcohol 1

(c) C5H

10O (1)

cyclopentanol (1) (also accept cyclobutylmethanol or cyclopropylethanol) 2

(d) Carbonyl/C=O 1

(e)

1

(f) Compound X or Y or unreacted 2,4-dinitrophenylhydrazine or water (any one). 1

(8)

12. (a) Separate ether layer and distil (evaporate) off the ether. 1

(b) Advantage – lack of reactivity/good solvent/low boiling point (1)Disadvantage – highly flammable/anaesthetic/toxic/forms peroxides (1)

2



ORGANIC CHEMISTRY

(c) Ethanoic acid/CH3COOH (1)

1700 to 1725 cm–1 (C=O stretch) (1)2

(d) In Grignard reagent C– Mg+ Br therefore the C– acts a nucleophile. (1)In alkyl halide C+ Br – therefore the C+ acts as an electrophile. (1)(This difference is due to electronegativity differences.) 2

(7)

13. (a) (i) Carbon to carbon double bond 1(ii) Geometric isomerism 1

(b) (i)

1

(ii) C=C does not allow rotation about this bond whereas C–C does allow rotation. 1

(c) If the molecule has changed from cis- to trans- then the enzyme will not recognise it. 1

(5)

14. (a) (i)

2

(ii) Pentane is a longer molecule and will have more van der Waals’ forces between the molecules (1)and thus will have a higher boiling point. (1)

2

(b) (i) Maleic acid is cis-butenedioic acid; fumaric acid is trans-butenedioic acid. (1+1)

2



ORGANIC CHEMISTRY

(ii) The trans isomer will be able to form hydrogen bonds with neighbouring molecules but the cis isomer will form hydrogen bonds with the other acid group on the same molecule. (1) The trans isomer will have the higher melting point. (1)

2(8)

15. (a) The two carbon atoms at either end of the molecule, i.e.

1

(b) Three 1(2)

16. (a) (i) 2,3-dimethylbutane 1(ii) Hexane 1

(b) No (1)Four of the C atoms have more than one H atom attached, and two have methyl groups attached therefore none of them has four different groups attached. (1) 2

(c)

1(5)

17. (a) 2–hydroxypropanoic acid 1

(b)

2



ORGANIC CHEMISTRY

(c) It is a racemate (racemic mixture), i.e. it consists of equal amounts of the two enantiomers. 2

(5)

18. (a)

1

(b) (i) No difference 1(ii) No difference 1

(c) +10°. 1

(d) It could be a racemate (racemic mixture) of the two enantiomers. 2

(6)

19. (a) Optical isomerism 1

(b) (i) a C=C 1(ii) an asymmetric C atom 1

(c) (i) B or H 1(ii) D or F 1

(5)

20. (a) Molecule A reacts with acidified dichromate solution/hot copper oxide/Benedict’s solution/Tollens’ reagent, but B does not. (1)Molecule A decolourises bromine solution but B does not. (1)

2

(b)

2



ORGANIC CHEMISTRY

(c) (i)

(1)

The chiral C has four different groups attached (1)2

(ii) A racemate (racemic mixture) of the two isomers will be formed. 1

(7)

21. (a) (i) Structural formulae (ii) CH3CH

2CH

2CH

3

(iii) C=C (iv) A chiral carbon

6

(b)

1

(c)

1

(d)

1(9)



ORGANIC CHEMISTRY

22. (a) Mass of C = 12/44 0.630 = 0.172 g Mass of H = 2/18 0.258 = 0.0287 g Mass of O = 0.315 – (0.172 + 0.0287) = 0.1143 g (–1 for each error) 2

Mole ratio Moles of C = 0.172/12 = 0.0143 2Moles of H = 0.0287/1 = 0.0287 4Moles of O = 0.1143/16 = 0.00714 1 (1)The empirical formula is C

2H

4O (1)

2(4)

23. (a) Mass of water = 0.610 g moles of water = 0.61/18 = 0.034 moles of hydrogen = 2 0.034 = 0.068 mass of hydrogen = 1 0.068 = 0.068g(1) Volume of CO

2 = 610 cm3 moles CO

2 = 610/24000

= 0.0254 moles of carbon = 0.0254 mass of carbon = 12 0.0254 = 0.305 g(1) Mass of oxygen = 0.508 – (0.068 + 0.305) = 0.135 g (1)

Mole ratio Moles of C 0.305/12 = 0.0254 3 Moles of H 0.068/1 = 0.068 8 (1)Moles of O 0.135/16 = 0.0084 1 Empirical formula is C

3H

8O,

CH

3CH

2CH

2OH (1)

5(b) To improve the reliability of the results 1

(6)

24. (a) CO+ 28; COH+ 29; CH3O+ 31; CH

3OH+ 32. 2

(b) Relative abundance of each ion formed. 1

(c) At m/z = 14; 14.5; 15.5; 16 1(4)



ORGANIC CHEMISTRY

25. (a)

2

(b) A (1)Molecule B cannot be fragmented into all of these parts. (1)

2(4)

26. (a)

1

(b) The peaks at 112 and 114 correspond to the two isotopes of chlorine, 35Cl and 37Cl, being present. 1

(c) A molecular ion is formed from the molecule with the loss of only one electron. (1)C

6H

5Cl+ (1)

2

(d) (i) It is part of the molecule with a positive charge. 1(ii) There is only one peak at 77 therefore the two

isotopes of Cl are not present. 1(iii) C

6H

5+ has a mass of 77. 1

(7)

27. (a) Vibration in the molecule, i.e. specific bonds lengthen and shorten rapidly. 1

(b) The presence of different types of bonds, e.g. C=O, O–H, etc. 1

(c) Compare the infra-red spectrum of the distilled sample with that of a pure sample of the expected distillate. (1)They should be identical. (1)

2(4)



ORGANIC CHEMISTRY

28. (a) The sample is not used up nor changed during analysis and can be recovered and used again. 1

(b) X–ray or infra-red of liquid sample 1

(c) No, because the sample is used up. 1(3)

29. NMR – radio; infra-red – infra-red; colorimetry – visible (3 1)(3)

30. (a) It produces a magnetic field. 1

(b) (i) Lined up with or against the magnetic field. 1(ii) The one that is lined up against the magnetic field. 1

(c) The nucleus changes orientation from high to low spin, i.e. changes from being aligned against to being aligned with the magnetic field. 1

(d) Absorption as energy is required to bring about the change. (1+1) 2

(e) (i) The chemical environment, i.e. the other atoms close to the hydrogen atom. 1

(ii) The number (proportion) of hydrogen atoms in the same chemical environment. 1

(8)

31. (a) 12 1

(b) All the H atoms are in CH3 groups attached to the central

silicon. 1

(c) 0.0 1(3)

32. (a) Tetramethylsilane (1)It is used as a reference and is given the chemical shift value of 0. (1)

2



ORGANIC CHEMISTRY

(b) The chemical shift of the peak is where the CH3 peak is

found. (1)The area under the curve/integral shows three H atoms. (1)

2

(c) Peak A has a chemical shift of 3.6 and has an integral of 2, therefore it corresponds to CH

2. 1

Peak B has a chemical shift of 2.8 and has an integral of 1, therefore it corresponds to the H in the OH group. 1

(6)

33. (a) Peak D is the reference peak (TMS). 1

(b) (i) Integral 1(ii) It gives the relative number of hydrogen atoms in

the same environment. 1

(c)

The phenyl group has five H atoms; the CH2 has two H

atoms; the CH3 has three H atoms and this agrees with

the integral values of 5; 2; 3. (1)Also, the chemical shift values from the Data Booklet agree. (1)

3(6)

34. (a) CH3CH

2CH

2OH (1)

CH3CH(OH)CH

3 (1)

2

(b) (i) CH3CH

2CH

2OH 1

(ii) 3 1(iii) 6 (2 CH

3 ): 1 (CH): 1 (OH) 1

(c) CH3CH

2OCH

3, therefore expect three peaks. 1

(6)



ORGANIC CHEMISTRY

35. (a) (i) CH4; C

2H

6(1+1)

2

(ii) All the H atoms are in the same chemical environment for CH

4 and for CH

3CH

3. (1)

For the CH3CH

2CH

3 and the CH

3CH

2CH

2CH

3

expect two peaks from the CH3 and the CH

2

groups. (1)2

(b) CH3CH(CH

3)CH

32 peaks – one for the CH

3 groups

and one for the CH group (1)CH

3CH

2CH

2CH

32 peaks – one for the CH

3 groups

and one for the CH2 groups (1)

2

(c)

2(8)

36. (a) The atom has only one electron. 1

(b)

1

(c)

1(3)



ORGANIC CHEMISTRY

37. (a) X–ray crystallography 1

(b) Helps to find mode of action or allows vitamin B12 or related compounds to be synthesised 1

(c) (i) A carbon atom has fewer electrons and would appear smaller. 1

(ii) They have only one electron and so their electron density is too small for them to be easily seen. 1

(d) (i)

1

(ii) At the centre of the concentric circles. 1

(iii)

1(7)

38. (a) A receptor is a large protein molecule situated in the cell membrane. 1

(b) Receptors bind the body’s chemical messengers, e.g. neurotransmitters or hormones, and this produces a response in the cell. 1

(c) An agonist has the same effect on the cell as the body’s own chemical messengers and produces a biological effect. 1

(d) An antagonist has no effect on the cell when it binds to the receptor. (1)



ORGANIC CHEMISTRY

However, by binding to the receptor, it blocks the body’s own chemical messengers from binding and thus prevents them having their usual effect on the cell (this could then lead to a biological effect). (1)

2(5)

39.

(2)

40. (a) Pharmacologically active – a substance that alters the biochemical processes of the body. (1)Derivative – another molecule with a significant part of its structure the same as the chosen molecule, and which is found to be pharmacologically active. (1)

2

(b) The pharmacophore is the structural fragment of the molecule, that confers pharmacological activity (alters the biological processes in the body). (1)Look for a common structural fragment. (1)

2

(c) Many tests are carried out in the laboratory on animals then on humans – checking toxicity, side effects, checking all isomeric forms of the derivative, test effectiveness of medicine, etc. (some appreciation of testing shown). 1

(5)

41. (a) Moulds, land plants and seaweeds, etc. 1

(b) (i) A pharmacophore is the structural fragment of the molecule that confers pharmacological activity (alters biological processes in the body). 1

(ii) The structure of a (lead) compound can be altered to give a large number of derivatives. (1)As long as the pharmacophore is retained, there is a good chance these derivatives will be biologically active. (1)

2(4)



ORGANIC CHEMISTRY

42. (a)

1

(b) Tests on both optical isomers are carried out. 1(2)

43. (a) X–ray crystallography 1

(b) (i) Pharmacophore 1(ii) It has to fit into the receptor site in the cell

membrane. 1

(c) Computers can have a large database of molecules. They can examine many molecular structures quickly looking for a possible pharmacophore. 1

(4)

44. (a) They will only work as an enzyme for one specific biological reaction. 1

(b) (i) The surface shape at the active site of the enzyme (1)is such that only a specific molecule with a definite shape can fit into the active site. (1)

2

(ii) Once the reaction has taken place at the active sites, the products leave the site. The enzyme is now available, leaving it free for another reactant to fill the site. 1

(4)



EXTRA QUESTIONS

EXTRA QUESTIONS

1. (a) (i) Only a small part of the diagram in the question represents the visible spectrum (around 3 10 6 m–1

to 2 10 6 m–1). 1

(ii) Balmer series 1

(b) An electron is excited from its ground state to a higher energy level. (1)On its return it emits energy (1)equal to the E between the two energy levels involved. (1)

3

(c) E = hc/ = hc for one photon E for 1 mol = Lhc

(Note: the question reminds you ‘mol –1’, so L is needed.) (1)

= 6.02 1023 6.63 10–34 3 108 11 106 (1)

= 1317.1 103 J mol–1

(Check the units by ‘fitting’ them in the above equation.)

= 1317 kJ mol–1 (1)3

(d) The first ionisation energy 1(9)

2. (a) 1s2 2s22p6 1

(b) N3– or Al3+ 1

(c) When electrons occupy degenerate orbitals, these orbitals are filled singly, keeping spins parallel, before pairing occurs. 1

(d) Consider three factors for each of these three elements (taken in the order C, N, O) and predict the effect each factor has on ionisation energy (IE).



EXTRA QUESTIONS

Nuclear charge: increases regularly; regular increase in IE Atomic size: decreases regularly; regular increase in IE Shielding effect: constant; no effect on IE

The net effect of these three factors is to predict that IEs should increase regularly. (1)

This does not agree with the given values. The increased stability provided by a half-filled energy level leads to a higher value of IE for N than expected. (1)

2 (5)

3. (a) (i) The principal quantum number indicates the shell to which the electron belongs (counting from the nucleus). 1

(ii) Degenerate orbitals are those with exactly the same value of energy. 1

(b) 3s fills completely before 3p starts to fill, because 3p is of higher energy. (1)The three degenerate orbitals are half-filled (singly filled) before doubly filling as pairing of electrons requires more energy. (1)

2 (4)

4. (a) Ti3+ 1s2 2s2 2p6 3s2 3p6 4s0 3d1 1(The second last term may be omitted.)

(b) Al [ Ne ] 3s2 3p1

Ti [ Ar ] 4s2 3d2 (1)

The 4 th IE for Al represents the removal of an electron from the stable octet, hence the dramatic rise after the 3rd value. The 4 th IE for Ti removes the last 3d electron only and does not involve breaking into a stable octet arrangement. (1)

2 (3)



EXTRA QUESTIONS

5. (a) p-type 1

(b) n-type 1 (2)

6. (a) Fossil fuels: pollution or an example of pollution or CO2

linked with greenhouse effect or finite resources, etc. (1) Nuclear fuel: risk of (large-scale) accident or finite resources or disposal of dangerous waste, etc. (1)

2

(b) Si, As, P, B, Al, Ga, Sb (any 4 ½) 2

(c) (i) An n–type semiconductor consists of Si doped with As, P or Sb. (1)A p–type semiconductor consists of Si doped with Al, B or Ga. (1)These are arranged as shown to create a p–n junction. (1)

3

(ii) Electron-hole pairs are created at the p–n junction by solar radiation. Electrons migrate towards the n-type and holes towards the p-type, (½ + ½) i.e. a potential difference is set up. (1)

2

(iii) Photovoltaic effect 1 (10)

7. (a)

(1)



EXTRA QUESTIONS

(b) (i) 8:8 in CsCl 1(ii) Each Cs+ ion is surrounded by 8 Cl– ions and each

Cl– ion is surrounded by 8 Cs+ ions. 1

(c) The relative sizes of the positive ion and the negative ion are very different. or The Cs+ ion is much bigger than the Na+ so it is more stable with more Cl– ions around it (and vice versa ). 1

(d) Na+/Cl– = 95/181 Cs+/Cl– = 174 /181 K+/F– = 133/133= ~ 0.5. = ~ 1 (0.96) = 1 (1)

KF may take a CsCl but not an NaCl structure. (1)2

(6)

8. (a) (i) Covalent (or polar covalent) (1)(ii) Ionic (1)

2

(b) Relative formula mass for AlCl3 = 27 + 106.5 = 133.5

This is exactly half the measured value, so a dimer is indicated, i.e. the formula is Al

2Cl

6 in the vapour phase. 1

(c) Al2Cl

6 + H

2O 2Al(OH)

3 + 6HCl (–½ per error)

2 (5)

9. (a)

Lithium hydride Hydrogen bromide

Types of bonding Ionic (½) Covalent (½)

Effect on moist pH paper

Reacts to form an alkali (½)

Ionises to form an acid (½)

2

(b) (i) 2H– H2

+

2e– (1)(ii) 2Br– Br

2 + 2e– (1)

2 (4)



EXTRA QUESTIONS

10. (a) (i) No two electrons can have the same four quantum numbers, i.e. exactly the same electron address. The nearest any two can come is to differ only in their spin, e.g. in 1s or 2s or 2p. 1

(ii) In degenerate orbitals, e.g. 2p, the electrons fill all the orbitals singly first before doubly filling. 1

(b) (i) All three orbitals are of exactly the same energy. 1

(ii) They are orientated at right angles to each other. 1

(c) 1s2 2s22p6 3s23p6 4s0 3d5 (The 4s term may be omitted.) 1 (5)

11. (a) The ligand causes to be of a value that absorbs the red/orange (1)so green is transmitted. (1)

2

(b) CN– creates a larger split in the d orbitals, i.e. is greater. (1)This means absorption would probably include yellow and green as well as the red/orange (1)transmitting blue or violet. (1)or The value of may be even greater still (1)absorbing say yellow/green/blue/indigo/violet (1)and the red/orange might then be transmitted. (1)

3

(c) Fe(II) has the electronic arrangement [Ar] 4s 0 3d6 in both complex ions. (1)In the hexacyano complex the value for is larger than the hexaaquo complex. (½)The larger value of the less likely electrons are to be promoted, (½)i.e.

(1)



EXTRA QUESTIONS

The hexaaquo complex is probably paramagnetic (with four unpaired electrons) but not the cyano complex (with no unpaired electrons). (1)

4(9)

12. (a) (i) ICl3

= 0I + [(–1) 3] = 0

I = +3 1

(ii) HIO3

= 0(+1) + I + [(–2) 3] = 0

I = +5 1

(b) (i) Reaction A (The oxidation numbers of I and of Cl have not changed.) 1

(ii) Hydrolysis 1

(c) (i) Tetrachloroiodate (ii) Square planar 1

1 (6)

13. (a) Each has a lone pair to donate in the formation of a dative covalent bond. 1

(b) (i) Tetrachlorocobalt(II) 1(ii) Hexaamminecobalt(II) 1



EXTRA QUESTIONS

(c)

(Clues: intense red; high % absorbance intense red; blue end of spectrum is absorbed.)

The Data Booklet, page 14, reminds you how to match colour to wavelength.) 2

(5)

14. (a) Adsorbed 1

(b) (i) [Ar] 4s2 3d6 1

(ii) The d orbitals of Fe form easily reversible bonds with reactants. (1)This loosens other bonds within the reactants. or This holds one or more reactants in a suitable orientation for a successful collision. (1)

2

(c) (i) The surface area is extremely large. 1

(ii) Smaller surface area per unit mass 1(6)

15. Metal oxides are basic/alkaline and non–metal oxides are acidic. (1)Al

2O

3 is amphoteric, i.e. it shows both basic and acidic

properties. (1)(2)

16. (a) 2Fe3+(aq) + 2I–(aq) 2Fe2+(aq) + I2(aq) 1



EXTRA QUESTIONS

(b) n(I–) = c v = 0.1 40/1000 = 0.004 mol (1)

so n(I2) = 0.002 mol (1)

2

(c) 1 mol I2 2 mol S

2O

32–

n(S2O

32–) = 0.004 mol (1)

c = n/v = 0.004/15 1000 mol l–1

= 4/15 = 0.267 mol l–1 (1)2

(d) Starch added as an indicator would lose its blue colour at the end-point. 1

(6)

17. (a) Fe3+ or +3 or III 1

(b) Ti3+(aq) Ti2+(aq) + e– (1)Fe3+(aq) + e– Fe2+(aq) (1)

2

(c) (i) 22.63 cm3 (½ for number + ½ for units)The rough titre should be ignored (½) and an average of three concordant titres is our best estimate. (½)

2

(ii) n(Ti3+) = c v = (0.10 22.63)/1000 = 2.263 10–3 mol (1)

so n(Fe3+) = 2.236 10–3 mol= 55.8 2.263 10–3 g Fe= 0.0126 g Fe in 25 cm3 aliquot = 0.126 g in 250 cm3 (1)

so %Fe = 0.126/2.65 100 = 47.65% (1)3

(8)

18. (a) K = [NH3]2/[N

2][H

2]3 1

(b) (i) 0.4 mol NH3

requires 0.2 mol N

2, leaving 1.8 mol unreacted (1)

0.6 mol H2, leaving 1.4 mol unreacted (1)

2



EXTRA QUESTIONS

(ii) K = 0.42/(1.8 1.43) = 0.16/4.94 = 0.0324 l2 mol–2 1

(c) K will decrease (1) because raising T favours the endothermic reaction, i.e. [reactants] increases lowering K (1)

2 (6)

19. (a) K = [H2][I

2]/[HI]2 1

(b) K = (0.11 0.11)/0.782 (1 + 1)= 0.02 (1)

(–½ if units given)3

(c) (i) Increases (Le Chatelier) 1(ii) No change (same number of gaseous molecules

on each side of the equation) 1

(d) Less than that for HI 1(7)

20. (a) The solute is distributed between the two solvents in a definite ratio called the partition coefficient. 1

(b) Yes 1

(c) 0.081/0.00026 = 311.50.11/0.00035 = 314.3 average = 310.70.16/0.0052 = 307 (Note: this is a ratio and0.31/0.001 = 310 therefore has no units.) 2

(d) By shaking, say, equal volumes of, the two solvents with

varying amounts of ammonia (1)and by titrating aliquots of these two solvents with a standard solution of a strong acid. (1)

2

(e) Water 1

(f) Water is a polar solvent capable of forming H-bonds with ammonia. (1)Ether is a non–polar covalent solvent with no such strong attraction for ammonia solvent. (1)

2



EXTRA QUESTIONS

(g) 0.5 1

(h) Water and alcohol are miscible. 1(11)

21. (a) Pencil lead (a mixture of clay and graphite) is insoluble in the solvents and will not move. 1

(b) Rf = distance moved by spot distance moved by solvent

front For spot A R

f = 4/5 = 0.8; for spot B R

f = 3/5 = 0.6 1

(c) (i) Allow the chromatogram a longer run, e.g. by using descending chromatography. 1

(ii) Find a ‘better’ solvent mixture. 1

(d) No, since either spot may consist of two or more dyes with very similar R

f values 1

(e) Either less was spotted or the concentrations are less in sample 2. 1

(6)

22. (a) A base is a proton acceptor. 1

(b) CH3COOH is the base. 1

(c) NO3

–(aq) 1 (3)

23. (a) C 1

(b) Phenol red 1

(c) (i) The mixture consists of a weak acid (methanoic acid) and its sodium salt (sodium methanoate) – the ingredients for a buffer. 1

(ii) If H+(aq) is added it will be neutralised by reacting with the methanoate ions (available in large amountsfrom the salt) to form molecules of the weak acid, (1)

i.e. HCOO– + H+ HCOOH (1) 2(5)



EXTRA QUESTIONS

24. (a) 1 cm3 ammonia has a mass of 0.88 g Mass of ammonia in 1 cm3 = 0.88 28/100 = 0.2464 g (1)No. of moles of ammonia per 1 cm 3 = 0.2464/17

= 0.01449 mol (1)= 14.49 mol l–1 (1)

3

(b) Because it is so volatile, the concentrated solution loses NH

3 gas, so the concentration is constantly falling. 1

(c) Salt of a strong acid with a weak base has a pH below 7 (1)so bromocresol green is the indicator to use, with a pK

a value below 7. (1)

2 (6)

25. (a) (i) Voltage of the cell = 1.60 V 1(ii) G° = –nFEo (1)

= – (2 96500 1.6)/1000 (1)= – 308.8 kJ mol–1 (1)

3

(b) Conditions are not standard, e.g. [H2SO

4] > 1 mol l–1 1

(c) Lead sulphate (since Pb2+ is a product in both half reactions and the solution has a high concentration of sulphate ions). 1

(6)

26. (a ) Above 2100K ( 50K) 1

(b) Separation of gaseous products; building a furnace that can withstand such high temperatures; cost of maintaining such high temperatures; Mg could well be extremely reactive at this temp; cost of fuel/energy; etc. (any two earn two marks) 2

(c) (+850 – 670 ) kJ mol–1 (1)= +180 kJmol–1 ( 20 kJmol–1) (1)(–1 for wrong or missing sign) 2

(d) +88 kJ (Note: if incorrect accept an answer that is 92 kJ less than (c) above.) 1



EXTRA QUESTIONS

(e) Reverse reaction is slower if concentration of Mg is reduced (or nil if all Mg removed). or If equilibrium is disturbed by the removal of Mg then the equilibrium position moves in favour of making more Mg. 1

(7)

27. H for reaction = –46.2 + (–92.3) – (–315.0)= 176.5 kJ mol–1 (1)

S for reaction = +193 + 187 – 94.6 = 285.4 J K –1 mol–1 (1)But G = H – TS = 0 at minimum temperature (1)i.e. T = H/S

= 176.5 103/285.4 = 618.4K (1)(4)

28. (a) Rate = k [RCl] or rate [RCl] 1

(b) Rate = k [R3 CCl ][I–] or rate [(CH

3)

3CCl ][I–] 1

(2)

29. (a) The order with respect to HgCl2(aq) is first. (½)

The order with respect to C2O

42– is second. (½)

So rate = k [HgCl2]1 [C

2O

42–]2 (1)

2

(b) k = 1.82 10–4 (0.128 (0.304 )2) (½)(applying units to this formula: mol l –1 min–1 mol l–1 mol2 l–2) (½)

= 0.0154 mol–2 l2 min–1 (½ + ½)2

(c) Rate = 0.0154 0.1 0.1 0.1= 1.54 10–5 mol l–1 min–1 (½ + ½)

(remember the units!) 1



EXTRA QUESTIONS

(d)

The first mark is for a line graph that curves generally as shown, representing a decreasing rate of change of concentration. (1) The second mark is for a tangent to the curve at time zero. (1)

2(7)

30. (a)

(1)

Neither C atom has four different groups attached. (1)2

(b)

The methyl group could be R or any group. 1

(c) It will exist in two enantiomeric forms that are optically active. 1(4)



EXTRA QUESTIONS

31. (a) (i) By an addition reaction with hydrogen using a finely divided Ni catalyst. 1

(ii) To lower the melting point of the triglycerides in the oil so that the margarine can have the desired properties, e.g. spread straight from the fridge. 1

(b)

(½ + ½ for each cis-arrangement) (½ + ½ for double bond correctly placed on C

9 and C

12) 2

(c)

1

(d) trans-octadec-9-enoic acid 1

(e) There is no chiral centre. or There is no optically active carbon (½) so it is not optically active. (½)

1(7)

32. (a) Mass of C = 12/44 0.905 = 0.247 g (½)Mass of H = 2/18 0.369 = 0.041 g (½)Total mass of C + H = 0.288 g So mass of O = 0.507 – 0.288 = 0.219 g (1)



EXTRA QUESTIONS

Elements C H O

Masses 0.247 0.041 0.219

Relative atomic mass 12 1 16

Ratio 0.021 0.041 0.014

Allowing for experimental error

21

3

41

6

14

2

(–1 for each error) (2)

empirical formula = C3H

6O

2(1)

5

(b) C2H

5COOH 1

(6)

33. (a) Mass = 60 therefore isomers are C3H

6O

(i) CH3CH(OH)CH

3 and (ii) CH

3CH

2CH

2OH 2

(b) (i)

4

(c) Isomer B the has biggest intensity at m/z = 31. This corresponds to CH

2OH and is only possible with the primary alcohol,

structure (ii). 2 (8)



EXTRA QUESTIONS

34. (a) CH3–CH

2–OH (1) CH

3–O–CH

3 (1) 2

(b) (i) CH3–CH

2–OH 1

(ii) 0 is TMS; 1.0 is CH3; 3.5 is CH

2; 4.0 is OH (4 ½) 2

(c) (i) Integrals (1)(ii) The length of the lines indicates the relative areas

under the peaks. (1)In this case relative areas are 1:2:3 (A:B:C) 2

(d) The molecule is symmetrical with all the hydrogen atoms in the same environment. There will only be one peak.

1(8)

35. (a) The NMR operator sets the value to zero as it is the internal standard. 1

(b) There are only two ‘types’ of protons in compound G. 1

(c) From the integral ratio of the peaks of 2:1 no reasonable structure can be drawn for CH

3. Thus, try C

2H

6O but molecule must be

symmetrical.The structure must be HOCH

2CH

2OH – it is the only one

to fit the data. 3

(d) A broad peak due to OH stretch between 3600 and 3300 cm–1 1

(6)



EXTRA QUESTIONS

36. (a) (b)

CH3CH2CH2CH2OH five peaks; five different hydrogen environments

butan-1-ol

CH3CHCH2CH3 five peaks; five different hydrogen environments

OHbutan-2-ol

CH3CHCH2OH four peaks; four different hydrogen environments2 CH3 groups are identical

CH3

methyl propan-l-ol

CH3

CH3 C OH two peaks; two different hydrogen environments3 CH3 groups are identical

CH3

methyl propan-2-ol(correct number of peaks ½ each;

(structures and names ½ each) correct explanation ½ each)(6)

37. (a)

1

(b) (i) Cyclohexane has one peak (1)Cyclohexane-1,4-diol has three peaks (1)

2



EXTRA QUESTIONS

(ii) All hydrogen atoms in cyclohexane are in the same environment. (1)In cyclohexane-1,4-diol there are three different environments and three peaks, as shown.

(2) (–1 per error)3

(6)

38. (a)

2

(b) (i) CH3CH

2OH

(ii) CH3OCH

3 (2 ½ )

1



EXTRA QUESTIONS

(c)

Isomer Group Approximate value

Integral

CH3CH

2OH CH

3 0.8–1.3 3

CH2 3.5–4.0 2

OH 3.0–6.0 1

Isomer Group Approximate value

Integral

CH3OCH

32 CH

3O 3.0–4.0 6

(8 ½) (½ for approximate value and ½ for integral)

4

(d)

Both peaks in correct position, i.e. at correct approximate value (ignore integral) 2

(9)

39. (a) C 70.59/12 = 5.88 H 5.88/1 = 5.88 O 23.53/16 = 1.47 (1)(divide by 1.47) C 5.88/1.47 = 4 H 5.88/1.47 = 4 O 1.47/1.47 = 1 Therefore empirical formula = C

4H

4O (1)

(which equates to mass of 68) 2

(b) Molecular mass = 136 so molecular formula = C8H

8O

2 1



EXTRA QUESTIONS

(c) The infra-red absorption at 1710 indicates the presence of a carbonyl group. (1)Two oxygens present so ester or acid (1)but acid ruled out because no large peak between 3500 and 2500 cm–1. (1)NMR spectrum peak at 6–7 (integral 5) indicates a monosubstituted benzene ring. (1)Peak around 2 (integral 3) indicates a CH

3 attached

directly to C=O. (1)5

(8)

40. (a) (i) Agonist interacts with receptor to give response similar to the natural compound. 1

(ii) Antagonist interacts with receptor to produce no response and it prevents action by a natural compound. 1

(b) Agonist acts like a good copy of a car key and is able to switch the car engine on (1)but an antagonist is a poor copy of the key which will go in the lock but will not turn and so cannot switch theengine on. (1)It can also prevent the real key from being used if it sticks in the lock. (1)

3

(c) There are three chiral carbons at atom numbers 3, 5 and 6 (–1 for each wrong answer.) 2

(7)



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