electronic structure and the periodic table unit 1 · = 6.02 × 1023 × 8.35 × 10Œ19 j molŒ1 =...

ELECTRONIC STRUCTURE AND THE PERIODIC TABLE

ANSWERS TO ADDITIONAL QUESTIONS (AH CHEMISTRY) 1

UNIT 1

1. BrBr bond enthalpy = 194 kJ mol1 (1)E = Lhν for one mole of bonds (See units above.) (1)ν = E/Lh = 194000/6.63 × 1034 × 6.02 × 1023 J/Js

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

= 4.86 × 1014 s1 (or Hz) (1)(4)

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

(b) The energy gaps between energy levels decrease withincreasing energy, i.e. the higher energy levels getcloser and closer together. 1

(c) (i) ∆E = hν= 6.63 × 1034 × 1.26 × 1015 J= 8.35 × 1019 J (1)

But IE = L × ∆E= 6.02 × 1023 × 8.35 × 1019 J mol1= 50.27 × 104 J mol1= 502.7 kJ mol1 (1)

2

(ii) The first ionisation energy of the element.(Check page 10 of the Data Booklet to confirmthat this is a likely answer.) 1

(5)

3. (a) Line A has a longer wavelength than all the othersshown. (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)2

(b) E = Lhc/λ (1)(Note: The question tells you to include L in thisequation.)

= (6.02 × 1023 × 6.63 × 1034 × 3 × 108 )/656 × 109 (1)= 0.1825 × 106 kJ mol1= 182.5 kJ mol1 (1)

3(5)

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

(b) E = Lhc/λ (1)= 6.02 × 1023 × 6.63 × 1034 × 3 × 108/284 × 109 (1)= 0.422 × 10 (23 34 + 8 + 9) J mol1= 422 kJ mol1 (1)

3

(c) Energy from the spark excites some electrons to a higherenergy level. (1)When these electrons return to ground state a specificamount of energy is released and this shows up as aline of measurable wavelength (or frequency) in thespectrum. (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/λ = Rh (1/2

2 1/42)= 1.097 × 107 × (1/4 1/16)= 1.097 × 107 × 3/16= 2.06 × 106 (1)

λ = 1/(2.06 × 106) = 0.485 × 106= 485 nm (1)

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

3



(b) n1 = 1 (1)

All the jumps must be of shorter wavelength,i.e. of higher energy than those for n1= 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 aroundthe nucleus. (1)p orbitals are dumb-bell shaped and aresymmetrical around each axis. (1)

2

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

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

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

degenerate orbitals singly before any one isdoubly filled. (1)

The correct configuration is:

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)

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



8. (a) (ii) is wrong because the 2s electrons have identicaladdresses, i.e. all four quantum numbers the samefor both.or Electrons cannot have parallel spin in the sameorbital. (1)This violates Paulis exclusion principle. (1)

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

orbitals (with parallel spin). (1)Hunds rule (1)

2

(c) (vi) has violated the Aufbau principle since the ssublevel (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 likelyto be found. 1

(c) It signifies the second energy level. 1

(d) Of equal energy 1

zy

x

2s

2px

zy

x

zy

x

zy

x

2py 2pz



(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 mol1 (±20 kJ mol1 ) 1

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

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

(ii) Each new energy level means a larger radius (lessattraction for the outermost electron) and providesa greater shielding effect (again reduced attractionby the nucleus). 1



(d) Two factors apply: the steady increase in nuclear chargeand the slight decrease in atomic radius from Li to Nemakes the attraction of the nucleus for outer electronsgreater. 1

(e) (i) Be 1s2 2s2

B 1s2 2s2 2p1, i.e. B has started a new subshell soits outermost electron is relatively easier to removethan that of Be, where a complete subshell has to bebroken into. 1

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

(7)

13. (a) Electrons are excited by electric discharge to a higherlevel. (1)These electrons emit energy as they return to a lowerenergy level. (1)The quantity of energy emitted depends on the energyvalues of the two energy levels involved. (Many linesmay be produced and each will represent a specificelectronic jump.) (1)

3

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

2(5)

14. (a) Since E = hc/λ or E ∝ 1/λ, shorter wavelengthscorrespond to higher energy. (1)or It can be assumed that these lines represent a partof a series of lines which converge (at a continuum, notshown) at the higher energy end, i.e. to the left on thediagram shown in the question. Both arguments lead tothe conclusion than 393 nm represents the highestenergy value shown. (1)

2



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

(c) ν = 1/λ = 1/620 × 107 (1)= 16129 cm1 (Remember the units.) (1)

2

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

(6)

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

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

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

(ii) A spectral line on the spectrum for each jumpor 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 Ca2+(aq). (1)Use the solutions to make a graph of intensity ofradiation 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 aspecific metal. Particles of the metal in the sampleabsorb a measurable amount of this light in proportionto their concentration as electrons are promoted. 1

(ii) The quantity of energy absorbed, i.e. the differencein intensity, between the incident light and thetransmitted light. 1

(12)

16. (a)

1

(b) (i)

2

(ii)

1

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

so the negative charge is considered to be sharedequally between all three atoms and not to beresiding 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 inbonding pairs with fluorines. All four bonds areidentical. The bond angle will be the tetrahedral angleof 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 downwardsso the bond angle is slightly less than 109.5°. (1)

2(3)


ANSWERS TO ADDITIONAL QUESTIONS (AH CHEMISTRY)1 0

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 thebonding pairs hence they are pushed down below thetetrahedral angle, creating a pyramid: (1)

Boron has only three outermost electrons, so BCl3 hasonly three pairs of bonding electrons. They spreadthemselves symmetrically (or as far from each other aspossible), i.e. pointing to the corners of an equilateraltriangle. (1)

2(4)


ANSWERS TO ADDITIONAL QUESTIONS (AH CHEMISTRY) 1 1

20. (a) There are five pairs of electrons around the centralchlorine atom (since an atom of Cl has seven outermostelectrons and each F atom contributes one electron tothe 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 theoppositely charged ions. (1)

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

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

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

Na+/Cl = 95/181 = 0.52Cs+/Cl = 174/181 = 0.96Fe2+/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 iontherefore has six near neighbours ⇒ 6:6 co-ordination. 1

(b) Since Na and Cl have a valency of 1 they cannot havesix bonds.or In the sketch each Na+ has 6 Cl near neighbours andnone of these can be considered to be the partner forthe Na+ any more than any other therefore the linesrepresent the direction along which the forces of ionicbonding apply (and do not represent shared pairs ofelectrons as in covalent bonds of covalent molecules.) 1

(c) Molecules of sodium chloride do not exist. Onlycovalent 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 + SiHCl3 → 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 itconducts electricity by a mechanism that we associatewith 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) andthese extra electrons are the charge carriers. (1)

2

(ii) Boron 1

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

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

2

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

(ii) When a pn junction is irradiated with light,electron hole pairs are formed and electronsmigrate. (1)

The electrons migrate → n-type semiconductorand the holes → p-type. The n-type is thereforenow negative with respect to the p-type, i.e. apotential 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 resistanceat all to the flow of an electric current. 1

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

(d) Tc for some new materials was above the boiling point ofnitrogen a much cheaper coolant. 1

(e) (i) This would bring superconductivity within reach atambient temperatures, with no need for expensivecoolants. 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 networkstructure so covalent bonds have to be broken onmelting, i.e. the melting point is very high. (1)

N2 has a covalent molecular structure so only van derWaals 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 inelectronegativity decreases to zero. (1)

Polarity rises from N to F since the difference inelectronegativity increases again (with the polarityreversed). (1)

2(7)

30. The H+ ion would consist of only a nucleus and as suchcannot have a separate existence.(In fact it exists in aqueous solution attached to a watermolecule, i.e. H

3O+.) (1)

31. (a) AlCl3 = 27 + (3 × 35.5) = 133.5 but the relative molecularmass = 267, i.e. twice the value for the empirical formula.The structure must therefore be a dimer of AlCl3,i.e. Al2Cl6. 1

(b) Covalent molecular structure with dipoledipoleinteractions. 1

(c) Aluminium hydroxide. (1)

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

2(4)



32. (a) Li+H or Mg

2+(H)2

(1)Each reacts vigorously (at least) with water, releasingH2 gas, leaving the metal hydroxide in solution. (1)

2

(b) Al2O3 + 6HCl → 2AlCl3 + 3H2O (1)base + acid → salt + water (½)(Note: there is no need to show the dimeric nature ofthe salt since it will be in solution as separate ions.)

Al2O3 + 3H2O + 2NaOH → 2NaAl(OH)4 (1)acid + base → salt (½)

sodium aluminateor 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 tothe 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 simplydissolve in non-polar solvents without reaction, but arehydrolysed by water, since it is polar, giving fumes ofHCl(aq), as well as an oxyacid or an oxychloride bothseen as an oily droplet.) 2

(d) The hydrogen bond is the force of attraction betweenan H atom of one molecule (which is directly attachedto an element of high electronegative value) (1)and an atom of high electronegative value on anothermolecule. (1)For water (or HF or HCl) (1)the boiling points are higher than predictions basedonly on molecular size. (1)(This is also true of melting point, of viscosity, ofsurface tension and of specific heat capacity.Hydrogen bonding also causes water to have itsmaximum value of density at 4oC, (ice floats), by holdingneighbouring molecules a little further apart than liquidmolecules.) 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 thatis exactly half-filled. This is a more stable arrangementthan that with one more electron. 1

(3)

34. (a) +1 1

(b) 10 1

(c)

When the ion is irradiated with white light, energy isabsorbed and promotes one electron from a lower 3denergy 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 dativecovalent bond with the central metal ion, Fe2+. (1)

2(b)

(1) octahedral (1)

2

(c) When a water molecule acts as a ligand its OH bondswill become more polarised than before so H+ ionswill tend to form. (1)

However, the hydroxide ion donates a complete negativecharge to the central metal ion, effectively reducing thepositive charge on it and so reducing the polarisingeffect on the OH bonds. H+ ions are therefore lesslikely to form, i.e. the solution is less acidic. (1)

2(6)

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

(b)

(Note: the 4s box may precede the 3d or it may beomitted 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 covalentbonds, not by ionic bonds. The ionic chloride ions arefree to form a precipitate with Ag+ ions whereas thecovalently bonded chlorides are not. (1)

4

(b) The degree of dd splitting is controlled by the ligandsinvolved. (1)NH3 on its own will therefore be different from NH3with 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 CrO42 is in oxidation state +6 (see above).

In Cr2O72 , Cr = ?

(2 × ?) + (7 × 2) = 2? = +12/2 = +6So Cr in Cr2O7

2 is also in oxidation state +6. (1)This is therefore not a redox reaction. (1)

2

(c) dd splitting can only apply when there is at least oneelectron in a d orbital (this theory similarly does notapply to a completely filled d sub-shell). 1

(5)



39. (a) Colour intensity is at a maximum when the concentration ofthe 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 theNi2+ ion. (1)

2

(c) The complex is purple, i.e. it absorbs in the greenregion. (1)The filter transmits green to get the best absorptionfrom 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 absorbedas electrons are promoted across the small energy gap, ∆,now existing in the d orbitals. Hence the green colour isseen. (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 l1 (1)(Remember to subtract 3 from the colorimeter readingfor the unknown.) 3

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

(d) Duplicates should be preparedor Large volumes should be used (to reduce error).or A blank should be carried out on the buffer and solventwithout any sample present. 1

(e) dd splitting or dd transitions 1(7)

42. (a) Five degenerate orbitals are split into two orbitals ofhigher energy and three of lower energy. Electrons canbe promoted across this gap by absorbing energy fromthe visible spectrum. (1)

The peak around 410 nm represents the wavelengthsabsorbed (and equals the energy value of the dd split). (1)

2



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

3(5)

43. (a) [VO2]+ V + (2 × 2) = +5

V = +5 (or V) (1)[V(H2O)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 thegreen 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 offrequencies) from the visible spectrum promoteselectrons across the gap. (1)Remaining frequencies are transmitted and theseproduce the colour. (1)

2

(c) The size of the dd split is altered by the presence ofdifferent ligands, so frequencies absorbed are differentand 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/λ (1)λ = Lhc/E = 6.02 × 1023 × 6.63 × 1034 × 3 × 108/239 × 103 (1)

= 119.74 ×10 6/239= 5.01 × 107m= 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/λ (1)= 6.02 × 1023 × 6.63 × 1034 × 3 × 108/540 × 109 J mol1(1)= 0.222 × 106 J mol1= 222 kJ mol1 (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 significantcatalytic properties.(any two) 2

(c) Fe Haber processPt 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 covalentbonds, often with d orbitals of the catalyst) (1)in a suitable orientation for a more probably successfulcollision.or in such a way that bonds within the reactant areweakened (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 × 103 × 0.1 = 2.48 × 103 1

(d) Moles of Cu2+ = 2.48 × 250/25 = 2.48 × 102 (1)Mass of Cu = 2.48 × 102 × 63.5 = 1.575 g (1)Percentage of Cu= 1.575/2.63 × 100 = 59.89% (1)

3(6)

2. Fe(NH4)2(SO4)2.6H2O = 392Moles of Fe2+ = 1.8/392 = 4.6 × 103 (1)From redox equation Fe2+: MnO4

= 5:1 (1)Moles of MnO4

= 4.6 × 103/5 = 9.2 × 104 (1)Concentration = 9.2 × 104/0.04 = 0.023 mol l1 (1)

(4)

3. (a) (i) AgCl =143.4Mass 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 = 121.6Mass 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 + MgCl2 → 2AgCl + Mg(NO3)2Moles of AgCl = 2.01/143.4 = 0.014 (1)Moles of MgCl2 = 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 × 103 × 1.0 = 1.5 × 102 (1)Moles of Na2CO3 = 1/2 × 1.5 × 10

2 = 7.5 × 103 (1)Mass of Na2CO3 = 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 Na2CO3 = 7.95/106 = 0.075 (1)Ratio of moles = 0.447/0.075 = 5.96 = 6 (i.e. 6H

2O) (1)

2(6)

6. (a) Moles of NaOH = 18.2/1000 × 0.1 = 1.82 × 103 (1)Moles of CH2(COOH)2 = 9.1 × 10

4 (1)2

(b) Mass of acid = 9.1 × 104 × 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 BaSO3 = 1.09/217 = 0.005 (1)

2

(b) Moles of Na2SO3= 0.005 × 250/50 = 0.025 (1)Mass of Na2SO3 = 0.025 × 126 = 3.15 g (1)%Na

2SO

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 sameequilibrium more quickly. (1)

2

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

2(6)

9. (a) (i) K = [SO3]/[SO2]2[O2] 1

(ii) No units 1

(b) (i) Le Chateliers principle states that if a system issubjected to any change, the system readjusts itself tocounteract the applied change. 2

(ii) Increased temperature favours ∆H +ve. Thus thebackward reaction is favoured and the equilibriumposition 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 smallervolume therefore an increase in pressure favours theproducts.

Reaction B goes from a small volume (no gas) to a largevolume therefore an increase in pressure favours thereactants.

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

Reaction E has no gas therefore pressure change has noeffect.

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

(b) It will have the same general shape but it will be lesssteep. 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 +vetherefore the products are favoured (1)therefore K decreases. (1)

2(d) 15 = [NO2]

2/[0.1]2 × [0.1][NO

2] = √0.015 (1)

= 0.12 mol l1 (1)2

(7)

12. (a) [Ba2+] = √1 × 1010 = 1 × 105 mol l1 1

(b) BaSO4 = 233.4 g mol1 (1)

Mass dissolving = 233.4 × 1 × 105 (1)= 2.334 × 103g (1)

3



(c) (i) Ksp

is a constant therefore stays the same. 1

(ii) The equilibrium moves to the left therefore [Ba2+]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)/100V = 0.106 gTotal extracted = 0.943 + 0.106 = 1.049 g (1)

2(d) Many organic compounds are nonpolar and dissolve

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

14. (a) Moles of acid originally = 2.36/118 = 2 × 102 1

(b) (CH2COOH)

2 + 2NaOH → (CH

2COONa)

2 + 2H

2O

Moles of NaOH = 34.8/1000 × 1 = 3.48 × 102 (1)Moles of acid = 1.74 × 102 (1)Moles in ether = (2 × 102) (1.74 × 102) = 2.6 × 103 (1)

3

(c) K = 2.6 × 103/0.1/1.74 × 102/0.1 = 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 maynot travel the same distance as glycine. 2

(v) Solvent front 1

(vi) Base line1

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

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

substance has the same Rf value for the sameconditions. 1

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

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

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

(vi) Yes (mobile and stationary phases) 1

(vii) The components are colourless and so must bereacted with a developing agent to make themvisible. 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 thecolumn from the point it was injected at to the pointwhere 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 mustbe an another isomer of pentane.Peak A is at about 1.7 and therefore the moleculemust be smaller than ethane. 3

(9)

17. (a) H2O 1(b) OH 1(c) NH3 1

(3)

18. (a) (i) The HSO3(aq) provides an H+(aq) ion. 1

(ii) The SO32(aq) is the conjugate base. 1

(b) HSO3(aq) + H+(aq) H2SO3(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 × 105 × 0.01) (1)= √1.7 × 107= 4.12 × 104

pH = log H+ = log 4.12 × 104 = 3.4 (1)3

(b) CH3COOH(aq) CH3COO(aq) + H+(aq) acid

CH3COONa+(s) → CH

3COO(aq) + Na+(aq) salt

(i) Add HCl the H+ (aq) ions added react withthe 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 ionisesto 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 × 1014, therefore [H+] = 7.16 × 107 andpH = 6.15 (1+1+1)

3

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

w and

therefore more ions form. Thus the ionisation isendothermic. 1

(5)

21. (a) NaH2PO4; Na2HPO4; Na3PO4 (any two, 1+1)2

(b) Ka = [H+][ H2PO4

]/[H3PO4] (1)7.08 × 103 = [H+]2/0.1[H+]2 = 7.08 × 104[H+] = 2.66 × 102 (1)pH = 1.6 (1) 3(You could use pH = ½pKa ½log[C].) (5)

22. Kw at 288K = 0.45 × 1014 (1)

[H+] = 6.71 × 108 (1)pH = log 6.71 x 108 = 7.17 (1)

(3)

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

= 0.017 mol l1 (1)[OH] = 2 × 0.017 = 0.034 mol l1 (1)

2

(b) [H+] = 1014/0.034 = 2.94 × 1013 (1)pH = log 2.94 × 1013 = 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 l1 1

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

2

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

(d)

2

(e) No suitable indicator 1(7)

Small point of inflection about pH 7.



26. (a) 1.6/64.1 = 0.025 mol in 250 cm3 (1)= 0.1 mol l1 (1)

2

(b) pH = ½ pKa ½logc (1)= (½ × 1.8) (½ log 0.1) (1)= 1.4 (1)

orK

a= [H+][HSO

3]/[H

2SO

4] (1)

H+ = √Ka × [H2SO4] = √(1.5 × 102) × (0.1) (1)

= 3.87 × 102pH = 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 isunstable. 1

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

(7)

27. Ka = [H+][A]/[HA] = [H+]2/[HA] (1)

[H+]= √Ka × [HA] = √1.3 × 105 × 0.1 (1)

= 1.14 × 103pH = 2.9 (1)orpH = 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 + ∆H1 + 2∆H2 (1)= (2326) + (1923) + 2(338) (1)= 273 kJ mol1 (1)

3

(b) MgCl2(s) + 6H2O(l) → MgCl2.6H2O(s)∆H = Σ∆Hp Σ∆Hr (1)

= 2500 (642) + 6(286) (1)= 142 kJ mol1 (1)

3(6)



29. (a) Ionic bonding becomes weaker going down the series. orHalide ion increases in size going down the series (mustmention ionic/ion). 1

(b) Enthalpy hydration = enthalpy solution lattice enthalpy(breaking) = 17.2 701 = 683.8 kJ mol1 1

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

2 (4)

30. (a) C(s) + HOH(g) → C~O(g) + HH(g) ∆H = 130 kJ mol1Bond breaking C(s) → C(g) = 715; two OH bonds =2 × 458 = 916; total = 1631 kJ (1)Bond making C~O; HH = 432; total = 432 + C~O; (1)C~O = 130 1631 + 432 = 1069 kJ mol1

Energy given out in forming one mole of C~O bonds= 1069 kJ (1)

3

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

(c) CO poisonous; H2 could be explosive; CH3OH is flammable(any two correct = 1+1) 2

(6)

31. (a) (i) 3C(s) + 4H2(g) → C3H8(g) 140 (1)(ii) C(s) + O

2(g) → CO

2(g) 394 (1)

(iii) H2(g) + ½O2(g) → H2O(g) 286 (1)3

(b) Find C3H8(g) + 5O2 (g) → 3CO2(g)∆H = ∆H(i) + 4 ∆H(ii) + 3∆H(iii)

= (104) + (4 × 286) + (3 × 394) (1)= 2222 kJ mol1 (1)

2 (5)



32. (a) (i) Sublimation enthalpy 109 kJ mol1

(ii) Bond dissociation enthalpy 248.5 kJ(iii) 1st ionisation energy 502 kJ mol1

(iv) 1st + 2nd electron affinity 703 kJ mol1

(v) Lattice formation enthalpy 2481 kJ mol1 (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 mol1 1

(8)

33. (a) (i) ∆S = ΣSp ΣS

r = 208 (201 + 2(187))

= 367 J K1 mol1 1(ii) ∆H = Σ∆Hp Σ∆Hr = (166 (227 + 2(92.3)))

= 209.4 kJ mol1 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 ∆Gθ ve (1)∆Hθ +ve or endothermic, therefore drop intemperature (1)

2



(c) ∆G = ∆H T∆S ∴∆ S = (∆H ∆G)/T = (92 (95))/298 (1)= 10 J K1 mol1 (1)

2(5)

35. (a) C8H

18 → C

8H

10 + 4H

2(1)

∆S = ΣSp ΣSr = (352 + 4(131)) 463 (1)= 413 J K1 mol1 (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 ifK



37. (a)Zn(s) + O2(g) → ZnO2

∆G/kJ mol1

2C(s) + O2(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 mol1 (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 + H2O 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 meansthat ∆S is more ve. 2

(e) Although the reaction is feasible at room temperature itdoes 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 mol1 (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 toproduce more CO2 (any one) 1

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

40. (a) Eθcell

= EθAg

EθM ∴ ∴ ∴ ∴ ∴ E

θM

= 1.03 + 0.8 = 0.23 V 1

(b) ∆G = nFE = 2 × 96500 × 1.03 (1)= 198.79 kJ mol1 (1)

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

(c) Maximum work would not be obtained from cell orthermodynamically reversible conditions do not applyor 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 mol1 (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 potassiumhydroxide (OH). 1

(c) H2 + ½O2 → H2O 1

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

(e) Fuel cells do not produce CO2, SO2 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 × 102 mol l1 (1)

2

(c)T1

Number (fraction) T2of molecules withgiven energy

(1)Energy Ea (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. 4and 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 mol2 l2 s1value (1) units (1) 2

(d)

Initial rate of

N2 /mol l1s1

0.0 0.01 0.02 0.03Initial [NO]/mol l1 1

(6)

45. (a)Experiment 1 2 3

Av. rate/mol l1 min1 0.1 0.2 0.4

Av. rate mol l1 s1 1.67 × 103 3.33 × 103 6.67 × 103

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 mol2 l2 min1 (1.67 mol2 l2s1) (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-determiningstep as both are in the rate expression. (1)

2

(b) (i) A blue/black colour (1)The Br2 displaces the I2 from the KI. The I2 reactswith 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 C2H5*; W Br*; X Br2; Y C4H10;

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 stepproduces more of these radicals to keep the reactiongoing. 1

(8)

2. (a) 2 1

(b) The 2s and 2p orbitals combine to form four hybridisedsp3 orbitals. (1)Each occupied by one electron. (1)

2

(c) The C=C bond consists of a σ bond and a π bond (1)but the CC is just a σ bond. (1)

2

(d)

1(6)

3. (a) Nuclophilic substitution 1

(b) They have a polar CX 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) CH3CHBrCH3 or full structural formula 1(b) Nucleophile 1(c) An ether (alkoxyalkane) 1(d) Any hexanol isomer, e.g. CH3CH2CH2CH2CH2CH2OH 1

(e)

1

(f) Use infra-red to detect the presence of C=O in R or itsabsence 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/ CH3OHesterification (concentrated H

2SO

4)

(6) Neutralisation NH3

(7) Chlorination PCl5/PCl3 /SOCl2

(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

31

(Note: phenols will react with HNO3 alone.)

(c) Reduction 1

(d) C6H4NH2OH → C6H4NHCOCH3OH111 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 hydrocarbonare rearranged to produce a different carbon skeletonby the action of heat and a catalyst. 1

(b) (FriedelCrafts) 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 = RCO2H.)

(3 × 1)3

(b) Reagent acidified dichromate or Tollens or BenedictsChange orange to blue-green or silver mirror formed

or blue to orangeName 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 cm1 and 1725 to 1700 cm1 (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

C5H10O (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-dinitrophenylhydrazineor water (any one). 1

(8)

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

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

2

ORGANIC CHEMISTRY


(c) Ethanoic acid/CH3COOH (1)

1700 to 1725 cm1 (C=O stretch) (1)2

(d) In Grignard reagent Cδ Mgδ+ Br therefore the Cδ acts anucleophile. (1)In alkyl halide Cδ+ Brδ therefore the Cδ+ acts as anelectrophile. (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 bondwhereas CC does allow rotation. 1

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

(5)

14. (a) (i)

2

(ii) Pentane is a longer molecule and will have morevan 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 acidis trans-butenedioic acid. (1+1)

2

ORGANIC CHEMISTRY


(ii) The trans isomer will be able to form hydrogenbonds with neighbouring molecules but the cisisomer will form hydrogen bonds with the otheracid group on the same molecule. (1)The trans isomer will have the higher meltingpoint. (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 atomattached, and two have methyl groups attached thereforenone of them has four different groups attached. (1)

2

(c)

1(5)

17. (a) 2hydroxypropanoic acid 1

(b)

2

ORGANIC CHEMISTRY


(c) It is a racemate (racemic mixture), i.e. it consists of equalamounts 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 twoenantiomers. 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/Benedicts solution/Tollens reagent,but B does not. (1)Molecule A decolourises bromine solution but B doesnot. (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 isomerswill be formed. 1

(7)

21. (a) (i) Structural formulae (ii) CH3CH2CH2CH3(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 gMass of H = 2/18 × 0.258 = 0.0287 gMass of O = 0.315 (0.172 + 0.0287) = 0.1143 g(1 for each error) 2

Mole ratioMoles 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 CO2 = 610 cm

3 ∴∴∴∴∴ moles CO2 = 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 ratioMoles of C 0.305/12 = 0.0254 3Moles of H 0.068/1 = 0.068 8 (1)Moles of O 0.135/16 = 0.0084 1Empirical formula is C3H8O, CH3CH2CH2OH (1)

5

(b) To improve the reliability of the results 1(6)

24. (a) CO+ 28; COH+ 29; CH3O+ 31; CH3OH

+ 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 theseparts. (1)

2(4)

26. (a)

1

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

(c) A molecular ion is formed from the molecule with theloss of only one electron. (1)C6H5Cl

+ (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) C6H5

+ has a mass of 77. 1(7)

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

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

(c) Compare the infra-red spectrum of the distilled samplewith 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 analysisand can be recovered and used again. 1

(b) Xray 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 alignedwith the magnetic field. 1

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

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

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

(8)

31. (a) 12 1

(b) All the H atoms are in CH3 groups attached to the centralsilicon. 1

(c) 0.0 1(3)

32. (a) Tetramethylsilane (1)It is used as a reference and is given the chemical shiftvalue 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 threeH atoms. (1)

2

(c) Peak A has a chemical shift of 3.6 and has an integral of 2,therefore it corresponds to CH2. 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 withthe integral values of 5; 2; 3. (1)Also, the chemical shift values from the Data Bookletagree. (1)

3(6)

34. (a) CH3CH2CH2OH (1)CH

3CH(OH)CH

3(1)

2

(b) (i) CH3CH

2CH

2OH 1

(ii) 3 1(iii) 6 (2 × CH3 ): 1 (CH): 1 (OH) 1

(c) CH3CH2OCH3, 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 chemicalenvironment for CH4 and for CH3CH3. (1)For the CH3CH2CH3 and the CH3CH2CH2CH3expect two peaks from the CH

3 and the CH

2

groups. (1)2

(b) CH3CH(CH3)CH3 2 peaks one for the CH3 groupsand 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) Xray crystallography 1

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

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

(ii) They have only one electron and so their electrondensity 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 cellmembrane. 1

(b) Receptors bind the bodys chemical messengers, e.g.neurotransmitters or hormones, and this produces aresponse in the cell. 1

(c) An agonist has the same effect on the cell as the bodysown chemical messengers and produces a biologicaleffect. 1

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

ORGANIC CHEMISTRY


However, by binding to the receptor, it blocks thebodys own chemical messengers from binding and thusprevents 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 thebiochemical processes of the body. (1)Derivative another molecule with a significant part ofits structure the same as the chosen molecule, andwhich is found to be pharmacologically active. (1)

2

(b) The pharmacophore is the structural fragment of themolecule, 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 animalsthen on humans checking toxicity, side effects, checkingall isomeric forms of the derivative, test effectivenessof 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 themolecule that confers pharmacological activity(alters biological processes in the body). 1

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

2(4)

ORGANIC CHEMISTRY


42. (a)

1

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

43. (a) Xray 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. Theycan examine many molecular structures quickly lookingfor a possible pharmacophore. 1

(4)

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

(b) (i) The surface shape at the active site of the enzyme (1)is such that only a specific molecule with adefinite 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 nowavailable, leaving it free for another reactant to fillthe site. 1

(4)


EXTRA QUESTIONS

EXTRA QUESTIONS

1. (a) (i) Only a small part of the diagram in the questionrepresents the visible spectrum (around 3 × 10 6 m1to 2 × 10 6 m1). 1

(ii) Balmer series 1

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

3

(c) E = hc/λ = hcν for one photonE for 1 mol = Lhcν

(Note: the question reminds you mol1,so L is needed.) (1)

= 6.02 × 1023 × 6.63 × 1034 × 3 × 108× 11 × 106 (1)

= 1317.1 × 103 J mol1(Check the units by fitting them in theabove equation.)

= 1317 kJ mol1 (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 orbitalsare filled singly, keeping spins parallel, before pairingoccurs. 1

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


EXTRA QUESTIONS

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

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

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

2 (5)

3. (a) (i) The principal quantum number indicates the shellto which the electron belongs (counting from thenucleus). 1

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

(b) 3s fills completely before 3p starts to fill, because 3p is ofhigher energy. (1)The three degenerate orbitals are half-filled (singly filled)before doubly filling as pairing of electrons requiresmore 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 4th IE for Al represents the removal of an electronfrom the stable octet, hence the dramatic rise after the3rd value.The 4th IE for Ti removes the last 3d electron only anddoes not involve breaking into a stable octetarrangement. (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 CO2linked with greenhouse effect or finite resources, etc. (1)Nuclear fuel: risk of (large-scale) accident or finiteresources or disposal of dangerous waste, etc. (1)

2

(b) Si, As, P, B, Al, Ga, Sb (any 4 × ½) 2(c) (i) An ntype semiconductor consists of Si doped with

As, P or Sb. (1)A ptype semiconductor consists of Si doped withAl, B or Ga. (1)These are arranged as shown to create a pnjunction. (1)

3

(ii) Electron-hole pairs are created at the pnjunction by solar radiation. Electrons migratetowards the n-type and holes towards thep-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 ionare very different.or The Cs+ ion is much bigger than the Na+ so it is morestable 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 = 133.5This is exactly half the measured value, so a dimer isindicated, i.e. the formula is Al

2Cl

6 in the vapour phase. 1

(c) Al2Cl6 + H2O → 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 Reacts to form Ionises to formpH paper an alkali (½) an acid (½)

2

(b) (i) 2H → H2 + 2e (1)

(ii) 2Br → Br2 + 2e (1)

2 (4)


EXTRA QUESTIONS

10. (a) (i) No two electrons can have the same four quantumnumbers, i.e. exactly the same electron address. Thenearest any two can come is to differ only in theirspin, e.g. in 1s or 2s or 2p. 1

(ii) In degenerate orbitals, e.g. 2p, the electrons fill allthe 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 thered/orange (1)so green is transmitted. (1)

2

(b) CN creates a larger split in the d orbitals, i.e. ∆ isgreater. (1)This means absorption would probably include yellowand 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] 4s0 3d6 inboth complex ions. (1)In the hexacyano complex the value for ∆ is larger thanthe hexaaquo complex. (½)The larger value of ∆ the less likely electrons are to bepromoted, (½)i.e.

(1)


EXTRA QUESTIONS

The hexaaquo complex is probably paramagnetic (withfour 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 Clhave 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 dativecovalent bond. 1

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


EXTRA QUESTIONS

(c)

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

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

(5)

14. (a) Adsorbed 1

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

(ii) The d orbitals of Fe form easily reversible bondswith reactants. (1)This loosens other bonds within the reactants.or This holds one or more reactants in a suitableorientation 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 nonmetal oxides areacidic. (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 S2O32

n(S2O

32) = 0.004 mol (1)

c = n/v = 0.004/15 × 1000 mol l1= 4/15= 0.267 mol l1 (1)

2

(d) Starch added as an indicator would lose its blue colourat 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 bestestimate. (½)

2

(ii) n(Ti3+) = c × v = (0.10 × 22.63)/1000= 2.263 × 103 mol (1)

so n(Fe3+) = 2.236 × 103 mol= 55.8 × 2.263 × 103 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/[N2][H2]

3 1

(b) (i) 0.4 mol NH3 requires0.2 mol N2, leaving 1.8 mol unreacted (1)0.6 mol H

2, leaving 1.4 mol unreacted (1)

2


EXTRA QUESTIONS

(ii) K = 0.42/(1.8 × 1.43) = 0.16/4.94 = 0.0324 l2 mol2 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][I2]/[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 adefinite 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 withvarying amounts of ammonia (1)and by titrating aliquots of these two solvents with astandard solution of a strong acid. (1)

2

(e) Water 1

(f) Water is a polar solvent capable of forming H-bondswith ammonia. (1)Ether is a nonpolar covalent solvent withno 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 inthe solvents and will not move. 1

(b) Rf = distance moved by spot ÷ distance moved by solventfront

For spot A Rf = 4/5 = 0.8; for spot B Rf = 3/5 = 0.6 1

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

(ii) Find a better solvent mixture. 1

(d) No, since either spot may consist of two or more dyes withvery similar Rf values 1

(e) Either less was spotted or the concentrations are less insample 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) theingredients for a buffer. 1

(ii) If H+(aq) is added it will be neutralised by reactingwith the methanoate ions (available in largeamounts from the salt) to form molecules of theweak acid, (1)

i.e. HCOO + H+ HCOOH (1) 2(5)


EXTRA QUESTIONS

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

= 0.01449 mol (1)= 14.49 mol l1 (1)

3

(b) Because it is so volatile, the concentrated solution losesNH3 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 apK

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 mol1 (1)

3

(b) Conditions are not standard, e.g. [H2SO4] > 1 mol l1 1

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

(6)

26. (a ) Above 2100K (± 50K) 1

(b) Separation of gaseous products; building a furnace thatcan withstand such high temperatures; cost of maintainingsuch high temperatures; Mg could well be extremely reactiveat this temp; cost of fuel/energy; etc. (any two earn twomarks) 2

(c) (+850 670 ) kJ mol1 (1)= +180 kJmol1 (± 20 kJmol1) (1)(1 for wrong or missing sign) 2

(d) +88 kJ (Note: if incorrect accept an answer that is 92 kJless 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 bythe removal of Mg then the equilibrium position moves infavour of making more Mg. 1

(7)

27. ∆H for reaction = 46.2 + (92.3) (315.0)= 176.5 kJ mol1 (1)

∆S for reaction = +193 + 187 94.6 = 285.4 J K1 mol1 (1)But ∆G = ∆H T∆S = 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 ∝ [(CH3)3CCl ][I

] 1(2)

29. (a) The order with respect to HgCl2(aq) is first. (½)The order with respect to C2O4

2 is second. (½)So rate = k [HgCl

2]1 [C

2O

42]2 (1)

2

(b) k = 1.82 × 104 ÷ (0.128 × (0.304 )2) (½)(applying units to this formula: mol l1 min1 ÷ mol l1mol2 l2) (½)

= 0.0154 mol2 l2 min1 (½ + ½)2

(c) Rate = 0.0154 × 0.1 × 0.1 × 0.1= 1.54 × 105 mol l1 min1 (½ + ½)

(remember the units!) 1


EXTRA QUESTIONS

(d)

The first mark is for a line graph that curves generally asshown, representing a decreasing rate of change ofconcentration. (1)The second mark is for a tangent to the curve at timezero. (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 opticallyactive. 1

(4)


EXTRA QUESTIONS

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

(ii) To lower the melting point of the triglycerides in theoil so that the margarine can have the desiredproperties, 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 activecarbon (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 gSo 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 21 41 14experimental error 3 6 2

(1 for each error) (2)

empirical formula = C3H6O2 (1)5

(b) C2H5COOH 1(6)

33. (a) Mass = 60 therefore isomers are C3H6O

(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. Thiscorresponds to CH2OH and is only possible with the primaryalcohol, structure (ii). 2

(8)


EXTRA QUESTIONS

34. (a) CH3CH

2OH (1) CH

3OCH

3 (1) 2

(b) (i) CH3CH2OH 1(ii) 0 is TMS; 1.0 is CH

3; 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 atomsin 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 internalstandard. 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 reasonablestructure can be drawn for CH3. Thus, try C2H6O butmolecule must be symmetrical.The structure must be HOCH2CH2OH it is the only oneto fit the data. 3

(d) A broad peak due to OH stretch between 3600 and3300 cm1 1

(6)


EXTRA QUESTIONS

36. (a) (b)

CH3CH

2CH

2CH

2OH five peaks; five different hydrogen environments

butan-1-ol

CH3CHCH

2CH

3five peaks; five different hydrogen environments

OHbutan-2-ol

CH3CHCH

2OH four peaks; four different hydrogen environments

2 × CH3 groups are identical

CH3

methyl propan-l-ol

CH3

CH3

C OH two peaks; two different hydrogen environments3 × CH

3 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 sameenvironment. (1)In cyclohexane-1,4-diol there are three differentenvironments and three peaks, as shown.

(2) (1 per error)3

(6)

38. (a)

2

(b) (i) CH3CH

2OH

(ii) CH3OCH3 (2 × ½ )1


EXTRA QUESTIONS

(c)

Isomer Group Approximate δδδδδ Integralvalue

CH3CH2OH CH3 0.81.3 3

CH2 3.54.0 2

OH 3.06.0 1

Isomer Group Approximate δδδδδ Integralvalue

CH3OCH3 2 × CH3O 3.04.0 6

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

4

(d)

Both peaks in correct position, i.e. at correct approximateδ 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 = 1Therefore empirical formula = C

4H

4O (1)

(which equates to mass of 68) 2

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

8O

21


EXTRA QUESTIONS

(c) The infra-red absorption at 1710 indicates the presenceof a carbonyl group. (1)Two oxygens present so ester or acid (1)but acid ruled out because no large peak between3500 and 2500 cm1. (1)NMR spectrum peak at 67 (integral 5) indicates amonosubstituted benzene ring. (1)Peak around 2 (integral 3) indicates a CH3 attacheddirectly to C=O. (1)

5(8)

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

(ii) Antagonist interacts with receptor to produce noresponse and it prevents action by a naturalcompound. 1

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

electronic structure and the periodic table unit 1 · = 6.02 × 1023 × 8.35 × 10Œ19 j molŒ1 =...

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