2008-Nov-GCE-A-CH-H2-soln-SBS.pdf

08N-1

2008 Nov (9746) Paper 1 1.

5.

[08N P1 Q01 Moles] B

[08N P1 Q05 Atomic Structure] [04J P1 Q05] [93J P4 Q02]

N–

C

1× 6.02 ×1023 24000

At r.t.p., volume of 1 mol of gas = 24 dm3 1 ⇒ mol of O2 in 1 cm3 = mol 24 ×103 ∴ no. of O2 molecules = nL 1 = × 6.02 × 1023 24 ×103 Hence, option B. (ans)

N– has 8 electrons. On losing an electron, it forms N which has 7 electrons (configuration 1s2 2s2 2px1 2py1 2pz1) and so, has a half-filled set of p orbitals. (ans) © Step-by-Step

6.

[08N P1 Q06 Bonding] N atoms numbered 1 and 3 2 and 4 ionic co-ordinate

© Step-by-Step

C 2.

[08N P1 Q02 Relative Masses] [99N P3 Q01] A

Chlorophyll is a chlorin pigment, and has a Mg2+ ion at the centre of the chlorin ring.

argon

Since density of 'atmospheric N2' is higher than that of chemically pure N2, the gas that causes this discrepancy would, therefore, be one of higher mass than N2. [Mr : N2 = 28; Ar = 39.9; He = 4; CH4 = 16; Ne = 20.0] (ans)

R

Element (X) in C has 3 valence electrons and would lose all 3 valence electrons most easily to achieve a stable octet configuration (forming X3+ ion). (ans) © Step-by-Step

4.

40

H H

C H C C

C C

C C

In the decomposition, a neutron is converted to a proton (⇒ mass number remains unchanged since there is no change in total no. of neutrons and protons) and an electron (⇒ atomic no. increases by 1). Hence, the process is 40 described by 40 19 K → 20 Ca . (ans) © Step-by-Step

R

C

R

O © Step-by-Step

7.

[08N P1 Q07 Chemical Equilibria] A

0.17

initial atm, p eqm atm, p–a

K → 40Ca

R

C

2NO2(g)

[08N P1 Q04 Atomic Structure] D

C C R

C

N Mg2+ – N N

C R

1s2 2s2 2p6 3s2 3p6 3d1 4s2

C

N–

C

H

C

C

H C

[08N P1 Q03 Atomic Structure] C

C C

N-1 and N-3 form ionic bonds with Mg2+, while N-2 and N-4 form co-ordinate (or dative covalent) bonds with Mg2+. (ans)

R

C

C

© Step-by-Step

3.

H

R

2NO(g) + O2(g) 0 a

0 ½a

Since pressure at equilibrium is 20 % greater than initial pressure, ⇒ a = 0.4 p (p – a) + a + ½ a = 120 100 p Hence, at equilibrium 2NO2(g) eqm atm,

p – 0.4 p = 0.6 p

2NO(g) + O2(g) 0.4 p

½ × 0.4 p = 0.2 p

∴ mole fraction of O2 in equilibrium mixture, 0.2 p x= = 0.17 (ans) 0.6 p + 0.4 p + 0.2 p © Step-by-Step

A-Level Solutions – Chemistry

08N-2 8.

12. [08N P1 Q12 Chemical Equilibria]

[08N P1 Q08 Solids] copper cations

D

C

iodine molecules

Copper has a giant metallic structure – lattice particles are metal cations. Iodine has a simple molecular structure – lattice particles are discrete I2 molecules. (ans) © Step-by-Step

9.

When the equilibrium constant is independent of temperature, it suggests that the reaction is neither exothermic nor endothermic; i.e. enthalpy change, ∆H = 0. (ans) © Step-by-Step

13. [08N P1 Q13 Kinetics]

[08N P1 Q09 Energetics] D A

burning an element in oxygen

When an element is burnt in oxygen, the enthalpy change (enthalpy change of combustion, ∆Hc) is always negative (i.e. exothermic). (ans) © Step-by-Step

The enthalpy change is zero.

1 210

Since half-life, t½, of iodine-131 is 8 days, 80 days ⇒ 10 t½ have elapsed. ∴ fraction of isotope that remained = ( 12 )10 or 110 . (ans) 2 © Step-by-Step

10. [08N P1 Q10 Entropy] 14. [08N P1 Q14 Periodicity] ∆H +

C

∆S +

∆G –

D

The phrase instant 'cold packs' suggests that the reaction is endothermic; i.e. ∆H is positive. When the pack is squeezed, NH4NO3(s) dissolves in the water suggests that the reaction is spontaneous; i.e. ∆G is negative. Dissolution of NH4NO3(s) is accompanied by an increase in entropy (less orderly); i.e. ∆S is positive. Hence, option C. (ans) © Step-by-Step

11. [08N P1 Q11 Ionic Equilibria] [93N P4 Q10] D Ka

0

V

HCl (produced when the chloride of R is hydrolysed) gives a white precipitate of AgCl with AgNO3(aq). Ag+(aq) + Cl –(aq) → AgCl(s) mol of Cl – from chloride of R = mol of Ag+ = 0.30 × 100 1000 = 0.030 mol ⇒ 0.010 mol of chloride of R ≡ 0.030 mol of Cl – or 1 mol of chloride of R ≡ 3 mol of Cl – However, the low b.p. (76 °C) suggests that the chloride of R is not an ionic compound, i.e. R is not a Group III element. Hence, R is a Group V element which forms two chlorides, RCl3 and RCl5. RCl3 gives 3 mol of HCl when mixed with water. (ans) © Step-by-Step

V

Ka, the acid dissociation constant, is only affected by changes in temperature. It is not affected by changes in volume, V. (ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-3 15. [08N P1 Q15 Group II] A

18. [08N P1 Q18 Transition Elements]

Some of the barium hydroxide had reacted with carbon dioxide in the air to from solid barium carbonate.

In the titration, the reaction is Ba(OH)2.8H2O + 2HCl → BaCl2 + 10H2O The titres were lower than expected suggests that the solution contains less Ba(OH)2 than expected. This is because when the prepared solution was left in an open beaker, some Ba(OH)2 has reacted with CO2(g) in the air to form solid BaCO3 (in an acid-base reaction). Ba(OH)2 + CO2 → BaCO3 + H2O (ans) © Step-by-Step

16. [08N P1 Q16 Group VII]

C •

•

boiling point increases

C

Fe2O3

The formation of chromium(VI) compounds from chromium(III) involves oxidation. Al2O3: Al 3+ + 3e– → Al E o = –1.66 V CuO: Cu2+ + 2e– → Cu E o = +0.34 V Fe2O3: Fe3+ + e– → Fe2+ E o = +0.77 V ZnO: Zn2+ + 2e– → Zn E o = –0.76 V From the list, the best oxidising agent is Fe2O3 (most positive E o value). (ans) © Step-by-Step

19. [08N P1 Q19 Hydrocarbons] A

It is an sp2–sp2 overlap.

Each carbon atom in buta-1,3-diene is sp2 hybridised. Hence, the covalent bond between C2 and C3 are formed by the overlap of sp2 hybrid orbitals. (ans)

electron affinity less negative

© Step-by-Step

Boiling point of Group VII elements increases from Cl2 to I2 due to stronger intermolecular van der Waals' forces as the number of electrons increases from Cl2 to I2. From Cl to I, electron affinity becomes less negative due to the increase in atomic size and hence, weaker attraction for the additional electron. (ans) © Step-by-Step

17. [08N P1 Q17 Transition Elements] B

20. [08N P1 Q20 Alkenes] B

HBr

Br2, hv

Option B will not give a good yield of 1,2dibromocyclohexane because the second step involves a free-radical substitution reaction which is not stereospecific and so, a mixture of substitution products is obtained. (ans) © Step-by-Step

1s2 2s2 2p6 3s2 3p6 3d6

21. [08N P1 Q21 Alkanes] 6

2

26Fe has electronic configuration: [Ar] 3d 4s (where [Ar] is 1s2 2s2 2p6 3s2 3p6). In forming ions, electrons are removed from the 4s orbitals first before the 3d orbitals.

∴ Electronic structure of Fe2+ is [Ar] 3d6 4s0. (ans) © Step-by-Step

C

CH3CH2CH3 + Cl •

The relevant bond energies (in kJ mol–1) are: C–Cl C–H Bond energy 340 410 C–Cl bond is weaker than C–H bond. Hence, homolytic fission of C–Cl bond occurs and products formed are CH3CH2CH3 and Cl •. (ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-4 22. [08N P1 Q22 Halogen Derivatives] A

26. [08N P1 Q26 Carbonyl Compounds]

BrCH2CH2CH2CH2Br B

Compound X is a halogenoalkane, BrCH2CH2CH2CH2Br.

CH3COCH3

HCN

CH3 OH C CH3 CN

Reaction of CH3COCH3 (a ketone) with HCN involves the use of a homogeneous catalyst, NaCN(aq) – the catalyst is in the same physical state as the reactants. (ans)

BrCH2CH2CH2CH2Br KCN in ethanol

NCCH2CH2CH2CH2CN

© Step-by-Step

reduction

27. [08N P1 Q27 Acid Derivatives]

(ans)

H2N(CH2)6NH2

OH

© Step-by-Step

OCH3

23. [08N P1 Q23 Halogen Derivatives]

C

[92J P4 Q24]

A

CH2–CH=CH2

0

PCBs contain only aryl Cl (i.e. Cl attached directly to benzene ring) which are resistant to hydrolysis. Hence, no Cl atoms will be removed by hydrolysis. (ans)

CH3COCl reacts with the phenol functional group to give the corresponding ester. OH OCH3

© Step-by-Step

24. [08N P1 Q24 Alcohols] C

CH3COCl

CH2–CH=CH2 (ans)

CH2–CH=CH2

© Step-by-Step

CH3(CH2)10O(CH2CH2O)10H

A possible formula of the non-ionic detergent made by the reaction of excess epoxyethane with a C11 alcohol is CH3(CH2)10O(CH2CH2O)10H C11 alcohol

excess epoxyethane

(ans) © Step-by-Step

25. [08N P1 Q25 Alcohols] [03J P1 Q29]

28. [08N P1 Q28 Esters] C

(CH3)2CHCO2CH2CH3

Acid hydrolysis of ester X gives a carboxylic acid and an alcohol. The yellow precipitate obtained with alkaline aqueous iodine (positive iodoform test) suggests that the alcohol contains a CH3CH(OH)– group. Hence, X is (CH3)2CHCO2CH2CH3. Hydrolysis of ester:

reagent Y

dilute NaOH

X (CH3)2CHC OCH2CH3

When ethanol is heated with an excess of concentrated H2SO4 (Y) at 170 oC, it undergoes dehydration to give ethene. H H

H H C

C

H

c.H2SO4 heat

O

solution Z

concentrated H2SO4

C

H

OCOCH3 OCH3

H

C

C

acid hydrolysis

O (CH3)2CHC

+ CH3CH2OH OH

H + H2O Iodoform test:

H OH The impure ethene is then bubbled into dilute NaOH (Z) to remove any excess acid. (ans) © Step-by-Step

CH3CH2OH + 4I2 + NaOH CHI3 + HCO2– Na+ + 5HI yellow ppt

(ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-5 29. [08N P1 Q29 Amino Acids]

C

33. [08N P1 Q33 Electrochemistry]

total number of positive charges

total number of negative charges

2

0

H+ + e–

At pH 2 (acidic), the basic –NH2 group would react with the H+ ions present to give a salt. The product formed is

CH2

CH

NH3+

NH3+

3+

Fe + e

D

1 2

3

HO2CCH2CH(CH3)NH2

1

cell

= ER – EL = +0.77 – 0.00 = +0.77 V

34. [08N P1 Q34 Kinetics] [96N P4 Q34] 1 2

The rate equation for the reaction can be written: rate = k [H2O2] [I–]. The reaction is zero order with respect to acid.

Comparing expt 1 and 2: [I–] and [H+] constant, when [H2O2] triples, the initial rate is also tripled. ∴ first order w.r.t. H2O2. Comparing expt 2 and 3: [H2O2] and [H+] constant, when [I–] doubles, the initial rate is also doubled. ∴ first order w.r.t. I–.

CO2, COS and COSe are all linear in shape. Down Group VI, electronegativity of the elements decreases from O to Se. i.e. electronegativity: O > S > Se

32. [08N P1 Q32 Energetics]

= +0.77 V

o

© Step-by-Step

The C=S bond is more polar than the C=Se bond. The C=O bond is more polar than the C=S bond.

© Step-by-Step

= 0.00 V

Under standard conditions, concentrations of Fe3+ and Fe2+ in the right hand solution should be 1.0 mol dm–3. (ans)

(ans)

∴ bond polarity: C=O > C=S > C=Se This accounts for the overall polarity of COS being less than that of COSe. (ans)

E

o

•

31. [08N P1 Q31 Bonding]

2

o

o

At the left hand electrode, oxidation takes place and so, it is the negative electrode. 1 H → H+ + e– 2 2

© Step-by-Step

1

Fe

2+

E

•

HO2CCH2CH(CH3)NH2

Proteins in the body are built from α - amino acids (or 2-aminocarboxylic acid). The amino acid in D is a 3-aminocarboxylic acids.

–

H2

E

© Step-by-Step

30. [08N P1 Q30 Amino Acids]

o

1 2

•

CO2H

(ans)

E o cell = 0.77 V. The left hand electrode is the negative electrode.

1 2

Comparing expt 3 and 4: [H2O2] and [I–] constant, when [H+] doubles, the initial rate remains unchanged. ∴ zero order w.r.t. H+. ∴ rate equation is rate = k [H2O2] [I–]. •

rate [H 2O 2 ][I- ] 2 ×10-6 From expt 1, k = 0.010 × 0.010 = 0.02 dm3 mol–1 s–1

rate constant, k =

Hence, k ≠ 2.0 × 10–1 dm3 mol–1 s–1 (ans)

NH3(g) + HCl(g) → NH4Cl(s)

© Step-by-Step

At 298 K (or 25 °C), carbon exists in the solid state, C(s), which therefore rules out option 2. Option 3 is ruled out because at 298 K, water exists as H2O(l) and not, H2O(g). (ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-6 35. [08N P1 Q35 Periodicity] 1

δ–

The cation has a greater nuclear charge than the anion.

Group II cation has charge 2+, while Group VII anion has charge 1–. Hence, configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 suggests that the Group II element has (36+2) = 38 electrons (or proton no. 38), while the Group VII element has (36 – 1) = 35 electrons (or proton no. 35). ∴ The smaller size of the cation is due to the greater nuclear charge which results in the same number of electrons being attracted more strongly by the nucleus. (ans)

3

–

X2(aq) + 2e Y2(aq) + 2e– Z2(aq) + 2e– •

–

2X (aq) 2Y –(aq) 2Z –(aq)

C OH + Hal– The energy difference between X and Z represents the activation energy. (ans) © Step-by-Step

38. [08N P1 Q38 Alcohols] 1 2

•

C

+1.36 V +1.07 V +0.54 V •

The reaction is an example of nucleophilic substitution. Between X and Y the C–Hal bond will be lengthening. The energy difference between X and Z represents the activation energy.

Halogenoalkanes undergo nucleophilic substitution. The energy profile shows a onestage process involving the formation of an intermediate Z. Between X and Y the C–Hal bond will be lengthening as C–OH is gradually formed.

OH

With excess hot conc. KMnO4, the 1° alcohol group, 2° alcohol group and C=C are oxidised to give the product: O

C CH3

O CH3

O

CO2H

OH

H

H

H

HO2C O

37. [08N P1 Q37 Halogen Derivatives] [02J P1 Q39] [93J P4 Q37]

3

CH3

CH2OH

O

© Step-by-Step

2

Hydrocortisone has seven chiral centres (indicated by * ): O

2Z –(aq) + Y2(aq) → Z2(aq) + 2Y –(aq) Ecell o = ER o – EL o = +1.07 – (+0.54) = +0.53 V (> 0, ∴ reaction is feasible) Hence the reaction occurs. (ans)

1

The hydrocortisone molecule has 7 chiral centres. When treated with an excess of hot concentrated acidified KMnO4, hydrocortisone will produce a compound with 2 carboxylic acid groups.

* * CH3 H * * * * H *H

E o value becomes less positive suggests a decrease in oxidising power : X2 > Y2 > Z2.

OH

C fast

HO

∴ reducing power increases: X – < Y – < Z –. •

Θ Hal

..

36. [08N P1 Q36 Group VII] There is an increase in reducing power in the sequence X –, Y – and Z –. The reaction 2Z –(aq) + Y2(aq) → Z2(aq) + 2Y –(aq) occurs.

slow

C

OH–

© Step-by-Step

2

δ+

Hal

(which has 2 carboxylic acid groups). •

With hot acidified K2Cr2O7, only the alcohol groups are oxidised. The product formed is: O C CH3

O CH3 H

CO2H OH

H H

O

(which has 3 carbonyl groups). (ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-7 39. [08N P1 Q39 Aldehydes] H OCH3

2 O

3 •

40. [08N P1 Q40 Proteins]

O

–CH2CONH2

2

–CH2CH2CH2NHC

3

–CH2

NH2

CH3CH=NNH

NH

Option 2 involves the reaction of aldehyde group with CH3O– nucleophile (from CH3OH). H O

–

OCH3

O

OH O

OH HO

N H

H OCH3

In each of the above, the amino group forms hydrogen bonding and hence, the cross-link to stabilise the tertiary structure of a protein. (ans)

- H2O

O

•

1

O

H OCH3

Option 3 involves the reaction of aldehyde group with phenylhydrazine. H

CH3

O + H2 N N

C H

- H2O H

CH3 C H

N

N

(ans) © Step-by-Step

© Step-by-Step

[08N P1 MCQ Key] Q.

Key

Q.

Key

Q.

Key

Q.

Key

1 2 3 4 5

B A C D C

11 12 13 14 15

D C D D A

21 22 23 24 25

C A A C C

31 32 33 34 35

B D B B D

6 7 8 9 10

C A D A C

16 17 18 19 20

C B C A B

26 27 28 29 30

B C C C D

36 37 38 39 40

C A B C A © Step-by-Step

A-Level Solutions – Chemistry

08N-8

2008 Nov (9746) Paper 2

2.

1.

(a) (i) Ksp = [Ag+] [Br–] (ans)

[08N P2 Q01 Hydroxy Compounds]

(a) (i)

(ii) [Ag+] = [Br–] = [AgBr]

H H

∴ Ksp = (7.1 × 10–7) × (7.1 × 10–7)

H O C C O H

= 5.0 × 10–13 mol2 dm–6 (1 d.p.) (ans)

(ans)

H H

(ii)

(b) (i) ∆G

H H δ– δ+ H O C C O H H H hydrogen bonding

[08N P2 Q02 Solubility Product]

o

δ– H H

(ans)

H H

(ii) ∆G

(b) (i) Gas A is ethene, C2H4. (ans)

∆S

o o

(ii) The purple acidified KMnO4 solution is decolourised. (ans) (c) The three functional groups are: • phenol • secondary alcohol, and • primary amine. (ans) (d) (i)

CH(OH)CH2NH2 Br

Br Br

(ii)

= ∆H o – T ∆S o o o = ∆H − ∆G T −66.0 − (−55.3) = − 10.7 = 298 298 = – 0.03591 kJ mol–1 K–1 = –35.9 J mol–1 K–1 (1 d.p.) (ans)

(iii) ∆S o is negative suggests that there is a decrease in entropy due to the formation of an ordered lattice as the precipitate forms. (ans) (c) ∆G

[Reaction of phenol ]

OH

o

ppt =

2.303RT log Ksp

= 2.303 × 8.31 × 298 × log 1.006 = +14.8 J mol–1 (1 d.p.) Since ∆G o ppt is positive, precipitation will not occur, i.e. AgF is soluble in water. (ans)

(ans)

OH

2.303RT log Ksp

= 2.303 × 8.31 × 298 × log (2.0 × 10–10) = –55310 J mol–1 = –55.3 kJ mol–1 (1 d.p.) (ans)

H O C C O H

δ+

ppt =

CH(OH)CH2NH2

© Step-by-Step

[Acid-base reaction of phenol ] O– Na+ O Na+ –

(iii)

O

C

CH2NH3+

[Oxidation of secondary alcohol ] OH

OH

(iv)

(ans)

(ans)

CH(OH)CH2NH3+ Cl –

OH OH

[Acid-base reaction of amine] (ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-9 3.

4.

[08N P2 Q03 Carbonyl Compounds]

(a) Products are CH3 CH3

(a) Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9 (ans)

CO2H

CHCO2H and CO2H

(b) (i) B is Ag(NH3)2Cl. (ans)

(ans)

(ii) CuCl2(aq): [Cu(H2O)6]2+ (ans)

(b) (i) Ecell o = ER o – EL o = +1.52 – (+0.70) = +0.82V (ans) (ii) 5 HO

5O

[08N P2 Q04 Transition Elements]

C: (c) (i) D

OH + 2MnO4– + 6H+

O + 2Mn2+ + 8H2O

E

(ans)

(ii) D E

[Cu(NH3)4(H2O)2]2+. (ans) CuCl2 + 2HCl → H2CuCl4 (ans) CuCl2 + Cu + 2HCl → 2HCuCl2 (ans) [CuCl4]2– (ans) [CuCl2]– (ans)

(iii) Ligand exchange occurs. (ans) (iii) To reduce quinone, the reducing agent must have a E o value less positive than +0.70 V. SO42– + 4H+ + 2e– → SO2 + 2H2O

E

o

= +0.17 V

∴ SO2(g) is a suitable reducing agent. (ans) (c) Reagent used is NaBH4. (ans) (d) (i) reagent: 2,4-dinitrophenylhydrazine observation: orange precipitate formed. (ans) (ii) reagent: neutral FeCl3 solution observation: violet/purple coloration. (ans) (iii) reagent: aqueous Br2 solution observation: brown Br2(aq) decolourised (with quinol, white ppt and steamy fumes of HBr are also observed) (ans) © Step-by-Step

(iv) tetrahedral in shape. (ans) (v) +1 oxidation state. (ans) (d) (i) reduction occurs. (ans) (ii) F is CuCl. (ans) (e) (i) mole ratio Cu : F : K = 21.5 : 38.7 : 39.8 63.5 19.0 39.1 = 0.338 : 2.036 : 1.018 = 1:6:3 ∴ empirical formula of G is CuF6K3. (ans) (ii) +3 oxidation state. (ans) (f) (i) Cu+ 1s2 2s2 2p6 3s2 3p6 3d10 Cu(I) in E and F has no empty or partially filled d-orbitals. Hence, d-d* electron transitions cannot occur which accounts for both E and F being colourless. (ans) (ii) Cu3+ 1s2 2s2 2p6 3s2 3p6 3d8 Cu(III) in G has empty d-orbitals. Hence, d-d* electron transitions can take place and the light energy not absorbed is seen as the colour of the complex. (ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-10

2008 Nov (9746) Paper 3 1.

[08N P3 Q01 Group VII / Halogenoalkanes]

(a) Bond energy (of dissociation) is the energy required to break one mole of a covalent bond between two atoms in the gaseous state. (ans) (b) (i) This reaction can be carried out by heating the hydrogen halides. (ans) [or by inserting a hot wire/glass rod into a test tube of the gas.] (ii) Down Group VII, the hydrogen halides, HX, is increasingly easy to decompose. (ans)

(d) (i) boiling point: C2H5Cl < C2H5Br < C2H5I Boiling point increases from C2H5Cl to C2H5I due to stronger intermolecular van der Waals' forces as the number of electrons increases from C2H5Cl to C2H5I. (ans) (ii) bond polarity: C–Cl > C–Br > C–I Bond polarity decreases from C–Cl to C–I due to decrease in electronegativity from Cl to I. (ans) (iii) reactivity towards nucleophilic reagent: C2H5Cl < C2H5Br < C2H5I Reactivity towards nucleophilic reagent increases from C2H5Cl to C2H5I due to the decrease in strength of C–X bond. (ans)

H2 + Cl2

(iii) 2HCl

∆H = 2 × EH–Cl – EH–H – ECl–Cl

(e) P is a tertiary bromide (chiral carbon indicated by *) CH

= (2 × 431) – 436 – 244 = +182 kJ mol–1 (ans) 2HI

3

CH3CH2 *C CHCH3

∆H = 2 × EH–I – EH–H – EI–I

The four alkenes are:

= (2 × 299) – 436 – 151 = +11 kJ mol–1 (ans)

H CH3

H

δ+ δ–

slow

C

H

C⊕

H

C

H

H

X

+ X–

H

H C⊕

H

C H

H

X

+ X–

fast

CH3

II

H

CH2CH3

CH3

CH3

C=C

C=C CH(CH3)2

CH3CH2

X

C

H

H

C

X

H

CH3

IV

I and II are geometrical isomers. (ans) NB. Q is a tertiary alcohol since it does not react with hot, acidified Na2Cr2O7(aq); i.e. not oxidised. Hence, P is a tertiary alkyl halide (chiral carbon indicated with *). HBr can be eliminated from P in three possible ways: H CH3 CH3 CH3C *C C CH3

carbocation

H

CH3

III

H X––X

C=C CH(CH3)2

H

Mechanism: electrophilic addition with the positive end of the polarised X δ+––X δ– as electrophile. C

CH(CH3)2

H

I

(c) Electrophilic addition. (ans)

H

CH3 C=C

(iv) Down the group, less energy is needed for the reaction. As the size of the halogen increases, the H–X bond becomes longer and weaker and so, breaks more easily. (ans)

•

(ans)

Br CH3

H2 + I 2

H Br

H

(i) (ii)

(iii)

(i)

(iii)

H

CH3 C=C

CH3 CH(CH3)2 exists as geometrical isomers

(ii)

(ans)

H

CH3

CH3 C=C

CH2CH3

CH3CH2

CH3

C=C H

CH(CH3)2 © Step-by-Step

A-Level Solutions – Chemistry

08N-11 2.

(e) (i) P2O5 + 3H2O → 2H3PO4 (ans)

[08N P3 Q02 Electrolysis / Alkanes]

(a) Products at the cathode: H2(g) and NaOH(aq) (ans) Overall reaction is: 2RCO2– Na+ + 2H2O → H2 + R–R + 2CO2 + 2NaOH (ans) (b) charge, Q = I × t = (2.0 × 40 × 60) C = 4800 C = 4800 F = 0.0497 F 96500

CO2 + 2NaOH → Na2CO3 + H2O (ans) (ii) mass of H2O = 1.55 g 1.55 mol of H2O = m = = 1.55 M r 2(1.0)+ 16.0 18.0 = 0.0861 mol (ans) •

2RCO2– Na+ → R–R + 2CO2 + 2e–

= 0.0689 mol (ans)

From equation, 2 F ≡ 1 R–R or 0.0497 F ≡ ( 12 × 0.0497) mol of R–R

(iii) mol of H in D = 2 × mol of H2O = 2 × 0.0861 = 0.172 mol mol of C in D = mol of CO2 = 0.0689 mol

∴ mol of C2H6 = ( 12 × 0.0497) mol Mr of C2H6 = (2 × 12.0) + (6 × 1.0) = 30.0 ∴ mass of C2H6 = ( 12 × 0.0497) × 30.0

∴ H : C ratio in D = 0.172 : 0.0689 = 2.5 : 1 = 5:2

= 0.746 g (ans) (c) Since A is a mono-carboxylic acid, mol of A = mol of NaOH = cV

•

molecular formula of D is C4H10. (ans)

(f) Possible structures:

11.4 = 0.100 × 1000 = 1.14 × 10–3 mol

∴ mass of 1.14 × 10–3 mol of A = 0.100 g mass of 1 mol of A =

mass of CO2 = 3.03 g 3.03 mol of CO2 = m = = 3.03 M r 12.0+ 2(16.0) 44.0

CH3CH2CH2CH2CH2CH3

CH3CH2CO2H

alkane E

acid B

0.100 = 87.7 g 1.14 ×10−3

(ans) (g) Structures of F, G and H:

Mr of A = 87.7 (ans) O

•

CH3

Possible structure of A: CH3CH2CH2C

F is CH3CH CHCH3 ; G is CH3CH C CH3

OH

CH3

(ans) (d) Using the ideal gas equation, (since n = m ) pV = nRT = m RT Mr Mr 0 . 20 × 8 . 31 × 380 mRT Mr of C = = 1.01×105 × 87 ×10−6 pV

CH3 Br CH3

H is CH3CH CHCH2Br * CH3

(ans)

•

F is a symmetrical alkane since it gives only two isomeric monobromo compounds when reacted with Br2 under u.v. light; i.e. H atoms in only two different chemical environments.

•

H has a chiral carbon atom indicated by *.

= 71.9 (ans) •

CH3

Molecular formula of C is C5H12. (ans)

(ans) © Step-by-Step

A-Level Solutions – Chemistry

08N-12 3.

[08N P3 Q03 Acid Derivatives]

(a) (i) Reaction I: HCN + trace amount of NaOH; 10 – 20 °C (ans) (ii) Reaction II: H2O/H+ [or dilute H2SO4(aq)]; heat (ans)

(e) (i) Reaction III: aq. NaOH, heat; followed by acidification (say, with aq. H2SO4). (ans) (ii) Structures for K and L: O CH3CHC

(b) HA H+ + A – (where HA ≡ lactic acid) From the above equation, [H+] = [A–] and since the degree of dissociation is very small, [HA]eqm = [HA]initial Ka =

[H + ][A − ] [HA]

[H + ]2 0.20 [ H+]2 = (0.20)(1.38 × 10–4)

Cl

[H+] = 5.25 × 10–3 mol dm–3

∴ pH = –log10 [H+]

•

(ii) pH = pKa + log10 [salt] [acid] = –log10 Ka + log10 [salt] [acid] = –log10 (1.38 × 10–4) + log10 00..30 20

NH2

N C2H5 H

L

(ans)

In J, the Cl atom attached to C=O (i.e. acyl chloride) is much more reactive due to the electron-withdrawing effect of C=O group. (ans)

(f) The two chemical tests are: 1. Add 2,4-dinitrophenylhydrazine to each compound separately and warm.

= –log10 (5.25 × 10–3) = 2.28 (ans) (c) (i) A buffer solution is a solution whose pH remains almost unchanged on dilution or when small amounts of acid or base are added to it. (ans)

N C2H5 H

K

Ka =

⇒

O CH3CHC

Compound M gives an orange precipitate. No such precipitate is observed with N. (ans) 2.

Add Tollens' reagent to each compound separately and warm. Compound M gives a silver mirror. No such silver mirror is observed with N. (ans)

[or warm with Fehling's solution – only M gives a red precipitate; or add Na2CO3 – only N gives brisk effervescence of CO2(g); or add Na(s) – only N gives effervescence of H2(g).] © Step-by-Step

= 4.04 (ans) (iii) CH3CH(OH)CO2– + H+ → CH3CH(OH)CO2H (added) (ans) (d) (i) Ester functional group is present in PLA. (ans) (ii) Hydrolysis might occur during biodegradation of PLA. (ans)

A-Level Solutions – Chemistry

08N-13 4.

[08N P3 Q04 Bonding / Energetics]

q+ ⋅q, (r+ + r- ) the lattice energies of the oxides of Group II elements decrease down the group as the size of the cation, M2+, increases down the group. (ans)

(ii) Since lattice energy ∝

(a) molecular formula of vapour: Al2Cl6 (ans) Cl

Cl

Cl Al

Al Cl

Cl

Cl

Al2Cl6 dimer

(ans)

(b) With a few drops of water, steamy white fumes of HCl(g) evolved and a white solid, Al2O3(s), remains. 2AlCl3(s) + 3H2O(l) → Al2O3(s) + 6HCl(g) [or Al2Cl6 + 6H2O → 2Al(OH)3 + 6HCl ] •

With a large amount of water, AlCl3 undergoes hydrolysis to give a weakly acidic solution. AlCl3 + 6H2O → [Al(H2O)6]3+ + 3Cl – [Al(H2O)6]3+ → [Al(H2O)5(OH)]2+ + H+ (ans)

(c) When Na2O(s) is added to a solution of litmus in water, a blue solution is obtained as Na2O(s) dissolves readily in water to give an alkaline solution (pH ≈ 13). (ans) Na2O(s) + H2O(l) → 2NaOH(aq) (ans) (d) (i)

3 Mg

2+

3–

2

N

(ans)

(ii) Mg3N2 + 6H2O → 2NH3 + 3Mg(OH)2 (ans) •

q+ ⋅q, (r+ + r- ) the lattice energy of Mg3N2 is larger than that of MgO due to the bigger charge of N3– ions (compared to O2– ions). (ans)

(iii) Since lattice energy ∝

The white insoluble solid is Mg(OH)2. Mr of Mg3N2 = (3 × 24.3) + (2 × 14.0) = 100.9 Mr of Mg(OH)2 = 24.3 + 2 (16.0 + 1.0) = 58.3 mol of Mg(OH)2 = 3 × mol of Mg3N2 = 3 × 2.0 = 0.0595 mol 100.9 mass of Mg(OH)2 = 0.0595 × 58.3 = 3.47 g (ans)

© Step-by-Step

5.

[08N P3 Q05 Group II / Acid Derivatives]

(a) Two reactions in which iron or its compounds behaves as a catalyst: 1.

N2(g) + 3H2(g) → 2NH3(g) with finely divided iron, Fe(s), as heterogeneous catalyst.

2.

2I–(aq) + S2O82–(aq) → I2(aq) + 2SO42–(aq) with Fe3+(aq) as homogeneous catalyst. (ans)

(b) Mg(NO3)2.6H2O → MgO + 2NO2 +

•

ionic charge – the bigger the ionic charge, the bigger (more exothermic) is the lattice energy;

•

ionic radius – the smaller the ionic radius, the bigger (more exothermic) is the lattice energy;

•

arrangement of ions in the crystal (crystal structure) – but this effect is small.

A-Level Solutions – Chemistry

O2

+ 6H2O (ans) (c) (i) HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+ (ans) (ii) Type of reaction: nitration via electrophilic substitution mechanism with NO2+ as the electrophile. Step 1: Production of electrophile, NO2+. HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+ Step 2: The NO2+ electrophile is attracted to the delocalised electron system in benzene to form an unstable intermediate. H +

NO2

[or Mg3N2 + 3H2O → 2NH3 + 3MgO in which case, mass of MgO = 2.40 g] (e) (i) Factors affecting magnitude of lattice energy:

1 2

slow

+

NO2

carbocation intermediate

Step 3: Expulsion of a proton from the intermediate and the stable delocalised system of benzene is retained. H +

NO2 + HSO4–

fast

NO2

+ H2SO4 (ans)

08N-14 (d) Synthesis of T from R:

(iii) Structural formulae of compounds formed: •

OCH2CH2CH3

cold HCl (aq) :

R O c. HNO3

O2N

C

CH2 HO2C

CH2

CH

CH N H

NH3+ Cl –

OCH2CH2CH3 NO2

•

S

OCH3 C O

(ans)

CH3COCl :

(i) Sn + HCl (small amount) (ii) aq. NaOH

O C

CH2 O2N

HO2C

OCH2CH2CH3

CH N

NH2

(ans)

T

(e) (i) The four functional groups in aspartame are: carboxylic acid, amine, amide and ester. (ans)

CH2 CH N H

H

OCH3 C O

C O

CH3

(ans) © Step-by-Step

(ii) The three compounds are: CH3OH, O Na+ –O2C

C

CH2 CH

O– Na+

and

NH2

CH2 O– Na+

CH H2N

C O

(ans)

A-Level Solutions – Chemistry