1990 Al Chemistry Paper I Marking Scheme

1990 A-Level Chemistry Paper I marking scheme There are some minor discrepancies with the original marking scheme. This is due to the corrections made to the original one. If there is anything inaccurate, please contact me. Some of the questions (or part of) are outside syllabus so the marking scheme of those parts are NOT PROVIDED. 1.(a) (i) Out of syllabus.

(ii) (iii) Out of syllabus. (iv) Out of syllabus.

(b) (i) or with a 5 membered ring, or FISHER PROJECTION

Note Any appropriate structure is correct which has the

group for reducing and

group for non-reducing. Rings can be 5 or 6; H‘s can be absent; any

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configuration of OH. (ii) Hemiacetal groups [1/2m] account for reducing properties. This is because they exist in equilibrium [1/2m] with the open chain form (aldehyde form) [1/2m]

The part of the structure having the hemiacetal MUST be shown in the answer. It must be stated or shown that the non-reducing form IS NOT A HEMIACETAL. [1/2m] (c)(i) Out of syllabus.

(ii) OR via an acid chloride:

[Other Nylons (e.g. Nylon 6) are possible, important is salt + heatà amide, or acid chloride + heat à amide] 2.(a)

[3m for correctly labeled diagram, reduce 1m for each WRONG / MISSING label, deduct 1/2m if Δplatinized‘ omitted] Conditions 298 K / 25 °C H2 gas at 1 atm. [1/2m]

[1/2m]

[H+] = 1 M [1/2m] How to maintain the stated conditions

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Thermostat / Water bath [1/2m] Pressure gauge [1/2m] Using a standard acid solution [1/2m] (b) ˚ H– / kJ mol-1 (1) Cu(s) + S(s) + 2O2(g) à CuSO4(s) -773 (2) 5H2(g) + 2.5O2(g) à 5H2O(l) 5(-286) (3) CuSO4(s) à CuSO4(aq) -66 . (4) CuSO4(aq) + 5H2O(l) à CuSO4 5H2O(s) -8 From (1) + (2) + (3) + (4) Cu(s) + S(s) + 5H2(g) + 4.5O2(g) à CuSO4.5H2O(s) ˚ H–f = -773 + 5(-286) + (-66) … 8 = 2277 kJ mol-1 [2m for correct working (cycle / equations / expression) 1m for correct answer (sign, value & unit) , -1/2m for wrong unit] (c) Metal having the hexagonal close packing has the ababO arrangement for atoms in different layers. [1m] Metal having the face-centred cubic close packing has the abcO arrangement for atoms in different layers [1m] Both arrangements have the same packing efficiency (or atoms in both close packing arrangements all have the coordination number of 12) [1m] (d)(i) Rate = k[A][B]2 (ii)

[Each graph: axis & label: 1m; each slope : 1/2m] (c)(i) Critical point [1/2m] At points on an isotherm above P, the gaseous states of the substance cannot be liquefied by the increase in pressure alone. [1m] [OUT] (ii) Out of syllabus. (iii) Out of syllabus.

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3.(a) * CO32- and HCO3- hydrolyze to a different extent. [1/2m] * In a solution of Na2CO3, the equilibrium CO32- + H2O ⇔ HCO3- + OH* Resulting in a pH sufficiently high (or [H +] sufficiently low) Such that [1/2m] * the equilibrium In + H+ ⇔ InH+ red colourless shifts to left and a pink / red colour is exhibited [1/2m] *In a solution of NaHCO3, whose pH is also above 7, [H +] is sufficiently high for the equilibrium to shift to the right. [1/2m] (b)(i) A white ppt. Of Al(OH)3 is formed. [1m] (ii) No ppt. Or less white ppt. of Al(OH)3 is observed. [1m] In the presence of an excess of ammnonia chloride, [NH4+] is high and the equilibrium position of the following reaction [1/2m] NH3(aq) + H2O(l) ⇔ NH4+(aq) + OH-(aq) Lies to the left so that [OH-] is low and [1m] the equilibrium position of Al(OH)3 ⇔ Al3+(aq) + 3OH-(aq) shift to right. [1/2m] (c) 2F2(g) + 2OH-(aq) à OF2(aq) + F-(aq) + H2O(l) [1m] Cl2(g) + 2OH-(aq) à OCl-(aq) + Cl-(aq) + H2O(l) [1m] OR 3Cl2(g) + 6OH-(aq) à BrO3-(aq) + 5Br-(aq) + 3H2O(l) 2Br2(l) + 6OH-(aq) à BrO3-(aq) + 5Br-(aq) + 3H2O(l) [1m] (d) Catalytic property: e.g. the use of a nickel catalyst [1m] in the hydrogenation of oils to make margarine. Complex ion formation: e.g. [Ni(H 2O)6]2+, [Ni(NH3)6]2+ [1m] Accept also Δcolours‘ (e.g. [Ni(H2O)]2+ is green), Δparamagnetism‘, or other properties can be illustrated, e.g. variable oxidation states: Ni 2O3, NiO, NiO2 etc. (e) (i) The copper foil dissolves to give a blue solution [1/2m] and brown fumes are evolved. [1/2m] 3Cu(s) + 8HNO3(aq) à 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l) 2NO(g) + O2(g) à 2NO2(g) [1/2m + 1/2m] (ii) The surface of the sodium becomes dull/less shinny/more white [1/2m] 4Na(s) + O2(g) à 2Na2O(s) [1m] The piece of sodium melts and burns with a yellow flame. [1m] 2Na(s) + 2H2O(l) à 2Na+ + 2OH-(aq) + H2(g) [1m] 2H2(g) + O2(g) à 2H2O(l) [1m] (f) The protective layer Al2O3 is destroyed when the piece of Al is dipped in a solution of HgCl 2. [1m] When withdrawn from the solution, the fresh surface of Al reacts highly exothermically [1m]

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with O2 (air) to form white Al2O3 powder. [1m] 4.(a) TWO methods are possible: (I) - Dissolves the mixture in diethyl ether, place in a separating funnel and shake with dil. HCl and separate. Call ether layer E. [1m] - To aqueous layer dil. alkali (NaOH), ether and separate - The ether layer on evaporation will yield the amine Y. [1m] - To ether layer E, add dil. alkali, and shake, separate. Ether layer will contain Z - Neutralize alkaline layer with acid (dil. HCl) and extract with ether which on evaporation will yield phenol X. [1m] (II) … Dissolve the mixture in diethyl ether using a separating funnel. Shake with dil. NaOH and separate. (1/2m) - Acidify (HCl) the aqueous layer and extract with ether to obtain X. (1m) -

Acidify (dil. HCl) the other layer (from NaOH extraction), shake and separate. The ether layer will contain Z. (1m) Add NaOH or alkali to this aqueous layer and extract with ether. The ether layer contains Y .(1m)

(b)(i) Add alcoholic FeCl3 solution. [1/2m] A colour change to red or blue denotes presence of (X). [1/2m] (ii) Add NaNO2 and dil. HCl to Y at 5°C [1/2m] followed by β-naphthol (naphthalen-2-ol). An orange ppt. denotes presence of a primary aromatic amine (Y). (iii) Out of syllabus. (c)

(i) (ii)

correct r.m.m. [1/2m] mass of Y = (4.0 x 93) / 140.5 = 2.65 g [1m] 100 % yield is (4 x 197) / 140.5 = 5.6085 g % yield = (4.0 x 100) / 5.6085 % = 71.3 % [1.5m]

5.(a)

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l l l l

Place the iron wire in a tube, add dilute sulphuric acid to completely cover the wire [1m] Heat the tube to complete the reaction, taking precaution to exclude air by, e.g. pass in N 2 or CO2. [1m] Leave the concentrated solution to cool in the absence of air (pass in N 2). [1m] Decant quickly and dry crystals with filter paper. [1m] Note: Alternative plausible steps are acceptable.

(b) l l l l l l l l

Weigh, accurately ( to at least 2 decimal places) a portion of the sample. [1/2m + 1/2m] Dissolve the weighed sample in dil. H 2SO4. [1/2m] Make up the solution to a standard volume [1/2m] in a volumetric flask. [1/2m] Pipette 25 ml of the Fe(II) solution into a flask. [1/2m] Titrate with the standardized KMnO4 solution from the burette. [1/2m] Swirling the flask to ensure thorough mixing. [1/2m] Titrate slowly near the end point. Stop at the first permanent pink colouration. [1m] Record the volume of KMnO 4 used, reading the bottom of the meniscus. [1/2m] Repeat the titration to obtain consistent results. [1/2m]

Essay type question 1. Essential points (1m / point) [max. 6m] 1. What is thermochemsitry … in terms of energy change [1/2m] giving rise to enthalpy changes that are exothermic and endothermic. [1/2m] 2. Definition of enthalpy changes [1/2m for per definition, max 1m] 3. Description of a practical method to determine ∆H 4. Description of an indirect method of measuring ∆H based on Hess‘s law. 5. Why measure thermochemical data? 6. Standard conditions [1/2m] and notations. [1/2m] Optional points (1m / point) [max. 4m]

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1. 2. 3. 4.

Expressing ∆H as ∆U + P∆V and explaining [OUT] Additional practical methods for determining ∆H. State clearly the experimental precautions. Additional examples of applications of using Hess‘s law / Born-Haber cycle to determine

∆H indirectly. 5. Definition & description of Hess‘s law. (if separate from point 4) 6. Calculation details using Hess‘s law. 7. Additional examples of applications of thermochemical data Organization [6m] [5 if exceptionally good CK] Look out for and mark the following points: (i) Use of reasonable subheadings O O [2m] If only implied not stated O O [1m] If stated out but not reasonable O O [1-0m] (ii) The essay should be easy to follow and not haphazardly put down O O [1m] If a jumble of reactions etc. [0m] (iii) Logical with the CORRECT MATERIAL under each heading O O [1m] Penalize incorrect or superfluous chemistry … award proportion out of the 3 marks [see (iv)]. All C.K. material would have to be relevant if 3/3 is awarded. Note … incorrect material is not penalized in C.K. (iv) Note also that a student who hardly write anything. Can‘t get the full 6 marks! The 6 marks assumes a full essay, so mark accordingly … proportionality with respective to C.K., as shown above. PRESENTATION [4m] (i) Scripts with very little useful material can‘t hope to get the full 4 marks, therefore award proportionately …

(ii) (iii)

Sugeest: For C.K. 8- 10 awards up to 4m 6-7 max. is 3m 4-5 max is 2m, etc. Diagrams / schemes: [1m] Note: This is not a neatness mark. Quality of English: This is not a spelling test so do not penalize the odd spelling error, but

(iv)

if particularly bad, or if meaning is lost, then penalize. [2m] Award marks for clear unambiguous precise explanations or statements. Continuity: Answer should be in ESSAY form, not short bits of answers. [1m] Check if there is an introduction or Δthread‘ to take interest.

2. Out of syllabus.

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3. Chemical Knowledge [normal 10m to outstanding 12m] At some point the type of compounds involved should be clarified: With R = Ph à Phenols With R = CnH2n+1 à Alcohol (R3COH is tertiary, R2CHOH is secondary & RCH 2OH is primary) [1/2m] in an introduction, but not necessarily. Since Chemistry involves PREPARATION and REACTION, these are expected as major sub-headings. PREPARATION PHENOL: [max. 1.5m] [OUT] ALCOHOL: By SUBSTITUTION: [max. 2.5m] e.g. RCH2Cl + OH- à RCH2OH + ClNaNO2 + HCl RCH2NH2

RCH2OH + N2

By ADDITION: H2SO4(l) Hydration of alkenes :

H2O(l)

>C=C<

By REDUCTION: Ketones : sec. alcohols Aldehydes : primary alcohol Methods / reagents : LiAlH4, Na + C2H5OH Each complete method carries a mark … do not mark secondary and primary as TWO. ONE MARK should be awarded for at least 4 unrelated facts, e.g. reagents, products. By GRIGNARD REAGENTS [OUT] OTHER METHODS also exists, but are generally less attractive (less mark), e.g. hydrolysis of ester. REACTION [max 6m] Some common …OH reactions for phenols and alcohols: ACYLATION to give esters:

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ROH + R‘COCl or (R‘CO)2O à R‘COOR ALKOXIDE formation: PhOH + Na à PhONa R-OH + Na à RONa Both are useful intermediate e.g. to give ALKYLATION (ether formation) : RONa + R‘I à R-O-R‘ PHENOL ONLY REACTIONS: Ring substitution:

Many other examples exists, e.g. nitration ALWAYS o- and p- substitution Kolbe Schmitt synthesis:

Polymerization with HCHO to give phenol-methanal

ALCOHOL ONLY REACTIONS: -OH SUBSTITUTION: ROH + PCl5 à R-Cl similarly for RBr ALCOHOL OXIDATION : RCH2OH à RCHO à RCOOH R2COH à RCOR (Ketone) Tertiary alcohol do not oxidize Reagents: chromic acid (CrO3 + H2O), K2Cr2O7, KMnO4 etc. -OH ELIMINATION: >CH-C-OH à >C=C<

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Ease of dehydration: 3°>2°>1° Reagents: Heat (about 400°C, Al2O3 cat.) Heat ( > 150°C, H2SO4(l) cat.) OTHER ASPECTS OF THE CHEMISTRY OF …OH COMPOUNDS [max. 1.5m] 1. Reason for the acidic nature of phenols: Resonance stabilization of the phenoxide ion

2. How to - distinguish between phenol and alcohol: neutral FeCl3(aq) colour for phenols. - Separate phenol and alcohol: solvent extraction with NaOH(aq) and water. 3. How to distinguish among 3°,2° and 1° alcohol : Oxidation of alcohol or Lucas test 4. Physical chemical properties: strength of acid (phenols). Note: ONE mark is for a very complete reaction … may include some mechanisms, a comparison, or in fact 2 reactions (reduction or both aldehyde and ketones for examples) Organization [6m] [5 if exceptionally good CK] Look out for and mark the following points: (i)

(ii) (iii)

(iv)

Use of reasonable subheadings O O [2m] If only implied not stated O O [1m] If stated out but not reasonable O O [1-0m] The essay should be easy to follow and not haphazardly put down O O [1m] If a jumble of reactions etc. [0m] Logical with the CORRECT MATERIAL under each heading O O [1m] Penalize incorrect or superfluous chemistry … award proportion out of the 3 marks [see (iv)]. All C.K. material would have to be relevant if 3/3 is awarded. Note … incorrect material is not penalized in C.K. Note also that a student who hardly write anything. Can‘t get the full 6 marks! The 6 marks assumes a full essay, so mark accordingly … proportionality with respective to C.K., as shown above.

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PRESENTATION [4m] (v) Scripts with very little useful material can‘t hope to get the full 4 marks, therefore award proportionately … Sugeest: For C.K. 8- 10 awards up to 4m 6-7 max. is 3m 4-5 max is 2m, etc. (vi) Diagrams / schemes: [1m] Note: This is not a neatness mark. (vii)

Quality of English: This is not a spelling test so do not penalize the odd spelling error, but

(viii)

if particularly bad, or if meaning is lost, then penalize. [2m] Award marks for clear unambiguous precise explanations or statements. Continuity: Answer should be in ESSAY form, not short bits of answers. [1m] Check if there is an introduction or Δthread‘ to take interest.

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