HCI Prelim C2 P2 MS

1 HWA CHONG INSTITUTION  2017 C2 CHEMISTRY PRELIMINARY EXAMINATIONS  PAPER 2 MARK SCHEME    1 (a) (i) H 2(g) +I 2(g) ⇌ 2HI(g) Initial conc/ mol dm¯3 0.05 0.05 0 Change / mol dm¯3 - 0.03 - 0.03 + 0.06 Eqm conc / mol dm¯3 0.02 0.02 0.06 2 22 [HI] [H ][I ] cK  [1] Kc =   2 2 0.02 0.06 = 9 [1] (ii) H 2(g) +I 2(g) ⇌ 2HI(g) Initial conc/ mol dm¯3 0.02 0.02 0.06 Change / mol dm¯3 +x +x -2x Eqm conc / mol dm¯3 0.02+x 0.02+x 0.06-2x Kc =   2 2 0.06 2 0.02 x x   = 0.36    0.06 2x 0.02 x   = 0.6 x = 0.01846 [1] [H2] = [I2] = 0.02+x = 0.0385 mol dm¯3 [HI] = 0.06-2x = 0.0231 mol dm¯3 shown [1]

Hwa Chong Institution 2017 9729 / 02 / C2 Prelim 2017 2 (iii) Correct axis labels [0.5] Appropriate scale [0.5] Correct shape of curves [0.5] Horizontal lines from 30 – 40 mins and 60 – 70 mins [0.5] *Correct concentration and time values for HI at t = 0, 30 and 60 min [1] *Correct concentration and time values for I2 at t = 0, 30 and 60 min [1] *-[0.5] for every wrong plot (iv) To prevent the position of equilibrium from shifting during the cooling process or when H I is removed when dissolved in water. [1] (b) IIIx xx xxxx xxxx xx xx or   [1] (c) Both I2 and HI have simple molecular structure / consist of simple discrete molecules [0.5] held together by dispersion forces. However, I2 has a larger number of electrons, therefore a larger electron cloud than HI, leading to stronger dispersion forces. [1] So, a larger amount of heat energy [0.5] is needed to separate the molecules, leading to a higher boiling point. 0 0.01 0.02 0.03 0.04 0.05 0.06 0 1 02 03 04 05 06 07 08 0 concentration / mol dm−3 t/ min I2 HI

Hwa Chong Institution 2017 9729 / 02 / C2 Prelim 2017 3 2    (a) (i) A ligand is an ion or molecule with one or more lone pairs of electrons available to be donated into the vacant orbitals of transition metal atom or ion.         (ii)  From the graph, VFe2+:Vphen = 2.5:7.5 therefore you can deduce the following reacting ratio - Fe2+:phen is 1:3 Formula of the complex: [Fe(phen)3]2+             (iii)   Fe 2+ has an incomplete/ partially filled 3d subshell (insufficient to just give electronic configuration)  In the presence of ligands (phen), the degenerate 3d orbitals of Fe 2+ split into two different energy levels with an energy gap ΔE  ΔE falls within the visible region of the electromagnetic spectrum  An electron in a lower energy d-orbital can absorb energy from the visible spectrum and be promoted to a higher energy d orbital that is vacant  The orange-red colour seen is the complement of the blue light absorbed.         (iv)  The energy gap, ∆E, is of a different magnitude in both complexes. Hence wavelength of light absorbed by the ferrozine complex is different from that absorbed by the phen complex and different colours are observed. (no need details on the exact colour of wavelength).                02468 1 0 Absorbance 10                             8                              6                              4                              2                              0  VFe 2+  / cm 3 Vphen / cm 3

Hwa Chong Institution 2017 9729 / 02 / C2 Prelim 2017 4 (b) (i) Reagent: CH3CH2Cl ; condition: AlCl3, warm     (ii) Electrophilic substitution   (c) (i)    [1]     (ii) Accepted answers:  2,4-dinitrophenylhydrazine. Orange precipitate observed for vanillin but no precipitate for 4-VG  Tollen’s reagent with heating. Silver mirror observed for vanillin but no silver mirror for 4-VG  Hot acidified KMnO4. Solution turns from purple to colourless for both but only 4-VG gives an effervescence that formed white ppt when passed through limewater  Hot acidified K 2Cr2O7. Orange solution turned green for vanillin but solution remained orange for 4-VG. 3 (a) (i) hybridisation: sp3 [1] typical bond angle: 109.5° [1] (ii) Forcing the bond angle in the epoxide ring to 60° brings electron pairs in the covalent bonds closer and they experience increased repulsion, weakening the C–O bonds and making them easier to break. [1] or

Hwa Chong Institution 2017 9729 / 02 / C2 Prelim 2017 5 After opening the ring, the resulting product is able to attain an optimal bond angle of 109.5° around the carbon atoms, minimising electron repulsion and eliminating the ring strain. (b) (i) curly arrow from benzylic C–O+ bond to O+ [½], correct carbocation [½] curly arrow from lone pair on O in H 2O to C bearing the positive charge [½], correct intermediate formed [½] curly arrow from O +–H bond to O+ and H+ regenerated [1] (ii) [1] (iii) The other C–O bond was broken instead, forming a primary carbocation intermediate. H218O was then able to attack the primary carbocation to form isotopic isomer A. [1] The primary carbocation formed is much more unstable than the secondary carbocation (which is particularly stable as it is resonance stabilised). The reaction mechanism is therefore much less likely to proceed via the 1° carbocation intermediate to form A. [1] (c) step 1: conc HNO3, conc H2SO4 [½], maintained at 55 °C or < 55 °C [½] step 2: KMnO 4(aq), H2SO4(aq), heat [1] step 3: CH3CH2OH (accept ethanol), few drops of conc H2SO4, heat (under reflux) [1] [1] each

Hwa Chong Institution 2017 9729 / 02 / C2 Prelim 2017 6     4 (a) Carboxylic acid, aldehyde, ketone, alkene 4 x [1] (b) (i)    [1]  (ii) 2 x [1] (iii) 1 mark for correct structure for oxidative cleavage of C = C bond 1 mark for correct structure for the oxidation of –CHO to –CO2H (d) [1] Extensive hydrogen bonding exists between PEG and water molecules. Energy released from this intermolecular hydrogen bonding with water is able to compensate for dispersion and p.d.-p.d. interactions between PEG, and intermolecular hydrogen bonding in water. [1]

Hwa Chong Institution 2017 9729 / 02 / C2 Prelim 2017 7 (c) Elenolic acid is a stronger acid / has higher acidity than compound M, which is a primary alcohol. [1] The negative charge on the carboxylate ion (conjugate base of elenolic acid) is delocalised equally over two highly electronegative oxygen atoms. The negative charge is dispersed and the carboxylate ion is greatly stabilised. [1] Whereas the electron-donating alkyl group intensifies the negative charge on the alkoxide ion (conjugate base of compound M) and the alkoxide ion is destabilised. [1]   5 (i) Cu (s) → Cu2+ (aq) + 2e– [1] (ii) Q = 3100 x 0.700 = 2170 C [1] Amount of electrons transferred = (0.714/63.5) x 2 = 0.02249 mol [1] Since Q = e x L x e 2170 = 0.02249 x L x 1.60 x 10–19 L = 6.03 x 1023 mol–1 [1] (b) (i) Ni2+(aq) + 2e– Ni(s) –0.25V Cu2+(aq) + 2e– Cu(s) +0.34V [1] (correct quotation) The reduction potential of the Cu 2+/Cu half cell is more positive than that of Ni 2+/Ni half cell and therefore Cu2+ will be preferentially reduced at the cathode. [1] (ii) 1s22s22p63s23p63d8 [1] (iii) Upper x y z dxy x z y dxz z y x dyz OR OR [1] for any of the three being drawn + correct label (shading not required; ignore if drawn) Lower x y z dx2 - y2 x y z dz2 OR [1] for any of the two being drawn + correct label (shading not required; ignore if drawn)

Hwa Chong Institution 2017 9729 / 02 / C2 Prelim 2017 8 (c) (i) From the graph, since the plot of kobs against OH – is a straight line that passes through the origin, m=1. [1] (ii) Rate = k[OH–][isocyanide] [1] When [OH–] = 0.74 mol dm–3, kobs = 5.6 x 10–3 s–1 rate = kobs [isocyanide] = 5.6 x 10–3 x (5.0 x 10–4) = 2.8 x 10–6 mol dm–3 s–1 [1] OR Gra