HCI H2 CHEM P3 ANSWERS Prelim

2014 HCI H2 Chemistry Prelims Paper 3 Mark Scheme 1 (a) Description: Cl2 reacts explosively with H2 under direct sunlight. Br2 reacts slowly with H2 upon heating. I2 reacts partially/reversibly with H2 upon heating. Explanation: Down the group, the bond strength for HX decreases in order of HCl > HBr > HI. The enthalpy of reaction to become less exothermic, thus the reaction is less vigorous down the group. [2] (b) (i) Name of mechanism: Free radical substitution Initiation: Propagation: + Br + HBr + Br2 Br + Br Termination: + Br Br + [3] (ii) Large excess of alkane / limited amount of bromine. [1] (c) (i) The bromine is added on the least substituted carbon atom. [1] (ii) Bromine radical is electron deficient, while C=C bond is electron rich. OR The  bond of the C=C bond is weaker than the  bond of the C-H bond. Hence, the radical will attack the alkene preferentially. [1] (iii) Br The intermediate formed in the reaction ha s more electron donating alkyl group s attached to the carbon radical compared to that drawn above . Hence it is more stable, which will lead to the formation of the major product. [2] (iv) [3] (d) (i) Unit of k: s-1 (or h1 or min-1) [1]

(ii) At higher temperatures, the proportion of molecules with kinetic energy greater than or equal to activation energy increases . Thus, frequency of effective collision increases, rate constant increases. [3] (iii) Chlorine radical acts as homogeneous catalyst. It exists as same phase as reactants and is regenerated. [2] (iv) C-Cl bond in CFCs can be broken down by UV light to form chlorine radicals, which react with ozone. [1]

2 (a) Longifolene is a non–polar molecule. It is does not form favourable interactions with water. [1] (b) In order of increasing basicity: Hydrazine, ammonia, ethylamine Each nitrogen atom on hydrazine is bonded to an electron withdrawing nitrogen atom, which reduces the availability of the lone pair of electrons to accept a proton . Hence, it is a weaker base compared to ammonia. Ethylamine has an electron donating alkyl group bonded to the nitrogen atom, increasing the availability of the lone pair of electrons to accept a proton compared to ammonia and hydrazine. Hence it is the strongest base. [3] (c) (i) I: LiAlH4/ dry ether or H2/Ni or NaBH4/ methanol II: KMnO4 or K2Cr2O7/ H2SO4(aq) / heat IV: SOCl2 or PCl3 or PCl5 or conc HCl V: Ethanolic NaOH, heat [4] (ii) Add 2,4–dinitrophenylhydrazine to the sample. Orange precipitate observed with A. No orange precipitate with B. Accept other tests, i.e. PCl3, PCl5, oxidising agent tests, etc. [2] (d) (i) OR Trigonal pyramidal [2] (ii) HCl(aq) or HNO3(aq) or H2SO4(aq) or H2O(l) [1] (iii) Name: Nucleophilic addition R R O CH3 R CH3 R O  slow R CH3 R O H R CH3 R OH fast [3] (e) (i) The folding of the farnesyl carbocation results in a locked ring structure which reduces the possible conformations of the molecule. There is hence less ways to distribute energy in the system, so ∆S is negative. [2] (ii) 4 [1] (iii) Laboratory synthesized longifolene contains one of the optical isomer which smells differently from extracted longifolene which contains the other optical isomer. OR

Laboratory synthesized longifolene could be a racemic mixture while extracted longifolene contains only one optical isomer. Each optical isomer would smell differently, hence the samples have different smells. OR Laboratory samples may contain a mixture of the different stereoisomers but the natural sample only contains one optical isomer. Each optical isomer would interact differently with the body and smell differently, hence the 2 samples have different smell. [1]

3(a) (i) [2] O–H bond is polar as O is more electronegative than H. H2O molecule is bent. There is a net dipole (or overall dipole) for the molecule. (ii) [2] H2O forms stronger hydrogen bond than NH3 because O is more electronegative than N. H2O forms more hydrogen bonds than NH3 per molecule. Boiling water involves overcoming more hydrogen bonds and stronger hydrogen bonds, which need more energy. (b) (i) A complex ion is an ion that contains a central metal cation or atom, dative bonded by surrounding anions or molecules, known as ligands. [2] (ii) [1] Transition metal ions have high charge density and polarising power. They have a strong tendency to form dative covalent bonds with ligands. or Transition metal ions have vacant 3d orbitals (as well as 4s and 4p orbitals) of low energy level (or suitable energy) that can accommodate lone pair of electrons donated by ligands, resulting in dative bond formation. (iii) [2] Cu(OH)2 + 4NH3  Cu(NH3)4(OH)2 or Cu(OH)2 + 4NH3  [Cu(NH3)4]2+ + 2OH– A dark blue (or deep blue) solution is formed. (c) [2] 2NH3 ⇌ NH4 + + NH2 – KNH3 = [NH4 +][NH2 –] = 1  10–30 At equilibrium, [NH4 +] = [NH2 –]  [NH4 +] = NH3K = 1.00  10–15 mol dm–3 (d) [2] pH = 9  pOH = 5  [OH–] = 10–5 mol dm–3 [salt][OH–] / [base] = Kb = 1.8  10–5  [salt]/[base] = 1.8 = 18/10  he has to convert 0.18 mol of NH3 into NH4Cl so that the resultant buffer contains 0.10 mol of NH3 and 0.18 mol of NH4Cl, which gives the 18/10 ratio. n(HCl) needed = 0.18 mol Volume of HCl needed = 0.18 dm3 or 180 cm3 (no s.f. penalty  exact ans) (e) [4] For phenylamine: Br2(aq) For benzene: Br2, FeBr3 (or AlBr3) Comment on two differences: For benzene to undergo electrophilic substitution, a (Lewis acid) catalyst like FeBr3 is needed and only mono-substitution occurs. For phenylamine, no catalyst is needed and tri-substitution occurs. Explain: Lone pair of N of NH2 group delocalises into the benzene ring. This increases the electron density in the benzene ring, making the benzene ring more susceptible towards electrophilic substitution (or electrophilic attack).

(f) [3] ClCH2 C N H CH2CH2NH2 O C: The COCl carbon is attached to two highly electronegative atoms, Cl and O, as compared to only one electronegative Cl in the case for the CH2Cl carbon. Hence, the COCl carbon has a higher partial positive charge than the CH2Cl carbon OR The COCl carbon has trigonal planar geometry while the CH2Cl carbon has tetrahedral geometry. The planar geometry of the COCl carbon makes it less hindered for the nucleophile to attack.

4 (a) (i) PCl5 + H2O  POCl3 + 2HCl [1] (ii) Step 2: PCl4OH + H2O  PCl3(OH)2 + HCl Step 3: PCl3(OH)2  POCl3 + H2O [2] (b) (i) PCl5 + 4H2O  H3PO4 + 5HCl Hydrolysis pH = 1 (accept any value from 0 – 3) [3] (ii) +5 [1] (iii) [1] (iv) [1] (v) H6P4O13 [1] (or H4P4O12 if cyclic) (c) (i) [3] (ii) The high bond energy / strength of P=O bond that is formed drives this reaction to completion. [1] (iii) To ethanal: K2Cr2O7, dilute H2SO4, heat with immediate distillation To ethanoic acid: K2Cr2O7 OR KMnO4, dilute H2SO4, heat (with reflux) [2] (d) Test 1 : To all three unknowns, add dilute NaOH and heat. (hydrolyse the ester and amide bonds) The sample which gives off a gas that turned moist red litmus paper blue is Y, whilte the samples which did not give off any gas is X or Z. Test 2: To the hydrolysed products of the other two unknowns, add I2(aq), NaOH(aq) and heat. (test for presence of CH3CH(OH)− group)

The sample which gives the yellow ppt of CH I3 is X, while the sample which does not give the yellow ppt is Z. Alternative answer: Test 1: To all three unknowns, add dilute H 2SO4 and heat. Then add I2(aq), NaOH(aq) and heat. The sample which