ASR 2020 J2Prelim H2Chem P3 MS

ASRJC JC2 PRELIM 2020 9729/03/H2 1 Anderson Serangoon Junior College Preliminary Examinations H2 Chemistry (9729) Paper 3 Mark Scheme Section A Answer all questions. 1 Many naturally occurring organic compounds contain either acidic groups or basic groups or both. The strength of the acid or the base depends on the structure of the molecule. (a) The pKa values of ethanoic acid and two amino acids are given in Table 1.1 below. Table 1.1 name structure pKa1 pKa2 pKa3 ethanoic acid CH3COOH 4.7 – – serine (ser) H2N CH CO2H CH2 OH 2.2 9.2 – aspartic acid (asp) H2N CH CO2H CH2 CO2H 1.9 3.7 9.6 (i) Suggest two reasons why the p Ka1 value of serine is so much less than the p Ka1 of ethanoic acid. [2] Electron withdrawing –NH3+ (or –NH2) group is in close proximity to the negative charge of the carboxylate anion of deprotonated serine, hence dispersing the negative charge and stabilizing the anion, and –OH of the R group of serine is able to form intramolecular hydrogen bond with the O atom of –COO–. This stabilizes the carboxylate anion. Hence, carboxylate anion of serine is more stabilised than ethanoate ion (ii) Draw a skeletal structure for the dipeptide, asp–ser. [2] N OH O OH O H2N O HO H

ASRJC JC2 PRELIM 2020 9729/03/H2 2 Electrophoresis is a technique used to separate charged particles placed in an electric field. The system consists of two oppositely –charged electrodes connected by a conducting medium, typically a gel. The separation of the charged particles is based on their electrical charge and Mr. A sample consisting of serine and aspartic acid was analysed by electrophoresis using a gel buffered at pH 5 .7. The small quantity of the mixture was placed at the centre of the gel, at equal distance from the two electrodes. (iii) Using the information given in Table 1 .1, suggest the structures of the major species present in the buffer solution. Label the structures clearly. [2] asp ser (iv) Hence, describe the relative positions of the two species in the gel after the separation by the electrophoresis process. Explain your answer. You may find it helpful to draw a diagram to illustrate your descriptions. [2] Ser has no net charge / electrically neutral / exists as a zwitterion. Hence it will not move and remain at the centre of the strip. Asp has an overall negative charge, and so it will migrate towards the anode. asp Anode (+) Cathode (–) ser (b) A student prepared a buffer solution of pH 5.7 by adding solid sodium hydroxide to 200 cm3 of 0.200 mol dm–3 ethanoic acid. (i) With reference to the information in Table 1.1, calculate the mass of sodium hydroxide the student would have used. [2] pKa (CH3COOH) = 4.7 Ka = 10–4.7 mol dm–3 Initial n(CH3COOH) = 200 x 10–3 x 0.200 = 0.04 mol Let the no.of mol of NaOH needed be a mol. CH3COOH + NaOH  CH3COO–Na+ + H2O a mol a mol Alternative: Ka = [H+]([A–]/[HA]) [A–]/[HA] = 10 Let x be amt of HA left 11x = 0.04 x = 3.636 x 10–3 mass of NaOH added = (0.04 – 3.636 x 10–3) x 40.0

ASRJC JC2 PRELIM 2020 9729/03/H2 3 Ka = [HA] ]][A[H  10–4.7 (0.04 – a) = 10–5.7 x a a = 0.03636 mol mass of NaOH = 0.036359 x 40.0 = 1.45 g = 1.45 g A buffer solution generally loses its effectiveness when one component of the buffer pair is less than about 10% of the other. (ii) Calculate the percentage of ethanoic acid component present in the buffer solution in (b)(i) above and hence comment on the effectiveness of the buffer. [2] To determine the percentage of ethanoic acid component present in the buffer solution Alternative: from previous part [HA]/[A–] = 0.1 %HA in buffer = (0.1)/1.1 x 100% = 9.09% To comment on the effectiveness of the buffer , need to find ratio of one component to the other component of the buffer. That is, ratio [acid] [salt]. From the above calculation, [acid] [salt] = 0.1 % [HA] with respect to the salt = 0.1  100% = 10% Comment and mention of how the above percentage is determined:  Buffer is useful/within limits of effectiveness as the % of one component is equal to 10% of the other component.

ASRJC JC2 PRELIM 2020 9729/03/H2 4 (c) Nicotinic acid and its amide, nicotinamide, are present in Vitamin B3 and are used to increase good cholesterol levels. N O OH N O NH2 1 2 Nicotinic acid Nicotinamide The structure of nicotinamide consists of a single delocalised system of electrons which includes:  an electron in the unhybridised p–orbital of the nitrogen atom in the six–membered ring  electrons from the carbon atoms of the six–membered ring  two electrons in the  bond of the >C=O group  two electrons in lone pair on the nitrogen atom of the amide group (i) State the number of delocalised electrons in one molecule of nicotinamide. [1] 10 (ii) Using the information provided, suggest the H –N–H bond angle in the NH 2 group in nicotinamide. [1] 120⁰ High tendency for the lone pair of electrons on N atom to delocalised and reduces it availability. Lone pair–bond pair repulsion is not as significant and thus greater chance of it forming a trigonal planar structure about the N atom resulting in bond angle to go close to 120o. (iii) Predict and explain the relative basicity of the two N atoms, labeled N1 and N2, in the nicotinamide molecule. [2] N2 is neutral because the lone pair on nitrogen atom is delocalised into the π bond of the adjacent C=O by resonance, hence not available for donation to a proton. N1 is more basic as the lone pair of electrons on the nitrogen are not in the same plane as the unhybridised 2p orbitals, thus making it more available for dative bond to a proton.

ASRJC JC2 PRELIM 2020 9729/03/H2 5 2 (a) A lithium–iodine electrochemical cell can be used to generate electricity for a heart pacemaker. The cell is non –rechargeable and it consists of a lithium electrode and an inert electrode immersed in bodily fluids. When current flows, lithium is oxidised and iodine is reduced. (i) Write two half–equations for the reactions taking place at the two electrodes. Hence , write the overall equation. [2] Oxidation: Li → Li+ + e– Reduction: I2 + 2e– → 2I– Overall equation 2Li + I2 → 2Li+ + 2I– (ii) Use the Data Booklet to calculate the Eocell for this cell. [1] Eocell = +0.54 – (–3.04) = +3.58 V (iii) A current of 2.5 × 10–5 A is drawn from this cell. Calculate how long a pacemaker will last when 0.1 g of lithium electrode is remaining in the cell. Assume the current remains constant throughout this period. Give your answer to the nearest day.