HCI 2021 Prelim Paper 3 Solutions

Updated 27 Sep 2021 2021 HCI C2 H2 Chemistry Prelim / Paper 3 1 1 (a) [1] – correctly drawn curve When temperature is increased, the reactant particles have greater average kinetic energy. They move more quickly and so collide more frequently. The proportion of particles that have kinetic energies greater than or equal to the activation energy increases. [1] Frequency of effective collisions hence rate of reaction and rate constant increases. [1] The question requires that you use the Boltzmann distribution to explain why an increase in temperature will affect the rate constant of a reaction. In order to do this, you must sketch the Boltzmann distribution curve and show how the kinetic energy of the particles in the reaction mixture changes with an increase in temperature. When drawing the graph, please ensure that the axes label are correct. Also, ensure that the shape of the curve is correct where it flattens out at higher kinetic energy. It is not sufficient to state in your answer that the average kinetic energy of the reactant particles increase. In order for the rate of the reaction to increase, the particles need to have kinetic energy greater than or equal to activation energy. This phrase is essential. Because of this, we can then conclude that the frequency of effective collision between the reactant particles increase and hence the rate of reaction would increase. For this question, we ask how temperature would affect the rate constant of the reaction. You need to explicitly tell us that the rate constant will increase with increasing temperature. HWA CHONG INSTITUTION 2021 C2 H2 CHEMISTRY PRELIM PAPER 3 SUGGESTED SOLUTIONS

Updated 23 Sep 2021 2021 HCI C2 H2 Chemistry Prelim / Paper 3 2 (b) (i) 1 mark for each step (deduct ½ for missing arrows, partial charges, lone pairs, slow fast step) The question states that Reaction 1 follows the SN1 mechanism. As such, it is not necessary to name the mechanism. Please ensure that the arrows that represent the movement of electrons start from a bond or a lone pair, and end at the atom where the new bond is formed. (ii) Rate = k[C6H5CHBrCH3] [1] For the SN1 mechanism, the only reactant in the rate determining step is the halogen derivative. Please ensure that the name of the compound is correct if you choose to use the name of the compound when writing the rate equation. (iii) Order of reaction with respect to a reactant is the power on its concentration term in the experimentally determined rate equation. [1] This is a definition question. Please study your definitions. (iv) [1] Show constant half-life on the graph [1]

Updated 27 Sep 2021 2021 HCI C2 H2 Chemistry Prelim / Paper 3 3 Based on the design of the experiment given in the question, the variables being monitored are the concentration of the bromoalkane at various times. As such, the logical graph to plot is the [bromoalkane] against time. The question requires that you show clearly on the graph how the results obtained can be used to confirm that the reaction is first order with respect to the [bromoalkane]. The graph need not be drawn to scale, but this means that you need to label both axes with values to illustrate that the half-life are constant. Written explanations are not required. No credit is given for answers that only consist of written explanations without any annotations on the graph. (vi) Reaction 2 is slower as the CCl bond is a stronger bond than CBr and requires more energy to break. [1] The rate determining step in the SN1 reaction involves the breaking of the C-Halogen bond. A direct answer on the relative strength of the bonds (you could even quote the bond energy values from the Data Booklet) would suffice here. (c) (i) 1-bromobut-2-ene [1] When naming a molecule, start by finding the longest chain of carbon atoms. For compound P, this would be a 4 carbon chain. Since the C=C is on C2 and C3, this is a but-2-ene. There is a bromo group on the terminal carbon. This terminal carbon could either be considered C1 or C4. Since IUPAC nomenclature requires that we use the smallest number to name the compound, we should assign the carbon connected to the bromine atom as C1. The name of compound P is therefore 1-bromobut-2-ene. Please note that hyphens are used only between number and letter. As the C=C in this compound shows the trans configuration, the more complete naming of the compound is trans-1-bromobut-2-ene. (ii) The carbocation formed when P undergoes SN1 mechanism is stabilised due to the delocalisation of the positive charge over C=C OR due to the delocalisation of the pi electrons of the C=C to the C+. [1] Due to the delocalisation, the positive charge shifts to the third carbon to form OR the position of the C=C can shift to form that forms R. [1] The question directs you to think about why this primary bromoalkene is able to undergo the SN1 reaction. Given that it is a primary halogen derivative, arguments for the R group causing steric hindrance and not favouring the SN2 will be invalid. This R group is not very big and would not do much to block the pathway of attack of the nucleophile. Instead, you should be thinking about how the carbocation can be stabilised. The question directs you to think about the concept of delocalisation in your

Updated 23 Sep 2021 2021 HCI C2 H2 Chemistry Prelim / Paper 3 4 answer. When you inspect the carbocation formed, you will realise that the carbocation is adjacent to the C=C, so there are 3 adjacent p orbitals where delocalisation can occur, hence the electrons in the C=C can delocalise to disperse the positive charge on the carbocation. (iii) Q: enantiomerism [1] [1] for correctly drawn pair of enantiomers When drawing the enantiomers, please make sure that you have drawn the 3D representation (this means using the solid wedge and hash lines in your answer) on the chiral carbon. While it is not necessary to draw the displayed formula here, you should ensure that you have written out clearly the different substituent groups around the chiral carbon, showing correctly how each group is connected to the chiral carbon. (iv) [1] OR From the earlier parts, we know that P can undergo nucleophilic substitution. Hence when P is reacted with ethylamine, ethylamine which has available lone pair on N can act as a nucleophile to substitute the Br atom on P. The subsequent loss of H+ produces S which fits C6H13N.

Updated 27 Sep 2021 2021 HCI C2 H2 Chemistry Prelim / Paper 3 5 Just like how R is formed, ethylamine can also attack the carbocation that has the positive charge on C3 after the position of C=C shifts. This is how the alternative answer above is derived. Please read the question carefully. The question is asking for the displayed formula of the product formed from the reaction. So after you figured out how the product is formed, you need to draw it as a displayed formula. (d) PBr3 is trigonal pyramidal [1/2] as there are 3 bond pairs and 1 lone pair around P [1/2] ClF3 is T-shaped [1/2] as there are 3 bond pairs and 2 lone pairs around Cl [1/2] Diagram with correct number of bond pairs and lone pairs [1/2] each, showing shape [1/2] each Bond angle labeled between two bond pairs [1/2] each The question requires that you do a few things: 1. Use VSEPR to predict the shape of each molecule. This means that you need to count and state the number of bond pairs and lone pairs around the central atom of the molecule, and use it to deduce the shape of the molecule. You need NOT state the principles of VSEPR as it is not asked for in the question. 2. Draw diagrams to illustrate the shape This means th