2022 EJC Prelims Paper 3 (solution with comments)

© EJC [Turn Over a EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2022 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME CIVICS GROUP 2 1 – INDEX NUMBER CHEMISTRY Paper 3 Free Response 9729/03 21 September 2022 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, civics group and registration number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 / 20 2 / 20 3 / 20 Section B 4 / 20 5 / 20 Total / 80 This document consists of 32 printed pages.

2 © EJC 9729/03/J2Prelims/22 Section A Answer all the questions in this section. 1 (a) Elements in Period 3 of the Periodic Table form chlorides and oxides with varying properties. (i) Equal amounts of MgC l2 and PC l5 dissolve in equal volumes of water to form solutions with pH 6.5 and 1.0 respectively. Construct two equations to account for the different pH values. [2] Comments: Many students forgot that the hydrolysis of the hydrated magnesium ion is a reversible reaction since it is only partial hydrolysis. (ii) Explain why the first ionisation energies of aluminium and sulfur are lower than the element before it. [2] Comments: Many students stated that the p orbital of Al i s higher in energy than the s orbital, without stating the principal quantum number. This makes it ambiguous, because it is only true if it is 3p vs 3s. If it was 2p vs 3s, then the 3s would be of higher energy. Thus, it is important to specify the principal quantum shell. Some students thought that the valence shell of Al and Mg contains 2p and 2s orbitals. (iii) Element L is an element in Period 3 . L forms a chloride with a simple molecular structure but an oxide with a giant molecular structure. Identify element L and construct an equation for the reaction of its chloride with NaOH(aq). [2] Comments: This was very badly attempted. One of the most common errors for this question was seeing HCl appear in the products. There is no chemical reaction known to mankind where adding a strong aqueous base, such as NaOH, will result in the formation of a strong aqueous acid, such as HCl. If this could happen, then none of us would be able to perform strong acid-strong base titrations. Some students did not read the question and constructed reactions of SiCl4 with water. Some students thought that the formula for the chloride of Si is SiCl2 / SiCl3. The most common error was that students thought the identity of L was aluminium. This shows that many students did not read the question FULLY. It is stated in the question that the chloride of L has a simple molecular structure (which is true for both aluminium and silicon), but an oxide with a GIANT MOLECULAR STRUCTURE! Al2O3 is an ionic compound! Only SiO2 has a giant molecular structure! (i) [Mg(H2O)6]2+  [Mg(OH)(H2O)5]+ + H+ PCl5 + 4H2O → H3PO4 + 5HCl (ii) The first I.E. of Al is lower than that of Mg since the electron is removed from the higher energy 3p orbital of Al, hence less energy is required. The first I.E. of S is lower than that of P since the electron removed comes from a fully-filled 3p orbital, where the paired electrons experience inter -electronic repulsion, hence less energy is required. (iii) Element L is silicon. SiCl4 + 4NaOH → SiO2 + 2H2O + 4NaCl or SiCl4 + 4NaOH → Si(OH)4 + 4NaCl or SiCl4 + 6NaOH → Na2SiO3 + 3H2O + 4NaCl

3 © EJC 9729/03/J2MYE/22 (b) 4-(butan-2-yl)phenol is an important intermediate in organic synthesis. It must be handled carefully as it is corrosive to the eyes and skin. 4-(butan-2-yl)phenol can be synthesi sed by heating phenol and butan -2-ol in the presence of acid. The mechanism of this synthesis is thought to proceed via an acid - catalysed Friedel-Crafts alkylation, where butan -2-ol is first protonated by H +, before forming a carbocation. (i) Draw the structure of the carbocation formed from intermediate X. [1] Comments: Many students do not seem to understand what a carbocation is. As the self - explanatory name suggests, it must carry a positive charge on a CARBON! (ii) Hence, describe the mechanism for the formation of 4-(butan-2-yl)phenol from the carbocation in (b)(i) and phenol. [3] (i) (ii) Electrophilic substitution

4 © EJC 9729/03/J2Prelims/22 Comments: Many did not write down the name of the mechanism. The arenium ion was poorly drawn by many students. In the arenium ion, there is one sp3 carbon atom and five sp2 carbon atoms in the ring. The delocalisation of pi electrons must be drawn from the first sp2 carbon atom to the last sp2 carbon atom. It cannot be any less or any more than that. The positive charge must also sit within this region of delocalisation. Many students drew the positive charge protruding from the region of delocalisation. Many students also missed out the regeneration of H + in the final step of the mechanism. (iii) Describe a simple chemical test that could distinguish between butan -2-ol and 4-(butan-2-yl)phenol. [2] Comments: Generally well done. Students are reminded to include the conditions such as room temperature if they are using Br2(aq) and neutral FeCl 3(aq). Students are reminded to state the observations for BOTH compounds, not just the positive test observations when the distinguishing test is conducted on both. For the negative test observations, students are expected to state the ‘opposite’ of the positive test observations for instance: No violet complex observed for butan -2-ol when using neutral F eCl3(aq)/r.t. Students were penalised for stating phrases like ‘no observations’, ‘no reaction’, ‘no visible change’ etc. (iv) Compounds M and N are two minor products in the reaction. Briefly explain why each of them is the minor product. [2] Comments: Generally well done. Candidates are reminded that Compound M is the position 3 isomer, not 3-directing group. Many candidates are penalised for omitting that M is the 3-isomer. This is because merely stating that the pre -existing side chain –OH is 2,4 directing does not adequately explain why M is minor product. Candidates were also penalised for merely stating inter electronic repulsion without further elaboration when attempting to explain why N is minor product. (iii) Add NaOH(aq), followed by I2(aq) and heat. Butan-2-ol will give a yellow ppt of CHI3, while 4-(butan-2-yl)phenol will not give yellow ppt. Accept Br 2(aq)/r.t., neutral FeCl 3(aq)/r.t., PCl 5/r.t. K 2Cr2O7(aq)/heat. With correct observations, Reject KMnO4(aq) as both give the same observations. (iv) Compound M, the 3-isomer, is a minor product as the phenolic OH group is a 2,4- directing group. Compound N, the 2-isomer, is a minor product due to steric hindrance between the phenolic OH group and the incoming carbocation.

5 © EJC 9729/03/J2MYE/22 Candidates are reminded the importance of reading question carefully as many were penalised for explaining why either M or N is the minor product and not both are minor products as required by the question. Several can