Mock chem paper 2 suggested solutions (corrected)

IDK Junior College CANDIDATE NAME CHEMISTRY SOLUTIONS 9476/02 2 hours Additional Material(s): None READ THE FOLLOWING FIRST I do hope you have enjoyed/learnt something from this paper. It is meant to be a tad unconventional and unusual, do take heart in having tried your very best! For organic structures, it will be penalised on the front page under structures if bonds are not bonded to the correct atom, e.g. C–HO, C–H 2 N et cetera. This is the first paper I have ever made, so any feedback/inquiries will be greatly appreciated! Despite best attempts to ensure accuracy, some mistakes may arise and if so, do email ian321714@gmail.com. In general, answers are written in blue with their marks denoted next to them. Diagrams are in black because my software does not support other colours unfortunately :( Information written in red are deductions/potential pitfalls that you must look out for, or just general helpful tips. This document consists of 16 pages 1

Answer all questions 1 Squaric acid is a dibasic organic acid. Fig. 1.1. shows the skeletal formula of squaric acid. Fig. 1.1 (a) State the IUPAC name of this compound. You are told that priority goes to the carbonyl groups. (suffix) [2] 3,4-dihydroxycyclobut-3-en(e)-1,2-dione - 1 mark for correct numbering - 1 mark for correct name - The (e) is optional. (b) Fig. 1.2 shows the delocalisation of the divalent squarate anion. Draw, on Fig 1.2, curly arrows and lone pairs to show how this resonance stabilisation is achieved, from left to right. [1] Fig. 1.2 (c) Hence, suggest why the divalent squarate anion is a perfect square. [1] The electrons are delocalised equally across the entire molecule , and so the electron density is equally distributed, so the bond lengths of the bonds are equal , resulting in a perfect square. - 1 mark for both parts (d) Given that the 1st pKa of squaric acid is 1.5, draw the predominant species present at pH 3 [1] [1] For correct species drawn (only 1 H is deprotonated) with a negative charge 2

(e) You are given squaric acid, ethanoic acid and ethanol at equal concentrations. Describe their relative acidities and explain. [3] From most acidic to least acidic: squaric acid, ethanoic acid, ethanol. Squaric acid is most acidic due to the negative charge on the oxygen atoms on the divalent squarate anion/conjugate base (if discussion is on squaric acid instead of the c.base, no marks awarded) being able to be delocalised throughout the entire molecule, dispersing the negative charge to the greatest extent compared to the other two molecules. (1 mark for describing squaric acid) For ethanoic acid, the negative charge on the ethanoate anion (if discussion is on ethanoic acid here, no marks awarded) is delocalised over two equally electronegative atoms but not the entire molecule (needed in some form to draw comparison), so the negative charge is dispersed to a smaller extent than in the conjugate base of squaric acid, so the ethanoate ion is less stabilised, hence less acidic. (1 mark for describing ethanoic acid) For ethanol, the negative charge on the oxygen of the conjugate base of ethanol is intensified by the inductively electron-donating alkyl group, destabilising the conjugate base , hence making ethanol the least acidic of the three. (1 mark for describing ethanol) - Structures of the chemical compounds instead of words are accepted so long as the structures are correct. (f) A notable derivative of squaric acid is dibutyl squarate. (Fig. 1.3) In order to form C-O-C (ether) bonds, an S N 2 reaction takes place where the conjugate base of an alcohol acts as a nucleophile and reacts with a suitable halogenoalkane. Hence, state the reagents needed to form dibutyl squarate from squaric acid. [2] Fig. 1.3 Step 1: sodium metal (Na (s)), step 2: 1-bromobutane/1-chlorobutane/1-iodobutane - If steps are not numbered, maximum mark = 1m - If “chlorobutane/bromobutane/iodobutane” is given, reject. State symbols are not needed (g) The divalent squarate anion is an example of the oxocarbon anion, which is a negative ion consisting solely of carbon and oxygen atoms. Each oxocarbon anion can also form a corresponding hydrogenated anion, H k C x O y m⁻ . Find the average oxidation number of carbon in this ion, leaving your answer in terms of k, x, y, z and m. Let z be the average oxidation number. [1] –m = zx+k(1) – 2(y) [1] z = 2 𝑦 − 𝑚 − 𝑘 𝑥 3

2 Alkynes are a class of organic compounds with the general formula C n H 2n⁻2 . Table 2.1 shows the carbon-hydrogen bond length in ethane, ethene and ethyne. Molecule Carbon⁻hydrogen bond length /Å Ethane 1.14 Ethene 1.09 Ethyne 1.06 Table 2.1 (1Å = 10 ⁻ 10 m) (a) Use the concept of hybridisation to explain the difference above. [2] The sp hybridised carbon atom in ethyne has the highest percentage s character , followed by the sp 2 hybridised carbon atom in ethene and lastly, sp 3 hybridised carbon atom in ethane. [1] Hence, the extent of orbital overlap between the sp hybridised carbon atom and the s orbital of H atom is the largest, resulting in the shortest bond length. [1] (b) Interestingly, terminal alkynes are more acidic than alkenes or alkanes, with a pKa of 25. By drawing the conjugate base of ethyne, suggest why this is so. [1] When C-H is deprotonated, the resulting carbanion is held in an orbital with 50% s-character . Since s -orbitals are closer to the nucleus than p -orbitals, this means that the electrons experience greater stabilisation from the positively charged nucleus than the conjugate bases of alkenes and alkanes. [1] - reject if ethyne is not drawn with linear geometry 4

(c) Using the information in Table 2.2, calculate the standard enthalpy change of formation of C 2 H 2 . [3] Table 2.2 By Hess’ Law, ⊖ f = 401 - 572/2 + 221/2 = +225.5 kJ mol -1 ∆ 𝐻[1] for correct energy cycle/working [1] correct calculation WITH units and positive sign (d) Suggest the type of reaction alkynes tend to undergo, briefly explain. [2] Electrophilic addition . [1] The C ≡ C bond is electron-rich and hence attract electrophiles , (which are electron-deficient species) [1]