HCI 2021 Prelim Paper 2 Solutions

2021 HCI C2 H2 Chemistry Prelim / Paper 2 1 (a) (i) The carboxylic acid functional group is acidic because the negative charge on its conjugate base anion is delocalized equally over two highly electronegative O atoms [1], stabilizing the conjugate base [1]. This is a recall question. Refer to Section 4.1.1 of Carboxylic Acid & Derivatives lecture notes. Do read the question carefully. You were asked to explain the acidity of the carboxylic acid functional group. You should not be explaining the effect of the NH2 group on the acidity of the carboxylic acid group Also, be very specific in the use of terms. You should explain why the conjugate base is stable by considering how well the negative charge is dispersed due to delocalization of negative charge of the conjugate base over 2 highly electronegative atoms. It is incorrect to say that the dispersion of the charge is through inductive effect. (ii) alkene [1] primary amine [1] QRWH>ò@IRU³SULPDU\´>ò@IRU³DPLQH´ This question was well done but many students did not state the type of amine. Do note that N atom is only attached to one carbon chain which makes it a primary amine. (iii) Hypoglycin A does not show cis-trans isomerism because one of the C=C carbon is attached to two identical groups / two hydrogen atoms. [1] You should apply the criteria for cis-trans isomerism and explain clearly the structural feature of hypoglycin A that makes it unable to exhibit cis-trans isomerism. In this case, there is a C=C that has restricted rotation but the molecule does not exhibit cis-trans isomerism as one of the carbon atoms is attached to two hydrogen atoms. Be direct in your answer and avoid stating what is missing. For instance, explanation such as “there is no two different groups on each side of C=C” is WRONG and confusing. You should be considering the two groups attached to each C of C=C, i.e. each end of C=C and not each side. It is also incorrect to say “hypoglycin A does not show cis- trans isomerism as there is no restricted rotation of bonds” because one of the criteria for cis-trans isomerism is in fact the presence of such bonds which is present in hypoglycin A (there is restricted rotation about C=C). Some students also recognize that there is restricted rotation about the 3 -membered ring. However, one of the carbon atoms of this 3-membered ring has 2 H atoms attached to it and hence cis-trans isomerism is not possible. Do note that “the presence of chiral carbon” does not give rise to cis -trans isomerism. Instead, it may exhibit enantiomerism. HWA CHONG INSTITUTION 2021 C2 H2 CHEMISTRY PRELIM PAPER 2 SUGGESTED SOLUTIONS

(b) P [1] Q [1] R [1] When hypoglycin A is heated with CH3OH in the presence of conc. H2SO4 to form P, many students were able to recognize that esterification took place and used the carboxylic acid group in hypoglycin A to form an ester with CH3OH. The presence of conc. H2SO4 means the basic NH2 group in hypoglycin A will be protonated to form NH3+. When hypoglycin A was reacted with LiAlH4 in dry ether to form Q, reduction takes place. Only carboxylic acid in hypoglycin A is reduced to form a primary alcohol (CO2H o CH2OH). Alkene is not reduced by LiAlH4 in dry ether. When hypoglycin A was reacted with cold KMnO4 in NaOH(aq) to form R, this set of reagent and condition is for mild oxidation of alkene to form a diol. The alkaline medium will also cause the deprotonation of carboxylic acid to form carboxylate (CO2). (c) (i) HBr [1] By comparing the structures of the reactants and product in stage I, most students were able to see that the small molecule produced is HBr. (ii) addition [1] In stage II, the S bond of one of the C=C bonds was broken and two new V bonds were formed. This type of reaction is addition. (iii) hydrolysis [1] In stage III, there are two CO2C2H5 groups in the reactant. You were told that one of the CO2C2H5 groups was lost and asked to state the type of reaction for the other CO2C2H5 group. From the product, you should be able to see that the other CO2C2H5 group was converted to CO2H. Hence the ester is hydrolysed to form carboxylic acid and the type of reaction is hydrolysis. (iv) [1] Generally well done. You were told to draw the dot -and-cross diagram of molecular HCO2H. Do remember to put in the lone pairs on the oxygen atoms. (d) dilute NaOH, heat (with reflux)

2021 HCI C2 H2 Chemistry Prelim / Paper 2 and OR dilute H2SO4, heat (with reflux) and [1] for correct reagent and condition GRQRWDFFHSW³HQ]\PH´ [1] for each correct product (must be correctly protonated or deprotonated) Hypoglycin B contains an amide functional group that can be hydrolysed. The other nitrogen atom on the right is not directly attached to C=O. This nitrogen-containing group is a primary amine and it is next to a carboxylic acid. Hydrolysis of the amide should be conducted in an acidic or alkaline medium and with heating. In acidic medium: In alkaline medium:

(e) [1] You were told that the group is retained. Hence there should not be any substituent attached to C=C in MCPA, just like hypoglycin A and hypoglycin B. This group accounts for 4 carbon atoms. Since MCPA gives effervescence with NaHCO3, it contains CO2H functional group which accounts for the 2 oxygen atoms. There is one last carbon atom to be accounted for and there can only be one chiral centre in the molecule. Hence MCPA has a CH2CO2H group attached to as shown in the answer (the chiral centre is the carbon on the cyclopropane ring bonded to the CH2CO2H group.) 2 (a) More energy [1] is required to break covalent C−C bonds [0.5] in giant covalent [0.5] lattice of carbon than to break metallic bonding [0.5] between tin cations and sea of delocalised electrons in giant metallic [0.5] lattice of tin. It was indicated under Group 14 on page 52 of the Data Booklet that carbon is covalent whereas tin is metallic. As covalent compounds can be either giant covalent or simple covalent, candidates were expected to know that carbon has a giant covalent structure due to a relatively high melting point. The question did not specify carbon as graphite or diamond, candidates were expected to simply state the general structure and bonding of carbon: giant molecular and covalent C−C bonds. The question asked for “structure and bonding”, candidates should address the question. There should be two “structures” and two “bondings” for C and Sn, so in total four points should be mentioned, but some candidates missed out at least one out of the four. A number of candidates did not show understanding of the definition of covalent bonding, incorrectly stating that covalent bonding occurs between the carbon “molecules”. In fact, the covalent bonding occurs between the carbon atoms, which is the electrostatic forces of attraction between the positively charged nucleus of both the bonded atoms and their shared pair of electrons. A number of candidates did not show understanding of the definition of intermolecular forces of attraction, which is the forces of attraction between simple covalent molecules. Melting or boiling involves breaking covalent bonding in

2021 HCI C2 H2 Chemistry Prelim / Paper 2 substances with giant covalent structure. For simple covalent molecules, it involves overcoming intermolecular forces of attraction between the molecules. (b) (i) H2SO4 is a stronger Bronsted acid / proton donor [1] than HNO3, hence the protonation of HNO3 by H2SO4 eventually produces a strong electrophile NO2+. As part of the learning outcomes under Topic 12 Arenes of the H2 Chemistry syllabus, candidates are required to recognise that concentrated sulfuric acid acts as a Bronsted-Lowry acid catalyst in the nitration of arenes with