NJC 2026 Organic Chemistry Revision Booklet (MCQ) Worked Solutions v2

Organic Chemistry Revision Booklet (MCQ) 2026 National Junior College 1 Worked Solutions for Organic Chemistry Revision (MCQ) Answer Key: 1 D 28 A 55 A 82 B 109 B 136 A 163 D 2 D 29 C 56 C 83 A 110 A 137 D 164 B 3 A 30 B 57 C 84 A 111 D 138 C 165 D 4 C 31 B 58 C 85 B 112 C 139 D 166 A 5 C 32 D 59 B 86 A 113 B 140 A 167 A 6 B 33 C 60 B 87 A 114 B 141 D 168 B 7 C 34 B 61 C 88 D 115 D 142 C 169 B 8 C 35 B 62 D 89 D 116 B 143 C 170 D 9 C 36 B 63 C 90 A 117 B 144 D 10 C 37 C 64 D 91 C 118 C 145 A 11 B 38 A 65 D 92 C 119 B 146 C 12 D 39 B 66 C 93 C 120 B 147 D 13 A 40 C 67 C 94 B 121 D 148 D 14 C 41 C 68 C 95 D 122 A 149 D 15 C 42 D 69 B 96 B 123 B 150 A 16 B 43 C 70 D 97 D 124 D 151 D 17 D 44 D 71 C 98 D 125 D 152 A 18 C 45 C 72 A 99 C 126 A 153 A 19 D 46 A 73 D 100 A 127 D 154 A 20 A 47 B 74 B 101 B 128 B 155 B 21 C 48 A 75 B 102 A 129 A 156 C 22 C 49 A 76 A 103 D 130 D 157 B 23 C 50 B 77 D 104 D 131 D 158 A 24 C 51 D 78 C 105 A 132 C 159 D 25 D 52 D 79 C 106 B 133 B 160 B 26 B 53 A 80 D 107 A 134 C 161 B 27 C 54 D 81 C 108 B 135 C 162 B

Organic Chemistry Revision Booklet (MCQ) 2026 National Junior College 2 General properties of organic compounds 1. D Structure Implication on Reactivity Benzene consists of delocalised π electrons above and below the plane. Susceptible to attack by electrophiles The ring of delocalised π electrons gives benzene additional aromatic stability. Reaction that destroys the aromatic ring requires an additional input of energy and is thus unfavourable. ● As such, even though benzene is highly unsaturated, it does not undergo addition reactions. ● Instead, benzene mostly undergoes substitution reactions. 2. D Option A: A catalyst is used to generate strong electrophile E+. (e.g. AlCl3 + Cl2 → AlCl4− + Cl+ conc HNO3 + conc H2SO4 → NO2+ + HSO4− + H2O) Option B: The carbon atom that forms a bond with the incoming E+ becomes sp3 hybridised. Option C: There is a loss of H hence the relative molecular mass of the organic product is 34.5 greater than that of benzene. Option D: During the rate -determining step (SLOW STEP), two of the six π electrons are used to form a dative bond with E +. This leaves four π electrons delocalized over the remaining five carbon atoms in the positively-charged arenium carbocation intermediate. 3. A Both butanal and pentane are simple covalent molecules. Both have instantaneous dipole - induced dipole interactions between molecules and Butanal, being polar, has additional permanent dipole- dipole interactions due to the net dipole moment in (C=O). More energy is thus required to overcome the stronger intermolecular forces in butanal resulting in higher boiling point. *Only intermolecular forces of attraction are broken during boiling of simple covalent molecules. Butanal is not able to form H -bonding with another butanal molecule due to lack of protonic H.

Organic Chemistry Revision Booklet (MCQ) 2026 National Junior College 3 4. C In crystalline form, amino acids exists as zwitterions which are held by strong ionic bonds. Large amount of energy is required to overcome the strong ionic bonds between the zwitterions, hence it has a higher melting point. 5. C To reduce pollution from motor cars, catalytic converters containing rhodium, platinum and or palladium are fixed onto the motor car exhaust pipes. These catalysts catalyse the redox reactions that convert the more harmful C, CO, NO, NO2 and unburnt hydrocarbons to less harmful or harmless compounds of CO2, H2O, N2 and O2. E.g. hydrocarbons + oxides of nitrogen → carbon dioxide + water + nitrogen 2CO + 2NO → 2CO2 + N2 (CO is oxidised to CO2 while NO is reduced to N2) 2NO2 → 2O2 + N2 (CO is oxidised to CO2 while NO is reduced to N2) *CO cannot react directly with CxHy in a redox reaction to produce CO2 6.

Organic Chemistry Revision Booklet (MCQ) 2026 National Junior College 4 7. 8. C Percentage of s character of hybridised orbitals are in order sp > sp2 > sp3. The higher the percentage of s character, the shorter the bond formed as the hybridised orbital will be nearer to nucleus and undergoes a more effective overlap. Option A has sp3-sp3 overlap. Option B and D has sp3-sp2. Option C has sp2-sp2 overlap. 9. C The 7 sp2 hybridised carbons are shown in red, each with 3  bonds and 1  bond.

Organic Chemistry Revision Booklet (MCQ) 2026 National Junior College 5 10. C As shown A, B and D are possible, but a molecule with an alkene and two alcohol groups would need 2 more H atoms, hence molecular formula C3H6O2 (one more degree of saturation) Combustion of organic molecules 11. B In alkane, y = 2x + 2 ∴ n(O2) = x + ( 2𝑥+2 4 ) = 1.5x + 0.5 So the graph should be a straight line with a positive gradient which is option B. Alternatively alkane = CH4 ; n(O2) = 2 C2H6; n(O2) = 3.5 C3H8; n(O2) = 5 C4H10; n(O2) = 6.5 As the number of carbon increases, the increase in n(O 2) is increasing at a constant rate. So the relationship should be a straight line with a positive gradient which is option B. 12. D CxHy O2 → CO2 H2O Reacting vol. / cm3 15 90 60 - Reacting molar ratio 1 6 4 - x =4 hydrocarbon contains 4 C atoms y = 8 hydrocarbon contains 8 H atoms Hence, hydrocarbon is C4H8.

Organic Chemistry Revision Booklet (MCQ) 2026 National Junior College 6 13. A Avogadro’s law: V ∝ n at constant p and T CH4 + 2O2 → CO2 + 2H2O CH3CH2CH3 + 5O2 → 3CO2 + 4H2O 𝑦 2 cm3 of CH4 will produce 𝑦 2 cm3 of CO2 𝑦 2 cm3 of CH3CH2CH3 will produce 3𝑦 2 cm3 of CO2 because 1 mole of propane give 3 moles of CO2. Total v(CO2) = 𝑦 2 + 3𝑦 2 = 2y cm3 14. C Amount of hydrocarbon = 100/ 24000 = 0.00417 mol Amount of water produced = 0.225/ 18.0 = 0.0125mol Amount of CO2 produced = 200/ 24000 = 0.00833mol CxHy : CO2 = 1 : 2 Hence x = 2 CxHy : H2O = 1 : 3 Hence y = 6 15. C 1st reduction in the total volume of gaseous products due to cooling  hot H2O(g) has condensed to form H2O(l) at r.t.p.  thus, vol of H2O(g) = 90 – 50 = 40 cm3 2nd reduction in the total volume of gaseous products due to reaction with KOH(aq) CO2(g) is an acidic gas and reacts with KOH(aq) via acid-base reaction  vol of CO2(g) = 40 cm3 Thus, CO2(g) : H2O(g) Thus, C : H = 40 : 40 = 1 : 2 (Every 1 mole of H2O contains 2H) = 1 : 1 Thus the organic compound must have a ratio C : H = 1: 2. 1 CH2CH2 C:H = 1:2 correct 2 CH3CO2H C:H = 1:2 correct 3 CH3CH2CH3 C:H = 3:8 incorrect 4 CH2CHCH2OH C:H = 1:2 correct

Organic Chemistry Revision Booklet (MCQ) 2026 National Junior College 7 Calculations involving Organic Molecules 16. B 0.01 mol of Compound X requires 0.03 mol of NaOH for hydrolysis. Hence 0.02 mol of NaOH is left for neutralisation with HCl. 0.01 mol of Compound Y requires 0.01 mol of NaOH for hydrolysis of the CN group. Hence 0.04 mole of NaOH is left for neutralisation with HCl. 0.01 mol of Compound Z requires 0.02 mol of NaOH for nucleophilic substitution reaction. Hence 0.03 mol of NaOH is left for neutralisation with HCl. 17. D Option A and D will not be further oxidised hence solution will remain orange in colour. CH3CH2CH2CH3 + 6.5O2 → 4CO2 + 5 H2O 0.05mol 0.325mol 0.20mol 0.25mol CH3CH2CH2COOH + 5O2 → 4CO2 + 4H2O 0.05mol 0.250mol 0.20mol 0.20mol 18. C Test Deductions Reacts with hot acidified KMnO4 Oxidative cleavage of C=C Product reacts with 2,4-DNPH Presence of carbonyl compound Product does not give white ppt with limewater No