RI 18. Carbonyl Compounds Tutorial (Suggested Answers to Practice Questions)

-1- RAFFLES INSTITUTION YEAR 6 H2 CHEMISTRY 2026 Tutorial 18 Carbonyl Compounds Reaction Mindmaps ❑ Aldehydes

-2- ❑ Ketones

-3- Answers to self-check questions Q1 Q2(a) LiAlH4 OR H2(g), Ni catalyst Q2(b) Heat propan-1-ol with acidified K2Cr2O7 carefully with distillation of the product formed to collect mainly propanal as distillate. Q2(c) Observations: Orange ppt is formed. Answers to MCQ: 3) B 4) C 5) B 6) D 7) B

-4- Suggested Solutions to Practice Questions: 1 (a) (i) Mechanism: Nucleophilic addition KCN(aq) ⎯→ K+(aq) + CN−(aq) Step 1 Step 2 (ii) KCN provides the initial CN– ions for the nucleophilic attack on the carbonyl carbon. HCN is used in the second step as a (Bronsted) acid to protonate the anionic intermediate. Note: Do NOT merely say that KCN helps to increase the rate of reaction and HCN is the reagent, as the question is about the roles of KCN and HCN in the mechanism. (iii) The carbonyl carbon in CH 3CHO is bonded to one alkyl group while that in (CH 3)2CO is bonded to two alkyl groups. Hence, the carbonyl carbon in (CH3)2CO is less electron deficient (less +) due to the additional electron donating alkyl group . There is also greater steric hindrance about the carbonyl carbon in (CH 3)2CO which hinder the approach of the attacking nucleophile. Therefore, propanone , (CH3)2CO, reacts at a slower rate than ethanal, CH3CHO. (b) Hydrolysis reaction OH O OH (c) (i) Lactic acid has a chiral carbon and no plane of symmetry through the molecule. Lactic acid in milk rotates the plane of polarised light because only 1 enantiomer is present (i.e. all the molecules present are of the same chirality). (ii) In the slow step, the CN− nucleophile can attack the planar carbonyl carbon of ethanal from either side with equal probability, resulting in the formation of a racemic mixture of two enantiomers, thus optical activity cancels out. Checklist for nucleophilic addition mechanism:  Name of mechanism  δ+ on C, δ– on O in Step 1 and δ+ on H, δ– on C of HCN in Step 2  Lone pair of electrons on Nu  Curly arrows to show flow of electrons - Lone pair on Nu to δ+ C - C=O bond to δ– O - Lone pair on O− to H (of HCN) - H−C bond to C  Label slow/fast steps NOTE: HCN is a weak acid and hence a poor source of CN− ion. The CN− ion comes initially from the catalytic amount of KCN(aq) added. δ– δ+

-5- 2 (a) (b) Note that the second step cannot be carried out using H2 and Ni, as the alkene group will also be reduced. Alternatively, you may carry out reduction in the first step followed by elimination in the second step. (c) (d) Note: The reagents and conditions for step 1 and step 2 cannot be interchanged as shown below.

-6- 3 (a) Test Add I2(aq) and NaOH(aq) to each compound in a test-tube and heat in a hot water bath. Observation gives yellow ppt of CHI3 while gives no yellow ppt of CHI3 IMPORTANT NOTE: Use of hot acidified KMnO 4 is not accepted because both compounds will undergo side-chain oxidation. will give benzoic acid while will give 1,2-benzenedicarboxylic acid. (b) Test Add Tollens’ reagent to each compound in a test-tube and heat in a hot water bath. Observation gives a silver mirror. gives no silver mirror. OR Test Add K2Cr2O7 (aq), H2SO4 (aq), heat (DO NOT heat under reflux) Observation will turn orange K2Cr2O7 green while will not produce a colour change.

-7- 4 (a) Evidence/Information Deductions A has molecular formula C8H8O C to H ratio = 1:1  A likely contains a benzene ring A orange ppt Condensation (or addition-elimination) reaction A is an aldehyde or ketone. A Ag A (no reaction) A undergoes oxidation with Tollens’ but not with Fehling’s. A is an aromatic aldehyde. A Oxidation reaction. A has 2 substituents on the ring in the 1 and 4 position.  A is (b) Evidence/Information Deductions C orange ppt Condensation (or addition-elimination) reaction C is an aldehyde or ketone. B (C3H8O) C (C3H6O) Oxidation reaction. B is a secondary alcohol and C is a ketone. (since there is no increase in number of oxygen atoms)  B is , C is . Note: • If B was a primary alcohol, C would be a carboxylic acid with two oxygen atoms. • Also, C cannot be an aldehyde as any aldehyde group formed (from oxidation of a primary alcohol using KMnO4) would be further oxidised to a carboxylic acid. Tollens’ 2,4 DNPH acidified KMnO4 heat Fehling’s 2,4 DNPH acidified KMnO4 heat

-8- 5 (a) Evidence/Information Deductions A B (orange ppt) Condensation (or addition-elimination) reaction A is an aldehyde or ketone. A (C5H10O) C (C5H10O2) Compound A undergoes oxidation to give compound C. A is an aldehyde and C is a carboxylic acid. (Note: 1 oxygen atom in A to 2 oxygen atoms in C) A D Compound A undergoes reduction to give compound D, which is a primary alcohol. D E Elimination of water from the primary alcohol to form a terminal alkene, E E CH3CH(CH3)COOH Oxidation of the C=C in alkene E. Loss of one C atom as CO2. Since CO2 and 2-methylpropanoic acid are products of the oxidative cleavage of an alkene, E is H C H C H CH(CH3)2 Hence, alcohol D is OH (a primary alcohol) A is H O B is C is OH O acidified K2Cr2O7 heat LiAlH4 2,4 DNPH conc H3PO4 heat acidified KMnO4 heat

-9- 6 (a) Molecular mass of K is 86. 86 – 16 = 70 = 12n + 2n ⇒ n = 5 (b) Evidence/Information Deductions J K Since there is no change in number of O atom  J is a secondary alcohol and K a ketone. K yellow ppt K undergoes oxidation and contains CH3CO- group J 2 alkenes (no cis-trans) Elimination of H2O J is K is For explanation (not required in answer): From the first evidence, J could be CH3CH(OH)CH2CH2CH3 or The two alkenes from the elimination of J are: (no cis-trans isomerism in both compounds) [O] alkaline I2(aq) conc. H3PO4 catalyst heat

-10- 7 1,3-butadiene does not undergo nucleophilic addition reaction with HCN due to the absence of an electron deficient C (+ C) to attract the CN− nucleophile. (Note: Both C=C and C=O  bonds are electron rich. Hence it is insufficient to state that the C=C  electron cloud repel the nucleophile CN − as carbonyl with C=O group can undergo nucleophilic addition reaction). In 4-methyl-1-penten-3-one, the nucleophilic attack on the + carbonyl C is sterically hindered by the –CH(CH3)2 group. Also, the p orbitals of the sp2 C in C=C and C=O overlap to form a delocalised  electron cloud. Due to the highly electronegative O atom, the delocalised  electron cloud is pulled towards the O atom and the terminal alkene C becomes +. Hence, the nucleophile will attack the + terminal alkene C instead, resulting in the nucleophilic addition occurring at the alkene group to form C instead of at the carbonyl group to form B. Resonance structure and mechanism (not required in answer, only for further understanding)