RI Carbonyl Compounds Tutorial - Answers to Discussion Questions

-1- RAFFLES INSTITUTION YEAR 6 H2 CHEMISTRY 2025 Tutorial 18 Carbonyl Compounds Reaction Mindmaps  Aldehydes R C H O aldehyde RC OH H H R C OH O R C O O + Ag R C O O + Cu 2O RC OH H CN H C O O + CH I3 Na N O2N NO2N H C R H K2Cr2O7(aq) / H2SO4(aq), heat or KMnO4(aq) / H2SO4(aq), heat oxidation silver mirror Tollens' reagent, heat oxidation reddish- brown ppt Fehling's solution, heat oxidation yellow ppt I2(aq), NaOH(aq), heat oxidation orange ppt 2,4-DNPH condensation LiAlH4 in dry ether or NaBH4 in methanol or H2/Ni, heat K2Cr2O7(aq), H2SO4(aq) heat with immediate distillation oxidation reduction *Ethanal is the only aldehyde that has positive test (R=CH 3) HCN,trace amt of KOH(aq) or HCN,trace amt of KCN(aq) or KCN(aq), H2SO4(aq) RC OH H CH2NH2 nucleophilic addition LiAlH4 in dry ether or H2/Ni, heat reduction RC OH H COOH H2SO4(aq), heat acidic hydrolysis *Positive test for ALL aldehydes *Positive test only for aliphatic aldehydes 1o alcohol 2,4-dinitrophenylhydrazone tri-iodomethane cyanohydrin

-2-  Ketones

-3- Answers to self-check questions Q1 Q2(a) LiAlH4 in dry ether, room temperature OR H2(g), Ni(s), heat OR NaBH4, room temperature Q2(b) Heat propan-1-ol with acidified K 2Cr2O7 carefully with immediate 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 NaCN(aq)  Na+(aq) + CN(aq) Step 1 Step 2 (ii) NaCN 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 NaCN helps to increase the rate of reaction and HCN is the reagent, as the question is about the roles of NaCN and HCN in the mechanism. (iii) The carbonyl carbon in CH3CHO is bonded to one alkyl group while that in (CH 3)2CO is bonded to two alkyl groups. Hence, the carbonyl carbon in (CH 3)2CO is less electron deficient (less +) due to the additional electron d onating 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, (CH 3)2CO, reacts at a slower rate than ethanal, CH3CHO. (b) Acidic hydrolysis reaction OH O OH (c) (i) Lactic acid has a chiral carbon and no pl ane 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, result ing 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 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 NaCN(aq) added. If trace amt of NaOH is used as catalyst: HCN + OH → CN + H2O δ–δ+

-5- 2 (a) CH3 C CH2CH3 O CH3 C CH2CH3 OH CN CH3 C CH2CH3 OH CH2NH2 HCN, trace amt of KCN LiAlH4 in dry ether (b) CH2 CH2 CHO Br H2CC H CHO H2CC H CH2OHethanolic KOH heat NaBH4 Note that the second step cannot be carried out using H2 and Ni with heat, 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) O Br benzene FeBr3 O I2(aq), NaOH(aq) heat O Na+ -O Note: The reagents and conditions for step 1 and step 2 cannot be interchanged as shown below. I2(aq), NaOH(aq) O Br O Na+ -O OH heat benzene FeBr3 electrophilic substitution does not occurX 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. O gives no silver mirror. O

-6- OR Test Add K2Cr2O7 (aq), H2SO4 (aq), heat (DO NOT heat under reflux) Observation will turn orange K2Cr2O7 green while O will not produce a colour change. 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 KMnO 4) would be further oxidised to a carboxylic acid. Tollens’ 2,4 DNPH acidified KMnO4 heat Fehling’s 2,4 DNPH acidified KMnO4 heat

-7- 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 Strong oxidation of the C=C in alkene E. Loss of one C atom as CO2. Since CO 2 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 (b) H is more electronegative than A l and B. As the electronegativity difference between A l and H is greater than that between B and H , the A l-H bond is more polar than B-H bond. This makes the H in Al–H more electron rich (more –) and more nucleophilic than the H in B–H. This allows LiAlH4 to be a stronger reducing agent. OR Since Al is larger than B, the valence orbital used in bonding is more diffuse for Al than B. The orbital overlap is less effective for the Al-H bond than the B-H bond, hence the A l-H bond is weaker, losing the H– nucleophile more readily. Therefore, LiAlH4 is the stronger reducing agent. acidified K2Cr2O7 heat NaBH4 2,