NJC Organic Chem Structured revision set Answer

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 1 Worked Solutions to Organic Chemistry Revision Booklet (Structured Questions) 1) Organic Synthesis 1. (a) Cl C O H D Cl C O OH E (b) Step 2: NaOH(aq); heat Step 3: KMnO4,H2SO4 (aq), heat or K2Cr2O7, H2SO4 (aq), heat (c)(i) (ii) NH2 OH CH3 CH3 and HO C O O− K + *remember to check for acid-base reaction after hydrolysis 2. (a) *Hydrolyse the amide, S, to deduce the structures of R and Q. *Displayed formula shows all the bonds in the correct shape (i.e. trigonal planar, bent)

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 2 (b) Step 1: Oxidation (1o alcohol to aldehyde) [controlled oxidation] Step 2: Nucleophilic addition (of HCN) Step 3: Reduction (of nitrile) (c) C O O C H CH2NH2 C H H Cl Cl Acyl chloride, R, can undergo condensation reaction with alcohol in Q to form an ester. 3. (a) cis-trans isomerism OH H H H H OH cis, cis- H H H H OH OH cis, trans- ,H2SO4(aq), cold cold

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 3 OHH H H H OH trans,cis- HH H H OH OH trans,trans- (b) Q (C11H12O2) Ester – undergoes acidic hydrolysis with aq HCl CH3CH2CO2H R (C3H6O2) S (undergoes reduction of alkene to give T) T (Phenol undergoes electrophilic substitution with Br2(aq) – multiple substitutions) U (C8H7OBr3) 4 (i) 1. Primary amine 2. Tertiary alcohol 3. amide 4. ester

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 4 (ii) cold HCl(aq) only acid-base reaction takes place (ester & amide hydrolysisrequire heat) C CH2NH3 + CH2 OH N C CH2CH3 H O OC O CH3 Hot NaOH(aq) alkaline hydrolysis followed by acid-base reaction of the products. CH3CO2Na CH3CH2CO2Na OH CH2NH2 CH2NH2 +Na-O 5 (i) AlBr3 or FeBr3 or AlCl3 or FeCl3 , anhydrous (ii) Benzene, although electron rich, is stabilised by resonance. Hence, benzene requires a strong electrophile for the electrophilic attack to take place. The presence of a lewis acid catalyst (eg. AlBr3) is necessary for the formation of the strong electrophile, (CH3)2CHCH2+. (iii) CH3 C CH3H C HH S O CCl T O O C U (iv) V is an aliphatic aldehyde as it gives a reddish brown ppt with Fehling’s solution. CH3 C CH3 H CH2Br CH3 C CH3 H C O H NaOH(aq) heat CH3 C CH3 H CH2OH K2Cr2O7, dil H2SO4 heat with immediate distillation V Cl –

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 5 6 (a) Condensation (Nucleophilic acyl substitution) (b) CH3CO2CH3 + CH3CH2OH → CH3CO2CH2CH3 + CH3OH (c) Methyl ethanoate should be the limiting reagent. Ethanol should be in excess so that all the methyl ethanoate can be converted. (d) More energy is required to overcome the stronger h ydrogen bond s between methanol molecules during boiling than the permanent dipole -dipole interactions between methyl ethanoate molecules, leading to a higher boiling point. The instantaneous dipole-induced dipole interactions between ethyl ethanoate molecules are stronger than the hydrogen bonds between methanol molecules due to greater ease of distortion of larger electron cloud size of ethyl ethanoate molecules, requiring more energy to overcome during boiling, leading to a higher boiling point. (e) Fractional distillation (particularly used when the boiling points of the liquids are quite similar) (f) CH3(CH2)5CH=CHCH2CH2CO2CH3 CH2(OH)CH(OH)CH2OH (g) Amt of HCl reacted = 1.00 × 15.20 × 10–3 = 0.0152 mol Total amt of NH3 = 1.00 × 25.0 × 10–3 = 0.0250 mol Amt of NH3 that reacted with ester = 0.0250 – 0.0152 = 0.0098 mol = amt of ester Mr of the ester = 1.00 / 0.0098 = 102.0 Thus, 15 + 12 + 32 + 14x + 15 = 102.0 x = 2 7 (i) CH3CH2OH + 3O2 → 2CO2 + 3H2O (ii) CH3CH2OH + 4I2 + 6NaOH → HCO2Na + CHI3 + 5NaI + 5H2O **No of C atoms decrease by 1. (iii) (iv) C O Cl OCH2CH3 OR C O CH3CH2O OCH2CH3

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 6 8 compound reagent(s) organic product O N CH3 CH3 OH O Y KMnO4, dilute H2SO4 [O] of 2º alcohol, acid-base reaction of 3 º amine, oxidative cleavage of alkene, acidic hydrolysis of ester followed by [O] of its 1 º alcohol product) O C N+ CH3 CH3 O H HO O O N CH3 CH3 OH O Y HBr(g), heat (N.S. of 2º alcohol, E.A. of alkene, acid-base reaction of 3º amine) HO OH NH2 Z propanoyl bromide (condensation of phenol, alcohol & amine with propanoyl bromide; yield of phenyl ester is enhanced if Na is used to convert phenol to phenoxide first) CH3CH2CO2 OCOCH2CH3 NHCOCH2CH3

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 7 9 (i) Step I: 1. cold concentrated H2SO4 2. warm with H2O Step III: HCN with trace amount of NaCN, cold (ii) CH3COCH2CH3 B CH3CCH2CH3 CH2NH2 OH D O N CH2CH3H3C O O H E (iii) The electron deficient C centre in the carbonyl group is sp2 hybridised / the geometry about the carbonyl C is trigonal planar. Hence, there is equal probability for the −CN nucleophile to attack from either side of the plane containing the C=O bond, leading to the formation of a pair of enantiomers in a 1:1 ratio. 10(i) step 1: acid-metal reaction / redox reaction step 3: alkaline hydrolysis (ii) CO2H CO2H Since 1 mol of X reacts with exactly 1 mol of sodium carbonate => 1 mol of X contained 2 -COOH groups. 11(i) HO CN A +Na -O OH B +Na -O CO2- Na+ C HO O D (ii) Step I: NaOH(aq)/KOH(aq); heat Step II: H2SO4(aq), K2Cr2O7(aq); heat with immediate distillation

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 8 Step III: NaOH/KOH in ethanol; heat Step IV: I2, NaOH(aq), warm 12(a) (-NO2 must be substituted at the 4-position relative to phenol in order form the following products in the given reaction scheme) (b) Stage 1: Sn, conc. HCl; heat Stage 2: NaOH(aq) (to liberate phenylamine) (c) (i) CH3COCl (ii) OR (consider condensation between phenol and acyl chloride) (d) (i) (acid-base reaction of phenol by aq NaOH) OH NO2 OCOCH3 NH2 OCOCH3 NHCOCH3 O-Na+ NHCOCH3

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 9 (ii) and CH3COO–Na+ (acid-base reaction of phenol by aq NaOH, alkaline hydroysis of amide) 13 X CH3CH2CH2OH Y CH3CH2CO2H Z CH3CHClCO2H Reagents and conditions for conversion of Y to Z: limited Cl2(g), uv light 14 CH2CH2CH2OHBrCH2CH2OH ethanolic KCN heat under reflux NCCH2CH2OH LiAlH4 in dry ether H2N 15(a)(i) CO + HCl + FeCl3 → +CHO + FeCl4– (ii) FeCl3 hydrolyses in water to yield [Fe(H2O)5(OH)]2+ and Cl− ions / dissolve in water to form [Fe(H2O)6]3+. Fe in [Fe(H 2O)5(OH)]2+ or Fe in [Fe(H 2O)6]3+ has no more vacant orbitals in the valence shell to accept the lone electron pair from Cl– to generate the strong electrophile. O-Na+ NH2

Organic Chemistry Revision Booklet (Structured Qn) 2024 National Junior College 10 (b) O CO/ HCl ,FeCl3; heat under reflux CHO LiAlH4 in dry ether CH2OH CH2Cl PCl5(s) / SOCl2(l) Heat under reflux O- (step 3 requires anhydrous condition; step 4 is N.S where the phenoxide is acting as nucleophile) 16(a)(i) HCl (g) + BaSO4 (ii) (E.A. of alkene with HCl)