RI 2026 Hydroxy Cpds Tutorial answers

-1- Tutorial 18 – Hydroxy Compounds (Suggested Solutions) Practice Questions 1 Relative acidity: methylpropan-2-ol < ethanol < water < 2-methylphenol < phenol < 2-nitrophenol Methylpropan-2-ol and ethanol are alcohols while 2- methylphenol and 2- nitrophenol are substituted phenols. The more stable the conjugate base, the more acidic the compound. Alcohols are less acidic than water. • Alkoxide ion, RO−, is less stable than hydroxide ion, OH–, due to the electron-donating alkyl group which intensifies the negative charge on the oxygen atom. Methylpropan-2-ol is less acidic than ethanol. • (CH3)3CO− has more electron- donating alkyl groups than CH3CH2O− that further intensifies the negative charge on the oxygen atom. • Thus (CH3)3CO− is less stable than CH3CH2O− Phenols are more acidic than water. • The p-orbital containing the lone pair of electrons on the oxygen atom of the phenoxide ions overlaps with the π electron cloud of the benzene ring and the lone pair of electrons is delocalised into the ring. • This results in the delocalisation of negative charge on oxygen into the ring, i.e. dispersal of the negative charge over the ring. • Thus, the phenoxide ion is resonance–stabilised and more stable than hydroxide ion. 2-methylphenol is less acidic than phenol. • Presence of electron-donating group, −CH3, on the benzene ring intensifies the negative charge on the oxygen atom. • Thus O CH3 is less stable than O . 2-nitrophenol is more acidic than phenol. • Presence of electron-withdrawing group, −NO2, on the benzene ring disperses the negative charge on the oxygen atom. • Thus O NO2 is more stable than O . 2 Reagents Condition Structure Type of reaction (a) sodium room temp O Na+ CH=CHCH2O − Na+ Redox (b) sodium hydroxide room temp O Na+ CH=CHCH2OH Acid-Base (c) sodium carbonate NA NO REACTION NA

-2- (d) phosphorus (V) chloride room temp, (anhydrous) OH CH=CHCH2Cl Nucleophilic Substitution (e) hydrogen bromide dry HBr OH C C H H CH2Br Br H Nucleophilic substitution and Electrophilic Addition Note: • Benzylic carbocation intermediate is more stable. (f) ethanoic acid conc. H 2SO4, heat OH C C H H CH2O C O CH3 Condensation (esterification) (g) ethanoyl chloride room temp. O C C H H CH2O C O CH3 C CH3 O Condensation (acylation) (h) aqueous bromine room temp. OH C C H H CH2OH OH Br BrBr Br Electrophilic Substitution and Electrophilic Addition Note: • Benzylic carbocation intermediate is more stable. (i) potassium manganate (VII) KMnO4(aq), H2SO4(aq), heat OH CO2H Vigorous oxidation (Oxidative cleavage of the C=C)

-3- 3 (a) Test Add Br2 (in CCl4) dropwise to each sample in a test-tube at room temperature, in the absence of uv light. Observations • For C6H5CH=CH2, there is (rapid) decolourisation of orange-red Br2. C6H5CH=CH2 + Br2 → C6H5CHBrCH2Br • For C6H5CH2CH2OH, there is no decolourisation of orange-red Br2. OR Test Add two drops of acidified K2Cr2O7(aq) to each sample in a test-tube, and heat the mixture in a hot water bath. (Note: DO NOT heat under reflux) Observations • For C6H5CH=CH2, the solution remains orange. • For C6H5CH2CH2OH, orange acidified K2Cr2O7(aq) turns green. C6H5CH2CH2OH + 2[O] → C6H5CH2COOH + H2O (b) Test Add I2(aq), followed by NaOH(aq) to each sample in a test -tube and warm each mixture in a hot water bath. (Note: DO NOT heat under reflux) Observation • For butan-1-ol, no yellow ppt is formed. • For butan-2-ol, yellow ppt of CHI3 is formed. CH3CH2CH(OH)CH3 + 4I2 + 6NaOH → CHI3 + CH3CH2COO–Na+ + 5NaI + 5H2O (c) Test Add Br2(aq) dropwise to each sample in a test-tube at room temperature, in the absence of uv light. Observation • For cyclohexanol, there is no decolourisation of orange Br2 and no white ppt is formed. • For phenol, there is decolourisation of orange Br2 and a white ppt is formed. OH + 3Br2 OH + 3HBr Br BrBr OR Test Add PCl5(s) to each sample in a test-tube at room temperature. Observation • For cyclohexanol, steamy white fumes of HCl are evolved. OH + PCl5 Cl + POCl3 + HCl • For phenol, no steamy white fumes are evolved. OR Test Add two drops of K2Cr2O7(aq), acidified with H 2SO4(aq), to each sample in a test -tube, and heat the mixture in a hot water bath. (Note: DO NOT heat under reflux)

-4- Observation • For cyclohexanol, orange acidified K2Cr2O7 turns green. OH + [O] O + H2O • For phenol, solution remains orange. 4 (a) OH propan-1-ol O O propyl propanoate heat O OH O Cl r.t OH conc. H2SO4 heat OH KMnO4(aq), H2SO4(aq) PCl5 step 1 step 2 step 3 room temperature (b) OH dilute HNO3 Br2 (aq) OH NO2 OH NO2 Br NaOH(aq) room temp. O − Na+ NO2 Br room temperature CH3COClO O2N Br C CH3 O step 1 step 2 step 3 step 4 room temperature room temperature Br Br Br 5 (a) Method 1 (using volume of gases only and Avogadro’s law) CxHyOH(l) + Na(s) → ½H2(g) + CxHyO−Na+(l) Vol of H2 = 10.9 cm3 CxHyOH(l) + 41 4 +−xy O2(g) → xCO2(g) + 1 2 +y H2O(l)

-5- ( ) ( ) 22 2 22 2 Change in volume Final volume of gases In itial volume of gases 54.4 Vol. of O left Vol. of CO produced Ini tial vol. of O 54.4 Initial vol. of O Vol. of O left Vol. of CO produced 54.4 Initial vol = − −= + − = −+ =( )22 2 22 22 3 . of O Vol. of O left Vol. of CO produced 54.4 Vol. of O reacted Vol. of CO produced Vol. of O reacted 54.4 Vol. of CO produced 54.4 109 163.4 cm −− = − = + = + = Since the same amount of C xHyOH(l) is used in both experiments, ½H2(g) ≡ 41 4 +−xy O2(g) ≡ xCO2(g) (to find x) = = 2 2 Vol of CO Vol of H 0.5 109 10.9 0.5 x x x = 5 ∴J is C 5H11OH Method 2 (using amt of gases formed) CxHyOH(l) + Na(s) → ½H2(g) + CxHyO−Na+(l) 4 2 10.9Amount of H (g) 4.54 10 mol24000 −= = × 44Amount of C H OH 2 4.54 10 9.08 10 molxy −−= ××=× CxHyOH(l) + 41 4 +−xy O2(g) → xCO2(g) + 1 2 +y H2O(l) 3 2 163.4Amount of O (g) 6.81 10 mol24000 −= = × 3 2 109Amount of CO (g) 4.54 10 mol24000 −= = × 22 4 33 Amount of J:Amount of CO (g):Amount of O ( g) 9.08 10 : 4.54 10 : 6.81 10 1 : 5.00 : 7.50 411 : : 4 − −−××× +−xyx ∴x = 5 and 41 7.54 11 +− = = xy y ∴J is C5H11OH (to find y) xy xy +− = +− = 2 2 41 Vol of O 4 Vol of H 0.5 41 163.4 4 10.9 0.5 where x = 5 y = 11

-6- (b) Evidence / Information Deduction and explanation J reacts with acidified K2Cr2O7. Oxidation of alcohol occurred. J is either a primary or secondary alcohol. J H 2O K Molecular formula of K is C5H10. K, C 5H10 + CH3 C OH O C CH3 CH3O excess of hot acidified conc KMnO4 Vigorous oxidation (oxidative cleavage of C=C) occurred. K is CH3 C H C CH3 CH3 and J is CH3 C H C CH3 CH3 HOH (c) concentrated H3PO4, heat (d) K cannot exhibit cis-trans isomerism as there are two identical –CH3 groups bonded to one of the carbon atoms of the C=C bond. 6 (a) • The –CH(OH)CH3 group is absent in A. • A contains one C=C bond. Examiner Comments • Students should not describe the –CH(OH)CH3 group as “methyl alcohol” as it is ambiguous. No credit will be given for ambiguous descriptions. • It is important to state that there is ONE C=C bond given that ONE mole of Br 2 reacts with ONE mole of A. • Answers that just stated A has a double bond or a doubly bonded carbon was not given credit as this could also imply a C=O group. • 1 degree of unsaturation was also not accepted as C=O is also unsaturated. (b) OH A B O HO OH O C O D Examiner Comments • Many students did not realise that C and subsequently B can be obtained independently without having to obtain the structures of A and B. • The thought process for deducing the structures is as follows: Molecular formula (C8H16O) and 1 mol of A reacts with 1 mol of Br2 in the dark. ⇒ A undergoe