RI 2025 Hydroxy Cpds Tutorial - Anwers to Discussion Questions

-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.  (CH 3)3CO has more electron-donating alkyl groups than CH 3CH2O that further intensifies the negative charge on the oxygen atom.  Thus (CH 3)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  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 is less stable than . 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 is more stable than . 2 Reagents Condition Structure T ype of reaction (a) sodium room temp Redox (b) sodium hydroxide room temp Acid-Base (c) sodium carbonate NA NO REACTION NA

-2- (d) phosphorus (V) chloride room temp, (anhydrous) Nucleophilic Substitution (e) hydrogen bromide dry HBr, heat Nucleophilic substitution and Electrophilic Addition Note:  Benzylic carbocation intermediate is more stable.  Without heating, only electrophilic addition of C=C occurs. (f) ethanoic acid conc. H2SO4, heat Condensation (esterification) (g) ethanoyl chloride room temp. Condensation (acylation) (h) aqueous bromine room temp. Electrophilic Substitution and Electrophilic Addition Note:  Benzylic carbocation intermediate is more stable. (i) bromine in CCl4 room temp. (or 2- or 4-substituted product) Electrophilic Substitution and Electrophilic Addition (j) potassium manganate (VII) KMnO 4(aq), H2SO4(aq), heat 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 C 6H5CH=CH2, there is (rapid) decolourisation of orange-red Br2. C6H5CH=CH2 + Br2  C6H5CHBrCH2Br  For C 6H5CH2CH2OH, there is no decolourisation of orange-red Br2. OR Test Add two drops of K 2Cr2O7(aq) acidified with dilute 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) Observations  For C 6H5CH=CH2, the solution remains orange.  For C 6H5CH2CH2OH, 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 heat 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 CH I3 is formed. CH3CH2CH(OH)CH3 + 4I2 + 6NaOH  CHI3 + CH3CH2COO–Na+ + 5NaI + 5H2O (c) Test Add neutral FeCl3(aq) to each sample in a test-tube at room temperature. Observation  For cyclohexanol, no violet colouration is observed.  For phenol, violet colouration is observed. (structure of violet complex is not in syllabus) OR 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 Br 2 and no white ppt is formed.  For phenol, there is decolourisation of orange Br 2 and a white ppt is formed. OR Test Add PCl5(s) to each sample in a test-tube at room temperature. Observation OOH6Fe3+(aq) + 6 Fe 3 + 6 H+

-4-  For cyclohexanol, steamy white fumes of HC l are evolved.  For phenol, no steam y white fumes are evolved. OR Test Add two drops of K 2Cr2O7(aq) acidified with dilute 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) Observation  For cyclohexanol, orange acidified K 2Cr2O7 turns green.  For phenol, solution remains oran ge. 4 (a) OH propan-1-ol O O propyl propanoate heat / heat under reflux O OH O Cl r.t OH conc. H2SO4 heat / heat under reflux OH KMnO4(aq), H2SO4(aq) SOCl2, PCl5 or PCl3 step 1 step 2 step 3 room temperature (b) OH dilute HNO3 Br2 in CCl4 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

-5- 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)   22 2 22 2 Change in volume Final volume of gases Initial volume of gases 54.4 Vol. of O left Vol. of CO produced Initial 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   (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- 22 43 3 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 (b) Evidence / Information Deduction and explanation J reacts with acidified K2Cr2O7. Oxidation of alcohol occurred. J is either a primary or secondary alcohol. Molecular formula of K is C5H10. Vigorous oxidation (oxidative cleavage of C=C) occurred. K is and J is (c) excess concentrated H2SO4, 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)CH 3 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 Br2 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) Examiner Comments  Many students did not realise that C and subsequently B can be obtained independently without having to obtain the structures of A a