Name: CG: Class: SERANGOON JUNIOR COLLEGE H2 9647 CHEMISTRY 2015 JC2 Hydroxyl CompoundsTutorial 6Compulsory Questions [To be discussed during tutorial. Do all questions on foolscap paper. Show all relevant workings.] 1.Using the table below to compare and contrast the reactions of ethanol and phenol witheachofthefollowingreagents.Includeinyouranswer,thetypeofreaction, balanced equations and all products formed. Reagents CH3CH2OHOH Comments (Any differences in reactions? Reasons for reaction /no reaction) a. SodiumRedox CH3CH2O-Na+ + H2 Redox O-Na++ 1/2H2 1 mol of ethanol or phenol produced only mol of H2 b. Aqueous potassium hydroxideNo reactionNeutralisation O-K+ + H2O Alcohols are weaker acids than water considered to be neutral. c.Phosphorus pentachloride Nucleophilic substitution CH3CH2Cl + POCl3 + HCl No reactionStrong C-O bond in phenol, no nucleophilic sub. d. Aqueous bromine No reactionElectrophilic substitution OHBrBrBr + HBr -OH group is a strong activating group (2,4,6-) Question:Whatis thedifferencein productifliquid bromine is used? e. Conc H2SO4 with ethanoic acid, reflux esterification/ nucleophilic substitution of CH3COOH CH3COOCH2CH3 + H2O No reactionPhenolrequires acylchlorideto form ester. f.Acidified hot Potassium manganate(VII) / potassium dichromate(VI) Oxidation CH3COOH + H2O noreaction as oxidation willresultindestruction ofthearomatic character 2.Identifyallthefivefunctionalgroupspresentinthecompoundbelow.Statethetypeof reaction and draw the organic product when the structure below reacts with: OHCH CHCH2OH C ClCH2OHH 2(a) sodium, rtp; ReagentType of reaction(s)(Circlethefunctionalgroup(s) involved in the reactions) Product (a) Sodium, rtpOHCH CHCH2OH C ClCH2OHH 1.Redox/acidmetal displacement O-Na+CH CHCH2O-Na+C ClCH2O-Na+H (b) aqueous sodium hydroxide, heat; OHCH CHCH2OH C ClCH2OHH 1.Nucleophilic substitution 2.NeutralisationCH=CHCH2OH C HOCH2OHHO-Na+ (c) thionyl chloride, SOCl2 OHCH CHCH2OH C ClCH2OHH 1.Nucleophilic substitution OHCH CHCH2Cl C ClCH2HCl Functional Groups: 1.1o alcohol 2.2o alcohol 3.Phenol 4.Alkene 5.Halogenoalkane (d) ethanoic acid, conc. H2SO4, reflux OHCH CHCH2OH C ClCH2OHH 1.Nucleophilicsubstitution/ esterification CH CHCH2O C ClCH2HOHO CCH3OCCH3O Note:Phenoldoesntreactwith carboxylic acids to form esters. (e) ethanoyl chloride, rtp OHCH CHCH2OH C ClCH2OHH 1.Nucleophilicsubstitution/ acylation CH CHCH2O C ClCH2HOO CCH3OCCH3OCCH3O (f) aqueous iodine, rtp OHCH CHCH2OH C ClCH2OHH 1.Electrophilic addition 2.Easy electrophilic substitution OHCH CHCH2OH C ClCH2HIOH III OHCan have alternative product (EA) 3. Outline, with equations, reagents and essential conditions, how the following conversions could be efficiently carried out: (a)Butan-1-ol to butan-2-ol I : + H2O Reagents & Conditions: excess concentrated H2SO4, heat, 170C Type of reaction: Elimination I II 1 alcohol2 alcohol C C HHHCHHCHHHHOHC C HHHCHHCHOHHHHC C HHHCHHCHHHC C HHHCHHCHHHHOHC C HHHCHHCHHH II : +H2O Reagents & Conditions: steam, H3PO4, heat at 300C, 70 atm Type of reaction: Electrophilic addition (b)cyclopentanol to cyclopentylmethanol (hint: 4 steps) I :+ PCl5+ POCl3+ HCl Reagents & Conditions: PCl5 , room temperature Type of reaction: Nucleophilic substitution II :+ KCN+ KCl Reagents & Conditions: alcoholic KCN, heat Type of reaction: Nucleophilic substitution III :+ HCl+ 2 H2O+NH4Cl Reagents & Conditions: dilute HCl, heat Type of reaction: Acidic hydrolysis C C HHHCHHCHHHC C HHHCHHCHOHHHHI IV 5 carbon alcohol 6 carbon1 alcohol OHCl COOH CNII III OH ClClCNCNCOOHCH2OH IV :+ 4[H]+H2O Reagents & Conditions: LiAlH4 in dry ether, rtpType of reaction: Reduction (c) OHCH2CH2OHOCCH2Oto (hint: 3 steps) OHCH2CH2OHOCCH2OOHCH2COOHOHCH2COCl I : OHCH2CH2OHOHCH2COOH+ 2[O] + H2O Reagents & Conditions: K2Cr2O7 in H2SO4 (aq), heatType of reaction: Oxidation III II I *** cannot use KMnO4, otherwise side chain oxidation will result in benzoic acid formed COOHCH2OHAlc. KCN, heat LiAlH4 in dry ether, rtp II : OHCH2COOHOHCH2COCl + + PCl5POCl3HCl+ Reagents & Conditions: PCl5 (s), rtpType of reaction: Nucleophilic substitution III :Reagents & Conditions: rtp Type of reaction: Nucleophilic substitution 4. Cyclohexanol can be synthesised from cyclohexene as follows: OH BrH

CH2NH2 (a)State the type of reaction and essential reagents and conditions for step I and II. I:Reagent/Conditions:HBr (g), r.t.pType of rxn: Electrophilic AdditionII:Reagent/conditions: Aqueous KOH, reflux Type of rxn: Nucleophilic Substitution (b) Suggest a suitable synthesis pathway for step III using the minimum number of steps. OH ClCN CH2NH2 (c)Explain why phenol could not be synthesised from bromobenzene. Br OH The porbital of the bromine atom can overlap with the -electroncloud of the benzenering.Asaresult,thelonepairofelectronsontheBratomcan delocalise into the benzene ring. Concepts: 1. cyclic ester2. Acyl chloride +phenol ester Step IStep II Electrophilic substitution Step III PCl5, rtp ThisresultsinpartialdoublebondcharacterinC-Brbondandthe strengthening of the C-Br bond. does not undergo nucleophilic substitution/ substitution by a nucleophile / OH- is difficult (d)DrawthestructuralformulaofalltheorganicproductsformedwhencompoundZ reacts with the following reagents.CH3CH CH2OH ClOH (d)(i) Hot, aqueous sodium hydroxide O-Na+CH3CH CH2OHOH (d)(ii) Br2 in CCl4, absence of ultraviolet light (d)(iii) Hot acidified KMnO4 Br5.Describe simple chemical tests to distinguish butanol, butan-2-ol and 2-methylbutan-2-ol.Youmustincludechemicalreagentsandconditionsandwritedownappropriate balanced chemical equations in your answer. Observations butanolbutan-2-ol2-methylbutan-2-ol Step 1Add potassium dichromate(VI), H2SO4(aq)to each of the compound separately and heat. Orange turns greenOrange turns green Remains orange Step 2Add 2,4-DNPH into the resulting mixtures from step 1 at rtp. No orange pptOrange ppt formed No orange ppt Equations:CH3CH2CH2CH2OH + 2[O] CH3CH2CH2COOH + H2O CH3CH2CH(OH)CH3 + [O] CH3CH2COCH3 + H2O CH3CH2COCH3 + NO2NO2N N H HH NO2NO2N N H CCH3CH2CH3 + H2O (will be covered in carbonyl compound) 6.CompoundCisadiolcommonlyusedinperfumes.Itisalsoausefulintermediatefor the synthesis of other organic compounds as illustrated by the following scheme. Given that compound C contains a tertiary alcohol, suggest structures for compoundsC, D, E and F. H3C C COHCH3OHHCH3H3C C COHCH3OO-Na+H3C C COHCH3OCH3EH3C C CO-Na+CH3OCH3CDF 7. AlcoholBformsesters whichareresponsible forthe flavoursofvariousfruitsandhas the molecular formula C5H12O. ReactionofBwithacidifiedpotassiumdichromate(VI)producesacompoundC, C5H10O2.Heating B over Al2O3 produces D, C5H10.Vigorousoxidation of D forms 2-methylpropanoic acid as one of the products. E, a structural isomer of B, decolourises purple potassium manganate (VII) to produce a compound F, C5H10O.E forms a yellow precipitate G, when warmed with alkaline aqueousiodine.WhenEisheatedwithconcentratedsulfuricacid,itproduces CH3CH2CH=CHCH3 as one of the products. Suggest the structures for B, C, D, E, F and G and explain the reactions involved. Analysis (flowchart) Analysis (flowchart) AlcoholB(C5H12O)undergoesoxidationwithhotacidifiedK2Cr2O7toformC (C5H10O2) B is a primary (1) alcohol C is a carboxylic acid (C5H10O2) B undergoes elimination of H2O with hot Al2O3 to form D D is an alkene (C5H10) D undergoes oxidation to form 2-methylpropanoic acid as one of the products. D is an alkene with a terminal double bond D: C CHHCHOCH3HOH2D has total 5 carbon:4 Carbon+ 1 Carbon C CHHCHCCH3HHHHC CHHCHCCH3HHHHH OH1 2 3 H CHHCCH3HCHHCOHOB: C:

(1 alcohol) (carboxylicacid) *** Chemical Equations are not required in this question.Oxidation of B: CH3CH(CH3)CH2CH2OH + 2[O] CH3CH(CH3)CH2COOH +H2O Elimination of B: CH3CH(CH3)CH2CH2OH CH3CH(CH3)CH=CH2 + H2O Oxidation of D: CH3CH(CH3)CH=CH2 + 5[O] CH3CH(CH3)COOH + CO2 + H2O E (C5H12O), a structural isomer ofB, undergoes oxidation with hot acidified KMnO4 to form F (C5H10O) F has only one O atom it is a ketone (C5H10O) E is a secondary (2) alcohol E undergoes oxidation with alkaline aqueous iodine to form G E is an must contain structure G is yellow ppt, CHI3 E undergoes elimination to form as one of the products. E :F :(C5H12O) (C5H10O) *** Chemical Equations are not required in this question. Oxidaton of E to F: 4 5 6 C CH3HOHC CHHCHCHHHCH3HC CHHCHCHHHCH3HH OHC CHHCHCHHHCH3H OC CHHCHCHHHCH3HH OHC CHHCHCHHHCH3H O + [O]+ H2O Positive iodoform test of E to give G: + 4I2 + 6 OH + 5 I + 5H2O+ CHI3

Elimination reaction of E: CH3CH2CH2CH(OH)CH3

CH3CH2CH=CHCH3 + H2O 8N2008/III/Q1(e)The molecules of compound P, C7H15Br, are chiral. On treatment withNaOH (aq), P produces alcohol Q, C7H15OH, which does not react with hot, acidified Na2Cr2O7 (aq). TheeliminationofHBr fromcompoundPproducesa mixtureof4differentisomeric alkeneswiththeformulaC7H14, onlytwoofwhicharegeometricalisomersofeach other. Suggest the structural formulae of compound P and the four alkenes. Analysis:Q must be a 3 alcohol since it cannot be oxidised by hot acidified Na2Cr2O7 Compound PCompound Q

H3C CCH3HCBrCH3C CH3HH

H3C C3CH3HC2OHC4H3C1CH3HH CH3CH2CH2COO-C CHHCHCHHHCH3HH OHH3C C3CH3HC2BrC4H3C1CH3HHH3C CCH3HCCH2C CH3HHH3C CCH3HC CCH3HCH3H3CC CCH3CH3CHCH3HC CCH3H3C CCH3HCH3HC CCH3H3C CCH3HHCH3 Remove H from C4 and Br from C2 Remove H from C1 and Br from C2 Remove H from C3 and Br from C2 cis isomertransisomer 2012/CJC/P2/Q5 9(a)A reaction scheme involving compound B is shown below. (i)Based on the above information, draw three possible structural isomers of B, which are labelled as B1, B2 and B3. B1 B2 B3 (ii)Based on the structural isomer B1, draw the structural formulae of C, D and E. C D E (b) A structural isomer of D, C8H8O, which is labelled as F, contains a C-O-C bond.F does not react with HBr(g). (i)Suggest a structural formula of F.

(ii)Although F does not react with HBr(g), it can react with concentrated HBr(aq). The reaction of F with concentrated HBr(aq) is similar to the reaction of primary alcohols with concentrated HBr(aq). The process involves the following two stages:

SuggestamechanismfortheStageIIprocessinthereactionofFwithconcentrated HBr(aq),includingcurlyarrowstodenotemovementofelectrons,andallcharges.You do not need to draw the 3-dimensional representation of the molecules involved. or OR [arrows, lone pair on Br- indicated, SN2 mechanism] 10.Astudentwas given5unlabelledbottlesandeachbottlecontainsoneof the following organic substances. CH3CH2CH2Br CH3CH2CH(OH)CH3 C6H5CHCH2 C6H5OH C6H5CH2OH Outlineasequenceofsimplechemicaltestswhichthestudentcouldusetoidentify eachoftheaboveorganicsubstances.Foreachtest,statethereagentsand conditions used, type of reaction and describe the expected observations. TestProcedureObservation 1Add NaOH(aq) to each compound separately and heat. Acidify with excess dilute HNO3and then add 2-3 drops of AgNO3(aq).Cream ppt. is observed in the test tube that contains CH3CH2CH2Br. No cream ppt is observed in the other test tubes 2Add I2(aq) to each of the remaining compound separately, followed by NaOH(aq) and warm. Decolourisation of brown solution is seen. Yellow ppt of CHI3 is formed for propan-2-ol No yellow ppt. observed in other test tubes. 3Add neutral FeCl3 (aq) to each of the remaining compound separately at rtp. Purple/violet complex observedfor C6H5OH. No purple/violet complex observed in other test tubes. 4Add Br2(aq) in the dark to each of the remaining compound separately at rtp. Decolourisation of reddish-brown solution observedfor C6H5CHCH2 No decolourisation of reddish-brown solution observed in other test tubes. 5To the last organic compound (C6H5CH2OH), add KMnO4(aq) in H2SO4 (aq) and heat. OrTo the last organic compound (C6H5CH2OH), add K2Cr2O7(aq) in H2SO4 (aq) and heat. Alternative Test for C6H5CH2OH. To the last organic compound (C6H5CH2OH add PCl5 at rtp and test gas evolved with damp blue litmus paper. Purple KmnO4 decolourised Orange K2Cr2O7 turns green White fumes of HCl that turns damp blue litmus paper red END Name: Date:Class: Score:/20 Assignment Hydroxyl Compounds [These questions are for submission.] Deadline for submission __________________ 1 (a) DrawthestructureoftheorganicproductwhencompoundZreactswiththe following reagents.CHCH2OHClCHCCH3CH2CH3 Z(i)Acidified KMnO4, heat COOHCOOH[1] (ii)Br2 (aq), room temperature and pressure CHCH2OHClHCBrCOHCH3CH3CH2[1] (iii)CH3COOH, concentrated H2SO4, heat CHCH2ClCHCCH3CH2CH3O C CH3O[1] (b)Outline, with reagents and essential conditions, how the following conversion could be efficiently carried out. CH3 toCHO CH3 COOH

CH2OH CHO [2] for all reagents and conditions correct [5] N11/A levels/P2/Q6 (modified) 2 CompoundWhasamolecularformulaofC7H7OCl.Itisanaromaticcompoundwhich contains two functional groups. Data about the reactions of W are given in the table. ReactionReagentResult 1AgNO3 (aq), warmWhitesolidformedwhichissolubleinan excess of NH3 (aq) 2Br2 (aq) in excessWhite solid formed which has an Mr = 379.2 3MnO4-/OH- heatunderreflux then acidify MnO4- is decolourised; One organic product formed with Mr =138 4NaColourless gas evolved; White solid formed which is soluble in H2O 5NaOH(aq)atroom temperatureColourless solution formed (i)Identify the white solid formed in reaction 1, and hence name the functional group that shows to be present in W. White solid: AgCl (s)Functional group: halogenoalkane [1] for both correct ans (ii)Based on reaction 4, two different functional groups could be present in W. Which of these functional group is confirmed by reaction 5? Explain your answer. Two functional groups that could be present: alcohol and phenol Functional group confirmed by reaction 5: phenol [1] Phenols are slightly acidic and can undergo neutralisation with NaOH (aq) at KMnO4, H2SO4 (aq), heat 1.LiAlH4 in dry ether 2.H2O, heat K2Cr2O7, H2SO4 (aq),heat with distillation r.t.p. [1] Alcohol is considered to be neutral thus it will not undergo acid-base reaction with NaOH (aq). (iii)Usingyouranswersin(i)and(ii),aswellasinformationfromthetableabove, draw the fully displayed structure of W. Explain clearly why you have placed each of the two functional groups in their particular positions.CH HClO H[1] Chlorineatommustnotbeattachedtothebenzeneringtoform chlorobenzene.white ppt of silver chloride will not be formed as the porbital of the Cl atom can overlap with the -electroncloud of the benzene ring. As a result, thelonepairofelectrons on the Cl atom candelocaliseinto the benzene ring. -ThisresultsinpartialdoublebondcharacterinC-Clbondandthe strengthening of the C-Cl bond. -Hence, the Cl atom is difficult to substitute. [1] 3 bromine atoms must be substituted into compound W when it reacted withBr2(aq)thusthesubstituentsmustbeinthe1and3positions respectively. [1] [6] SRJC 2008 Prelim/II/Section B Q7(a) modified 3When treated with colddilutealkalinepotassiummanganate(VII), P,C10H12forms compoundQ.However,onheatingPwithacidifiedpotassiummanganate(VII),2 products, compound R, C8H8O and compound S, C2H4O2 are formed.WhenRwaswarmedwith2,4-dinitrophenylhydrazine,anorangeprecipitatewas obtained. OnrefluxingQwithethanoicacidinthepresenceofconcentratedsulphuricacid, compound U, C14H18O4 is obtained.Deduce, with reasoning, the structures of P, Q, R, S, T and U. [9] P undergoes (mild) oxidation to form Q. P contains C=C Q contains diol / alcohol [1] P undergoes oxidationto form compounds R and S. P contains C=C R is a ketone S is a carboxylic acid [1] R undergoes condensation[1] to form orange precipitate. R is a ketone Q undergoes esterification/ nucleophilic substitution [1]to form compound T. Q contains diol T is a diester[1] Max 3 marks for reasoning Structure: [1] each (total 6 marks) C CC H3CH3H C CC H3CH3HO H OH P Q CCH3O

C H3COHO R S C CC H3CH3HO OCCCH3OCH3O T Max 7 out of 10 END