Chuyên đề Hóa học hay: Este - Lipit

Chuyên đề: Este LipitBiên soạn: Thầy PVC (Phạm Văn Cử)Email: pvc.chhk21gmail.comSĐT: 097.4472015Este Lipit là chuyên đề trọng tâm trong hóa học hữu cơ 12. Kiến thức chương este liên quan nhiều đến các chương ancol, anđehit, ... chúng ta đã học ở lớp 11. Nên để học tốt chuyên đề này, chúng ta cần phải ôn lại kiến thức đã học của năm trước. Chúc các em học tốt và đạt kết quả cao trong các kì thi.

CHUYấN:ESTEưLIPIT c H a I.KHINIMCHUNG ú I.1 K/N ESTE H 201 ưEsteldnxutcaaxitcacboxylicthuckhithaythnhúmưOHtrongbngOR C OH C V O VD: CH C OCH P 44 o O iỏ 97 g I.2.NGNGưNGPHNưDANHPHP y T: I.2.1.ngng h S T CTTQ: C H O (n 2,k1,a2) m A.LíTHUYT A.1 ESTE 3 n 2n + - 2k a co l i ma *Esteno,nchc,mchh(k=1,a=2):CnH2nO2hocRCOOR(R,RlgcHCno) g *Estekhụngno,1lkC=C,nchc,mchh(k=2,a=2):CnH2n - 2O2hocRCOOR(RhocRcúC=C) @ 21 k *Esteno,2chc,mchh(k=2,a=4):CnH2n - 2O4 h c H a ú HO VD: XỏcnhsngphõnestecúCTPTlC H 201 Ligii Cỏch 1: C V HCOOCH -CH HCOOCH-CH P -CH o ỏ i CH g : y CHTCOOCH -CH CH -CH COOCH h T S Cỏch 2:Sngphõneste=2 (n < 5) = =4ngphõn I.2.2.ngphõn p il: a Em ch c v c H a ú H 015 C V P 44 Cỏch 3:GiC H O cúcụngthcdngRCOOR o m iỏ 97 co g => R + R = 3C = i+l = + = + = + 0 a : y m T g Peste Paxit @ Th S = 1.2 + 1.1 + 1.1 + 1=6ngphõn(4peste+2paxit) 2 3 3 n-2 m :p ail v 3 4-2 3-2 hk h c.c 3-2 *Chỳý:SngphõngcCnH2n + (n < 6) = 2n - c D 29 H a I.2.3 Danh phỏp ú H Tờneste=tờngcHCancol+tờnaniongcaxit C V P 44 VD:Gitờncỏcestesau: o iỏ 97 4) C H COOC H g 1) HCOOCH : y 5) C H COOCH C H 2) CH =CHCOOCH=CH T h 6) CH OOC - CH - COOC H T 3) CHS =CCOOCHCH (Etylmetylmalonat) E VD:ShpchthuccúCTPTC 6H12O2tỏcdngcviNaOHl A 26 B 27 C 28 2 CH3 5 6 5 CH3 Biờnson:ThyPhmVnC ST:097.447.2015 Trang CHUYấN:ESTEưLIPIT II.TNHCHTCAESTE II.1.TNHCHTVTL Estekhụngcúliờnkthiro: + tos (este) < tos (ancol) < tos (axit) (CúMtngng) +Khụngtantrongnc Estenhhnnc,tantrongdungmụihuc Mtsestecúmựictrng: II.2.TNHCHTHểAHC II.2.1.Phnngcanhúmchc II.2.1.1.Phnngthyphõn Estecúththyphõnctrongmụitrngkimhocaxit R C OR + H OH H+/OH c H a ú H 015 C V P o ỏ i g : 09 y T h T S R C OH + R OH gm @ 21 om c ail O k h c h phnng vicỏcchttrongmụitrng Snphmthuctiptc chuynhúahoc c.c H v p a Mụitrngaxit il: ú a H 15 H E+m R OH R C OH R C OR + H0 OH C V O PO (Phnngthunnghchviphnngestehúa) c o ỏ i H g H a : COOCH + HOH y VD:TCH CH COOH + CH OH ú H h 201 T S Mụitrngkim(phnngxphũnghúa) C t V RCOONa + ROH RCOOR + NaOH P 44 o m ỏ OH 97 i co + NaOH t VD: CH COOCH CH COONa + CH g l : y m T Mtstrnghpthyphõnestecbit g h S @ T +Estecaphenol k O + + 3 3 o o 3 hhVD: HCOOC6H5 c vc + NaOH :p il+Estecacỏcancolkhụngbn a Em to HCOONa + C6H5ONa + H2O c CH C H (anehit) C=CH H a O OH ú Ancolkhụngbn H 201 C C = C R (xeton) CH C R V P o O ỏ OH i g :0 y VD: T CH COOCH=CH + NaOH CH COONa + CH CH=O h T S CH3COOC = CH2 CH3 Biờnson:ThyPhmVnC ST:097.447.2015 3 + NaOH CH3COONa + CH3 C CH3 O Trang CHUYấN:ESTEưLIPIT c LiAlH H R-CH OH + R OH R C OR a ú O H 015 LiAlH C CH CH OH + CH OH V VD: CH COOCH P 44 o II.2.2.TớnhchtcagcHC iỏ 97 g II.2.2.1.Phnngcngvogckhụngno :0 y,HA)vogcHCkhụngnoto ưTngtanken,estekhụngnocngthamgiaphnngcng(H , Br T thnhhpchtno Th S II.2.1.2.PhnngkhnhúmchcCOO(bngLiAlH4) 4 3 3 VD: CH2=CHCOOCH3 + Br2 II.2.2.2.Phnngtrựnghp VD: to, xt, p c nCH3COOCH=CH2 H a Vinylaxetat ú H C t , xt, p V nCH =CH-COOCH P o ỏ i g : 09 CH y TMetylmetacrylat h T ưPMMcsdngsnxut thytinhhuc S o 3 II.2.3.Phnngtchỏy ưGicụngthctngquỏtcaEste: om CnH2n + 2kOa c il a CnH2n + 2m g 2kOa + @ 21 k III.IUCHVNGDNG hh c III.1.IUCH vc p : III.1.1.Estecaancol ail m E 3n + - k - a O2 ( CHCH2 ) n OOCCH3 pv Poli vinylaxetat (PVA)ail: hh c.c Em COOCH3 c CH H a Poli metylmetacrylat (PMM) ú H 015 C V P 44 o iỏ 97 g nCO +y (n + 1T - :k)H O Th S ( CH2CH )n 2 c H SO c Este + H O H H SO c a VD: CH COOH + CúH OH CH COOC H + HO H III.1.2.Estecaphenol C V Phenol+cloruaaxit/anhiritaxit Estecaphenol P o VD:7C H4 OH + CH COCl CH COOC H + HCl ỏ i g : 09 y III.1.2.Estecaancolkhụngbn T h T S VD: CH COOH + CHCH t , xt CH COOCH=CH Axit + Ancol 2 k gm @ 21 om c ail 3 5 o 3 III.2.NGDNG(SGK) Biờnson:ThyPhmVnC ST:097.447.2015 Trang CHUYấN:ESTEưLIPIT c I.KHINIMCHUNG H a I.1.K/NCHTBẫO ú Chtbộoltriestecaglixerolvcỏcaxitmonocacboxyliccúschnnguyờntcacbon(t12ư24C), H 201 khụng phõn nhỏnh (axit bộo) C CH -OOCR V P 44 o =>CTTQcachtbộo: CH-OOCR iỏ 97 g CH -OOCR y T: Mtsaxitbộothnggp: Th S m A.2.CHTBẫO 2 Loiaxit Cụngthc Tờngi CH3[CH2]14COOH Axit bộo no Axit panmitic c h Axit stearic ch c v :p l i a EmAxit oleic CH3[CH2]16COOH H a ú [CH ]5 CH H Axit bộo 201H C=C khụng no C V 4] P CH 4[CH CH o ỏ C=C i g H H H : y T h T S I.2.CCHGITấNCHTBẫO [CH2]7COOH H k g @ 21 co l i ma c H H a ú H 015 C V P 44 Tờnchtbộo=Tri+tờnaxit(ic in) o m ỏ 97 i co (gcaxitgingnhau) g l : y m T g @ H CH -OOCC Th S 15 2 [CH2]7COOH C=C Axit linoleic 31 VD: hCH-OOCC hk 15H31 c vc CH2-OOCC15H31 p : ail Tripanmitin II.TNHCHTCACHTBẫO Em II.1.TNHCHTVTLí Chtbộonhhnncvkhụngtantrongnc,tantrongcỏcdungmụihucnh:benzen,xng,ete, c H a ú no:5l chtrnt thng(mngvt) H Chtbộochanhiugcaxitbộo C khụng no: l chtlngt thng(duthcvt,ducỏ,) V P 44 o II.2.TNHCHTHểAHC iỏ 97 II.2.1.Phnngthyphõn g y T: +Mụitrngaxit Th S H ,t C H (OH) + 3RCOOH (RCOO) C H + 3H O o o + 3 Biờnson:ThyPhmVnC ST:097.447.2015 o Trang CHUYấN:ESTEưLIPIT c H C H (OH) + 3RCOONa (RCOO) C H + 3NaOH a ú X phũng H 201 II.2.2.Phnnghirohúa C Dun+H ắắđ M V P 44 o Chtbộolng Chtbộorn iỏ 97 g B.MTSDNGBITPTHNGGP y T: DNG1:BITONTCHYESTE Th S +Mụitrngkim(xphũnghúa) to 3 5 to Ni GHINH PTPTngquỏt: Cn H 2n + 2-2k Oa + c 3n +1- k -a O ắắ đ nCO + (n + - k)H O gm 1@ hk h c.c n H2O - n CO2 om c ail => n = H v 1- k a :p l i ú a Dóyesteno,nchc,mchh: H C H O ắắắ đ n = n Em (k = 1, a = 2) 20 C V x=2C H O :Estekono,1C=C,n,h P c 4đ n = n - n C H O (k = 2) ắắắ o ỏ i x=4C H O :Esteno,haichc,h H g a y T: ú VD1:tchỏyhontonesteno,nchcthuc n CO = n O P.Tờncaestel H Th S A metylfomiat B etyl axetat C n-propylaxetat D Cmetylaxetat V Ligii P 44 o m ỏ 97 i PTP: co C H O + nO nCO + nH O g l : y m T Vỡ n CO = n O Pg =>Hscõnbngphnngbngnhau h 1@ T S k2 este +O2 n 2n CO2 H2O n 2n - 2 n 2n - +O2 n 2n - x este CO2 H2O n 2n 2 2 2 TheonhlutbotonnguyờntO,tacú: 2+2n=2n+n=>n=2 hh c => CTPT: C H O HCOOCH (metylfomiat) => ỏpỏnA vc :p l i a VD2 (A/11):tchỏy0,11gamesteX(tobiaxitnchcvancolnchc).Saukhiphnngxyra Em hontonthuc0,22gamCO v 0,09 gam c H2O.SestengphõncaXl C D H a ú Ligii H ưTa cú: X20+1 O CO + H O Theobotonk/lg,tacú: C m 7+ m = m + m V => m = 0,2 gam => n = 0,00625 (mol) P o Theobotonnguyờnt,tacú: n + n = n + n ỏ i g : 0 n + 2n = 2n + n y T => n = 0,0025 (mol) h n T Xlestenchc=>n S = 0,00125 (mol) => M = 88 => CTPT este l C H O = A B X O2 O2 CO2 O (X) O (O2) H2O O2 O (X) O2 CO2 O (CO2) O (H2O) H2O O (X) O(X) X X => ỏpỏnA Biờnson:ThyPhmVnC ST:097.447.2015 Trang CHUYấN:ESTEưLIPIT VD3 (A/11):tchỏyhonton3,42gamhnhp(a.acrylic,vinylaxetat,metylacrylat,a.oleic)rihpth snphmvobỡnhngdungdchCa(OH)2d.Sauphnngthuc18gamkttavdungdchX KhilngdungdchXsovikhilngcadungdchCa(OH)2banuthayinhthno? A.Gim7,74gam B.Tng7,92gam C.Tng2,7gam D.Gim7,38gam Ligii Nhnxột:C4chtucúk=2=>CTTQ: C n c H a ú H 015 C : a (mol) V P o ỏ i n=6 g : => y a = 0,03 T h T S H 2n - 2O2 (14n + 30).a mhh = 3,42 = Theogithuyt,tacú:=> n CO2 = n CaCO3 = 0,18 = n.a PT chỏy: 2C6H10O2 + 15O2 12CO2 + 10H2O 0,18 mol 0,15 mol 0,03 mol mdungdchX = mCa(OH)2 + mCO2 + mH2O m => mdungdchX mCa(OH)2 = 0,18.44 + 0,15.18 18 = 7,38 (gam) c hh c.c k gm @ 21 om c ail v H a :p l i ú a H 015 VD4(C/10):tchỏy2,76gamhnhpXgm:C H COOH, C H COOCH Em , CH OHthuc2,688lớt CO (ktc)v1,8gamH C O.Mtkhỏccho2,76gamXtỏcdngvavi30mlNaOH1Mthuc V 0,96 gam CH OH.CụngthccaC H COOH l P 4B.4CH COOH c o A C H ỏCOOH C C H COOH D C H COOH i H Ligii g a y T: ú H h n = 0,12 mol > n = 0,1 mol Nhnxột: 201 T S C Nuaxitlno,nchc,h(ỏpỏnB,D): X ắắắ đ n > n (vụ lớ) => LOIB,D V P 44 o m =>Axitlaxitkhụngno,1lkC=C,nchc,h ỏ 97 i co g l Theobi,tacú: n = 0,03 (mol) = n + n : y m => n = n T ng = 0,03 (mol) = n + n (k = 2) 1@ Th(k = 0) S => mdungdchXgim7,38gamsovimCa(OH)2banu=> ỏpỏnD x y x y 3 3 x y CO2 H2O + O2 H2O CH3OH este ancol NaOH este axit k2 Nhli: Bitoỏntchỏy2cht(k=2,k=0): hh p c vc => Ta icú: l: Axit, ancol a m E neste => + O2 ắắắ đ ồn CO = ồn ồn CO CO2 ancol = ồn axit H2O H2O c => n = n = 0,01 (mol) Hn = 0,01 mol Nubti1H O (axit + ancol) a=> ú 0,03 5mol =>HnhpXchcúC H COOCH H 201 2,76 - 0,18(m ) = 86 C 7.M = 0,03 V P => CT este: C H COOCH 4=> CT axit C H COOH => ỏpỏnC o iỏ 97 g y T: Th S = ồn CO - ồn H2O = 0,02 (mol) axit ancol H2O x y H2O este 3 Biờnson:ThyPhmVnC ST:097.447.2015 Trang CHUYấN:ESTEưLIPIT c H a ú GHINH H 015 ắắđ RCOOH + ROH PTPTQ: RCOOR+HOH ơắắ C V P RCOOR + NaOH RCOONa + ROH Nhnxột: Nu: n = n =>Estenchc iỏo g : 09 cbit:NuRCOORcúRtrựngviC H hocvũngbenzencúnhúmth y T h => n = 2n vsnphmchora2mui T S m DNG2:BITONVPHNNGTHYPHN/XPHềNGHểAESTE 1.Bitoỏnthyphõnestenchc H+ este NaOHP NaOH este o c VD1 (B/07):Xlestenchc,tkhicaXsoviCH4=5,5.unnúng2,2gamesteXvidungdch il a NaOHdthuc2,05gammui.CụngthccutocaXl gm @ A HCOOCH2CH2CH3 B C2H5COOCH3 21 k C HCOOCH(CH3)2 D CH3COOC2H5 h c ch c H v VD2:ThchinphnngxphũnghoỏchthucXnchcvidungdchNaOHthucmtmui a :p l i ú YvancolZ.tchỏyhonton2,07gamZcn3,024lớtO (ktc)thuclngCO nhiuhnkhi a H m E lngncl1,53gam.NungYvivụitụixỳtthuckhớTcútkhisovikhụngkhớbng1,03 20 CTCTcaXl C V A C H COOCH B CH COOC H P H 44 c o C C H ỏCOOC D C H COOC H i H Ligii g a y T: ú H h + NaOH/CaO T (d = 1,03) 201 T S + NaOH MuiY t C X(nchc) V + 3,024 lớt O (ktc) m P = m 4+41,53 Ancol Z; 2,07 gam Z o m ỏ o i c g + ROH GicụngthccaXlRCOOR: RCOOR+NaOH RCOONa il a : y (Y) T(Z) gm h @ T CaO, t RH + Na CO S Y + NaOH/CaO: k2 RCOONa + NaOH 2 5 3 2 5 T/kk o CO2 H2O o :p l i a Z + O2 Em v hh c c (T) =>RCOONa C2H5COONa MT = 1,03.29 30 => T l C2H6 CO2 + H2O Theo BTKL: 2,07 + 32.0,135 = mH2O + 1,53 + mH2O => mH2O = 2,43 gam => mCO2 = 3,96 gam c H a Theo BTNT O: n ú + 2n = 2n H => nư = 0,045 (mol) => M C => ỏpỏnD V P 44 o iỏ 97 g y T: Th S ancol ancol Biờnson:ThyPhmVnC ST:097.447.2015 O2 + nH2O ancol = 46 => ROH C2H5OH CO2 Trang CHUYấN:ESTEưLIPIT VD3:un20,4gammtchthucAnchcvi300mldungdchNaOH1MthucmuiBvhp chthucC.ChoCphnngviNadthuc2,24litH2(ktc).NungBvihnhpvụitụixỳtthu ckhớDcútkhiiviO2 bng0,5.KhioxihúaCbngCuOcchthucEkhụngphnngvi AgNO3/NH3.XỏcnhCTCTcaA? A CH3COOCH2CH2CH3 B CH3COOCH(CH3)2 C C2H5COOCH2CH2CH3 D C2H5COOCH(CH3)2 Ligii c H a ú H 015 C V P o + NaOH/CaO ỏ i D (d = 0,5) MuiB g : 09 t y T +Nad 2,24 lớt T Hh (ktc) S C D/O2 o 20,4 gam A + 0,3 mol NaOH + CuO E(KhụngphnngviAgNO3/NH3) Tỏpỏn=> Alesteno,nchc,mchh => GiCTcaAlRCOOR PTP: RCOOR+NaOH RCOONa + ROH (B) (C) c gm @ 21 om c ail (1) hh c.c k v (2) H + NaOH CaO, t RH + Na lCO a :p i ú a (D) H m E Ta cú: d =ư0,5 => M0 = 32.0,5 = 16 => D l CH => RCOONa CH COONa C V (3) + Na RONa + 1/2H P 4ROH C + Na: c o ỏ i 0,2 mol 0,1 mol H g a : y Tathy:n ú T = 0,3 mol > n =0,2mol=>(1)NaOHd,Estephnnght H h 201 T STheo (1): n = n = n = 0,2 mol C => M = 102 => M = 102 59 = 43 => R C H V P 44 o Vỡ C + CuO EkhụngPAgNO /NH => C l (CH ) CHưOH(ancolbcII) m ỏ 97 i co g =>ỏpỏnB l : y m T g VD4(C/08):EstenchcXcúd =6,25.Cho20gamXtỏcdngvi300mldungdchKOH1M 1@ Th S B + NaOH/CaO: o RCOONa D/O2 D NaOH RCOOR ROH NaOHP RCOOR ROH R 3 X/CH4 k2 Saukhiphnngxyrahontonthuc28gamchtrn.CTCTcaXl h h A CH B CH2=CHCOOCH2CH3 2COOCH3 .c2=CH-CH c v C.pCH3COOCH=CH-CH3 D CH3-CH2COOCH=CH2 il: a Ligii Em d = 6,25 => M = 100 => n c= 0,2 mol H a ú H C V P 44 => ỏpỏnD.iỏo g y T: Th S X/CH4 X X RCOOR + KOH RCOOK + ROH Vỡ nX < nKOH=>KOHd,estephnnght nKOHd = 0,3 0,2 = 0,1 mol =>28gamchtrngm:KOH(0,1mol)RCOOK(0,2mol) => 56.0,1 + (MR + 83).0,2 = 28 => MR = 29 => R C2H5 Biờnson:ThyPhmVnC ST:097.447.2015 Trang CHUYấN:ESTEưLIPIT c H a THYPHNMTSESTECBIT ú H 015 EstenX+NaOH 2muihoc n = 2n =>Xlestecaphenol C V P H /OH o (khụngbn) RCOOH + R C=CHR + HO RCOOC=CHR ỏ i g : 09 OH R y T (Estecúgcancolkhụngbn) h T MtmuddBr S m GHINH NaOHP + X 2 1 2 R2 l H: Anehit R2C=CHR1 + AgNO3/NH3 R2CCH2R1 OH Ag g co l i ma R2lgcHC: Xeton (khụngcútớnhchtnhAnehit) @ O hk c ch c Ag HCOOX H v a :p (X:H,Na,gcHC) l i ú a H 015 Em C VD5 (B/11):Cho0,15molestenchcXtỏcdngvadungdchcha12gamNaOH.Saukhicỏc V P 44 phnngxyrahontonthuccỏcsnphmhuccútngkhilngl29,7gam.Sngphõncu c o tocaXl iỏ H A 4.g B C D a y T: ú Ligii H h 201 T S + 0,3 mol NaOH C n = 2.n =>Xlestecaphenol 0,15molestenchcX V P 44 o GicụngthccaXlRCOOCmH ỏC H ONa97+ H O i co PTP: RCOOC H + 2NaOH RCOONa + g l : y m 0,15 mol 0,15 mol 0,15 mol T g h S @ T 29,7 gam k + AgNO3/NH3 NaOH X 6 h =>Theoubi,tacú:(M ch R + 67).0,15 + 116.0,15 = 29,7 => MR = 15 => R CH3 c v p CỏccụngthccutocaXl l: m E c H a úCH H HCOO C V P 44 => ỏpỏnA o iỏ 97 g y T: Th S CH3COO CH3 HCOO Biờnson:ThyPhmVnC ST:097.447.2015 HCOO CH3 Trang CHUYấN:ESTEưLIPIT c H a ú H 015 C V P GiCTcaXlRCOOR o ỏ + NaOH i RCOONa RCOOR g : 09 y 3,4 gam gam T h Theo PT, ta cú: n =n = 0,05 mol => M T = 68S=> M = => R H VD6 (A/09):ChthucXcúCTPTlC5H8O2.Cho5gamXtỏcdnghtvidungdchNaOH(va) Saukhiphnngxyrahontonthuc3,4gammuivmthpchthuckhụnglmmtmu dungdchBr2.CTCTcaXl A CH3COOC(CH3)=CH2 B HCOOC(CH3)=CH-CH3 C HCOOCH2-CH=CH-CH3 D HCOOCH=CH-CH2-CH3 Ligii RCOONa RCOOR RCOONa R => CụngthccaXcúdng:HCOOC4H7 Vỡ X + NaOH HCOONa+Hpcht (khụnglmmtmuBr2) =>CTCTcaX:HCOOC(CH3)=CH-CH3 gm @ 21 om c ail + NaOH HCOONa + CH3-CH=CCH3 (khụngbn) hk c ch OH c H v a :p CH CH CCH (KhụnglmmtmuddBr CH -CH=CCH l ) i ú a H m E O OH Butan-2-on (xeton) C V => ỏpỏnB P 44 c o ỏ 2.Bitoỏnthyphõnhnhpestenchc i H g a y T: VD7 (A/09):Xphũnghúa1,99gamhnhp2estebngdungdchNaOHthuc2,05gammuica ú H h 1axitcacboxylicv0,94gamhnhp2ancollngngktipcanhau.Cụngthcca2estel T A HCOOCH S v HCOOC H B C H COOCH v C H COOC H 20 C C CH COOC H v CH COOC H D CH COOCH v CHVCOOC H P 44 Ligii o m ỏ 97 i co g l Gicụngthcchungca2estel RCOOR' : y m T g h + SR' @ PTP: RCOOR' + NaOH RCOONa OH T k 1,99 gam 2,05 gam 0,94 gam HCOOC(CH3)=CH-CH3 3 3 hh c Theo vc BTKL, ta cú: p il: a Em 3 3 5 1,99 + mNaOH = 2,05 + 0,94 => mNaOH = gam => nNaOH = 0,025 mol M = 82 => M = 15 => R CH c Theo PT, ta cú: n =n =n =0,025 mol H M = 37,6 => M = 20,6 => CH v C H a ú H Cỏch 2: Nhnthy: C m = 2,05 gam > m = 1,99 gam V P 4=>4 M < M = 23 =>2gcancoltngngl CH v C H o iỏ 97 g => ỏpỏnD y T: Th S RCOONa RCOONa R NaOH R'OH R'OH RCOONa R' Biờnson:ThyPhmVnC ST:097.447.2015 R' RCOOR' Na Trang 10 CHUYấN:ESTEưLIPIT c H a ú H 015 C V P o ỏ i Hnhp3mui g : 09 y 896mlkhớ(ktc) T h m gam ancol Y +bỡnhNad T S VD8(THPTQucgia2015):HnhpXgm3estenchc,tothnhtcựngmtancolYvi3axit cacboxylic(phõntchcúnhúmCOOH)trongú,cúhaiaxitnolngngktipnhauvmtaxit khụngno(cúngphõnhỡnhhc,chamtliờnktC=Ctrongphõnt).Thyphõnhonton5,88gamX bngdungdchNaOH,thuchnhpmuivmgamancolY.ChomgamYvobỡnhngNad,sau phnngthuc896mlkhớ(ktc)vkhilngbỡnhtng2,48gam.Mtkhỏc,nutchỏyhonton 5,88gamXthỡthucCO2 v 3,96 gam H2O.PhntrmkhilngcaestekhụngnotrongXl A 29,25 % B 38,76 % C 40,82 % D 34,01 % Ligii + NaOH mbỡnhtng2,48gam 5,88 gam X + O2 Ancol + Na: CO2 v 3,96 gam H2O ROH + Na RONa 0,08 mol c + 1/2H2 0,04 mol H a ú H C 2H 0O1 (x mol) v C H O (y mol)Em GicụngthccỏcestetrongX: C (2 este no) (este khụng no) V Pra, ta cú: 44 x + y = n = 0,08 mol o Theo bi iỏ 97 g : X + O CO + H O X + O:y T m + m = m +m Th Theo SBTKL: => 44n 32n = 1,92 (*) hh c.c k gm @ 21 om c ail v Theo bi ra, ta cú: mbỡnh tng=2,48=mROH mH2 => mROH = 2,56 : p gam l i => MROH = 32 => ROH CH3OH a n m 2n 2m - c H a ú H 015 C V P 44 o (**) iỏ 97 g y T: 0,24 - 0,22 = 0,02 mol T-hồ n S= CH3OH 2 X O2 CO2 CO2 2 H2O O2 Theo BTNT O: 2nX + 2nO2 = 2nCO2 + nH2O => 2nCO2 om 2nO2 = 0,06 c T(*)v(**)=>nCO2 = 0,24ilmol; nO2 = 0,27 mol + Og2m 1@ k2 Vỡ Cn H 2n O =>cnhChH vc O n 2n a nCO2 = nH2O => n Cm H2m - 2O2 = ồn = 0,06 mol => n.0,06 + m.0,02 = 0,24 n CO 0,24 l: p NC = = =3 iSCtrungbỡnhtrongX: a nX 0,08 m E c(**) => < n < T (*) => < m < H CO H2O (*) => Trong X cú HCOOCH3 v CH3COOCH3 a ú T(**)v(***)=>m=5=>CTcaestekhụngno:C H O 0,02 mol H C 7.H O X: 100.0,02 V PhntrmkhilngcaC %m = 100% = 34,01% P 5,88 o => ỏpỏnD.iỏ g :0 y T h T S Doaxitkhụngnocúngphõnhỡnhhc=>m 5(CH3-CH=CH-COOCH3) (***) 5 Biờnson:ThyPhmVnC ST:097.447.2015 8 C5 H O Trang 11 CHUYấN:ESTEưLIPIT c H a ú THYPHNESTEACHC H 201 nNaOH R(COONa) + nROH n = n.n C nNaOH nRCOONa + R(OH) n = n.n V nNaOH R(COONa) + R(OH) n o P =n iỏ 97 g n Snhúmchceste: n = y T: n Th S 3.Thyphõnesteachc GHINH R(COOR)n + (RCOO)nR + R(COO)nR + n n n n ancol mui mui ancol mui ancol NaOHP este VD9: Cho0,01molmtesteXcaaxithucphnngvavi100mldungdchNaOH0,2 M,sn om c phmtothnhchgmmtancolYvmtmuiZvismolbngnhau.Mtkhỏc,khixphũnghoỏ ail m hon ton 1,29 gam este úbngmtlngval60mldungdchKOH0,25 M,saukhiphnngkt g @ thỳcemcụcndungdchc1,665gammuikhan.CụngthccaesteXl 21 k A C2H4(COO)2C4H8 B C4H8(COO)2C2H4 hh C C2H4(COOC4H9)2 D C4H8(COOC2H5)2 c.c v Ligii l: p c H a ú H m n =2 E Theobi:Snhúmchceste:a= n C V n = n =>EsteXcúdngR(COO) R PX + NaOH c o ỏ i H Este X + gKOH: R(COO) R + 2KOH R(COOK) + R(OH) a y T: ú 0,0075 mol 0,015 mol 0,0075 mol H h T = 172 => M + M = 84ư (*) S Theobi:m = 1,29 gam => M m = 1,665 gam => M = 222 => M = 56 C (**) V 47 P o m M = 56 => R C H ỏ o T(*)v(**)=> => ỏpỏnB i { c g : 09 M = 28 => R C H il a y T gm h VD10: unnúng7,2gamesteXvidungdchNaOHd.Phnngktthỳcthucglixerolv7,9gam @ T hnhpmui.ChotonbhnhpmuiútỏcdngviH SO loóng S thu c3axithucno,n k2 NaOH este ancol mui 2 este mui R R(COO)2R R R(COOK)2 R R R 4 hh chc,mchhY,Z,T.TrongúZ,Tlngphõncanhau,ZlngngktipcaY.SlngH c vc X l p : B 24 C 26 D 18 ailA 16 m Ligii E c H a ú PTP: (C H COO) ưHC H20+13NaOH Cgam 7,2 V P 447,9 o 7,2.3 => Ta cú: iỏ42n + 17697= 14n + 68 => n = 2,67 g y T: C H COO-CH Th S =>CTCTcaX: GicụngthccaesteXl (Cn H 2n +1COO)3C3H n 3 2n +1 5 7,9 gam { C2H5COOH (Y) =>3axittngng: CH3-CH2-CH2COOH (Z) (CH3)2-CHCOOH (T) CH3-CH2-CH2COO-CH (CH3)2-CHCOO-CH2 Biờnson:ThyPhmVnC ST:097.447.2015 ắắ đ 3Cn H 2n +1COONa + C3H5 (OH)3 => ỏpỏnB Trang 12 CHUYấN:ESTEưLIPIT c H a ú H 015 C V P o 164,7 gam H O ỏ i g : 09 23,85 gam Na CO y + OT h 44,4gamhhc/rnZ 14,85 gam H O T S VD11(THPTQucgia2016):HpchthucX(chaC,H,O)chcúmtloinhúmchc.Cho0,15 molXphnngvavi180gamdungdchNaOH,thucdungdchY.LmbayhiY,chthuc 164,7 gam hi nc v 44,4 gam hn hp cht rn khan Z. t chỏy hon ton Z, thu c 23,85 gam Na2CO3; 56,1 gam CO2 v 14,85 gam H2O.Mtkhỏc,ZphnngvidungdchH2SO4loóng(d),thuc haiaxitcacboxylicnchcvhpchtT(chaC,H,OvMT mNaOH (trong dd) = 18 gam => mH2O (trong dd NaOH) = 162 gam g @ 21 SP: X+NaOH Z + H2O (1) k h c => mH2O (1) = 164,7 162 = 2,7 gam => nH2O (1) = 0,15 mol .ch c H pv Theo BTKL, ta cú: mX = 164,7ú+a 44,4 180 = 29,1 gam => MX = 194 il: a GiCTPTcaXlCxHyOưzH 15 Em 20 C V 4=7n + n n P+ n => n = 1,5 mol => m = 1,5 gam c o ỏ i => m 9= m (m + m ) = 9,6 gam => n = 0,6 mol H g a y =>Tx:: y : z = n : n : n = 1,5 : 1,5 : 0,6 = : : ú H h H O ) ; vỡ M =194=>n=2=>CTPTcaX: C H20O1 TVycụngthcthcnghimcaXl(C S C HO V P 44 + NaOH o => Xchcúthleste2chc m X ỏ 97 i co g l (1loinhúmchc) Hnhpc/rnZ : y m T g h S @ T n HCOOưCH C H -OOCCH = =>Xlestecúgcphenol=>CTcaX: M: k nC (X) = nC (CO2) + nC (Na2CO3) = 1,275 + 0,225 = 1,5 mol => mC (X) = 18 gam H (X) H (NaOH) O (X) X C H (Z) H (H2O (1)) C H H (X) H H (X) O (X) O 5 n X 10 10 NaOH nX hh c c HCOO-CH pv 2C6H4-OOCCH3 + NaOH HCOONa + CH3COONa + HOCH2C6H4ONa + H2O il: a Em } H+ H SO a ú H HOCH C HưONa C V => ỏpỏnB P 44 o iỏ 97 g y T: Th S HCOONa Z + H2SO4: CH3COONa Biờnson:ThyPhmVnC ST:097.447.2015 Z c HCOOH + CH3COOH + HOCH2C6H4OH + Na2SO4 T (MT = 124) Trang 13 CHUYấN:ESTEưLIPIT c H GHINH a ú BiToỏn:Cho2chthuctỏcdngviNaOHhocKOHtora: H 201 +2muiv1ancolthỡcúkhnng2chthucúl C RCOOR v RCOOR cú n =n V Hoc: RCOOR v RCOOH cú n >n P 44 o +1muiv1ancolcúnhngkhnngsau iỏ 97 g RCOOR v ROH y T: Hoc: RCOOR v RCOOH Hoc: RCOOH v ROH Th S DNG3:BITONHNHPESTEVCCCHTHUCKHC NaOH ROH NaOH ROH +1muiv2ancolthỡcúnhngkhnngsau om c RCOOR v RCOOR ail Hoc: RCOOR v ROH m g *cbit: Nunúichthucúchcúchcestethỡ khụng sao, nhngnunúicúchc @ este thỡchỳngtacnchỳýngoichcestetrongphõntcúthcúthờmchcaxithocancol! k2 h c ch c v VD1(C/08): MthnhpXgmhaichthucnchc.ChoXphnngvavi500mlddKOH H a :p l i ú 1M.SauphnngthuchnhpYgmhaimuicahaiaxitcacboxylicvmtancol.Chotonb a H 015 Em lngancolthuctrờntỏcdngviNa(d),sinhra3,36litH (ktc).HnhpXgm C A.mtestevmtancol B mtaxitvmteste V C.mtaxitvmtancol D hai este P 44 c o Ligii ỏ i H g n = 0,5 mol; n = 0,15 mol a y T: ú VỡXgm2chthucnchc=>Ancolnchc(ROH) H h 201 T S 2ROH + 2Na 2ROna + H C V 0,15 mol 0,3 mol P 44 o m ỏ 97 i R COOR v R COOR co g l X + KOH 2mui+1ancol=>Xgm: : y m RCOOR v RCOOH T g @ Th S Vỡ n k>2n =>XgmRCOORvRCOOH=> ỏpỏnB KOH H2 KOH ROH hh c VD2 (B/10):Hụ vc nhpMgụmaxitcacboxylicX,ancolY(ờunchc,sụmolXgõphailõnsụmolY)v p : esteZctoratXvY.ChomtlngMtỏcdngvaviddcha0,2molNaOH,tora16,4gam ail m muụ E iv8,05gamancol.CụngthccaXvYl c B CH COOH v CH OH H D CH COOH v C H OH a ú ưH 2015 Ligii 2x molCRCOOH + 0,2 mol NaOH V ROH47 16,4 gam RCOONa + 8,05 gam ROH xP mol o 2x + y mol x + y mol y mol 7RCOOR ỏ i g :0 y Ta cú:T2x + y = 0,2 => M = 82 => M = 15 => R CH h T nS = x + y < 2x + y = 0,2 => M > 8,05 = 40,25 => M > 23,25 => ỏpỏnB A HCOOH v CH3OH C HCOOH v C3H7OH 3 } RCOONa ROH Biờnson:ThyPhmVnC ST:097.447.2015 R ROH 0,2 R } Trang 14 CHUYấN:ESTEưLIPIT VD3(THPTQucgia2016):HnhpXgmmtaxitcacboxylicT(haichc,mchh),haiancoln chccựngdóyngngvmtestehaichctobiTvihaiancolú.tchỏyhontonagamX,thu c8,36gamCO2.Mtkhỏc,unnúngagamXvi100mldungdchNaOH1M,saukhiphnngxy rahonton,thờmtip20mldungdchHCl1MtrunghũalngNaOHd,thucdungdchY.Cụ cnY,thucmgammuikhanv0,05molhnhphaiancolcúphõntkhitrungbỡnhnhhn46 Giỏtrcaml A 7,09 B 5,92 C 6,53 D 5,36 Ligii { R(COOH)2 x mol y mol a gam X R'OH R(COOR') z mol c H a ú H 015 C V P o ỏ i +O g : 09 8,36 gam CO y T + 20 molhHCl 1M + 0,1 mol NaOH Dungdch DungdchY T THNaOHd S 2 Cụcn om c R(COONa) ail m 0,05 mol ancol (M < 46) m (g) g @NaCl 21 k Theobi,tacú: n R'OH = y + 2z = 0,05 mol h c ch nNaOHphnng = 2xH + 2z = 0,1 0,02 = 0,08 mol => x + z = c0,04 v a :p l i ú R(COONa)2 0,04 mol ma Trongmgammuicú: ưH E { { NaCl 0,02 mol C V 47 P C H (COOH) : 0,04 mol o (SCkhụngthayisoc CoihnhpXbanugmcúaxitvancol ỏ i H vihnhpbanu) g : 0,05 mol C H OH a y T: ú H h T S(n + 2).0,04 + n.0,05 = n = 0,19 Vỡ M < 46 (C H OH) ị1< n < 2C 201 0,19 -0,04n - 0,04.2 V < => 0,25 < n < 1,5 =>Pn = =>4C4H CH => < o ỏ om 0,05 i c g : 09 l (COONa) 0,04 mol iCH a y Vy,trongmgammuicú: => ỏpỏnA T => m = 7,09 gm h @ 0,02 mol NaCl T S k2 { n m n 2n +1 ancol CO2 n { m 2 h DNG4:BITONVCCCHSCACHTBẫO ch Em vc p : ail GHINH c H a ú H C V VD1: trunghũaaxitdcútrong5,6gamchtbộocn6mldungdchNaOH0,1M.Chsaxitcacht P 44 o bộoúl iỏ 97 B 2,4 A 6g C D 4,8 : y Ligii T h T =n = 0,6 (mmol) S n = 0,1.6 = 0,6 (mmol) => n Chsaxit (aaxit):lsmgKOHcntrunghoaxitbộotdocútrong1gamchtbộo Chsxphũnghoỏ (axp):lsmgKOHcnxphũng hoỏ glixerit v trung ho axit bộo tdocútrong1gchtbộo Chseste (aeste):lsmgKOHcnxphũnghoỏglixeritca1gamchtbộo Chsiot (aiot): lsgamiotcúthcngvoniụiC=Cca100gamchtbộo NaOH KOHcnthit NaOH THaxittdocútrong5,6gamchtbộocn0,6.56=33,6mgKOH =>THaxittdotrong1gamchtbộothỡcn1.33,6:5,6=6mgKOH=> ỏpỏnA Biờnson:ThyPhmVnC ST:097.447.2015 Trang 15 CHUYấN:ESTEưLIPIT c H a ú H 201 Gicụngthccatriestetrongchtbộol: (RCOO) C H C V (RCOO) C H + 3KOH ắắ đ 3RCOOK + C H (OH) CỏcPTP: P(1) 44 o RCOOH + KOH ắắ đ RCOOK + H O iỏ (2) 97 g y T: Chtbộocúchsaxit=7=>100gamchtbộocn7.100mgKOHTHaxittdo h S m = 700 mg => n =n =T 0,0125 mol VD2:phnngvi100gamchtbộocúchsaxitbng7phidựng17,92gamKOH.Tớnhkhilng muithuc A 108,265 B 150,256 C 120,265 D 103,256 Ligii 3 3 5 KOH (2) H2O (2) KOH (2) => nKOH (1) = 0,32 0,0125 = 0,3075 mol => nC3H5(OH)3 = 0,1025 mol Theo BTKL: 100 + mKOH = mmui + mH2O + mC3H5(OH)3 => mmui = 108,265 gam => ỏpỏnA c H a ú H C V 4C H P (RCOO) o + 90 ml KOH 0,1M ỏ 2,52gamchtbộo: i g : 0RCOOH y T h 5,04 gam T XS+ KOH: hh c.c k gm @ 21 om c ail v VD3:Xphũnghúa2,52gamchtbộoXcn90mldungdchKOH0,1M.Mtkhỏcxphũnghúahon :p l i a ton5,04gamchtbộoXthuc0,53gamglixerol.ChsxphũnghúavchsaxitcaXlnltl m E A 200 v B 196 v C 200 v D 196 v Ligii c H a ú H 015 C (RCOO) C H + 3KOH ắắ đ 3RCOOK + C H (OH) V(1) P 44 0,01728 mol 0,00576 mol o m ỏ 97 i co g RCOOH + KOH ắắ đ RCOOK + H O l (2) i a : y gm 0,01728 = 0,72.10 mol => m h= 0,04032T = 0,1.0,09.2 gam = 40,32 mg @ T S k2 40,32 3 3 0,1.90.56 ChsxphũngcaXl = 200 2,52 5 => nKOH (2) h =>ChsaxitcaX= ch p il: a Em vc 5,04 =8 KOH (2) => ỏpỏnA c H a ú H C V P 44 o iỏ 97 g y T: Th S Biờnson:ThyPhmVnC ST:097.447.2015 Trang 16 [...]... Snhúmchceste: n = y T: 0 n Th S 3.Thyphõnesteachc GHINH R(COOR)n + (RCOO)nR + R(COO)nR + n n n n ancol mui mui ancol mui ancol NaOHP este VD9: Cho0,01molmtesteXcaaxithucphnngvavi100mldungdchNaOH0,2 M,sn om c phmtothnhchgmmtancolYvmtmuiZvismolbngnhau.Mtkhỏc,khixphũnghoỏ ail m hon ton 1,29 gam este úbngmtlngval60mldungdchKOH0,25 M,saukhiphnngkt g @ thỳcemcụcndungdchc1,665gammuikhan.CụngthccaesteXl 21... Nunúichthucúchcúchcestethỡ khụng sao, nhngnunúicúchc @ 1 este thỡchỳngtacnchỳýngoichcestetrongphõntcúthcúthờmchcaxithocancol! k2 h c ch c v VD1(C/08): MthnhpXgmhaichthucnchc.ChoXphnngvavi500mlddKOH H a :p l i ú 1M.SauphnngthuchnhpYgmhaimuicahaiaxitcacboxylicvmtancol.Chotonb a ư H 015 Em lngancolthuctrờntỏcdngviNa(d),sinhra3,36litH (ktc).HnhpXgm C 7 2 A.mtestevmtancol B mtaxitvmteste V C.mtaxitvmtancol D hai este. .. trung ho axit bộo tdocútrong1gchtbộo Chseste (aeste):lsmgKOHcnxphũnghoỏglixeritca1gamchtbộo Chsiot (aiot): lsgamiotcúthcngvoniụiC=Cca100gamchtbộo NaOH KOHcnthit NaOH THaxittdocútrong5,6gamchtbộocn0,6.56=33,6mgKOH =>THaxittdotrong1gamchtbộothỡcn1.33,6:5,6=6mgKOH=> ỏpỏnA Biờnson:ThyPhmVnC ST:097.447.2015 Trang 15 CHUYấN :ESTE LIPIT c H a ú 5 H ư 201 Gicụngthccatriestetrongchtbộol: (RCOO) C H C 7 V (RCOO)... Theobi:Snhúmchceste:a= n C 7 V 4 n = n =>EsteXcúdngR(COO) R PX + NaOH c 4 o ỏ 7 i H Este X + gKOH: R(COO) R + 2KOH R(COOK) + R(OH) 9 a 0 y T: ú 5 0,0075 mol 0,015 mol 0,0075 mol H h 1 T = 172 => M + M = 84ư (*) 0 S Theobi:m = 1,29 gam => M 2 m = 1,665 gam => M = 222 => M = 56 C (**) V 47 P 4 o m M = 56 => R C H ỏ o 7 T(*)v(**)=> => ỏpỏnB i { c g : 09 M = 28 => R C H il a y T gm h VD10: unnúng7,2gamesteXvidungdchNaOHd.Phnngktthỳcthucglixerolv7,9gam... cú HCOOCH3 v CH3COOCH3 a ú T(**)v(***)=>m=5=>CTcaestekhụngno:C H O 0,02 mol ư H 2 0 1 5 C 7.H O trong X: 100.0,02 V PhntrmkhilngcaC %m = 100% = 34,01% 4 P 5,88 4 o => ỏpỏnD.iỏ 7 9 g :0 y T h T S Doaxitkhụngnocúngphõnhỡnhhc=>m 5(CH3-CH=CH-COOCH3) (***) 5 5 Biờnson:ThyPhmVnC ST:097.447.2015 8 2 8 2 C5 H 8 O 2 Trang 11 CHUYấN :ESTE LIPIT c H a ú 5 THYPHNESTEACHC H ư 201 nNaOH R(COONa) + nROH n = n.n... phnngthuc896mlkhớ(ktc)vkhilngbỡnhtng2,48gam.Mtkhỏc,nutchỏyhonton 5,88gamXthỡthucCO2 v 3,96 gam H2O.PhntrmkhilngcaestekhụngnotrongXl A 29,25 % B 38,76 % C 40,82 % D 34,01 % Ligii + NaOH mbỡnhtng2,48gam 5,88 gam X + O2 Ancol + Na: CO2 v 3,96 gam H2O ROH + Na RONa 0,08 mol c + 1/2H2 0,04 mol H a ú 5 H ư C 2H 0O1 (x mol) v C H O (y mol)Em GicụngthccỏcestetrongX: C 7 (2 este no) (este khụng no) V Pra, ta cú: 44 x + y = n = 0,08 mol o Theo bi iỏ 97... CH3COOH + HOCH2C6H4OH + Na2SO4 T (MT = 124) Trang 13 CHUYấN :ESTE LIPIT c H GHINH a ú 5 BiToỏn:Cho2chthuctỏcdngviNaOHhocKOHtora: H ư 201 +2muiv1ancolthỡcúkhnng2chthucúl C 7 RCOOR v RCOOR cú n =n V Hoc: RCOOR v RCOOH cú n >n P 44 o +1muiv1ancolcúnhngkhnngsau iỏ 97 g RCOOR v ROH y T: 0 Hoc: RCOOR v RCOOH Hoc: RCOOH v ROH Th S DNG3:BITONHNHPESTEVCCCHTHUCKHC NaOH ROH NaOH ROH +1muiv2ancolthỡcúnhngkhnngsau...CHUYấN :ESTE LIPIT c H a ú ư H 015 2 C 7 V 4 P o 7 4 ỏ i Hnhp3mui g : 09 y 896mlkhớ(ktc) T h m gam ancol Y +bỡnhNad T S VD8(THPTQucgia2015):HnhpXgm3estenchc,tothnhtcựngmtancolYvi3axit cacboxylic(phõntchcúnhúmCOOH)trongú,cúhaiaxitnolngngktipnhauvmtaxit khụngno(cúngphõnhỡnhhc,chamtliờnktC=Ctrongphõnt).Thyphõnhonton5,88gamX... = 40,25 => M > 23,25 => ỏpỏnB A HCOOH v CH3OH C HCOOH v C3H7OH 3 3 3 2 5 } RCOONa ROH Biờnson:ThyPhmVnC ST:097.447.2015 R ROH 0,2 3 R } Trang 14 CHUYấN :ESTE LIPIT VD3(THPTQucgia2016):HnhpXgmmtaxitcacboxylicT(haichc,mchh),haiancoln chccựngdóyngngvmtestehaichctobiTvihaiancolú.tchỏyhontonagamX,thu c8,36gamCO2.Mtkhỏc,unnúngagamXvi100mldungdchNaOH1M,saukhiphnngxy rahonton,thờmtip20mldungdchHCl1MtrunghũalngNaOHd,thucdungdchY.Cụ... thu c3axithucno,n k2 NaOH este ancol mui 2 2 2 este mui R R(COO)2R R R(COOK)2 R 4 R 2 R 8 2 4 2 4 hh chc,mchhY,Z,T.TrongúZ,Tlngphõncanhau,ZlngngktipcaY.SlngH c vc trong X l p : B 24 C 26 D 18 ailA 16 m Ligii E c H a ú 5 PTP: (C H COO) ưHC H20+13NaOH Cgam 7 7,2 V P 447,9 o 7,2.3 => Ta cú: iỏ42n + 17697= 14n + 68 => n = 2,67 g y T: 0 C H COO-CH Th S =>CTCTcaX: GicụngthccaesteXl (Cn H 2n +1COO)3C3H