Bồi dưỡng hóa học 12

^HCVNH VAN UT - HUYNH NHIEN DO QUYEN PHAM TH! TUfai - PHAM THj HONG THAIVI Giai thifdng Sach hay Viet Nam GV. Boi dif&ng hoc sinh gidi B6I DI/ONG HOA HOC • . r. _-__B NHA XUAT BAN DAI HOC QUOC GIA HA NOI NH^ XU^T BfIN Dfil HOC QUOC Gia Hfi NQI 16 Hang Chuoi - Hai Ba TrtCng - Ha Noi Bien thoai: Bien tap - Che ban: (04) 39714896 Hanh chinh: (04) 39714899: Tong Bien tap: (04) 39714897 Fax: (04) 39714899 Chiu trdch nhi^m xuat ban: Gidm doc - Tong bien tap: TS. PHAM THI TRAM Bien tap: QUOC THANG Sufa bai: NHA SACH SAO MAI Che ban: TlJCfNG VY Trinh bay bia: TUCfNG LINK Doi tdc lien ket xudt ban: NHA SACH SAO MAI SACH LIEN KET I DLfOtNG HOA HOC 12 so: 1L-323DH2012 2000 cuon, kho 16 X 24cm. T^i CTy TNHH MTV in diicyng sat Sai Gon. I chi: 136/lA TRAN PHU, P.4, Q.5, TP.HCM xuat ban: 1446-2012/CXB/07-231/DHQGHN ^et dinh xuat ban so: 328LK-TN/QD-NXB DHQGHN song va nop lu^u chieu quy I nam 2013. &i noi ddu Nham giup cac em hoc sinh c6 them W lieu de tiep can, nam vilng he thong kien thufc cof ban, nang cao va ren luyen ki nang giai bai tap theo yen cau doi mdfi. Chung toi xin tran trgng gidi thieu den cac em hoc sinh va cac ban dong nghiep bo sach "BOI Dl/OfNG HOA HOC 12" Noi dung sach "BOI DTJCfNG HOA HOC 12" difc/c bien soan bam sat chiicfng trinh sach giao khoa hien hanh va theo tiing chvfdng ijfng vdi tiing chu:dng trong sach giao khoa. Trong moi chifdng difgfc trinh bay gom 3 phan chlnh: A. KIEN THL/C CAN NHOf B. BAI TAP AP DUNG C. BAI TAP NANG CAO (CO lofi giai) Cac bai tap ap dung va nang cao difcfc giai chi tiet, Idi gidi phu hdp vdi moi do'i tifdng hoc sinh, nham giup cac em hoc sinh lam quen va khac sau kien thiJc thong qua cac bai toan va phufdng phap giai tifng bai toan. Quy en sach la tii lieu giiip cac em hoc sinh Idp 12 nang cao nang l\ic t\i hoc d nha cung nhu! ren luyen de tham gia vao cac doi tuyen hoc sinh gioi cac cap. Va la tai lieu tham khao cho cac giao vien tham gia giang day d cac tru:dng pho thong va giao vien tham gia boi ditdng hoc sinh gioi. Mac du da co' gang trong qua trinh bien soan song khong the tranh khoi nhGng thieu sot ngoai y muon. Tac giai xin chan thanh cam dn cac y kien dong gop, xay diing tijf phia ban doc de Ian tai ban sau cuon sach cd chat liidng tot hdn. Tdc gid CHl/dNG I. ESTE - LIPIT A. KIEN THLfC CAN NH6 I. ESTE. 1. Cong thijtc chung cua mot so este - Este tgo hdi R-COOH vdi R'OH: R-COO-R' Neu R va R' no thi este la CnHznOz (n >2) -, Este tg o bdi R-COOH vdi R '(OH),,: (RCOO)„R'. - Este tgo bdi R(COOH),„ vdi R'OH: R(COOR')^. - Este tgo bdi R(COOH)n, vdi R'iOH),,: Rn(COO)„„,R'„, 2. Ten ggi Ten niia he thong cua este diigc ggi nhii sau: 'i ' Ten este = Ten goc hidrocacbon cua ancol + ten goc axit (doi duoi ic at) Vi du: CH3COOC2H5 : etyl axetat CHz^^CH-COO-CHj : metyl acrylat CH3-OCO-[CH2]4-COO-CH3 : dimetyl adipat C17H35COO—CH2 C17H35COO—CH : glixerol tristearat C17H35COO—CH2 3. Tinh chat vat li. - Cdc este thiidng it tan trong niidc, nhe hdn niidc, di bay hdi. - Co mill thdm dgc triing. 4. Tinh chat hoa hoc a) Phan ihig thuy phan * - Trong dung dich axit: • RCOOR' + HOH :^i===±RCOOH + R'OH Phan ling theo chieu tii trdi sang phdi la phdn ling thuy phdn este, phdn ling theo chieu tiiphdi sang trdi Id phdn ijtng este hoa. Phdn ving thuy phdn este trong dung dich axit la phdn ling thugn nghich. - Trong dung dich baza: dun ndng este trong dung dich natri hidroxit, phdn ling tgo muoi cua axit cacboxylic vd ancol. Phdn ling nay la phdn ling mot chieu vd con diiOc ggi phdn ting xd phdng hoa. Bdi DI/ONG H(3A HCIC12 5 RCOOR' + NaOH > RCOONa + R'OH CH3COOC2HS + NaOH —> CHjCOONa + CzHsOH Tuy nhien, mot so triidng hap ngoai muoi cua axit cacboxylic khong tgo ra ancol ma tgo thanh andehit, xeton, muoi cua phenol, hogc chi tgo thanh mot sdn phdm duy nhdt. CH,COOCH=CH2 + NaOH —> CHjCOONa + CH,-CH=0 CHsCOOCiCHshCH^ + NaOH —> CHjCOONa + (CHjJzC =0 CH3COO—y + 2NaOH —> CHsCOONa + ^ ^^ONa ONa + NaOH > C ^OH b) Phdn thig khvT R-COO-R' —> R-CH2OH + R'OH c) Phan ihig cgng d goc hidrocacbon: Phdn ling cgng vd trtmg hap khi goc R chiia lien ket n. .0 COOCH3 n H2C=:C -^-^"-P ) CH, COOCH3 CH,—C CH2=C(CH3)COOCH3 + H2 ——> CHs-CHiCHaJCOOCHs ' d) Phdn iftig dot chay: Dot chdy este no dofn chv^c: CnH2n02 + (^^^) O2 > /jCO^ + /iH^O 2 Chu y: Ngoai cdc phdn ring tren, rieng este cua axit fomic, con c6 phdn iJtng trdng giiang giong nhii andehit. 5. Dieu che a) Este cua ancol - Thiic hien phdn ling este hda: RCOOH + R'OH < "^^"^'^ z> RCOOR' + HOH - W muoi vd ddn xudt halogen cua hidrocacbon RCOOAg + R'Cl > RCOOR' + AgCU BO'I oadNG HdA HOC 12 b) Este cua phenol Tif halogenua axit va phenolat: RCOCl + NaOCoHs > RCOOCeH^ + NaCl - Tir anhidrit axit vd phenol: (CH3CO)20 + HOCeHs > CHsCOOCgHs + CH3COOH 11. LIPIT. 1. Khdi niem va cd'u tgo - Chat beo (nguon goc dong vat, thitc vat] la este cua glixerol vdi axit beo (axit hHu ca mot idn axit mgch thdng, khdi liigng phdn tit Idn). Cdc chat beo nay diigc goi chung Id triglixerit. Cong thdc tong qudt cua chat beo: CH2—OCOR1 CH—OCOR2 CH2—OCOR3 Trong do: Ru R2, Rj cd the gio'ng nhau hogc khdc nhau. +) Mot so' axit beo thitdng gap: Axit panmitic : C15H31COOH Axit stearic : C^HssCOOH Axit oleic : CiyH33COOH Axit linoleic : CMiCOOH +] Thitdng gap cdc triglixerit pha tgp (R^ ^ R^ ^R^) - Trong chat beo, ngoai este cua glixerol vdi axit beo con cd mot liigng nhd axit d dgng tiJ do diigc dgc tritng hdi chi so axit. - Chi so axit cua mot chat beo la so miligam KOH can thiet de trung hda axit tit do cd trong 1 gam chat bdo. Vi du: Mot chat beo cd chi so axit bd-ng 9. Nghia la de trung hda 1 gam chat beo can 9 mg KOH. - Chi so este cua mot chat beo la so miligam KOH can thiet de thuy phdn hoan toan litgng este cd trong 1 gam chat beo. - Chi so xd phong hda cua mot chat beo Id so' miligam KOH can thiet de trung hda axit tii do vd thuy phdn hoan todn litgng este cd trong 1 gam chat beo. 2. Tinh chat hda hoc a) Phdn vtng thuy phdn - Trong mdi trifdng nitdc hogc axit: Chat beo khong bi thuy phdn bdi nitdc, khi cd xuc tdc axit hogc enzim chat beo bi thuy phdn thugn nghich cho glixerol vd axit beo: CH2—OCOR1 CH2-OH R1COOH CH—OCOR2 + 3H2O < ^ CH-OH + R2COOH CH2—OCOR3 CH2-OH R3COOH B(5i DI/SNG HOA HQC 12 7 - Trong moi triidng kiem (phan ling xa phdng hoa) pMn itng thuy phan xay ra hoan toan: CH2—OCOR1 CH2-OH RiCOONa CH—OCOR2 + 3H2O <===^==± CH-OH + R2C00Na CH2—OCOR3 CH2-OH RgCOONa triglixerit ghxerol xd phdng b) Phan vtng cong hidro: Bien glixerit chifa no thanh glixerit no (CM^COOJ^CsH, + 3H2 '° > (C.yHssCOOhCsHs chat long chat ran [II. XA PHONG vA CHAT GIAT RLTA TONG HOfP 1. Xa phdng la hdn hap cac muoi Na hoac muoi cua axit beo va mot so chat phu gia. 2. De san xuat xa phdng ngifdri ta c6 the: - Dun nong chat beo vdi dung dich kiem. - Oxi hoa ankan (parafin) cua ddu mo nhd oxi khong khi, d nhiet do cao CO muoi mangan xiic tdc roi trung hoa axit bhng NaOH hoQc KOH. 3. Chat giqt rvta tong hap la nhffng chat khong phdi Id muoi Na hoac K cua axit beo nhiing c6 tinh nang gigt riia nhiixd phdng. 4. Chat gidt rtia tong hap difcfc sdn xuat ti( cdc sdn phdm cua dau mo. 5. Xd phdng hi gidm tdc dung gigt rita trong niidc cvtng con chat gidt tdy riia tong hap khong bi gidm tdc dung gigt rifa. 6. Xd phdng cd liu diem Id khong 1dm hgi da, it gay 6 nhiSm moi triidng. Chat gigt riia tong hap gay hgi da do trong do cd chat tdy trdng, gay 6 nhiem moi triidng. BAI TAP AP DUNG lai 1. Ho^n th^nh cdc phiicfng trinh phan ufng theo scf do sau: CH4 )A —^B —i51_>C —> E + B lai 2. Viet phifcfng trinh phan ufng thifc hien day bien hda sau (viet cdc chat d\i6i dang cong thufc cau tao): C5H10O ^ CsHioBr^O ^ CsHsBra ^ C5H12O3 ^WCnHisOe CsHsOsNa CH4 Biet C5H10O la mot ancol bac ba. ai 3. Viet cong thufc cau tao cac dong phan dcfn chufc, mach hor c6 the c6 cua C4H6O2. ai 4. Viet cong thufc cau tao cdc chat c6 ten sau day: a) Isopropyl axetat b) Alylmetacrylat c) Phenyl axetat d) sec-Butyl fomiat BO'I DI/ONG HdA HQC 12 5- E^ot chdy ho^n toan 7,4 gam este X dcfn chufc thu di/ac 6,72 lit khi CO2 (dktc) va 5,4 gam nU6c. a) Xdc dinh cong thufc phan tuf cua X. b) Dun 7,4 gam X trong dung dich NaOH vCfa du den khi phan ufng ho^n toan thu diicrc 3,2 gam ancol va mot lugng muoi Z. Viet cong thufc cau tao cua X vd tinh khoi lufcfng cua Z. Bai 6. Dot chay hoan toan 1,76 gam mot este X thu diTOc 3,52 gam CO2 va 1,44 gam H2O. Xdc dinh cong thufc phan tijf cua X? Bai 7. E la este ciia mot axit don chufc va ancol don chufc. De thuy phan hoan toan 6,6 gam chat E phai dung 34,1 ml dung dich NaOH 10% (d = 1,1 g/ml). Lirgng NaOH nay dung di/ 25% so vdri lufong NaOH phan iJng. Hay de xuat cong thufc cau tao dung cua E? Bai 8. De xa phong hoa 17,4 gam mot este no don chufc c^n dung 300ml dung dich NaOH 0,5M. Tim cong thufc phan tuf cua este dem dung. Bai 9. 0,05 mol este E phan ufng vi/a du 100 gam dung dich NaOH 6%, ta thu difcfc 10,2 gam muoi va 4,6 gam rxiau. Biet rufou hoac axit tao thanh E don chufc. Hay xac dinh cong thufc cau tao cua E. Bai 10. Khi thiic hien phan ufng este h6a 1 mol CH3COOH va 1 mol 2 C2H5OH, Itfong este thu dufoc lorn nhat la - mol. De dat hieu suat CLTC tj dai la 90% (tinh theo axit) khi tien hdnh este hoa 1 mol CH3-COOH thi can so mol C2H5OH la bao nhieu? (biet cac phan ufng este hoa thiTc hien d cung nhiet do). Bai 11. Tron 1 mol axit axetic vdi 1 mol rUgu etylic. Khi s6' mol cac chat trong hon hop khong thay ddi nufa, nhan thay lUOng este thu dUOc \k - mol. Tinh hkng so can bkng (K). 3 Bai 12. Thuy phan hoan todn 444 gam m6t lipit thu difoc 46 gam glixerol (glixerin) va hai loai axit beo. Xdc dinh cong thufc cua hai loai axit beo do. Bai 13. Hay viet phi^ong trinh phan ufng cua chat beo c6 cong thufc phan tuf nhtf sau: CH2-OCO-(CH2X4CH3 CH - OCO - (CH2- CH = CH (CH2), CH3 CH2 - OCO - (CH2 \H = CH - CH2 - CH - CH (CH2 )^ CH3 a) Vdri dung dich KOH d nhiet do cao. b) Vdri I2 CO dU. c) Vdri H2 du, c6 Ni xiic tdc, d nhiet do vd dp suat cao. Bfii DUONG H6A HOC 12 . ^ tiii 14. Khi thiiy ph4n a gam m6t este X thu dUcfc 0,92 gam glixerol, 3,02 gam^ muoi linoleat CiTHsiCOONa vk m gam natri oleat CnHaaCOONa. Tinh gia tri cua a, m. Viet cong thijfc cau tao c6 the c6 ciia X. Bai 15. a) De trung h5a liicfng axit beo tif do c6 trong 14 gam mot mSu chat beo can 15ml dung dich KOH 0,1M. Hay cho biet chi so' axit cua mau chat beo tren. b) Tinh kho'i lacfng NaOH can thiet de trung h6a 10 gam mpt chfi't b6o c6 chi so' axit 1^ 5,6. Bai 16. Hay tinh chi so xk phong h6a cua m6t chS't b6o, biet rkng khi x^ ph6ng hoa hoan toan 1,5 gam ch§.'t b^o d6 can 50 ml dung dich KOH 0,1M. Bai 17. Hay tinh khoi lucfng NaOH can de trung hoa axit tii do c6 trong 5 gam chat beo v6i chi so axit bang 7. Bai 18. Hay tinh chi so' iot cua triolein. Hlf6NG DAN GIAI f Bai 1. Phan ufng: tach nhanh.^im knh nhanh ^ ^aHg + SHg C3H2 H- H,0 ) CH. CHO 1 ^ Mn^* CH3CHO + ^02 — > CH3COOH CH3COOH + CH-CH > CH3COOC=CH2 CH3COOC=CH2 + NaOH > CH3COONa + CH3CHO Bai 2. Phan ufng: CH3 CH3 (1) H2C=CH—C-OH + Br2 > H2C-CH—C-OH CH3 Br Br CH3 ^•^3 CH3 (2) H^C-CH-C-Br . HBr H^C-CH-i-Br . H2O Br Br CH3 Br Br CH I3 CH3 CH3 (3) H2C-CH—C-Br + SNaOH > H2C-CH—C-CHgi- 3NaBr III III Br Br CH3 OH OH OH 10 BO'I DUflNG H6A HOC 12 (4)H2C—CH—C—CH3 + 3CH3COOH OH OH OH CH3COO—CH2 -> CH3COO-CH + 3H2O CH3COO—C(CH3)2 (5) CH3COO-CH + 3NaOH ^ H2C—CH—C-CH3 + 3CH3COONa • OH OH OH CaO, t° ->CH4 + NaaCOg CH3COO-C(CH3)2 (6) CH3COONa + NaOH Bai 3. C4H6O2 c6 A = 2 va hai nguyen tuf oxi. => dong phan este dcfn chuTc, khong no c6 mot noi doi d gdc dong phan axit cacboxylic dcfn chufc khong no mot ndi doi d gdc. - DSng phdn este: HC00CH=CH-CH3 ; HCOOCH2-CH=CH2 CH3COOCH=CH2 ; CH2=CHCOOCH3 - Bong phdn axit cacboxylic: CH2=CH-CH2-COOH CH3-CH=CH-C00H CH2 - C - COOH Bai 4. a) CH3COOCH(CH3)2 C) CH3COOC6H5 I CH, b) CH2 = C - COOCH2 - CH = CH2 CH3 d) HCOO-CH(CH3)CH2-CH3 Bai 5. a) Ta c6: n^o, = = 0,3 (mol) va n^^^ = M = o,3 (mol). ' 22,4 ' 18 Vi khi ddt chay X thu dirge sd mol H2O bang sd mol CO2 nen X Ik este no, dcfn chufc. Goi cong thiJfc cua este no, dcfn chufc \k : CnH2n02 (n > 2) (3n-2] Phan lifng: (mol) I CnH2n02 + 0,3 0,3 O2 nCOz + nHaO (D <- 0,3 n Theo dl bai, ta c6: Mx = — (14n + 32) = 7,4 n = 3. n Vay cong thufc phan tijf cua X: C3H6O2. B6| DirdNG HOA H0C12 11 b) X&c dinh cong thijfc cau tao cua X kho'i luang ciia Z: Ta c6: nx = — = 0,1 (mol). 74 Phan ufng: RCOOR' + NaOH > RCOONa + R'OH. (mol) 0,1 -> 0,1 0,1 0,1 M^: IHR-OH = 0,1(R' + 17) = 3,2 R' = 15: CHg- Vay cong thufc cau tao dung cua X: CH3COOCH3. Va khoi lugng cua Z: 0,1 x 82 = 8,2 (gam). Bai 6. Ta c6: nco^ = = 0.08 (mol); nn^o = = 0.08 (mol) Do n^Og = i^HgO X c6 do bat bao h6a cua phan tuf A = 1 => X 1^ este no, dofn chufc => X dang CnH2n02 C„H2n02 ) nC02 nco2 0,08 1,76 . ^nx= = -— => Mx = 14n + 32 = ^ n = 4 n n 0,08 n Vay cong thufc phan tuf cua X: C4H8O2. m . 34,1 X 1,1 X 10 0 , Bai 7. Ta c6: m^^oHdemdtag = ^QQ = 3.751 (gam) 3,751 X 100 ^ , ^mOU phan .ng = (lOQ + 25) ^ ^ ^^^""^ 3 Mat khdc: IIE = nNaOH = — = 0,075 (mol) 40 => ME = 88 (gam) o R + 44 + R' = 88 =o R + R' = 44 -IQii R = 1=>R' = 43(C3H,) => CTCT (E): HCOOC3H7 (propyl fomiat) - Khi R = 15 ^ R' = 29 =^ CTCT (E): CH3COOC2H5 (etyl axetat) Bai 8. Gpi cong thufc cua este no, don chufc 1^ C^HanOg. Khi phbng h6a thi: n^^^^ = ^naou ne.te = 0,3 X 0,5 = 0,15 (mol) => M^^,^ =^ = ^16 14n + 32 = 116=>n = 6 vay cong thufc phan tuf cua este 1^ C^H^Pz • 2 BO'I Di/dNG HdA Hnr Bai 9. Ta c6: nNaOH = ^'^^^ = 0,15 (mol) = Sneste => este 3 chufc. 100.40 Theo de bai, se c6 1 trong 2 chat la dcfn chufc nen c6 2 truTcfng hop: Triidng h^p 1: axit don chufc rLfcfu ba chufc (RCOOsR' + 3NaOH > 3RC00Na + R'(OH)3 (mol) 0,05 -> 0,15 0,15 0,05 10 2 Ta c6: MRcooNa = ^r-^ = 68 (dvC) R + 67 = 68 => R = 1 (la H-) MR(OH) = =^2 (dvC) ^ R' + 51 = 92 R' = 41 ^ R' 1^ C3H5- y /x 0,05 Vay cong thufc cau tao la (HCOOgCgHs. Trildng hcfp 2: axit da chufc va rifgfu dcfn chufc. (RCOO R')3 + 3NaOH > R(C00Na)3 + 3R'0H (mol) 0,05^ 0,15 . 0,05 0,15 4 6 ^ = n »c o = 30,67: khong nguyen (loai) U, Uo.o Vay cong thufc cau tao ciia este (E) la: (HCOO)3C3H5 Bai 10. Can bang: CH3COOH + C,H,OH ^^"^ CH3C00C,Hs + H^O Ban ddu: IM IM 0.0 2 2 2 2 Phan ling: -M -M -M -M 3 3 3 o Can 6^71^ (1 - 0,9)M (a - 0,9)M 0,9M 0,9M /o^2 Cho V = 1 lit K = v3, = 4 I RCoi a la so mol CgH^OH can dCing, ta c6: Kc = j^^^^-4=>a = 2,925 (mol). ai 11. Goi the tich he phan ufng V Git) CH3COOH + C2H5OH ^ CH3COOC2H5 + H2O Ban ddu: 1 1 Phan ling: x x Sail phan iCng: 1-x 1-x x x I DUdNG HOA HOC 12 13 2 Do Heste = — => X = 1 3 3 3 K = "CH3COOC2H5" H2O" K = CH3COOH C2H5OH 73 V V V V J V = 4 Bai 12. Ta c6: n^,^^^„, 46 92 = 0,5 (mol) 3 Goi cong thijfc cua lipit c6 dang: C3H5(OCOR)3 v6i R = Phan ijfng: ^C3H5(OH)3+3RC00H - 0,5 C3H5(OCOR)3+3H2O: (mol) 0,5 Theo de b^i, ta c6: R = 238,333 = (*) "ihpit =0'5(41+ 132 +3R) = 444 o Ma: „ =239; „ =237 va M_ „ =213; „ =211 Ket hop vdi (*) Cap nghiem thich hop: C17H35- va CnHag Bai 13. CH2-OCO-(CH2)^,CH3 CH - OCO(CHa)^ - CH = CH(CH2)7 CH3 CH2 - OCO - (CH2), CH = CH - CH2 - CH = CH (CH2 \3 Chat beo tren Ik trieste cua glixerol vdi ba axit panmitic, axit oleic vk axit linoleic. a) Khi cho tac dung vdi dung dich KOH d nhiet do cao se xay ra phan LJfng xa phong hoa triglixerit va ba muoi kali cua ba axit beo tren. b) Khi cho tac dung vdi h c6 dtf thi c6 phan ufng cong I2 vao cac noi doi trong goc axit beo khong no. c) Khi cho tac dung vdri H2 diT, xiic tac Ni, t°, p thi c6 phan iJng cong vao noi doi trong g6c axit beo khong no. Bai 14. So d6: X + NaOH ^ CnHsiCOONa + CnHssCOONa + C3H5(OH)3 3,02 Ta c6: n •Ci,H3,COONa 302 = 0,01 (mol) 14 BO'I OUONG H6A HOC 12 => ne„„33cooNa = 0,02 (mol) => m,_^„^^,„„,^ = 0,02 x 304 = 6,08 (gam) 0 92 M^: nNaOH = 3ngiixeroi = 3 X = 0,03 (mol). => mNaOH = 0,03 X 40 = 1,2 (gam). Ap dung dinh luat bao to^n khoi luong, ta c6: = 3,02 + 6,08 + 0,92 - 1,2 = 8,82 (gam). Bai 15. a) Theo dinh nghia: chi so axit cua chat beo la so miligam KOH can dung de trung hoa het cac axit b^o tii do c6 trong 1 gam chat beo. Ta c6: mKon = 0.015 x 0,1 x 56000 = 84 (mg) 84 ^ Chi so axit la: — = 6. 14 b) Chi so axit la 5,6 nghia la de trung hoa 1 gam chat beo can 5,6mg KOH. Vay trung h5a 10 gam chat beo cdn 5,6mg x 10 = 56mg KOH. Hay = 0,001 mol KOH •^56 Vi NaOH la bazo don chiifc nhtf KOH nen can so mol bang nhau trong phan ufng trung hoa. Do vay so gam NaOH can c6 la: 40 x 0,001 = 0,04 (gam) NaOH Bai 16. So mol KOH: 0,050 lit x 0,1 mol/lit = 0,005 mol KOH So gam KOH: 0,005 mol x 56 g/mol = 0,280g hay 280mg KOH 280 Chi so xa phong hoa: = 186,67. 1,5 Bai 17. Theo dinh nghia, chi so axit cua chat beo bkng 7 c6 nghia la muo'n trung hoa lucfng axit beo tii do trong 1 gam chat beo phai dung 7mg KOH. Vay muon trung hoa axit beo i\i do trong 5 gam chat beo c6 chi so 7 thi phai dung 5x7 = 35mg KOH, hay ^^mol KOH ^ M^mol OH" => ^^^moi NaOH ^ khol lifong NaOH c^n dl 56 56 trung hoa axit i\i do trong 5 gam chat beo c6 chi so axit bkng 7 la: niNaOH = >^ 40g = 25 (mg) = 0,025g/5g chat beo. 56 Bai 18. Phan ufng: (C,,H33COO)3C3H5 + 3I2 (C,,H33COOl2)3C3H, (gam) 884 3 x 254 ^ Chi so' iot la i^^-^ x 100 = 86,2. . 884 Bfii DUONG HOA HQC 12 IS C. BAI TAP NANG CAO Bai 1. De thuy phan hoan toan 0,74 gam mot h6n hgp este dcfn chufc car, 7 gam dung dich KOH 8% trong nude. Khi dun nong h6n hgp este noj tren vdi axit H2SO4 80% sinh ra khi X. Lam lanh X, diia ve dieu kie^ thudng va dem can, sau do cho khi Igi tii tii qua dung dich brom dy trong nifdrc thi thay khoi lifgng khi giam 1/3, trong do khoi lifgng rieng cua khi gan nhu khong doi. | a) Tinh khoi liTgng mol cua hon hgp este, xac dinh thanh phan h6n hgp khi sau khi da lam lanh va tinh khoi lugng cua chung. b) Xac dinh thanh phan hon hgp este ban dau. c) Neu phan ufng de phan biet 2 este tren, viet phtfong trinh phan ufng. Bai 2. Lam bay hoi mot chat hufu co A (chufa cac nguyen to C, H, O) diJgc chat hoi CO ti kho'i doi vdi metan bang 13,5. Lay 10,8 gam chat A va 19,2 gam O2 cho vao binh kin, dung tich 25,6 lit (khong doi). Dot chay hoan toan A, sau do giuf nhiet do binh 163,8°C thi ap suat trong binh bang 1,26 atm. Lay toan bg san pham chay cho vao 160 gam dung dich NaOH 15% dirge dung dich B c6 chijfa 41,1 gam hon hgp hai muoi. Khi ra khoi dung dich c6 the tich Vi lit (dktc). a) Xac dinh cong thCfc phan tii, viet cong thufc cau tao cija A (biet rkng khi cho A tac dung vdi kiem tao ra mot ancol va 3 muoi). b) Tinh Vi va C% ciia cac chat trong dung dich B. c) Cho 10,8 gam A tac dung vCra du vdi V2 lit dung dich NaOH 3M thu dufgc a gam hSn hgp muoi. Tinh V2 va a. Bai 3. Thiiy phan hoan toan este A cua mot axit hufu co don chufc va mot ancol don chufc bang lufgng dung dich NaOH vifa du. Lam bay hoi hoan toan dung dich sau thiiy phan. Phan hoi do dugc dan qua binh 1 dgng CUSO4 khan di/, hoi con lai dugc ngUng tu het vao binh 2 dgng Na da thay CO khi G bay ra. DSn khi G qua binh dgng CuO dii, nung nong thi 6,4 gam Cu di/gc giai phong. Lugng este ban dau tac dung vdi dung dich brom dif thi co 32 gam brom tham gia phan ufng. Brom chiem 65,04% khoi lurgng phan tuf san pham sau khi cong hgp vao A. Hay: a) Xac dinh cong thufc phan tijf va cong thufc cau tao ciia A. b) Hoan thanh so do phan ufng: ^ trung htfp ^ g + NaOH ^ Q _I_ Bai 4. Dun 20,4 garri mot hgp chat hufu co A don chufc vdri 300 ml dung dich NaOH IM thu durgc muoi B va hgp chat hOu co C. C tac dung vdri Na dU cho 2,24 lit H2 (dktg). Biet rang khi nung muo'i B vdi NaOH thu dirge khi K CO ti khoi do'i vdi oxi bang 0,5. C la mot hgp chat don chufc khi hi oxi hoa bang khong khi tren Cu nong tao ra san pham D khSng phan ufng vdi dung dich AgNOa trong NH3. 16 - - • ' : • Bdl DI/SNG HOA HOC12 a) xac dinh cong thufc cau tao cua B, C, A va D. b) Sau phan ufng giffa A va dung dich NaOH thu dirge F. C6 can F dtfgc 1 cha't r^n. Tinh khoi lugng cha't r^n nay. c) Them vao 10,2,gam A mot cha't G don chufc cung chufc hoa hoc vdi A vdi so mol no = 0,5nA. Do't chay h5n hgp A, G thu dirge 33 gam CO2 va 12,6 gam H2O. Xac dinh cong thufc phan ttf va cong thufc cau tao bi§'t rkng khi dun G vdi dung dich NaOH ta thu difgc muoi B tren va m6t san pham co phan ufng vdi AgNOa/NHs. gai 5- Cho vao binh kin co dung tich 500 ml 2,64 gam mot este A roi dem nung binh den 273°C, toan bg este hoa hoi thi ap sua't b^ng 1,792 atfti. a) Xac dinh cong thufc phan tuf cua A. Tinh nong do mol ciia dung dich NaOH can thiet de thuy phan het lirgng este noi tren biet rang the tich dung dich NaOH la 50 ml. b) xac dinh cong thufc cau tao cua A va tinh khoi lifgng muoi thu dirge sau phan ufng (vdi hieu suat 100%) trong cac triTomg hgp sau: - San pham thu dirge sau phan ufng la h8n hgp 2 muoi va 1 rirgu. - San pham thu dirge la 1 muoi va 2 rifgu la dong d4ng lien tiep. Bai 6. Mot hon hgp gom hai este don chufc, co 3 nguyen to C, H, O. Lay 0,25 mol este nay phan ufng vdi 250 ml dung dich NaOH 2M dun nong thi thu dirge mot andehit no mach hd vk 28,6 gam hai muoi hufu cO. Cho biet kho'i lugng muoi nay hkng 4,4655 Ian khoi lugng muoi kia. De phan ufng het vdri NaOH con dU can dung 150 ml dung dich HCl IM, phan tram kho'i lugng cua oxi trong* andehit la 27,58%. Xac dinh cong thufc ca'u tao eua hai este. ' \ Bai 7. Mot h6n hgp 2 este don chufc dugc nung nong vqri -mot lugng NaOH viTa dij tao ra h6n hgp dong dang lien tiep va hon Hgp muo'ji. a) Dot chay 2 rUgu thu duge CO2 va hoi H2O theo ti le the tich 7 : 10. Tim cong thUe phan tuf va thanh phan phan tram theo s6' mol c^e rUgu trong hon hgp. b) Cho 2 muoi tac duijg vdi lugng H2SO4 viTa dii dugc hon hop 2 axit no. La'y 2,08 gam h5n hgp 2 axit (nguyen cha't) cho vao 100ml dung dich Na2C03 IM. Sau phan ufng lugng Na2C03 du tac djjng viraMij vdi''85 ml dung dich HCl 2M. Xac dinh cong thufc phan tuf eua 2 axit Va 2 este big't r^ng khi dot m6i este deu thu dugc mot the tich khi CO2 nho hon 6 Ian the tich hai este do d cung dieu kien t° va P. Bai 8. A va B la 2 chat hUu co don chufc eo ciing cong thUc phan tuf. Khi dot chay hoan toan 10,2 gam h6n hgp 2 chat nay^ean 14,56 lit O2 (dktc), khi CO2 va hoi nude jbae-^featth-reo-the- tichaxhu nliavrtdo «f cung dieu kien). BfiiDi/aNGH0AHgci2 THI/VIEN TI41NH THUAN ,7 Mat khdc khi cho A, B tac dung vdi dung dich NaOH ngUffi ta tha'y: - A tao dtrgc muo'i cua axit hOfu ca C va riigu D. Ti kho'i hoi cua C so vdi H2 Ik 30. Cho hoi rtrou D di qua Qng dUng Cu dun nong dtfcfc chat E khong tham gia phan uTng trang jgiiong. - B tao ra dirge chat C va D'. Khi cho C tac dung vdi H2SO4 di^oc E' tham gia phan ijfng trang gifong con khi D' tac dung vdi H2SO4 dac d t° thich hdp thi thu diidc 2 anken. Xac dinh cong thufc cau tao cua 2 chat A, B. ai 9. Mot hdp chat hiJu cd X mach hd chi chufa 1 loai nhom chufc difdc di4u che tii axit no A va rUdu no B. Biet: a) Khi dot chay a gam X thu difdc n^^ = n^j ^ + 0,2 mol, a gam X khi hda hdi chiem mot the tich bang the tich cua 5,8 gam khong khi d cung dieu kien (lay MKK = 29). b) Khi dot chay 1 mol riidu no B can 2,5 mol O2. c) a gam X tac dung vdi dung dich NaOH vCfa du tao ra 32,8 gam muoi khan. Hay cho biet cong thufc cau tao cua X. ai 10. Cho A la este cua glixerol vdi axit cacboxylic ddn chufc mach hd. Dun nong 7,9 gam A vdi NaOH cho tdi phan ufng hoan toan thu dtfdc 8,6 gam h6n hdp muo'i. Cho h6n hdp muoi tac dung vdi H2SO4 dii diidc hon hdp 3 axit X, Y, Z trong do X, Y la dong phan ciia nhau, Z la dong dang ke tiep cua Y. Lay mot phan hon hop axit do dem dot chay va cho CO2 thu difoc tac dung vdi dung dich Ba(0H)2 dii cd 2,561 gam ke't tua. a) Tim cong thi'jfc phan tuf va viet cong thufc cau tao cd the c6 cua A biet Z CO mach cacbon khong phan nhanh. b) ,Tinh khoi liidng hon hdp axit da bi dot chay. ai 11. Cho 5,7 gam hdn hop 2 este ddn chufc, mach hd dong phan cua nhau tac dung_vdi 50ml dung dich NaOH. Dun nhe, gia suf phan ufng xay ra hoan toan. De trung hoa liJdng NaOH d\i can 50ml dung dich H2SO4 0,5M ta thu dddc dung dich D. a) Tinh tong so' mol 2 este trong 5,7 gam hdn hdp biet rang de trung hoa 10 ml dung dich NaOH can 30 ml dung dich H2SO4 0,25M. b) ChiJng cat D diTOc h6n hop 2 rUdu c6 so' nguyen tuf cacbon trong phan tuf bang nhau. Hon hop 2 mau lam mat mau 6,4 gam Br2 trong dung dich. Ne'u cho Na tac dyng vdi h6n hdp 2 riidu thu difdc x lit H2 (dktc). Co can phan con lai sau khi chitog cat D roi cho tAc dung vdi H2SO4 thu dirge hSn hgp 2 axit. H6n hdp nay lam mat mau dung dich chufa y gam Br2. Tim cong thufc phan tuf cua cac este. c) Tinh X, y va kho'i lifgng mdi rqgu. } Bdi DirflNG HOA HOC 12 ai 12. A Ik axit hOfu ca mach th&ng, B \k ancol dcm chufc bac 1 c6 nh6nh. Khi trung hoa hoan toan A thi so mol NaOH can trung hoa gap doi so mol A. Khi dd't chay B tao ra CO2 vk H2O c6 ti le s6' mol tqong ufng Ik 4 : 5. Khi cho 0,1 mol A tac dung vdi B hieu suat 73,5 % thu ddgc 14,847 gam chat hufu CO E. a) Viet cong thufc cau tao cija A, B, E vagpi ten. b) Tinh khoi liTgng ciia A, B da phan ufng de tao ra lirgng chat E nhq tren. ai 13. Lay 100ml dung dich chufa 2 este A, B don chufc c6 nong do mol chung la 0,8M. Cho hon hgp nay tac dung vdi 150ml dung dich NaOH IM. Sau phan ufng thu dirge 2 san pham hflu cd la 2 muoi C, D c6 kho'i M 41 lirong la 10,46 gam (Ti le 2 phan tuf khoi —- = —) dong thdi thu dirge Mp 65 1 rirgu E CO khoi lirgng la 2,9 gam. Rirgu nay khong ben bien thanh andehit. Sau phan ufng phai diing 200ml dung dich HCl 0,2M de trung hoa NaOH dir. Xac dinh cong thufc phan tuf va cong thufc cau tao cua A va B. Bai 14. Dot chay 17,6 gam mot hon hgp 3 chat A, B, C dgn chufc la dong phan va cho san pham chay Ian lirgt qua binh 1 dirng P2O5, binh 2 dirng KOH dit thi khoi lirgng binh 1 tang 14,4 gam va binh 2 tang 35,2 gam. a) Xac dinh cong thufc phan tuf va cong thufc cau tao c6 the cd cua A, B, C biet r^ng ca 3 chat deu mach hd. b) Lay 17,6 gam hon hgp A, B, C va chia ra lam 2 phan bang nhau. Phdn 1: Bi trung hoa bdi 0,5 lit dung dich NaOH 0,1M d nhiet dp thirdng {phan ling thuc hien trong thai gian ngdn). Phan 2: Tac dung vtra du vdi 1 lit dung dich NaOH 0,1M {dun nong mot thai gian de phan ilng xdy ra hoan toan). Sau khi c6 can dirge chat ran D. Hgi E dugc lam ngirng tu va sau khi loai het nirdc chufa trong E con lai mot chat long c6 khoi lirgng la 2,58 gam. Xac dinh cong thufc cau tao dung cua A, B, C va thanh phan phan tram hon hgp theo khoi lirgng. c) Them NaOH dir vao chat rSn D va nung hon hgp nay dirge m6t hon hgp khi F. Tinh ti khoi cua F doi vdi H2. Bai 15. Mot hon hgp X gom 2 este A, B dong phan {khong chica chicc hoa hoc ndo khdc ngodi chdc este). Dot chay hoan toan X can 2,8 lit O2 (dktc). Cho ha'p thu toan bg san pham chay vao trong dung dich Ca(0H)2 diT tao thanh 10 gam ke't tua va kho'i lirgng dung dich giam 3,8 gam. a) Xac dinh cong thufc phan tuf va cac cong thufc cau tao cd the cd cua A va B. B6| DI/SNG HOA HOC 12 [...]... Neu cho 0,171 gam C12H22O11 + C a ( 0 H ) 2 > C12H22O11.CaO.H2O chat dUofng do tac dung vdi dung dich a x i t clohidric, r o i cho san pham C12H22O11.CaO.H2O + CO2 > C12H22O11 + C a C O g ^ + H2O glucozcf t h u difgc tac dung vcfi dung dich AgNOg t r o n g amoniac t h i t h u di/oc C12H22O11 + H2O 0,108 gam bac H a y cho biet cong thijfc p h a n ttf va cong thtfc cau tao cua > C6H12O6 + (Glucoza) •... chijfa 2 chat hOfu ccf la CH3COONH4 va CH2(COONH4)2 K h i dien phan d anot ta c6 phan ijfng oxi hoa 2 anion OOC - CH„ - 2e COO = 0 ,125 (mol) A 36 Kiem chtJng: a + b + c = 0 ,125 + 0,25 + 0 ,125 = 0,5 = — (phti hgp) /z 0 ,125 mol C H 2 = C H - C 0 0 H 0,25 mol H - C 0 0 C H = C H 2 0 ,125 mol C H O - C H 2 - C H O -> 4CO2 t + C2H4T 0,5c HSn h g p k h i F gom 2 c h a t hufu ccf CgHe ( - • C H 2 = C H - C 0 0 N... a c6 12x + y + 16z = 132 +) C h o z = 2 => 12x + y = 100 X +) C h o z = 4 5 6 7 8 40 28 16 4 C7H16O2 vi C7H16 thuoc C T T Q CnHgn ^ 4 5 (mol) ^ hOp cha't no 6 y 20 8 am 5CO2 + 5H2O C 5 H 8 O 4 thuoc C T T Q CnHan + 2 - 4O4 CO 2 li§n k e t n +) Cho z = 6 => 12x + y = 36 0,5 n^^^ = 0,75 - 0.5 = 0,25 ( m o l ) n ^ ^ ^ , ^ , = 0,7 - 0,5 = 0,2 DUONGlHdA HOCI2 Bdl ( m o ) 2 3 y = 0,7 2 12x + y = 68 X 12 0... Jai 10 P h a n ufng: CgHizOe 7,5 (CeHioOs),, + nHaO rnAg = 0 ,125 x 108 = 13,5 (gam) ) X X = 0,81 X 88n Bai 12 G o i x \k so m o l xenlulozcf: ^^ = 0 ,125 nAg = 4.nsaccaroz a 17,1% X 2nC2H50H + 2nC02 ^ 88n fructoza C a glucozcf vk fructozcf deu t r a n g bac => nAg = 2.(a + a) = 4a iA 162n C6H12O6 + C6H12O6 glucozcf , nC,U,,0, SnCO^ > (gam) Saccarozcf Khol ^^nraenruau... tao t h a n h diaxetat t h a n h t r i a x e t a t xenlulozo T i n h k h d i C l 2 H 2 2 0 i i + H2O lirgng dirge "'-^^ Mantoza ) C6H12O6 + C6H12O6 glucoza fructoza > 2C6H12O6 glucoza V a y cha't d u d n g c a n t i m la saccarozO Bdi OI/SNG H6A HOC 12 I DUflNG H6A HOC 12 59 B a i 2 a ) T a c 6 : mtinh bot = 1 t a n 6 5 % = 0 , 6 5 t a n = 6 5 0 k g (C6Hio05)„ + n H a O nCeHiaOe > kg 162n kg 180n 650kg... ™H20= 0 ,12 0,11 0 ,125 mo 22 mH " mc/o,o855 g a m x = 22 0 ,125 tham gia phan ufng t r a n g bac, san pham la glucozo C H 2 0 H [ C H O H ] 4 C H O + 2[Ag(NH3)2]OH > C H 2 0 H [ C H O H ] 4 C O O N H 4 + 2 A g i + SNHgt + H 2 O 180 0,9 216 Ta CO t i le Cong thufc cho glucozcr va fructozo Ket luan nay phu hop vdi duf k i e n de bai cho > C6H12O6 fructozcf (gam) (gam) 342 m ^ddCa(0H)2ban dlu 180 + C6H12O6... n ta c6 x = 12; y = 22 mH > Ca(HC03)2 (gam) Tong k h o i li^gng C O 2 : 0,044 + 0,088 = 0,132 (gam) 0,01 Theo de b a i : CaCOg^ + H 2 O + C O g t 162 (gam) K h o i liiong glucozcr h i len men = 0,75 x 180 = 135 (g) (moi) 0,1 Ca(HC03)2 Theo (1) so moi CeHiaOe = 0,75 (moi) CxHyO, 100 = 342 (gam) ^ 0 C6ng thufc phan tuf 0 012 = 0,0 012 (moi) M^^ „ o , = 342 ^ n = 1 *'- '12" 22'-'ll'n ciia X: C12H22O11 = ^... chufa 5 0 % x e n l u l o z o N e u m u o n d i e u che' 1 t a n ancol e t y l i c j 12; y = 22; z = 11 h i e u suat q u ^ t r i n h l a 70%, t h i k h o i l i / g n g n g u y e n l i e u m a n h a m a y V a y cong thufc p h a n tuf: C12H22O11 can d u n g l a bao n h i e u ? Bdl Ol/dNG HdA HOC 12 Bfil DU8NG H 6 A H O C 12 [ 11 T i n h k h o ' i l u g n g glucozcf d e m l e n m e n , b i e t rkng pham... u n d i c h nao c h u y e n m a u quy t i m t h a n h do l a a x i t axetic Sau do, ~ mo mjj 16z ,, 342 X 0 , 0 1 1 = 11; y = ^ 0,171 _ M m „^ = 22; 342 x 0,072 X = ^_ = 12 12x0,171 C o n g thufc p h a n tuf ciia dtfcfng: C12H22O11 C12H22O11 l a saccarozcf hoac m a n t o z o k h i tac d u n g v d i d u n g d i c h H C l - M a u thijf t a o d u n g d i c h m a u x a n h l a glucoza d u n n o n g v a... i c h t r o n g o n g n g h i e m k i a V a so m o l dtfcfng: k h o n g p h a n ufng l a saccarozof M I S ^ O N G HOA HOC 12 = i x 2 342 108 = 0,0005 ( m o l ) = 0,0005 (mol) = ngi^cozo 53 11 P h ^ n ufng: X 6 t phdn lifng thuy p h ^ n : CO2 + Ca(OH)2 ^^—^ > C6H12O6 + C6H12O6 C12H22O11 + H2O Saccarozd glucoza (mol) fructoza 5,5 Mantozcf (mol) glucozcf 2 V a y c h a t d u d n g c a n t i m l a saccarozcf . DLfOtNG HOA HOC 12 so: 1L-323DH2 012 2000 cuon, kho 16 X 24cm. T^i CTy TNHH MTV in diicyng sat Sai Gon. I chi: 136/lA TRAN PHU, P.4, Q.5, TP.HCM xuat ban: 1446-2 012/ CXB/07-231/DHQGHN. noi doi C=0. +) Cho z = 4 12x + y = 68 X 4 5 6 y 20 8 am C5H8O4 thuoc CTTQ CnHan + 2 - 4O4 CO 2 li§n ket n +) Cho z = 6 => 12x + y = 36 X 2 3 y 12 0 Loai tri/c;ng hcfp nay. x x I DUdNG HOA HOC 12 13 2 Do Heste = — => X = 1 3 3 3 K = "CH3COOC2H5" H2O" K = CH3COOH C2H5OH 73 V V V V J V = 4 Bai 12. Ta c6: n^,^^^„, 46