Bồi dưỡng kiến thức Hóa học 12: Phần 1

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Tài liệu Bồi dưỡng Hóa học 12 được biên soạn bám sát chương trình sách giáo khoa hiện hành và theo từng chương trong chương trình sách giáo khoa, mỗi chương gồm 3 phần chính: kiến thức cần nhớ, bài tập áp dụng, bài tập nâng cao có lời giải. Mời các bạn tham khảo phần 1 tài liệu.

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Nội dung Text: Bồi dưỡng kiến thức Hóa học 12: Phần 1

^ H C V N H 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^ X U ^ T B f I N Dfil HOC QUOC G i a H f i NQI &i noi ddu 16 H a n g Chuoi - H a i Ba TrtCng - H a N o i B i e n t h o a i : B i e n t a p - Che ban: (04) 39714896 N h a m g i u p cac e m h o c s i n h c6 t h e m W l i e u d e t i e p H a n h c h i n h : (04) 39714899: T o n g B i e n t a p : (04) 39714897 c a n , n a m v i l n g h e t h o n g k i e n thufc cof b a n , n a n g cao v a r e n Fax: (04) 39714899 l u y e n k i n a n g g i a i b a i t a p t h e o y e n c a u d o i mdfi. C h u n g t o i x i n t r a n t r g n g g i d i t h i e u d e n cac e m h o c s i n h v a cac b a n d o n g n g h i e p bo s a c h " B O I Dl/OfNG H O A H O C 1 2 " Chiu trdch nhi^m xuat ban: N o i d u n g s a c h " B O I DTJCfNG H O A H O C 1 2 " difc/c b i e n Gidm doc - Tong bien tap: TS. P H A M T H I T R A M s o a n b a m sat chiicfng t r i n h s a c h g i a o k h o a h i e n h a n h v a t h e o t i i n g chvfdng ijfng v d i t i i n g chu:dng t r o n g s a c h g i a o k h o a . T r o n g m o i c h i f d n g difgfc t r i n h b a y g o m 3 p h a n c h l n h : A . K I E N T H L / C C A N NHOf B. B A I T A P A P D U N G Bien tap: QUOC THANG C . B A I T A P N A N G C A O ( C O lofi giai) Sufa b a i : N H A SACH SAO MAI Cac b a i t a p a p dung v a n a n g cao difcfc g i a i c h i t i e t , I d i Che ban: TlJCfNG V Y g i d i p h u h d p v d i m o i do'i t i f d n g h o c s i n h , n h a m g i u p cac e m hoc s i n h l a m q u e n v a k h a c s a u k i e n thiJc t h o n g q u a cac b a i Trinh bay bia: TUCfNG LINK t o a n v a phufdng p h a p g i a i t i f n g b a i t o a n . Q u y e n sach l a t i i l i e u g i i i p cac e m h o c s i n h I d p 12 n a n g Doi tdc lien ket xudt ban: cao n a n g l\ic t\i h o c d n h a c u n g nhu! r e n l u y e n d e t h a m g i a NHA SACH SAOMAI vao cac d o i t u y e n h o c s i n h g i o i cac c a p . V a l a t a i l i e u t h a m k h a o cho cac g i a o v i e n t h a m g i a g i a n g d a y d cac tru:dng p h o thong v a giao v i e n t h a m gia b o i ditdng hoc sinh gioi. SACH LIEN K E T Mac d u d a co' g a n g t r o n g q u a t r i n h b i e n s o a n song I DLfOtNG H O A HOC 1 2 so: 1 L - 3 2 3 D H 2 0 1 2 k h o n g t h e t r a n h k h o i nhGng t h i e u sot n g o a i y m u o n . T a c 2000 cuon, kho 16 X 24cm. T ^ i CTy T N H H M T V i n diicyng sat Sai Gon. g i a i x i n c h a n t h a n h c a m d n cac y k i e n d o n g g o p , x a y d i i n g I chi: 1 3 6 / l A T R A N P H U , P.4, Q.5, T P . H C M tijf p h i a b a n d o c d e I a n t a i b a n s a u c u o n s a c h cd c h a t l i i d n g xuat b a n : 1446-2012/CXB/07-231/DHQGHN tot hdn. ^et d i n h xuat b a n so: 3 2 8 L K - T N / Q D - N X B DHQGHN Tdc gid song va nop lu^u chieu quy I n a m 2013.
CHl/dNG I . ESTE - LIPIT A. KIEN THLfC C A N N H 6 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
b) Este cua phenol RCOOR' + NaOH > RCOONa + R'OH Tif halogenua axit va phenolat: CH3COOC2HS + NaOH —> CHjCOONa + CzHsOH RCOCl + NaOCoHs > RCOOCeH^ + NaCl Tuy nhien, mot so triidng hap ngoai muoi cua axit cacboxylic khong - Tir anhidrit axit vd phenol: tgo ra ancol ma tgo thanh andehit, xeton, muoi cua phenol, hogc chi (CH3CO)20 + HOCeHs > CHsCOOCgHs + CH3COOH tgo thanh mot sdn phdm duy nhdt. 11. LIPIT. CH,COOCH=CH2 + NaOH —> CHjCOONa + CH,-CH=0 1. Khdi niem va cd'u tgo - Chat beo (nguon goc dong vat, thitc vat] la este cua glixerol vdi axit CHsCOOCiCHshCH^ + NaOH —> CHjCOONa + (CHjJzC =0 beo (axit hHu ca mot idn axit mgch thdng, khdi liigng phdn tit Idn). Cdc chat beo nay diigc goi chung Id triglixerit. C H 3 C O O — y + 2NaOH —> CHsCOONa + ^ ^^ONa Cong thdc tong qudt cua chat beo: CH2—OCOR1 ONa CH—OCOR2 .0 + NaOH > C CH2—OCOR3 ^OH Trong do: Ru R2, Rj cd the gio'ng nhau hogc khdc nhau. b) Phdn thig khvT +) Mot so' axit beo thitdng gap: R-COO-R' —> R-CH2OH + R'OH Axit panmitic : C15H31COOH Axit stearic : C^HssCOOH Axit oleic : CiyH33COOH Axit linoleic : CMiCOOH c) Phan ihig cgng d goc hidrocacbon: Phdn ling cgng vd trtmg hap +] Thitdng gap cdc triglixerit pha tgp (R^ ^ R^ ^R^) khi goc R chiia lien ket n. - Trong chat beo, ngoai este cua glixerol vdi axit beo con cd mot COOCH3 COOCH3 liigng nhd axit d dgng tiJ do diigc dgc tritng hdi chi so axit. CH,—C - Chi so axit cua mot chat beo la so miligam KOH can thiet de trung n H2C=:C -^-^"-P ) hda axit tit do cd trong 1 gam chat bdo. CH, 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. CH2=C(CH3)COOCH3 + H2 ——> CHs-CHiCHaJCOOCHs ' - Chi so este cua mot chat beo la so miligam KOH can thiet de thuy d) Phdn iftig dot chay: Dot chdy este no dofn chv^c: 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 CnH2n02 + ( ^ ^ ^ ) O2 > /jCO^ + / i H ^ O 2 de trung hda axit tii do vd thuy phdn hoan todn litgng este cd trong Chu y: Ngoai cdc phdn ring tren, rieng este cua axit fomic, con c6 1 gam chat beo. phdn iJtng trdng giiang giong nhii andehit. 2 . Tinh chat hda hoc a) Phdn vtng thuy phdn 5. Dieu che - Trong mdi trifdng nitdc hogc axit: a) Este cua ancol Chat beo khong bi thuy phdn bdi nitdc, khi cd xuc tdc axit hogc enzim - Thiic hien phdn ling este hda: chat beo bi thuy phdn thugn nghich cho glixerol vd axit beo: RCOOH + R'OH < "^^"^'^ z> RCOOR' + HOH CH2—OCOR1 CH2-OH R1COOH - W muoi vd ddn xudt halogen cua hidrocacbon CH—OCOR2 + 3H2O < ^ CH-OH + R2COOH RCOOAg + R'Cl > RCOOR' + AgCU CH2—OCOR3 CH2-OH R3COOH BO'I oadNG HdA HOC 1 2 B(5i DI/SNG HOA HQC 12 7
- Trong moi triidng kiem (phan ling xa phdng hoa) pMn itng thuy 5- E^ot chdy ho^n toan 7,4 gam este X dcfn chufc t h u di/ac 6,72 l i t k h i phan xay ra hoan toan: CO2 (dktc) va 5,4 gam nU6c. CH2—OCOR1 CH2-OH RiCOONa a) Xdc dinh cong thufc phan tuf cua X. C H — O C O R 2 + 3H2O < = = = ^ = = ± CH-OH + R2C00Na b) Dun 7,4 gam X trong dung dich NaOH vCfa du den k h i phan ufng ho^n toan thu diicrc 3,2 gam ancol va mot lugng muoi Z. Viet cong thufc CH2—OCOR3 CH2-OH RgCOONa cau tao cua X vd t i n h k h o i lufcfng cua Z. triglixerit ghxerol xd phdng Bai 6. Dot chay hoan toan 1,76 gam mot este X t h u diTOc 3,52 gam CO2 va b) Phan vtng cong hidro: Bien glixerit chifa no thanh glixerit no 1,44 gam H2O. Xdc dinh cong thufc phan tijf cua X? (CM^COOJ^CsH, + 3H2 '° > (C.yHssCOOhCsHs Bai 7. E la este ciia mot axit don chufc va ancol don chufc. De thuy phan chat long chat ran hoan toan 6,6 gam chat E phai dung 34,1 m l dung dich NaOH 10% [ I I . XA P H O N G vA CHAT G I A T R L T A T O N G HOfP (d = 1,1 g/ml). Lirgng NaOH nay dung di/ 25% so vdri lufong NaOH phan 1. Xa phdng la hdn hap cac muoi Na hoac muoi cua axit beo va mot iJng. Hay de xuat cong thufc cau tao dung cua E? so chat phu gia. Bai 8. De xa phong hoa 17,4 gam mot este no don chufc c^n dung 300ml 2. De san xuat xa phdng ngifdri ta c6 the: dung dich NaOH 0,5M. T i m cong thufc phan tuf cua este dem dung. - Dun nong chat beo vdi dung dich kiem. Bai 9. 0,05 mol este E phan ufng vi/a du 100 gam dung dich NaOH 6%, ta - Oxi hoa ankan (parafin) cua ddu mo nhd oxi khong khi, d nhiet do thu difcfc 10,2 gam muoi va 4,6 gam rxiau. Biet rufou hoac axit tao cao CO muoi mangan xiic tdc roi trung hoa axit bhng NaOH hoQc KOH. t h a n h E don chufc. Hay xac dinh cong thufc cau tao cua E. 3. Chat giqt rvta tong hap la nhffng chat khong phdi Id muoi Na hoac Bai 10. K h i thiic hien phan ufng este h6a 1 mol C H 3 C O O H va 1 mol K cua axit beo nhiing c6 tinh nang gigt riia nhiixd phdng. 2 4. Chat gidt rtia tong hap difcfc sdn xuat ti( cdc sdn phdm cua dau mo. C2H5OH, Itfong este t h u dufoc lorn nhat la - mol. De dat hieu suat CLTC 5. Xd phdng hi gidm tdc dung gigt rita trong niidc cvtng con chat gidt tj tdy riia tong hap khong bi gidm tdc dung gigt rifa. dai la 90% ( t i n h theo axit) k h i t i e n hdnh este hoa 1 mol CH3-COOH 6. Xd phdng cd liu diem Id khong 1dm hgi da, it gay 6 nhiSm moi t h i can so mol C2H5OH la bao nhieu? (biet cac phan ufng este hoa thiTc triidng. Chat gigt riia tong hap gay hgi da do trong do cd chat tdy hien d cung nhiet do). trdng, gay 6 nhiem moi triidng. Bai 11. T r o n 1 mol axit axetic vdi 1 mol rUgu etylic. K h i s6' mol cac chat BAI TAP AP DUNG trong hon hop khong thay ddi nufa, nhan thay lUOng este t h u dUOc \k - mol. T i n h h k n g so can bkng (K). lai 1. Ho^n t h ^ n h cdc phiicfng t r i n h phan ufng theo scf do sau: 3 Bai 12. T h u y p h a n hoan todn 444 gam m6t l i p i t t h u difoc 46 gam CH4 )A — ^ B —i51_>C — > E + B glixerol (glixerin) va hai loai axit beo. Xdc d i n h cong thufc cua hai lai 2. Viet phifcfng t r i n h phan ufng thifc hien day bien hda sau (viet cdc loai axit beo do. chat d\i6i dang cong thufc cau tao): Bai 13. Hay viet phi^ong t r i n h phan ufng cua chat beo c6 cong thufc phan C5H10O ^ CsHioBr^O ^ CsHsBra ^ C5H12O3 tuf nhtf sau: ^WCnHisOe CsHsOsNa CH4 CH2-OCO-(CH2X4CH3 Biet C 5 H 1 0 O la mot ancol bac ba. C H - OCO - ( C H 2 - CH = C H ( C H 2 ) , C H 3 ai 3. Viet cong thufc cau tao cac dong phan dcfn chufc, mach hor c6 the c6 CH2 - OCO - ( C H 2 \H = CH - C H 2 - C H - C H ( C H 2 )^ C H 3 cua C4H6O2. ai 4. Viet cong thufc cau tao cdc chat c6 ten sau day: a) Vdri dung dich K O H d nhiet do cao. a) Isopropyl axetat b) Alylmetacrylat b) Vdri I2 C O dU. c) Phenyl axetat d) sec-Butyl fomiat c) Vdri H2 du, c6 N i xiic tdc, d nhiet do vd dp suat cao. BO'I DI/ONG HdA HQC 12 Bfii DUONG H6A HOC 12 . ^
tiii 14. K h i thiiy ph4n a g a m m6t este X thu dUcfc 0,92 gam glixerol, 3,02 gam^ m u o i l i n o l e a t C i T H s i C O O N a vk m gam n a t r i o l e a t CnHaaCOONa. Tinh (4)H2C—CH—C—CH3 + 3CH3COOH -> C H 3 C O O - C H + 3H2O gia t r i cua a, m. V i e t c o n g thijfc cau tao c6 the c6 ciia X. OH OH OH CH3COO—C(CH3)2 B a i 15. a ) De t r u n g h5a liicfng axit beo tif do c6 trong 14 gam mot mSu chat beo can 15ml dung dich K O H 0,1M. Hay cho biet chi so' axit cua CH3COO—CH2 mau chat beo t r e n . (5) C H 3 C O O - C H + 3NaOH ^ H 2 C — C H — C - C H 3 + 3CH3COONa b) T i n h kho'i lacfng NaOH can t h i e t de trung h6a 10 gam m p t chfi't b6o • OH OH OH CH3COO-C(CH3)2 c6 c h i so' axit 1^ 5,6. B a i 16. H a y t i n h chi so xk phong h6a cua m 6 t chS't b6o, b i e t rkng k h i C a O , t° - > C H 4 + NaaCOg (6) CH3COONa + N a O H x^ ph6ng hoa hoan t o a n 1,5 gam ch§.'t b^o d6 can 50 m l dung d i c h B a i 3. C 4 H 6 O 2 c6 A = 2 v a h a i n g u y e n tuf o x i . K O H 0,1M. => d o n g p h a n e s t e dcfn chuTc, k h o n g n o c6 m o t n o i d o i d gdc dong B a i 17. Hay t i n h khoi lucfng NaOH can de trung hoa axit tii do c6 trong p h a n a x i t c a c b o x y l i c dcfn chufc k h o n g n o m o t n d i d o i d g d c . 5 gam chat beo v6i chi so axit bang 7. - DSng phdn este: HC00CH=CH-CH3 ; HCOOCH2-CH=CH2 B a i 18. Hay t i n h chi so' iot cua triolein. CH3COOCH=CH2 ; CH2=CHCOOCH3 Hlf6NG DAN GIAI f - Bong phdn axit cacboxylic: CH2=CH-CH2-COOH B a i 1. Phan ufng: CH3-CH=CH-C00H tach n h a n h . ^ i m k n h n h a n h ^ ^aHg + SHg C H 2 - C - COOH I C 3 H 2 H- H , 0 ) CH.CHO CH, 1 ^ Mn^* B a i 4. CH3CHO + ^ 0 2 — > CH3COOH b) C H 2 = C - C O O C H 2 - C H = C H 2 a ) CH3COOCH(CH3)2 CH3COOH + C H - C H > CH3COOC=CH2 CH3 CH3COOC=CH2 + NaOH > CH3COONa + CH3CHO C) CH3COOC6H5 d) HCOO-CH(CH3)CH2-CH3 B a i 2. Phan ufng: B a i 5. a ) Ta c6: n^o, = = 0,3 (mol) va n^^^ = M = o,3 (mol). CH3 CH3 ' 22,4 ' 18 V i k h i ddt chay X thu dirge sd mol H 2 O bang sd mol C O 2 nen X Ik este (1) H 2 C = C H — C - O H + Br2 > H2C-CH—C-OH no, dcfn chufc. CH3 Br Br CH3 Goi cong thiJfc cua este no, dcfn chufc \k : C n H 2 n 0 2 (n > 2) ^•^3 CH3 (3n-2] CnH2n02 + O2 nCOz + nHaO (D Phan lifng: (2) H ^ C - C H - C - B r . H B r H ^ C - C H - i - B r . H2O Br Br CH3 Br Br CHI 3 0,3 < - 0,3 (mol) I n CH3 CH3 0,3 Theo d l bai, ta c6: Mx = — (14n + 32) = 7,4 n = 3. (3) H 2 C - C H — C - B r + SNaOH > H2C-CH—C-CHgi- 3 N a B r n I I I I I I Br Br CH3 OH OH OH Vay cong thufc phan tijf cua X: C3H6O2. 11 10 BO'I DUflNG H6A HOC 12 B6| DirdNG HOA H0C12
b ) X&c d i n h c o n g thijfc cau t a o cua X kho'i l u a n g ciia Z: B a i 9 . T a c6: nNaOH = ^'^^^ = 0,15 ( m o l ) = Sneste => este 3 chufc. 100.40 T a c6: n x = — = 0 , 1 (mol). T h e o de b a i , se c6 1 t r o n g 2 c h a t l a dcfn chufc n e n c6 2 truTcfng h o p : 74 Triidng h^p 1: a x i t d o n chufc rLfcfu b a chufc P h a n ufng: RCOOR' + N a O H > RCOONa + R'OH. (RCOOsR' + 3NaOH > 3 R C 0 0 N a + R'(OH)3 (mol) 0,1 - > 0,1 0,1 0,1 (mol) 0,05 -> 0,15 0,15 0,05 M ^ : IHR-OH = 0,1(R' + 17) = 3,2 R' = 15: C H g - 10 2 Vay c o n g thufc cau t a o d u n g cua X : CH3COOCH3. T a c6: MRcooNa = ^ r - ^ = 6 8 ( d v C ) R + 6 7 = 6 8 => R = 1 ( l a H - ) V a k h o i l u g n g cua Z: 0,1 x 8 2 = 8,2 (gam). B a i 6. T a c6: nco^ = = 0.08 ( m o l ) ; nn^o = = 0.08 ( m o l ) MR(OH) = = ^ 2 (dvC) ^ R' + 5 1 = 9 2 R' = 4 1 ^ R' 1^ C 3 H 5 - y /x 0,05 D o n^Og = i^HgO X c6 do b a t bao h 6 a cua p h a n tuf A = 1 V a y c o n g thufc cau t a o l a ( H C O O g C g H s . => X 1^ este no, dofn chufc => X d a n g C n H 2 n 0 2 Trildng hcfp 2: a x i t d a chufc v a rifgfu dcfn chufc. ( R C O O R')3 + 3 N a O H > R(C00Na)3 + 3R'0H C„H2n02 ) nC02 (mol) 0,05^ 0,15 . 0,05 0,15 nco2 0,08 1,76 . 4 6 ^nx= = - — => M x = 1 4 n + 3 2 = ^ n = 4 ^ = n »c o = 30,67: k h o n g nguyen (loai) n n 0,08 U, Uo.o n V a y c o n g thufc cau t a o ciia este ( E ) l a : (HCOO)3C3H5 V a y c o n g thufc p h a n tuf cua X : C4H8O2. B a i 10. C a n b a n g : m . 3 4 , 1 X 1,1 X 10 0 , CH3COOH + C , H , O H ^ ^ " ^ CH3C00C,Hs + H^O Bai 7. T a c6: m ^ ^ o H d e m d t a g = ^QQ = 3.751 (gam) Ban ddu: IM IM 0 . 0 2 2 2 2 3,751 X 100 ^ , Phan ling: - M - M - M - M ^mOU phan .ng = (lOQ + 2 5 ) ^ ^ ^^^""^ 3 3 3 o I 3 Can 6^71^.- (1 - 0,9)M (a - 0 , 9 ) M 0,9M 0,9M M a t khdc: IIE = nNaOH = — = 0,075 ( m o l ) /o^2 40 => M E = 8 8 ( g a m ) o R + 4 4 + R ' = 88 =o R + R ' = 4 4 v3, Cho V = 1 l i t K = = 4 - I Q i i R = 1 = > R ' = 43(C3H,) => C T C T ( E ) : H C O O C 3 H 7 ( p r o p y l f o m i a t ) R C o i a l a so m o l C g H ^ O H c a n dCing, t a c6: - K h i R = 15 ^ R ' = 2 9 =^ C T C T ( E ) : CH3COOC2H5 ( e t y l a x e t a t ) Kc = j ^ ^ ^ ^ - 4 = > a = 2,925 (mol). B a i 8. G p i c o n g thufc cua este no, d o n chufc 1^ C^HanOg. a i 1 1 . G o i t h e t i c h h e p h a n ufng V Git) Khi p h b n g h 6 a t h i : n^^^^ = ^naou C H 3 C O O H + C2H5OH ^ CH3COOC2H5 + H 2 O n e . t e = 0 , 3 X 0 , 5 = 0 , 1 5 ( m o l ) => M^^,^ = ^ = ^16 Ban ddu: 1 1 14n + 32 = 1 1 6 = > n = 6 v a y c o n g thufc p h a n tuf cua este 1^ C^H^Pz • 13 2 BO'I Di/dNG HdA Hnr PhanHOA Sail I DUdNG phan ling: HOC 12iCng: 1x -x 1 x- x x x
2 1 => ne„„33cooNa = 0 , 0 2 (mol) => m,_^„^^,„„,^ = 0 , 0 2 x 3 0 4 = 6 , 0 8 (gam) X = Do Heste = — => 3 3 3 0 92 M^: nNaOH = 3ngiixeroi = 3 X = 0,03 ( m o l ) . 73 V V => = 0,03 X 40 = 1,2 (gam). mNaOH "CH3COOC2H5" H2O" K = = 4 Ap dung d i n h luat bao t o ^ n khoi luong, ta c6: CH3COOH C2H5OH V V = 3,02 + 6,08 + 0,92 - 1,2 = 8,82 (gam). V J Bai 15. a ) Theo dinh nghia: chi so axit cua chat beo la so miligam K O H 46 B a i 12. Ta c6: n^,^^^„, = 0,5 (mol) can dung de trung hoa het cac axit b^o t i i do c6 trong 1 gam chat beo. 92 Ta c6: mKon = 0.015 x 0,1 x 56000 = 84 (mg) Goi cong thijfc cua l i p i t c6 dang: C3H5(OCOR)3 v6i R = 84 3 ^ Chi so axit la: — = 6. Phan ijfng: 14 C3H5(OCOR)3+3H2O: ^C3H5(OH)3+3RC00H b) Chi so axit la 5,6 nghia la de trung hoa 1 gam chat beo can 5,6mg (mol) 0,5 - 0,5 KOH. Vay trung h5a 10 gam chat beo cdn 5,6mg x 10 = 56mg K O H . Hay = 0,001 mol K O H Theo de b ^ i , ta c6: •^56 Vi NaOH la bazo don chiifc nhtf K O H nen can so mol bang nhau trong "ihpit = 0 ' 5 ( 4 1 + 132 + 3 R ) = 444 R = 238,333 = (*) o phan ufng trung hoa. Do vay so gam NaOH can c6 la: 40 x 0,001 = 0,04 (gam) NaOH Ma: „ =239; „ =237 Bai 16. So mol K O H : 0,050 l i t x 0,1 mol/lit = 0,005 mol K O H va M _ „ =213; „ =211 So gam K O H : 0,005 mol x 56 g/mol = 0,280g hay 280mg K O H Ket hop vdi (*) Cap nghiem thich hop: C17H35- va CnHag-. 280 Chi so xa phong hoa: = 186,67. B a i 13. 1,5 CH2-OCO-(CH2)^,CH3 Bai 17. Theo dinh nghia, chi so axit cua chat beo bkng 7 c6 nghia la muo'n trung hoa lucfng axit beo t i i do trong 1 gam chat beo phai dung 7mg CH - O C O ( C H a ) ^ - C H = C H ( C H 2 ) 7 CH3 KOH. Vay muon trung hoa axit beo i\i do trong 5 gam chat beo c6 chi CH2 - OCO - ( C H 2 ) , CH = CH - CH2 - C H = CH (CH2 \3 so 7 t h i phai dung 5 x 7 = 35mg K O H , hay ^ ^ m o l K O H Chat beo t r e n Ik trieste cua glixerol vdi ba axit panmitic, axit oleic vk axit linoleic. ^ M ^ m o l O H " => ^ ^ ^ m o i NaOH ^ k h o l lifong NaOH c^n d l a) K h i cho tac dung vdi dung dich K O H d nhiet do cao se xay ra phan 56 56 LJfng xa phong hoa triglixerit va ba muoi kali cua ba axit beo tren. trung hoa axit i\i do trong 5 gam chat beo c6 chi so axit bkng 7 la: b) K h i cho tac dung vdi h c6 dtf t h i c6 phan ufng cong I2 vao cac noi niNaOH = >^ 40g = 25 (mg) = 0,025g/5g chat beo. 56 doi trong goc axit beo khong no. Bai 18. Phan ufng: c ) K h i cho tac dung vdri H2 diT, xiic tac N i , t°, p t h i c6 phan iJng cong (C,,H33COO)3C3H5 + 3I2 (C,,H33COOl2)3C3H, vao noi doi trong g6c axit beo khong no. B a i 14. So d6: X + NaOH ^ CnHsiCOONa + CnHssCOONa + C3H5(OH)3 (gam) 884 3 x 254 3,02 Ta c6: n•Ci,H3,COONa = 0,01 (mol) ^ Chi so' iot la i ^ ^ - ^ x 100 = 86,2. . 302 884 14 BO'I OUONG H6A HOC 12 Bfii DUONG HOA HQC 12 IS
C . BAI TAP N A N G C A O a ) x a c d i n h c o n g thufc cau t a o cua B , C, A v a D . B a i 1 . D e t h u y p h a n h o a n t o a n 0,74 g a m m o t h 6 n h g p este dcfn chufc car, b ) Sau p h a n ufng giffa A v a d u n g d i c h N a O H t h u dirge F . C6 c a n F dtfgc 7 g a m d u n g d i c h K O H 8 % t r o n g nude. K h i d u n n o n g h 6 n h g p este noj 1 cha't r ^ n . T i n h k h o i l u g n g cha't r ^ n n a y . t r e n v d i a x i t H2SO4 8 0 % s i n h r a k h i X . L a m l a n h X , diia ve d i e u k i e ^ c ) T h e m v a o 1 0 , 2 , g a m A m o t cha't G d o n chufc c u n g chufc h o a hoc v d i A t h u d n g v a d e m c a n , sau do cho k h i I g i tii tii q u a d u n g d i c h b r o m dy v d i so m o l n o = 0,5nA. Do't c h a y h 5 n h g p A , G t h u dirge 33 g a m CO2 v a t r o n g nifdrc t h i t h a y k h o i l i f g n g k h i g i a m 1/3, t r o n g do k h o i l i f g n g r i e n g 12,6 g a m H2O. X a c d i n h cong thufc p h a n t t f v a c o n g thufc cau t a o bi§'t cua k h i g a n nhu k h o n g d o i . | r k n g k h i d u n G v d i d u n g d i c h N a O H t a t h u difgc m u o i B t r e n v a m 6 t a ) T i n h k h o i liTgng m o l cua h o n h g p este, xac d i n h t h a n h p h a n h 6 n hgp s a n p h a m co p h a n ufng v d i A g N O a / N H s . k h i sau k h i da l a m l a n h v a t i n h k h o i l u g n g cua c h u n g . g a i 5- C h o v a o b i n h k i n co d u n g t i c h 5 0 0 m l 2,64 g a m m o t este A r o i d e m b ) X a c d i n h t h a n h p h a n h o n h g p este b a n dau. n u n g b i n h d e n 2 7 3 ° C , t o a n bg este h o a h o i t h i ap sua't b ^ n g 1,792 a t f t i . c ) N e u p h a n ufng de p h a n b i e t 2 este t r e n , v i e t phtfong t r i n h p h a n ufng. a ) X a c d i n h c o n g thufc p h a n tuf cua A . T i n h n o n g do m o l ciia d u n g d i c h B a i 2. L a m b a y h o i m o t c h a t hufu co A (chufa cac n g u y e n t o C, H , O) diJgc N a O H c a n t h i e t de t h u y p h a n h e t l i r g n g este n o i t r e n b i e t r a n g t h e t i c h c h a t h o i CO t i kho'i d o i v d i m e t a n b a n g 13,5. L a y 10,8 g a m c h a t A va d u n g d i c h N a O H l a 50 m l . 19,2 g a m O2 cho v a o b i n h k i n , d u n g t i c h 25,6 l i t ( k h o n g doi). D o t chay b ) x a c d i n h c o n g thufc cau t a o cua A v a t i n h k h o i l i f g n g m u o i t h u dirge h o a n t o a n A , sau do giuf n h i e t do b i n h 163,8°C t h i a p suat t r o n g b i n h sau p h a n ufng ( v d i h i e u suat 100%) t r o n g cac triTomg h g p sau: b a n g 1,26 a t m . L a y t o a n bg s a n p h a m chay cho v a o 160 g a m d u n g dich - S a n p h a m t h u dirge sau p h a n ufng l a h 8 n h g p 2 m u o i v a 1 rirgu. N a O H 1 5 % dirge d u n g d i c h B c6 chijfa 4 1 , 1 g a m h o n h g p h a i m u o i . K h i - S a n p h a m t h u dirge l a 1 m u o i v a 2 rifgu l a d o n g d 4 n g l i e n t i e p . r a k h o i d u n g d i c h c6 t h e t i c h V i l i t ( d k t c ) . B a i 6. M o t h o n h g p g o m h a i este d o n chufc, co 3 n g u y e n t o C, H , O. L a y a ) X a c d i n h c o n g thCfc p h a n tii, v i e t c o n g thufc cau t a o cija A ( b i e t r k n g 0,25 m o l este n a y p h a n ufng v d i 250 m l d u n g d i c h N a O H 2 M d u n n o n g k h i cho A t a c d u n g v d i k i e m t a o r a m o t ancol v a 3 m u o i ) . t h i t h u dirge m o t a n d e h i t n o m a c h hd vk 28,6 g a m h a i m u o i hufu cO. Cho b ) T i n h V i v a C % ciia cac c h a t t r o n g d u n g d i c h B . b i e t kho'i l u g n g m u o i n a y hkng 4,4655 I a n k h o i l u g n g m u o i k i a . D e c ) C h o 10,8 g a m A t a c d u n g vCra d u v d i V2 l i t d u n g d i c h N a O H 3 M t h u p h a n ufng h e t vdri N a O H c o n d U c a n d u n g 150 m l d u n g d i c h H C l I M , dufgc a g a m h S n h g p m u o i . T i n h V2 v a a. p h a n t r a m k h o ' i l u g n g cua o x i trong* a n d e h i t l a 2 7 , 5 8 % . X a c d i n h cong B a i 3. T h i i y p h a n h o a n t o a n este A cua m o t a x i t hufu co d o n chufc v a m o t thufc ca'u t a o eua h a i este. ' \ ancol d o n chufc b a n g lufgng d u n g d i c h N a O H v i f a d u . L a m b a y h o i h o a n B a i 7. M o t h 6 n h g p 2 este d o n chufc dugc n u n g n o n g vqri -mot l u g n g N a O H t o a n d u n g d i c h sau t h i i y p h a n . P h a n h o i do dugc d a n q u a b i n h 1 d g n g viTa dij t a o r a h 6 n h g p d o n g d a n g l i e n t i e p v a h o n Hgp muo'ji. C U S O 4 k h a n d i / , h o i c o n l a i dugc ngUng t u h e t v a o b i n h 2 d g n g N a d a a ) D o t c h a y 2 rUgu t h u duge CO2 v a h o i H2O t h e o t i l e t h e t i c h 7 : 10. t h a y CO k h i G b a y r a . D S n k h i G qua b i n h d g n g C u O dii, n u n g n o n g t h i T i m c o n g thUe p h a n tuf v a t h a n h p h a n p h a n t r a m t h e o s6' m o l c^e rUgu 6,4 g a m C u di/gc g i a i p h o n g . L u g n g este b a n d a u t a c d u n g v d i d u n g d i c h b r o m dif t h i co 3 2 g a m b r o m t h a m g i a p h a n ufng. B r o m c h i e m 6 5 , 0 4 % trong hon hgp. k h o i lurgng p h a n tuf s a n p h a m sau k h i c o n g h g p v a o A . H a y : b ) Cho 2 m u o i t a c d u i j g v d i l u g n g H 2 S O 4 viTa d i i dugc h o n h o p 2 a x i t no. a ) X a c d i n h c o n g thufc p h a n tijf v a cong thufc cau t a o ciia A . La'y 2,08 g a m h 5 n h g p 2 a x i t ( n g u y e n cha't) cho v a o 1 0 0 m l d u n g d i c h Na2C03 I M . Sau p h a n ufng l u g n g Na2C03 d u t a c d j j n g v i r a M i j vdi''85 m l b ) H o a n t h a n h so do p h a n ufng: ^ trung htfp ^ g + NaOH ^ Q _I_ dung dich H C l 2 M . Xac d i n h c o n g thufc p h a n tuf eua 2 a x i t V a 2 este big't r ^ n g k h i d o t m 6 i B a i 4. D u n 20,4 garri m o t h g p c h a t hufu co A d o n chufc vdri 3 0 0 m l d u n g d i c h este d e u t h u dugc m o t t h e t i c h k h i CO2 n h o h o n 6 I a n t h e tich hai este N a O H I M t h u durgc m u o i B v a h g p c h a t hOu co C. C t a c d u n g vdri N a dU cho 2,24 l i t H2 ( d k t g ) . B i e t r a n g k h i n u n g muo'i B v d i N a O H t h u dirge do d c u n g d i e u k i e n t ° v a P. k h i K CO t i k h o i do'i v d i o x i b a n g 0,5. C l a m o t h g p c h a t d o n chufc k h i h i B a i 8. A v a B l a 2 c h a t hUu co d o n chufc eo c i i n g cong thUc p h a n tuf. K h i d o t o x i h o a b a n g k h o n g k h i t r e n Cu n o n g t a o r a s a n p h a m D k h S n g p h a n c h a y h o a n t o a n 10,2 g a m h 6 n h g p 2 c h a t nay^ean 14,56 l i t O2 ( d k t c ) , k h i ufng v d i d u n g d i c h AgNOa t r o n g N H 3 . CO2 v a h o i nude jbae-^featth-reo-the- t i c h a x h u n l i a v r t d o «f cung d i e u k i e n ) . 16 - - • .. ' : • Bdl DI/SNG HOA HOC12 BfiiDi/aNGH0AHgci2 T H I / V I E N T I 4 1 N H THUAN ,7
M a t khdc k h i cho A, B tac dung vdi dung dich NaOH ngUffi ta tha'y: ai 12. A Ik axit hOfu ca mach th&ng, B \k ancol dcm chufc bac 1 c6 nh6nh. K h i - A tao dtrgc muo'i cua axit hOfu ca C va riigu D. T i kho'i hoi cua C so vdi trung hoa hoan toan A t h i so mol NaOH can trung hoa gap doi so mol A. H2 Ik 30. Cho hoi rtrou D di qua Qng dUng Cu dun nong dtfcfc chat E K h i dd't chay B tao ra CO2 vk H2O c6 t i le s6' mol tqong ufng Ik 4 : 5. khong t h a m gia phan uTng t r a n g jgiiong. K h i cho 0,1 mol A tac dung vdi B hieu suat 73,5 % thu ddgc 14,847 gam - B tao ra dirge chat C va D'. K h i cho C tac dung vdi H2SO4 di^oc E' chat hufu CO E. tham gia phan ijfng t r a n g gifong con k h i D' tac dung vdi H2SO4 dac d t° a) Viet cong thufc cau tao cija A, B, E v a g p i ten. thich hdp t h i thu diidc 2 anken. b) Tinh khoi liTgng ciia A, B da phan ufng de tao ra lirgng chat E nhq tren. Xac dinh cong thufc cau tao cua 2 chat A, B. ai 13. Lay 100ml dung dich chufa 2 este A, B don chufc c6 nong do mol ai 9. M o t hdp chat hiJu cd X mach hd chi chufa 1 loai nhom chufc difdc di4u chung la 0,8M. Cho hon hgp nay tac dung v d i 150ml dung dich NaOH che tii axit no A va rUdu no B. Biet: I M . Sau p h a n ufng thu dirge 2 s a n p h a m hflu cd la 2 muoi C, D c6 kho'i a) K h i dot chay a gam X thu difdc n^^ = n^j ^ + 0,2 mol, a gam X k h i M 41 hda hdi chiem mot the tich bang the tich cua 5,8 gam khong k h i d cung lirong la 10,46 gam (Ti le 2 phan tuf khoi — - = — ) dong t h d i thu dirge dieu k i e n (lay MKK = 29). Mp 65 1 rirgu E CO k h o i lirgng la 2,9 gam. Rirgu nay khong ben bien thanh b) K h i dot chay 1 mol riidu no B can 2,5 mol O2. andehit. Sau p h a n ufng phai diing 200ml dung dich H C l 0,2M de trung c) a gam X tac dung vdi dung dich NaOH vCfa du tao ra 32,8 gam muoi hoa NaOH dir. Xac dinh cong thufc p h a n tuf va cong thufc cau tao cua A khan. Hay cho biet cong thufc cau tao cua X. va B. ai 10. Cho A l a este cua glixerol vdi axit cacboxylic ddn chufc mach hd. Dun nong 7,9 gam A vdi NaOH cho t d i phan ufng hoan toan thu dtfdc Bai 14. Dot chay 17,6 gam mot hon hgp 3 chat A, B, C dgn chufc la dong 8,6 gam h6n hdp muo'i. phan va cho s a n pham chay Ian lirgt qua binh 1 dirng P2O5, binh 2 dirng Cho h6n hdp muoi tac dung vdi H2SO4 dii diidc hon hdp 3 axit X, Y, Z K O H dit t h i khoi lirgng binh 1 tang 14,4 gam va b i n h 2 tang 35,2 gam. trong do X, Y la dong phan ciia nhau, Z la dong dang ke tiep cua Y. a) Xac d i n h cong thufc p h a n tuf va cong thufc cau tao c6 the cd cua A, B, Lay mot phan hon hop axit do dem dot chay va cho CO2 thu difoc tac C biet r ^ n g ca 3 chat deu mach hd. dung v d i dung dich Ba(0H)2 dii cd 2,561 gam ke't tua. b) Lay 17,6 gam hon hgp A, B, C va chia ra lam 2 p h a n bang nhau. a) T i m cong thi'jfc phan tuf v a viet cong thufc cau tao cd the c6 cua A biet Phdn 1: B i trung hoa bdi 0,5 l i t dung dich NaOH 0,1M d nhiet dp Z CO mach cacbon khong phan nhanh. thirdng {phan ling thuc hien trong thai gian ngdn). b) ,Tinh khoi liidng hon hdp axit da bi dot chay. Phan 2: Tac dung vtra du vdi 1 l i t dung dich NaOH 0,1M {dun nong mot ai 11. Cho 5,7 gam hdn hop 2 este ddn chufc, mach hd dong phan cua thai gian de phan ilng xdy ra hoan toan). Sau k h i c6 can dirge chat ran D. nhau tac dung_vdi 50ml dung dich NaOH. Dun nhe, gia suf phan ufng Hgi E dugc lam ngirng tu va sau k h i loai het nirdc chufa trong E con lai xay r a hoan toan. De trung hoa liJdng NaOH d\i can 50ml dung dich mot chat long c6 khoi lirgng la 2,58 gam. Xac dinh cong thufc cau tao H2SO4 0,5M t a thu dddc dung dich D. dung cua A, B, C va thanh phan phan tram hon hgp theo khoi lirgng. a) T i n h tong so' mol 2 este trong 5,7 gam hdn hdp biet rang de trung c) Them NaOH dir vao chat rSn D va nung hon hgp nay dirge m6t hon hoa 10 m l dung dich NaOH can 30 m l dung dich H2SO4 0,25M. hgp k h i F. T i n h t i khoi cua F doi vdi H2. b) ChiJng cat D diTOc h6n hop 2 rUdu c6 so' nguyen tuf cacbon trong phan Bai 15. Mot hon hgp X gom 2 este A, B dong phan {khong chica chicc hoa hoc tuf bang nhau. H o n 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 l i t H2 (dktc). ndo khdc ngodi chdc este). Dot chay hoan toan X can 2,8 l i t O2 (dktc). Co can phan con l a i sau k h i chitog cat D r o i cho tAc dung vdi H2SO4 thu Cho ha'p thu toan bg s a n p h a m chay vao trong dung dich Ca(0H)2 diT dirge hSn hgp 2 axit. H6n hdp nay lam mat mau dung dich chufa y gam tao t h a n h 10 gam ke't tua va kho'i lirgng dung dich giam 3,8 gam. Br2. T i m cong thufc phan tuf cua cac este. a) Xac dinh cong thufc p h a n tuf va cac cong thufc cau tao cd the cd cua A c) T i n h X, y va kho'i lifgng mdi rqgu. va B. } Bdi DirflNG HOA HOC 12 B6| DI/SNG HOA HOC 12
b ) G o i Y , Z I k 2 d 6 n g p h d n m a c h hd c6 c u n g s6' nguy§n tuf C v k O v6i A, NhiX v a y cA 2 I j h a n S n g diu c6 1 este f o m a t , k h i d u n n 6 n g v d i H2SO4 b i B n h u n g i t hcfn 2 H . Cho h 5 n horp p h a n drng v6i N a O H difcfc 2 m u d i v a p h a n h u y t a o r a C O ( M = 28), n g o k i r a con m o t k h i b i h a p t h u b d i nude • rUcfu. Cac a x i t t a o r a 2 muo'i n a y l a 2 d o n g d ^ n g ke" t i e p v a c6 m a c h b r o m , k h i d6 p h a i Ik a n k e n s i n h r a k h i p h a n ancol t r o n g este b i tkch cacbon k h o n g p h a n n h a n h . Cho 2 a x i t n a y p h a n uTng v d i 6 0 0 m l d u n g ntfdc. M a t k h a c , k h o i Itfgng r i e n g h 6 n h o p k h i k h o p g d o i , tufc l a k h i d6 d i c h Na2C03 I M ( a x i t t h e m t i f tU) t h i k h o n g c6 CO2 bay r a . Sau p h a n p h a i c6 p h a n tuf k h o i = 28, do 1^' C2H4. ijfng de tac d u n g h e t v d i l u g n g cacbonat p h a i d u n g 8 0 0 m l d u n g d i c h H C l I M . D o t chay h e t 2 a x i t t h u dirge 32,48 l i t CO2 ( d k t c ) . C2H4 + Bra — — ^ C2H4Br2 Xdc d i n h c o n g thufc p h a n tuf v a cong thufc cau tao cua 2 a x i t . N e u t r o n g h 6 n h o p c6 H - C O O - C 2 H 5 > C O + C2H4 + H2O T i n h t h a n h p h a n p h a n t r S m theo k h o i Itfgng cua h 6 n hcfp Y , Z. t h i sau k h i d i q u a n i f d c b r o m k h o ' i l i i o n g k h i p h a i g i a m d i 1/2 (trdi 1 1 6 . M o t h 6 n h o p X g o m 2 c h a t hufu ccf A , B k h o n g tac d u n g vdi dung gia t h i e t ) . d i c h Br2 v a deu tAc d u n g v d i d u n g d i c h N a O H . T i k h o i cua X do'i v d i H2 V a y ckc gdc H C O O - v a C2H5- p h a i thuoc ve 2 este k h d c n h a u . b a n g 35,6. H 8 n h o p chufa H - C O O - C H 3 (x m o l ) v ^ R - C O O - C 2 H 5 (y m o l ) . Cho X tac d u n g h o a n t o a n vdfi d u n g d i c h N a O H t h i t h a y p h a i d i i n g T a CO : x + y = 0,01 ; x = 2y (do C O = 2 x C2H4 ) 4 g a m N a O H , p h a n ufng cho t a m o t r u g u don chiifc v a 2 m u o l ctia a x i t y = 0,01/3 v a X = 0,02/3 hufu cof dcfn chufc. N e u cho t o a n bo li/cfng ri/cfu t h u di/gfc tac d u n g v d i N a d a CO 6 7 2 m l k h i ( d k t c ) t h o a t r a . T a c 6 : 6 0 x ^ + (R + 73) x Ml o,74 3 3 Xac d i n h c o n g thufc p h a n tijf v a cong thufc cau t a o cija A , B. => R = 2 9 ^ C o n g thufc p h a n tuf 1^ : C 2 H 5 - C O O - C 2 H 5 , 1 7 . M o t h6n h o p X g o m 3 d6ng p h a n A , B , C m a c h h d deu chuTa C, H , K h o i l u g n g h d n h o p k h i sau p h a n ufng v d i H2SO4 Ik : 0 . B i e t 4 g a m h 6 n hcfp X d 136,5°C va 2 a t m c6 c u n g t h e t i c h v d i 3 g a m 28 X 0,01 = 0,28 (gam) p e n t a n d 273°C v a 2 a t m . => 0,02/3 X 60 = 0,4 (gam) H C O O - C H 3 54,1% a ) Xac d i n h c o n g thufc p h a n tuf cua A , B , C. b) Cho 36 g a m h o n hap tac d u n g viTa dij v d i d u n g d i c h N a O H c6 chufa 0,34 (gam) C2H5-COO-C2H5 45,9% m g a m N a O H . Co can d u n g d i c h diicfc c h a t r a n Y v a h 6 n h o p Z. Z tac P h a n b i e t 2 este b a n g p h a n ufng v d i d u n g d i c h A g N 0 3 t r o n g NH3 : d u n g vira d i i v d i d u n g d i c h AgNOs/NHa t a o r a 108 g a m A g v a d u n g d i c h H C O O - C H 3 + 2Ag(Nli3); + 2H2O >(NH4)2C03 + 2 N H ; Z' chufa 2 c h a t hufu ccf. D i e n p h a n d u n g d i c h Z' v d i d i e n chiic t r o , c6 + C H 3 O H + 2Ag>i. m a n g ng&n dugc h5n hcfp k h i F b e n a n o t . N u n g c h a t r d n Y v d i N a O H Bki 2. a ) Theo de b a i , t a c6: M A = 13,5 x 16 = 216 (dvG) diT dagc h6n h,ap k h i G. D u n G v d i N i xiic tac dUgfc h 6 n h o p k h i F ' g o m 2 k h i CO so m o l b k n g n h a u . Taco: n^ = — ^ = 0,05 ( m o l ) ; = = 0,6 ( m o l ) T r g n Ikn F v d i F ' r o i cho qua ni/dc Br2 dif t h i k h o i li/cfng d u n g d i c h t a n g l e n 1,75 g a m . K h i t r o n I a n F , F ' cac chat k h o n g c6 p h a n ufng v d i n h a u . n^o^ = 0,45 (mol) 1 . X^c d i n h c o n g thufc cau t a o cua A , B , C, b i e t r S n g m o i c h a t c h i chufa => So m o l h o n hop sau p h a n ufng chky \k 0,6 m o l . m o t l o a i n h o m chufc. K h i cho CO2 v^o d u n g d i c h N a O H 2. T i n h t h a n h p h a n p h a n t r a m cua A , B , C t r o n g hSrt h o p X. CO2 + N a O H > NaHCOg 3. T i n h m ( k h o i l i r g n g N a O H ) . HlfCfNG DAN GIAI CO2 + 2 N a O H — ^ NaaCOs + H2O 1. P h a n d n g : R-COO-R' + KOH > R - C O O K + R'OH Do't ch^y A : C^HyO, + (x + | - | ) 0 2 xCOa + |H20 => So m o l 2 este = 0,01 (mol) v ^ M = 74 (mol) 0,05 ^ 0,05x Co 2 k h a n S n g x a y r a : Ma: nj,o= o,05x = 0,45 X= 9 - Ca 2 este deu c6 p h a n tuf k h o i Ik 74 (HCOO-C2H5 v ^ CH3-COO-CH3) => C o n g thufc cua A di/gc v i e t l a i : CgHyO^ =i> y + 16z = 216 - 108 = 108 - M o t t r o n g h a i este c6 p h a n tuf k h o i < 74 do l a H C O O - C H 3 . ^ Cap n g h i e m p h i i hop: z = 6 v k y = 12 => C T P T ciia A 1 ^ C9H12O6 B6| Dl/dNG HOA HOC 1 2 BAI mfflMR uA« unr 4«
Theo de ra A tdc dung vdi NaOH cho ancol vk 3 muoi. Yky ancol nky c6 ba Ho'i bay ra gom ancol Ian hai nLfcfc. Dung C U S O 4 de ha'p thu hcfi nifdc, c6n nhom chufc - O H va la glixerol, ba axit khdc nhau c6 tong so' cacbon la 6, lai hoi ancol R'OH. Ngung tu hcfi ancol roi cho tac dung vdi Na du vay CO mot axit c6 1 nguyen tijf cacbon, mot axit c6 2 nguyen tCr cacbon va R'OH + Na -> R'ONa+ ^HaT (2) mot axit c6 3 nguyen ttf cacbon. 2 Cong thdfc cau tao cua A: H2C—OCOH K h i G la H2. Do tang khoi iLfgng binh dLfng Na: m^.^^ - m^^ = 6,4 (3) HC-OCOCH3 H2 + CuO — > Cu + H2O (4) H2C-OCOCH=CH2 6,4 (mol) 0,1 = 0,1 b)TinhVi: 64 Phanilng: CgHiaOe + 9O2 — > 9CO2 + 6H2O TCf (3) =^ mR.0H = 6,2 + 2 X 0,1 = 6,4 (gam) (mol) 0,05 9 X 0,05 9 x 0,05 6 x 0,05 TCf (2) => nR.oH = Sn^ = 0,2 (mol) ^ = 0,6 - 0,45 = 0,15 (mol) ^ V i = 0,15 x 22,4 = 3,36 (lit) 64 R' = 15 (CH,,-) Nong do phan trSm cua dung dich B: 0,2 ^ d u n g dich B ~ " ^ d u n g dich NaOH "^00^ "^H^O Cong thiJc phan tuf cua ancol la C H 3 O H = 160 + 0,45 X 44 + 0,3 x 18 = 185,2 (gam) Ta c6: nA = nancoi = 0,2 (mol) = n Br^ phan ilng A chufa 1 lien ket 71 (C=C) Phan Lfng: CO2 + NaOH > NaHCOg RCOOCH3 + Bra > RBr2COOCH3 (mol) X X X (mol) 0,2 -> 0,2 0,2 C02 + 2 N a O H - )• NasCOg + H2O 160 65,04 => % B r = R = 27 (C2H3-) (mol) y 2y y R + 219 100 x + y = 0,45 f x = 0,3 => C o n g thufc p h a n tLf cua A l a : C4H6O4 Theo de bai, ta c6 he: 84x + 106y = 41,1 |y = 0,15 Cong thurc cau tao cua este A : C H 2 = C H - C O O - C H 3 b ) P h a n ifng: Vay: C%,,„,,^ = X 100% = 13,6%; nH2C=:CH H2C—CH C%^ CO = ^ ' ^ ^ X 100% = 8,58% COOCH3 COOCH3 c) Phan Lfng: H2C—CH + nNaOH > nCHaCOONa C9H12O6 + 3 N a O H > C3H5(OH)3 COOCH3 - H 2 C — C H - (mol) 0,05 0,15 HCOONa + CHgCOONa + C H 2 = C H C 0 0 N a 0,05 0,05 0,05 I OH =^ Va = 0,05 (lit) B a i 4. a ) A l a e s t e v i k h i d u n A v d i d u n g d i c h N a O H t a t h u dLfcfc m u o i B v a =^ a = n^HCOONa + "lcH3COONa + "^C,H3C00Na = 12,2 (gam) 1 rifau C. a i 3. a ) Lap cong thufc phan tuf vk cong thufc cefu tao cua este A: RCOOR' + NaOH RCOONa + R'OH Goi cong thOfc phan tiuf ciia este A 1^: RCOOR' R'OH + Na R'ONa+ - H 2 Phan Lfng: RCOOR' + NaOH - — — > RCOONa + R'OH (1) 2 B6| DlJdNG HOA H0C12 Bfil DI/8NG H(5A HOC 12 23
2 24 Tac6: H A = nR.QH = 2njj = 2 x = 0,2 ( m o l ) CxHyOz + X + ^ - 1 O2 -> xCOa . |H20 , ' 22,4 4 Vay MA = ^ = 102 0,05y U, 2 (mol) 0,05 0,05x 2 K h i K l a 1 h i d r o c a c b o n c6 M K = 32.0,5 = 16. 0,05y Vay, K l a CH4 v a m u o i B 1 ^ C H s C O O N a do n CO, = 0,05x = 0,25 =:> X = 5 v a n^^ = = 0,20 y = 8 ^ay C T P T cua G l a C 5 H 8 O 2 thuoc C T T Q C„H2„ + 2 - 4O2 (co 2 l i e n k e t n CHgCOONa + N a O H CH4t + NaaCOg \k este k h o n g no). A : CH3COOR' ^ M A = 15 + 44 + R' = 102 Sir x a p h o n g h o a G cho r a m u o i C H s C O O N a . V a y G l a m 6 t a x e t a t mu R' 1 ^ CJiy =^ R' = 43 = 12x + y =^ N g h i e m h o p l i : x = 3, y = 7 r:> G 1^ ( C H 3 C O O ) 3 C 3 H 5 , g6c C 3 H 5 CO n d i doi C=C V a y R \k C3H7 c o n g thufc cua rifgu C l a C3H7OH San p h a m G l a 1 a n d e h i t (ttf 1 rifcfu (e m o l ) k h o n g b e n c h u y e n t h ^ n h ) Ydi C3H7OH t a CO 2 d o n g p h a n rtfcru V a y cong thufc cau t a o cua (G) l a C H 3 - C O O - C H = C H - C H 3 CH3-CH2-CH2OH C H 3 - C H 2 - C H O + H2O CH3-COO-CH=CH-CH3 + NaOH CHgCOONa + C H 3 - C H = C H 0 H Ri/gu n a y k h o n g b e n c h u y e n t h a n h a n d e h i t : C H 3 - C H 2 - C H O . CH3-CHOH-CH3 CH3-CO-CH3 + H2O Bat 5. a ) C o n g thufc p h a n tuf cua este A V i s d n p h a m o x i h o a D k h o n g cho p h a n ijfng vdri A g N 0 3 / N H 3 , D 1^ xeton Phiicfng phap: A c6 cong thijtc la CJiyO^ (Vdi z = 2, z ^ 4 (dieste), z = 6 (tri CH3-CO-CH3 vk rirgu B l a C H 3 - C H O H - C H 3 . este) v.v...) 4 an (x, y, z, nA) va 2 phiidng trinh (2,64 va so mol) thieu 2 C 6 n g thufc caiu t a o cua este A : C H 3 - C O O - C H ( C H 3 ) 2 (axetat i s o p r o p y l ) . phiiOng trinh nen sau khi c6 MA tif cho z = 2,4 tinh x, y, z, v.v... b ) C h a t r ^ n t h u di/cfc sau p h a n ufng PV 1,792 x 0,5 So' m o l este nA = = 0,02 ( m o l ) CH3COOC3H7 + N a O H > C H 3 C 0 0 N a + C3H7OH RT 22,4 X 2 X 273 273 (mol) 0.^ 0,2 0,2 9 fi4 => nNaOH ban diu = 0,3' X 1 = 0,3 (mol) Vay MA = =^ = 132 0,02 V a y dif 0,3 - 0,2 = 0,1 ( m o l ) N a O H . N e u A CO c o n g thufc p h a n tuf l a C^HyO, t a c6 12x + y + 16z = 132 Sau k h i c6 can d u n g d i c h F t a t h u dUgrc c h a t r ^ n g o m 0,1 m o l N a O H +) C h o z = 2 => 12x + y = 100 0,2 m o l C H g C O O N a (Ri/gu C3H7OH 1^ c h a t l o n g bay d i ) . X 5 6 7 8 => tCLrin = 0,1 X 40 + 0,2 X 82 = 20,4 (gam). y 40 28 16 4 c ) T a c6: H A = ^ =± 0,1 ( m o l ) ; no = 0,05 ( m o l ) L o a i c 6 n g thufc C7H16O2 vi C7H16 thuoc C T T Q CnHgn ^ 2 hOp cha't no 1U2 t r o n g k h i este p h a i c6 1 n o i d o i C = 0 . G c u n g l a 1 dcto este n e n t a c6 c6ng thufc l a CxHy02. +) C h o z = 4 12x + y = 68 T a t i n h n ^ o ^ n ^ ^ ^ d l suy r a x v& y X 4 5 6 C o n g thufc cua A v i e t g o n 1^ C5H10O2 y 20 8 am C5H10O2 + ^ 0 2 —> 5CO2 + 5H2O C 5 H 8 O 4 thuoc C T T Q CnHan + 2 - 4O4 CO 2 li§n k e t n (mol) 0,1 0,5 0,5 +) Cho z = 6 => 12x + y = 36 33 X 2 3 y 12 0 T a c6: n^^^ = —= 0,75 ( m o l ) ^ n^^^ = 0,75 - 0.5 = 0,25 ( m o l ) L o a i tri/c;ng hcfp nay V a y c h i c6n 1 nghiem C5H8O4 Va n„^o(A,G, = ^ = 0,7 (mol) ^ n ^ ^ ^ , ^ , = 0,7 - 0,5 = 0,2 ( m o lHdA Bdl DUONG ) HOCI2 Bfii DI/SNG H6A HOC 12 25
N6ng dp dung dich NaOH K h d i liiOng muo'i Este nay c6 2 chuTc este vay phan ufng vdi NaOH theo t i le mol 1 : 2 0,02 mol este A can 0,04 mol NaOH nnr—r\A COONa uuy ^2NaOH-> I + C2H5OH + CH3OH CM(NaOH) = ^ = 0,8M. oo(:-C2H5 ^oof^^ 0, 05 (mol) 0,02 0,02 b ) Cong thufc cau tao cua A => ^N.,c,o, = 0.02 X 134 = 2,68 (gam), Trifcfng i i t f p 1: San pham thu dirge gom 1 rirgu, 2 muoi t h i este phai phat xuat tCr 1 mgu 2 chiifc 2 axit khdc nhau pal 6. Ta c6: nNaOH = 0,5 (mol) va nHci = 0,15 (mol) RiCOOH, R2COOH, R3(OH)2 Phan iJng: NaOH + HCl > NaCl + H2O (mol) 0,15 0,15 Este A . ^ ^ R o HNaOH thay phan = 0,5 - 0,15 = 0,35 (mol) > 0,25 (mol) => M o t trong hai R2COO / mudi 1^ do hop chat phenol tao t h a n h tiep tuc b i trung hda bdi NaOH. Andehit no, mach hd xuat phat bdi este don chufc n e n cung don chufc Tdng so cacbon trong R i , R2, R3 la 5 - 2 = 3. Ri/Ou 2 chufc vay R3 t d i Cong thufc tdng qudt la: CnH2nO thieu phai c6 2 nguyen tuf cabon CH2OH-CH2OH: Ta c6: ^ = n = 3 (C3H6O) ^ CTCT: CH3-CH2-CHO R i , R2 con l a i 1 nguyen tuf cacbon vay R i = H (axit 1^ HCOOH) v ^ R2 la 14n 72,42 CH3- (axit la CH3COOH) Dat cong thufc tdng quAt cua 2 este la: Vay cong thufc cau tao cua este la: HCOO—CH2 R C 0 0 - C H = C H - C H 3 (a mol) va RCOO-R' (b mol) H3C—COO—CH2 => a + b = 0,25 (mol) (D Chii y: Loai triTomg hop R3 c6 3 cacbon vay R i , R2 khong c6 cacbon n^o RC00CH=CH-CH3 + NaOH ^ RCOONa + CH3-CH2-CHO ca, R i , R2 la H => chi c6 1 mudi la HCOONa (trai vdfi de) (mol) a a a Kho'i liiong 2 muo'i RCOOR' + NaOH > RCOONa + R'OH HCOO—CH2 (mol) b b b b , u n nr^r^ i u + 2NaOH -> HCOONa + CHgCOONa + (CH20H)2 R'OH + N a O H > R'ONa + H2O (chufa n h a n benzen) (mol) b b b (mol) 0,02 0,02 0,02 HNaOH = a + 2b = 0,35 (2) m 2 muoi = 0,02(68 + 82) = 3 (gam) TCr (1) va (2) suy ra: a = 0,15; b = 0,1 Trifcfng hcfp 2: San pham thu difOc gom 1 muo'i vk 2 rufOu. Vay este p h ^ t => m . = 0,25(R + 67) + (R' + 39).0,1 = 28,6 => 5R + 2R' = 159 (3) xuat tCr axit 2 chufc va 2 rirou khac nhau. muoi ' A x i t : Ri(C00H)2, RLTOU: R2OH, R3OH Xet hai triTcfng hop sau: Ri, R2 va R3 CO t a t ca 3 nguyen tuf cacbon R2, R3 phai c6 chufa cacbon +) Trifcfng hcfp 1: Neu mRcooNa = 4,4655 x mR-pNa vay chi c6 the R2 la CH3- va R3 la C2H5-. Vay Ri se kh6ng c6 cacbon => 0,25(R + 67) = 4,4655 x 0,1 x (R' + 30) => 2,391R' = R + 220,691 (4) vay axit la COOH Giai (3) va (4), ta difgc: R = 1 ( - H ) ; R' = 7 (CeHg-) . +; Trifcfng hap 2: Neu m ^ o N a = 4,4655 x mRcooNa => V6 nghiem COOH v a y CTCT cua 2 este: HC00-CH=CH-CH3 V ^ 2 rirgu \k CH3OH, C2H5OH do do A c6 c6ng thiJc cau tao: OOC—CH3 OOC—C2H5 B6| D I / S N G »6k HQC 12 _Bfl| D U S N G HOA HOC 1 2
+) P h a n ufng giufa 2 mu6'i vdri H 2 S O 4 Bai 7. a) X^c d i n h cong thuTc p h a n t t f cua 2 rUcfu G o i cong thufc cua 2 este A , B l a A : R i C O O R ' i (a m o l ) 2 R i C O O N a + H2SO4 -> 2 R 1 C O O H + Na2S04 B: RaCOOR'a (b m o l ) (mol) a a E s t e A , B d a n chufc n d n a x i t R i C O O H , R2COOH 2 r u a u R i ' O H , Rg'Oli 2 R 2 C O O N a + H2SO4 -> 2 R 2 C O O H + Na2S04 c u n g d a n chufc (mol) b b ^co, •• VH^O = 7 : 10 v a y n.^^ : n^^^ = 7 : 10 +) P h a n ufng giufa 2 axit vori Na2C03 "coj < "H^O "^^y g i o n g n h a trife/ng horp a n k a n 2 r i / a u R ' l O H R'20Ii 2 R 1 C O O H + NaaCOa -> 2 R i C 0 0 N a + CO2 + H2O a deu la r u g u no. (mol) 2 D l xAc d i n h c o n g thufc p h a n tuf cua 2 rifgu n a y t a d u n g 1 r u g u duy n h a t 2 R 2 C O O H + NaaCOg 2R2COONa + CO2 + H2O C-H - - O H de t i n h n n 2ii + 1 (mol) b ^ 2 a + b n N a j C O j p h a n ling (mol) (mol) (a + b) (a + b ) n ( n + l ) ( a + b) n N a , C O , ban dau ,„ = 0,1 X 1 (mol) '^H.o n + 1 10 - 7 „ „ „ => — — = - = — => n = - = 2,33 = > n < n < m = n + l ^cp, n 7 3 a + b '^Na,C03 da ~ 0,1 (mol) vay n = 2 R i ' O H m C2H5OH +) P h a n ufng giufa Na2C03 d u vdi H C l m = 3 =^ R'aOH 1& C3H7OH T h a n h p h a n % t h e o s6 m o l cua h 6 n h g p 2 r i i g u Na2C03 + 2HC1 > 2 N a C l + C 0 2 t + H2O 0,17 C2H5OH + 3O2 > 2CO2 + 3H2O nHci = 2 X 0,085 = 0,17 (mol) => nNa^COj du - = 0,085 (mol) (mol) a 2a 3a a + b a + b o 0,085 = 0,10 - = 0,015 ^ a + b = 0,03 C3H7OH + ^02 > 3CO2 + 4H2O V a a = 2b a = 0,02 ( m o l ) ; b = 0,01 (mol) (mol) b 3b 4b V a y c o n g thufc p h a n tuf cua 2 a x i t '^H,D 3a + 4b 10 a = 2b ma axit = 0 , 0 2 ( R i + 45) + 0,01(R2 + 45) = 2,08 n^^Q^ 2 a + 3b R2 + 2 R i = 73 V i 2 a x i t n a y deu no n e n Ri = CnH2n +1 v a R2 = CmHam + 1 vay: %C2H.0H = = = 66,67% a + b 3b V a y 2 ( 1 4 n + 1) + 14m + 1 = 73 o 2 8 n + 14m = 70 2n + m = 5 % C 3 H 7 0 H = 100% - 6 6 , 6 7 % = 3 3 , 3 3 % n 0 1 2 b ) Xac d i n h c o n g thufc p h a n tuf cua 2 a x i t m 5 3 1 RiCOOR'i + N a O H > RiCOONa + R'lQH Co 3 cap n g h i e m (mol) a a • n = 0 ^ H C O O H v a m = 5 => C g H n C O O H . n = 1=^ C H 3 C O O H va m = 3 C3H7COOH RsCOOR's'+NaOH ^^RgCOONa + R ' 2 0 H . n = 2 C 2 H 5 C O O H va m = 1 => C H 3 C O O H (mol) b b Bfil DUONG H6A HOC 12 B6\G HI^A Hnr 17
Vely cdc c o n g thufc c6 t h e c6 cua 2 este ijfng vdri m o i cfip n g h i § m t r e n Dg' c h o n c o n g thufc ca'u t a o d u n g cua A giufa 2 c o n g thufc n ^ y t a dUa t r e n H C O O C 2 H 5 va C5H11COOC3H7 c o n g thufc cau t a o cua 2 s a n p h a m oxi h d a cua rUOu D . CH3COOC2H5 v a C3H7COOC3H7 K h i x a p h o n g h o a este A , t a duoc vdri C3H7 m a c h t h ^ n g C2H5COOC2H5 v a CH3COOC3H7 CH3COOCH2CH2CH3 + N a O H - > CH3COONa + CH3CH2CH2OH K h i d o t chay mSi este CxHy02 t h i n^,^^ = xneste- V a y neu n^^^ < Sn^ste t h i Rifcfu E t h u dufoc l a rtfcfu bac 1 , c h a t n a y k h i b i o x i h o a t r e n C u n u n g X < 6, este chufa dvtdi 6 nguyen tijf cacbon. V a y c h i c6 cap n g h i e m cuoi n o n g cho a n d e h i t C2H5COOC2H5 v a CH3COOC3H" (so C = 5) l a t h o a m a n dieu k i e n t r e n . C H 3 C H 2 C H 2 O H + IO2 ^ — > C H 3 C H 2 C H O + H2O B a i 8. A chac c h a n l a este con B t a c d u n g vdfi N a O H t a o duoc 2 c h a t hOfu co C , D ' . V a y B c u n g l a este. A n d e h i t co p h a n t i n g t r a n g gifong, t r a i vofi de v a y l o a i trLfomg h o p n a y 1 K h i d o t c h a y A , B t a difgc n^^^ = n^^^^ +) N e u C3H7- p h a n n h a n h V a y cong thijfc cau t a o cua A , B deu chufa 1 l i e n k e t n (cua n h o m C = O). 1 H3C—COO—CH—CH3 + N a O H ^ C H 3 C 0 0 N a + C H 3 C H O H C H 3 V a y m a c h cacbon cua A , B k h o n g c6 l i e n k e t n, do do A , B l a este no, CH3 A x i t v a nfgfu t a o r a A , B c t i n g l a h o p c h a t n o . E l a rLfgru bac 2, k h i b i o x i hoa cho x e t o n +) X a c d i n h cong thu'c c u a A , B Phiidng phdp: Ta xdc dinh cong thu'c cdu tao cua axit C va rMu D, ti( CH3-CHOH-CH3 + IO2 ^ — > C H 3 - C O - C H 3 + H2O do viet cong thifc cdu tgo cua este. X e t o n k h o n g co p h a n ufng t r a n g giJofng, d u n g v d i de. V a y A co c o n g thufc T i k h d i d^n^ = 3 0 =^ M c = 2 x 3 0 = 6 0 ca'u t a o l a : H3C—COO—CH—CH3 Do C l a a x i t n o , C co c o n g thufc CnHgn + i - C O O H H a y 14n + 46 = 6 0 => n = 1 v a C l a C H 3 C O O H CH3 A , B CO c u n g c o n g thufc p h a n tiJf la CnH2n02 n e n k h i do't c h a y h 5 n h o p +) X a c d i n h cong thifc c a u tao ciia B A , B CO t h e x e m n h t f d o t c h a y 1 c h a t duy n h a t co c u n g c o n g thufc p h a n B k h i b i x a p h o n g h o a t a o r a C v a D'. D ' l a riJgu v i k h i b i khuf nifdfc (bcfi tuf vdri A , B H2SO4) cho r a a n k e n v a y C l a m u d i R C O O N a V(Ji H2SO4, muo'i n a y co p h a n ufng C„H2„02 + " O2 > nC02 + n H a O 2 R C 0 0 N a + H2SO4 > 2 R C 0 0 H + Na2S04 nA B = ' 10,2 14n + 32 . => n,, = (3n - 2 2) X 10,2 14n + 32 - 22,4 14,56 ^ ^_ = 0,65 A x i t R C O O H CO p h a n ufng t r a n g gifomg v a y a x i t n a y l a H C O O H B l a este cua H C O O H v a co cau t a o l a H C O O R , vcfi c o n g thuTc p h a n tii => n = 5 l a C5H10O2 t a t h a y ngay R l a C4H9 v a ri/cfu D ' l a C4H9OH. V a y A , B co c u n g c o n g thufc p h a n tuf l a C5H10O2. V d i C4H9OH t a CO 3 d o n g p h a n (rifcfu bac 1 , 2 v a 3) t a c h o n cong thufc +) X a c d i n h cong thu'c c a u tao c u a A ca'u t a o n a o m a k h i b i khuf nu'cfc t a t h u diicfc 2 a n k e n . A l a este cua C H 3 C O O H v a y A co cong thufc l a CH3COOR, C5H10O2 v a y H3C—CH2—CH—CH2 > C H 3 - C H 2 - C H = C H 2 + H2O R-C3H7.. H OH V a y A CO t h e co 2 cong thufc cau t a o t u y theo C3H7 m a c h t h ^ n g h a y C h i CO 1 d o n g p h a n a n k e n duy n h a t (loai) phan nhanh 1 2 3 4 CH3-COO-CH2-CH2-CH3 H3C—COO—CH—CH3 H2C—CH—CH—CH3 CH3 H OH H JO B6| Dl/flNG HflA HOC 12 Dl/flNO HOA HOCI2 31
Ttiy theo H t d c h r a v 6 i O H ( d l tab H g O ) 1^ d C i h a y C 3 , t a se diTgrc 2 anken: +) C 6 n g thtfc c i i a rtfofu B C H 2 = C H - C H 2 - C H 3 (khii H d C i ) l a r u g u n o c6 t h e dcfn h a y d a chufc C H 3 - C H = C H - C H 3 (khi^ H d C 3 ) B C O C T Ik C „ H 2 n + 2 O , . T r i i d n g hcfp n ^ y p h u hgrp vdri de. 3 n + 1 - z' CnH2n + 20z' + 02 -> n C 0 2 + ( n + D H a O OH 3 n + 1 - z' H3C C CH3 H 2 C — C CH3 Vofi 1 m o l B c a n mol O2. V a y CH3 CH3 3 n + 1 - z' = 2,5 ^ 3 n + 1 - z' = 5 => 3 n - z' = 4 V a y D ' 1^ C H 3 - C H O H - C H 2 - C H 3 C T C T cua este B Ik: z' 1 2 3 HOOC—CH—CH2—CH3 n 5 7 t = = 0,2 ( m o l ) (2) 29 m„,u6-i = 0,4(R + 67) = 32,8 R + 67 = 82 R = 15 G i a i (1) (2) ^ X - ^ = 1 => y = 2 x - 2 Vaiy R l a CH3 v a a x i t A 1^ C H 3 C O O H . -2 Do do c o n g thufc cau t a o cua X 1^: V a y X CO 2 l i e n ke't n (2 n o i d o i 0 = 0 ) H3C—COO—CH2 V a y X CO 2 chufc este. D o do riicfu B hoSc a x i t A da chufc. H3C—COO—CH2 2 . B6\G HdA HQC 12
Bai 10. a) Gid stjT X, Y, Z c6 cong thufc Ik RjCOOH, R2COOH vk R'COOH Ghep cong thufc cua 3 axit vdi cong thufc cua glixerol ta c6 cong thufc cau vdri R i = R2 = R (vi X, Y dong phan) va R' = R + 14 (do Z la dong dang tao cua este A: ke tiep cua Y (hcfn Y mot -CH2-) I ' Vay cong thijfc cau tao cua este A xuat phat tU 3 axit tren va glixerol la: H3C—CH—COO—CH2 Rl—COO—CH2 R^_COO—CH2 H3C—CH2—CH2-COO—CH R2—COO—CH ' R'—COO—CH H3C—CH2—CH2—CH2—COO-CH2 R'—COO—CH2 I CH3 R2—COO—CH2 K h i xk phong hoa A, ta thu dtfoc glixerol va 3 muo'i hoac H3C—CH—COO—CH2 Rl—COO—CH2 HO—CH2 RjCOONa I I H3C—CH2—CH2—CH2-COO—CH Ro—COO—CH +3NaOH ^ HO—CH + R^COONa R'COONa H3C-CH2—CH2—COO—CH2 R'—COO—CH2 HO—CH2 b) H5n hgfp 3 mudi vdri H2SO4 Goi a = nA, phan uTng phong hoa cho ta 3 muoi vdri so mol moi mudi 2C3H7-COONa + H2SO4 > 2C3H7-COOH + Na2S04 deu bhng a 2C3H7-COONa + H2SO4 > 2C3H7COOH + Na2S04 mA = a(Ri + R2 + R' + 173) = 7,9 hay a(2R + R' + 173) = 7,9 (1) 2C4H9COONa + H2SO4 > 2C4H9COOH + Na2S04 Kh6'i ItfOng 3 mudi Do so' mol 3 mudi bang nhau, so mol 3 axit X, Y, Z cung bang nhau. ma muo-i = a(2R + R' + 201) = 8,6 (2) Goi X sd mol m6i axit trong hon hop daoc dem dd't chay Lay (2) trCr (1), ta dugc: 28a = 0,7 =^ a = — mol C3H7-COOH + 5O2 ~ — > 4CO2 + 4H2O (mol) x • 4x 0,1 Thay vao (1): (2R + R' + 173) = 7,9 C3H7-COOH + 5O2 — > 4CO2 + 4H2O 2R + R' + 173 = = 316 2R + R' = 143 (mol) x 4x 0,1 C4H9-COOH + 6,502 ^—-^ 5CO2 + 5H2O V i R' = R + 14 (mol) x 5x 2R + R + 14 = 3R + 14 = 143 3R = 129 => R = 43 => n^Q^ = (4 + 4 + 5)x = 13x Neu R \k CxHy => 12x + y = 43 X 2 3 4 ^ n„„ = n„ I - ^'^^^ = 0,013 (mol) . CO2 BaCOg-L 197 J ^ ' y 19 7 am =^ 13x = 0,013 hay x = 0,001 (mol) Vay R 1^ -C3H7 va R' la -C4H9 K h d i lifong hon hop 3 axit dem dot chay A x i t Z CO mach th^ng vay c6 cong thufc cau tao la : 0,001(88 + 88 + 102) = 0,278 (gam) CH3-CH2-CH2-CH2-COOH Bai 11. a) Este A: RCOORi' (a mol) B: R2COOR2' (b mol) Y dong dang v d i Z nen cung c6 mach t h i n g CH3-CH2-CH2-COOH Tinh nNaOH trong 50 m l dung dich NaOH (diia t r e n 10 m l dung dich X dong phan vdri Y c6 mach phan nhanh: NaOH t r u n g hoa 30 m l dung dich H2SO4 0,25M) va nNaOH du H3C—CH—COOH Suy ra nNaOH phan ufng vdfi este CH3 HNaGH = n2 este => a + b = 0,05 (mol) B6| DliflNn H f i i H f i r 1 9 as
b) ma este = M ( a + b) = 5,7 => M A = M B = 1 1 4 V i ancol 1^ ancol bac hai c6 nhdnh n§n B c6 cong thufc cau tao l a : R i + R ' l = R2 + R'2 = 7 0 . H3C—CH—CH2—OH • ai^'^ol isobutylic 2 rugru vk axit d4u l ^ m m a t mau rnidc Br2 vSy trong 2 rUcru cung nhij CH3 trong 2 axit phai c6 i t nhat 1 chat khong no. 2 rtfou axit c6 cCing so nguyen tijf cacbon vay 2 axit cung c6 cung s6' h^n iJfng este \k: H,SO, nguyen tuT cacbon (do 2 este nay dong phan) R(C00H)2 + nC4H90H ^ > nH20 + R(COOH), _ ^ (COOC^H^) 9 'n Cong thuTc chung cua 2 este: C n H s n + 2 - 2kCOOCniH2m + 1 - 2k' 0,0735 (mol) 0,0735n (mol) 0,0735 (mol) (k, k ' so lien k e t n c6 t r e n gdc R , R ' ) ta c6 14 847 R + R ' = 70 7(n + m) - (k + k') = 34 M E = R + 45(2 - n) + l O l n = ' = 202 R + 56n = 112 k +k'=l=^n +m =5 0,0735 n , m phai > 2 (vi v d i 1 cacbon, khong the c6 2 riidu, 2 axit khac nhau) ) K h i n = 1 =^ R = 56 (~C4H8-) k = 0, k ' = 1 C2H5COOC3H5 (goc CO noi doi trong R ' O H phai t o i thieu Cong thufc cau tao cua A : HOOC-(CH2)4-COOH: axit adipic CO 3 nguyen ttf cacbon t h i riiOu mdi ben) => Cong thufc cSi'u tao cua E: k = 1, k ' = 0 ^ C2H3COOC3H7 HOOC—CH2—CH2-COO—CH2-CH-CH3 C2H5COO-CH2-CH=CH2 ; propionat propennil CH3 CH2=CH-COOC3H7 ; acrilat propil. Isobutylhidro adipat c) 2 rtrgu la C3H5OH hay C H 2 = C H - C H 2 0 H va C3H7OH chi c6 rifou dau + ) K h i n = 2 = > R = 0=> CTCT cua A: HOOC-COOH: axit oxalic '* lam mat mau n U d c Br2 vay Cong thufc ca'u tao c u a E: a = n^^H^oj, = ng^^ = 0,04 (mol) H3C—OH—CHo-OOC—COO—CH2—CH—CH3 3 I 2 2 I 3 ^=^o,n,ou =0,01(mol) CH3 CH3 = 2,32 (gam); m^^„ = 0,60 (gam) b) T i n h k h o i liTcfng cua A, B: +) K h i n = 1: m A = 0,0735 x (56 + 90) = 5,439 (gam) "H, = r.,. = 0>025 (mol) ^ = 0,56 (lit). mB = 0,0735 X 74 = 10,31 (gam) Chi CO axit C H 2 = C H - C 0 0 H l a m mat mau rnidc Brz +) K h i n = 2; mA = 0,0735 x 74 = 10,878 (gam); ^ n3_, = b = 0,01, (mol) => y = 1,60 (gam). me = 2 X 0,0735 x 74 = 6,615 (gam) +) X6t trifcfng hcfp 3: Loai v i nNaOH khong du thuy phan so mol este. Bai 12. Bai 13. a) ChiJtng minh rdng 1 trong 2 este A, B cd chvta goc benzen bang a) Theo de bai, t a c6: n A : nNaOH = 1 : 2 => A la axit hai chufc. each so sdnh so mol NaOH phan ling vdi tong so mol este. Goi cong thufc tong quat cua A la R ( C 0 0 H ) 2 Neu HNaOH > n2 este t h i 8 6 CO 1 este phan ijfng v6i N a O H theo t i le mol Ma n^^o^ : n^^^= 4 : 5 = ^ n^^< n^^^^B la ancol no 1 : 2, este n^y c6 dang R - C O O - C e H s Theo de r a B la ancol dcrn chufc => Goi cong thufc tong quat cija B la A: RiCOOR'i (a mol) CnH2n + l O H B: RgCOOR'a (b mol) vdi R'2=C6H5 Phan dfng giufa A, B v d i NaOH CnHzn. lOH + ^ 0 2 ^ — > nC02 + (n + D H g O RiCOOR'i + NaOH -> RiCOONa + R'lOH (mol) a a a a ^ "co ^ _ _n u _ ^ ^ _co^ 4 ^ 4 (C4H9OH) R 2 C O O R ' 2 + 2NaOH - > RgCOONa + R'20Na + H2O ^H,o n + 1 5 ^mol) b 2b b b )6 Bdl DlidNG HOA HOC 12 DU8NG HdA HOC 12 37
=> n 2 e s t e = a + b = 0,08 (1) pai 14. a) Cong thufc phan tuf cua A, B, C. (do n = C X V = 0,8 X 0,1 = 0,08 mol) A, B, C la dong phan nen c6 cung C T P T . P2O5 hut nude So mol NaOH ban dau = 0,15.1 = 0,15 (mol) (P2O5 + 3H2O > 2H3PO4) HNaOH du = H H C I = 0,2 X 0,2 = 0,04 (mol) 14,4 => n N a O H pt» vdi A,c = 0,15 - 0,04 = 0,11 (mol) N§n mjj^o = 14,4 (gam) => mn = = 1,6 (gam) a + 2b = 0,11 (2) K O H hut CO2 n§n: Tii (1) (2) ^ a = 0,05 (mol) A; b = 0,03 (mol) B 3.35,2 Chi c6 este A phan ufng cho ra rifcfu E m^Q = 35,2 (gam) => mc = = 9,6 (gam) 11 n E = H A = 0,05 (mol) M E = MR.QH = = 58 Kh6ng c6 nguyen t d n^o khac ngoai oxi. 0,05 Suy ra: mo = 17,6 - (1,6 + 9,6) = 6,4 (gam) Do mqu R'OH khong ben bien t h a n h andehit R"CHO ta c6 ME = R" + 29 = 58 =i> R" = 29 12x y 16z x _ y _ z hay Vay R" 1^ C 2 H 5 va andehit la CH3-CH2-CHO bien doi tii rUgfu khong 9,6 1,6 6,4 r96^ 16 r64^ b i n CH3-CH=CH0H. 112; [lej Phan ufng xa ph6ng h6a A, B chi cho ra 2 mudi vSy b^t bu6t trong 3 X mudi RiCOONa, RgCOONa va R'aONa c6 2 mudi gidng nhau. = — hay — = — = - = n C T N l a (C,H^O)„ ^ 8 = 16 4 - ^ 2 4 1 Yky R i t r u n g vdi R2 - Vi A, B, C dorn chufc, sd nguyen tuf oxi chi c6 the b^ng 1 (rLfgu, ete, K h d i iLfgtng 2 mudi: m2 muol = niRcooNa + ^n\om andehit, xeton) hay bkng 2 (axit, este). (a + b) (b) +) K h i z = 1 => Cong thufc phan tuf la C2H4O. Chi c6 the c6 1 c6ng thiJc 0,08(Ri + 67) + 0,03(R'2 + 39) = 10,46 cafu tao 1^ CH3-CHO (loai v i phai c6 3 dong phan). 8R1 + 3R'2 = 1046 - 653 = 393 (3) +) K h i z =.2 => C4H8O2. ^ M^^coo^^ ^ +67 ^ 4 1 . Vay c6 l i e n ket n. A, B, C chi c6 the la axit hay este. • M , - M,,^,,,^ - R; +39-65 Axit CH3CH2CH2COOH CH3-CH(CH3)-COOH 41R'2 + 1599 = 65Ri + 4355 41R'2 - 65Ri = 2756 (4) Este HCOOC3H7 CH3COOC2H5 C2H5COOCH3 Ttr (3) v^ (4) R i = 15 vay R i Ik CH3 Co 3 triidng hop: +) Trufcfng hcfp 1: 3 este R'2 = 91 R'2 la C 6 H 4 - C H 3 +) Cong thtfc cau tao cua A va B +) Triicfng hgp 2: 1 axit, 2 este A este cua axit C H 3 C O O H va rirgu C H 3 - C H = C H 0 H c6 cong thufc cau tao +) Trtrdng hop 3: 2 axit, 1 este. la: C H 3 - C O O - C H = C H - C H 3 PhiTOng phdp: de chon tritdng hcfp dung, ta difa tren so mol NaOH B este cua axit CH3COOH va phenol nSn B c6 cong thufc cau tao \k: phdn ling d nhiet do thitdng va khi dun nong va tren khoi liidng 2,58 gam cua chat long. H3C—COO HNaOH d t° thifdng: 0,5.0,1 = 0,05 (mol) nNaOH k h i dun nong: 1 x 0,1 = 0,10 (mol) +) Trififng hcfp 1: A, B, C deu 1^ este. Do este gan nhtr khSng phan ufng vdi NaOH trong t h d i gian ng^n, ta loai trifdng hcfp n^y. Bfii Di/flNG HOA HOC 12 39 B6I DUSNG HOA HOC 12