anken hay

L thuyt v bi tp anken

1L THUYT AN KEN (OLEFIN)

I. NG NG - C2H4 v cc ng ng ca n to thnh dy ng ng , gi chung l anken hay olefin - Anken l cc hiro cacbon khng no, mch h, trong phn t c 1 lin kt i C = C. - Cc anken c cng thc chung l CnH2n, n 2 II. NG PHN ng phn a. ng phn cu to - Cc anken C2, C3 khng c ng phn - T C4 tr i c ng phn mch C v ng phn v tr lin kt i. Cch vit ng phn ca anken: - Vit mch cacbon thng. t lin kt = vo cc v tr khc nhau. - B 1 cacbon lm nhnh. t nhnh vo cc v tr khc nhau trong mch. Sau mi mch C li t lin kt i vo cc v tr khc nhau. - Khi b 1 cacbon khng cn ng phn th b n 2 cacbon. 2 cacbon c th cng lin kt vi 1C hoc 2C khc nhau. Li t lin kt i vo cc v tr khc nhau. - Ln lt b tip cc Cacbon khc cho n khi khng b c na th dng. b. ng phn hnh hc a c - L ng phn v v tr khng gian ca anken - Gm 2 loi: ng phn cis (cng pha) v trans (khc pha) C=C * iu kin c ng phn hnh hc: - Trong phn t phi c lin kt i b d - Cc nhm th lin kt vi cng 1 nguyn t C mang ni i phi khc nhau III. DANH PHP 1. Tn thng thng - Mt s t anken c tn thng thng Tn thng thng = Tn ankan tng ng, thay ui an = ilen - Khi trong phn t c nhiu v tr lin kt i khc nhau th thm cc ch nh , , ... ch v tr ni i 2. Tn cc nhm ankenyl - Khi phn t anken b mt i 1 nguyn t H th to thnh gc ankenyl - Tn ca gc ankenyl c c tng t nh tn anken nhng thm ui yl VD: CH2 = CH2 CH2 = CH CH2 = CH2 CH3 CH2 = CH CH2 -H -H Eten Vinyl Propen anlyl (allyl) (Etenyl) (prop 2 en 1 yl ) 3. Tn thay th ca anken Tn anken = S ch v tr nhnh Tn nhnh + Tn mch chnh v tr lin kt i en - Mch chnh l mch c cha lin kt C = C v di nht, c nhiu nhnh nht . - xc nh v tr nhnh phi nh s cacbon trn mch chnh. nh s C trn mch chnh t pha C u mch gn lin kt C = C hn. Nu c nhiu nhnh ging nhau th phi nu y v tr ca cc nhnh v phi thm cc tin t i (2), tri (3), tetra (4) trc tn nhnh. Nu c nhiu nhnh khc nhau th tn nhnh c c theo th t ch vn ch ci. Lu : Gia s v s c du phy, gia s v ch c du gch IV. TNH CHT VT L

Mucdong_Tay Bac


2

- Trng thi: Anken t C2 C4 trng thi kh An ken t C5 tr ln trng thi lng hoc rn. - Mu: Cc anken khng c mu - Nhit nng chy, si: Khng khc nhiu so vi ankan tng ng nhng nh hn so vi xicloankan c cng s nguyn t C. Cc anken c nhit nng chy v nhit si tng dn theo khi lng phn t0 0 ng phn cis- anken c t nc thp hn nhng c t s cao hn so vi ng phn trans 0 0 Khi cu trc phn t cng gn th t nc cng cao cn t s cng thp v ngc li

- tan: Cc anken u nh hn nc, khng tan trong nc nhng tan nhiu trong cc dung mi hu c V. TNH CHT HA HC Nhn xt chung: - Do trong phn t anken c lin kt C = C gm 1 lin kt v 1 lin kt , trong lin kt km bn hn nn d b phn ct hn trong cc phn ng ha hc. V vy anken d dng tham gia cc phn ng cng vo lin kt C=C to thnh hp cht no tng ng 1. Phn ng cng a. Cng Hiro to ankan t 0 , Ni CnH2n + H2 CnH2n+2 b. Cng halogen X2 (Cl2, Br2) CnH2n + X2 CnH2nX2 VD: CH2 = CH2 + Br2(dd) CH2Br CH2Br (mu nu ) (khng mu) * Do anken lm mt mu dung dch Brom nn ngi ta dng dung dch Brom lm thuc th nhn bit ra anken. c. Cng axit HX (HCl, HBr) CnH2n + HX CnH2n+1X VD: C2H4 + HCl C2H5Cl * Quy tc cng HX: - Phn ng ch cng vo lin kt C = C - H cng vo C c nhiu H (C bc thp hn). Phn cn li cng vo C mang ni i cn li. d. Cng H2O trong mi trng axitH CnH2n + H2O CnH2n+1OH (ancol no n chc) H VD: C2H4 + HOH C2H5OH 2. Phn ng trng hp - Phn ng trng hp l phn ng cng hp nhiu phn t nh c cu to tng t nhau (gi l monome) thnh 1 phn t ln (gi l polime) .t , p, xt nA A n - n gi l h s trng hp - Phn trong ngoc gi l mt xch ca polime t , p, xt VD: nCH2 = CH CH3 CH CH2 0 0

+

+

CH3 n * iu kin monome tham gia phn ng trng hp l phn t phi c lin kt 3. Phn ng oxi ha a. Phn ng chy 3n t CnH2n + O2 nCO2 + nH2O 20

- Trong phn ng chy lun c: n CO2 = n H 2O b. Phn ng oxi ha khng hon ton - Dn kh C2H4 vo dung dch KMnO4 (mu tm) thy dung dch mt mu tm:

Mucdong_Tay Bac


3

3CnH2n + 2KMnO4 +4H2O 3CnH2n(OH)2 + 2MnO2 + 2KOH VI. IU CH 1. hiro ha ankan t 0 , xt CnH2n+2 CnH2n + H2 2. Phng php cracking crk CnH2n+2 CaH2a+2 + CbH2b 3. T ankin (l hp cht c ni ba C C), ankaien (c 2 ni i) t 0 , Pd CnH2n-2 + H2 CnH2n 4. T dn xut halogen ancol CnH2nX + KOH CnH2n + KX + H2O 5. T dn xut ihalogen t0 CnH2nX2 + Zn CnH2n + ZnX2 6. Tch nc ca ru no n chc170 C, H SO CnH2n+1OH CnH2n + H2O 2 4 0

BI TP P DNG L THUYT Loi 1. Bi tp l thuyt Bi 1. Hon thnh s phn ng CH3COONa CH4 C2H2 C2H4 C2H5Cl C2H5OH C2H4 C2H4(OH)2 C2H6 C2H5Cl C2H4 P.E Bi 2. Hon thnh s phn ng P.P +HCl +NaOH, t 0 C4H10 C3H6 A B +H 2 O, H H 2SO 4 , 170 C D E +KMnO 4 G + 0

Bi 3. Nhn bit cc dung dch ring bit a. CH4, C2H4, CO2, SO2, N2 b. Oxi, sunfur, cacbonic, etilen, butan, xiclopropan, nit. Bi 4. Hon thnh s phn ngH 2SO 4 , 170 C +H 2 O, H H 2SO 4 , 170 C +H 2 O, H 3-metylbut-1-ol A B D E Bi 5. Cho anken A c cng thc C6H12 tc dng vi HCl ch thu c 1 sn phm cng duy nht. Tm CTCT ng ca A. Bi 6. Thc hin phn ng tch H2 ca 2-metylpentan th thu c my anken (khng k ng phn hnh hc) Bi 7. Tch C2H4 tinh khit ra khi hn hp gm C2H4, CO2, SO2, CH4, N2, O2. Bi 8. Vit cc ptp iu ch cc cht sau t metan: Etilenglicol, P.E, ancol etylic, P.P, 1,2-ibrom propan. Bi 9. Cht X c CTPT l C4H8. X phn ng chm vi Brom nhng khng phn ng vi dung dch KMnO 4. Tm CTCT ca X.(metyl xiclopropan) Bi 10. Thc hin phn ng tch H2 ca isopentan th thu c ti a my anken Bi 11. Chng minh cng thc chung ca cc ankan l CnH2n+2 , cn ca an ken l CnH2n. Bi 12. Cho cc cht : 2-metylbut-1-en (1); 3,3-imetylbut-1-en (2); 3-metylpent-1-en (3); 3-metylpent-2-en (4); Nhng cht no l ng phn ca nhau ? Bi 13. Nhng hp cht no sau y c ng phn hnh hc (cis-trans) ? CH3CH=CH2 (I); CH3CH=CHCl (II); CH3CH=C(CH3)2 (III); C2H5C(CH3)=C(CH3)C2H5 (IV); C2H5C(CH3)=CClCH3 (V). Bi 14. Cho cc cht sau: CH2=CHCH2CH2CH=CH2; CH2=CHCH=CHCH2CH3; CH3C(CH3)=CHCH2; CH2=CHCH2CH=CH2; CH3CH2CH=CHCH2CH3; CH3C(CH3)=CHCH2CH3; CH3CH2C(CH3)=C(C2H5)CH(CH3)2; CH3CH=CHCH3. Nhng cht no c ng phn hnh hc? Bi 15. Khi tch nc t ru (ancol) 3-metylbutan-2-ol th sn phm chnh thu c l ?0 + 0 +

Mucdong_Tay Bac


4

Loi 2. Phn ng chy, cng Br2 Bi 1. Hn hp A gm 2 olefin l ng ng k tip. t 3 lit A cn 14 lit Oxi (o cng k). Tm CTPT ca 2 anken. Bi 2. Hn hp X gm 1 anken v 1 ankan l cht kh iu kin thng. t chy ht 2,688 lit X thu c 12,32 gam CO2 v 5,76 gam H2O. Tm CTPT ca A, B. Bi 3. Hn hp X gm CH4 v 2 anken l ng ng k tip. t chy ht 2,24 lit X thu c 4,14 gam nc v 8,36 gam CO2. Tm CTPT ca 2 anken. Bi 4. Mt hn hp A gm 2 hirocacbon X, Y lin tip nhau trong cng dy ng ng. t chy 11,2 lt hn hp X thu c 57,2 gam CO2 v 23,4 gam CO2. Tm CTPT ca X, Y v khi lng ca chng trong hn hp. Bi 5. em t chy hon ton 2,24 lit hn hp X gm 2 anken l ng ng k tip nhau thu c CO 2 v nc c khi lng hn km nhau 6,76 gam. Tm CTPT ca 2 anken. Bi 6. em t chy hon ton 3,36 lit hn hp X gm 2 anken l ng ng k tip nhau thu c CO 2 v nc c khi lng hn km nhau 8,45 gam. Tm CTPT ca 2 anken. Bi 7. t chy hon ton 1lit hirocacbon X cn va 6 lit kh oxi, sau phn ng thu c 4lit kh cacbonic. Bit X lm mt mu dung dch brom v c mch cacbon phn nhnh. Tm CTCT ca X Bi 8. Dn 3,36 lt hn hp X gm 2 anken l ng ng k tip vo bnh nc brom d, thy khi lng bnh tng thm 7,7 gam. Tm CTPT ca 2 anken ? Bi 9. Mt hn hp X c th tch 11,2 lt, X gm 2 anken ng ng k tip nhau. Khi cho X qua nc Br2 d thy khi lng bnh Br2 tng 15,4 gam. Xc nh CTPT % khi lng mi anken trong hn hp X. Bi 10. Hn hp X gm ankan A v anken B. Ly 11,2 lit X cho tc dng vi dung dch Brom th cn 0,5 lit dung dch Brom 0,4M. Nu t chy 5,6 lit X th thu c 13,44 lit CO2. Tm A, B. Bi 11. Hn hp kh X gm 1 ankan v 1 anken c cng s nguyn t C trong phn t (t l mol l 1;1). Ly m gam hn hp X lm mt mu va 80 gam dung dch Br 2 nng 20%. t chy hon ton m gam X thu c 0,6 mol CO2. Tm CTPT ca A, B. Bi 12. Dn 7,84 lit kh X gm ankan A v anken B qua bnh ng dung dch Brom thy dung dch Brom nht mu v khi lng bnh tng 4,2 gam. Kh thot ra khi bnh c khi lng 3,2 gam. Tm CTPT ca A, B. Bi 13. Cho 2,016 lit hn hp X gm anken A v ankan B li chm qua dung dch Brom d thy cn li 1,568 lit kh bay ra v khi lng kh gim i mt na. Tm CTPT ca A, B. Bi 14. Cho 5,04 lit hn hp X gm C2H4 v 2 ankan A, B l ng ng k tip li chm qua dung dch Brom d thy c 16 gam Brom phn ng v khi lng kh gim i mt na. Tm CTPT ca A, B. Bi 15. Hn hp X gm 1 ankan v 2 anken l ng ng k tip. S mol cc cht trong hn hp bng nhau. Cho X qua dung dch Brom d thy c 16 gam Brom tham gia phn ng. t chy X thu c 15,4 gam CO 2. Tm CTPT cc cht trong hn hp. Bi 16. Hn hp X gm etilen, propen v etan. t chy ht 2,016 lit X thu c 4,704 lit CO 2. Cn nu cho 2,016 lit X tc dng vi dung dch Brom d th thy c 9,6 gam Brom tham gia phn ng. Tm % khi lng tng cht c trong hn hp X. Bi 17. Hn hp A gm C2H4, C3H6, C4H8.

Mucdong_Tay Bac


5

- t chy m gam A thu c 2,016 lit CO2. - Nu hiro ha ht m gam A bng lng H2 va ri t chy ton b lng ankan th thu c 2,16 gam H2O. - Nu cho m gam A tc dng vi dung dch Brom d th thy c a gam Brom tham gia phn ng. Tm a. Bi 18. Cho 4,48 lit hn hp X gm etan, propan v propen qua dung dch brom d, thy khi lng bnh brom tng 4,2 gam. Lng kh cn li thot ra khi dung dch em t chy hon ton thu c 6,48 gam nc. Tnh % th tch cc cht c trong hn hp. Bi 19. Mt hn hp X gm 2 hirocacbon A, B c cng s nguyn t cacbon. A, B ch c th l ankan hay anken. t chy 4,48 lt (kc) hn hp X thu c 26,4 gam CO 2 v 12,6 gam H2O. Xc nh CTPT v s mol ca A, B trong hn hp X. Bi 20. Ba hirocacbon X, Y, Z k tip nhau trong dy ng ng, trong MZ = 2MX. a. Tm CTPT ca X, Y, Z. b. Vit phn ng cng H2O/H+, cng HBr, cng H2, cng KMnO4, phn ng trng hp ca Y. Bi 21. t chy hon ton 8,96 lt hn hp hai anken l ng ng lin tip thu c m gam H 2O v (m + 39) gam CO2. Tm CTPT ca hai anken. Bi 22. t chy hon ton 1,12 lit mt anken A thu c 4,48 lt CO2. Cho A tc dng vi dung dch HBr ch cho mt sn phm duy nht. Tm CTCT ca A Bi 23. Dn 2,24 lt anken A qua bt CuO d nung nng thy khi lng bt CuO gim 14,4g. a. Tm cng thc phn t ca A. b. Vit phng trnh phn ng cng H2, cng H2O, trng hp, cng KMnO4 ca A. c. Hn hp A vi mt ng ng B trn theo t l mol 1:1. t chy hon ton mt th tch hn hp cn 3,75 th tch oxi trong cng iu kin. Hy gi tn B. Bi 24. t chy hon ton a gam hirocacbon A cn dng 6,72 lt O2 thu c 3,6 gam H2O. a. Tm CTPT ca A, bit t khi hi ca A so vi nit l 2. b. Vit Tm CTCT ng ca A, bit A cng vi H2O cho 1 sn phm duy nht. Bi 25. Hn hp X gm ankan A v anken B (u th kh ). S nguyn t Cacbon trong B nhiu hn trong A. Th nghim1: t 0,12mol hn hp gm a mol A v b mol B thu c khi lng CO2 ln hn khi lng ca nc l 9,12gam. Th nghim2: t 0,12mol hn hp gm b mol A v a mol B thu c khi lng CO2 ln hn khi lng ca nc l 7,44gam. Tm CTPT ca A, B. Bi 26. Cho 3,36 lit hn hp X gm 1 ankan v 2 anken l ng ng k tip tc dng vi dung dch nc Brom thy dung dch nc Brom nht mu v khi lng bnh tng 3,5 gam. Kh i ra khi bnh c th tch 1,12 lit. Nu t chy ht 3 lit hn hp X th thu c 8 lit CO2 cng iu kin.Tm CTCT tng cht trong hn hp X. Bi 27. Cho m gam hn hp X gm 2 hidrocacbon mch h tc dng vi dung dch Brom thy dung dch nht mu, c 10 gam Brom phn ng v cn li 2,24 lit kh thot ra. Nu t chy ht 1,456 lit kh X ri cho sn phm chy qua bnh ng dung dch nc vi trong d thy trong bnh xut hin 7,5 gam kt ta v thu c dung dch A. un nng dung dch A li thu thm c 4 gam kt ta na. Tm CTPT ca 2 hidrocacbon. Bi 28. Cho hn hp X gm anken A v ankan B li chm qua 100 gam dung dch Brom 16% thy dung dch Brom mt mu, khi lng bnh tng 2,8 gam v c 3,36 lit kh D thot ra. t chy ht D thu c 8,8 gam CO 2 v 5,4 gam H2O. Tm CTPT ca A, B. Bi 29. Bn hn hp kh X gm 1 ankan v 1 anken vo bnh ng dung dch Brom d thy khi lng bnh tng thm 5,04 gam. t chy ton b kh thot ra khi bnh thu ri cho ton b sn phm chy hp th ht vo bnh

Mucdong_Tay Bac


6

ng dung dch Ba(OH)2 thy khi lng bnh tng thm 21,28 gam v trong bnh xut hin 63,04 gam kt ta. Tm CTPT ca A, B v % khi lng ca chng trong hn hp.

BI TP LUYN TP I. PHN NG CHY Cu 1. Hn hp X gm hirocacbon A va O2 (ti l mol tng ng 1:10). t chay hoan toan X c hn hp Y. Dn Y qua binh H2SO4 c d c hn Z co ti khi so vi hiro la 19. Tm CTPT ca A ? Cu 2. X, Y, Z la 3 hirocacbon k tip trong day ng ng, trong o MZ = 2MX. t chay hoan toan 0,1 mol Y ri hp thu toan b san phm chay vao 2 lit dung dich Ba(OH)2 0,1M c m gam kt ta. Tm m. Cu 3. Hn hp X gm 2 anken th kh l ng ng k tip phn ng va vi dung dch cha 48 gam brom. Mt khc t chy hon ton hn hp X dng ht 24,64 lt O2 (ktc). Tm CTPT ca 2 anken. Cu 4. t chy hon ton hn hp kh A gm 2 anken iu kin thng th nhn thy t l th tch gia A v oxi tham gia phn ng l 21/93. Bit anken c khi lng mol phn t cao c th tch chim khong 40% n 50% th tch hn hp. Xc nh CTPT ca 2 anken v tnh % th tch ca tng anken trong A. Cu 5. Hn hp X gm C3H8 va C3H6 co ti khi so vi hiro la 21,8. t chay ht 5,6 lit X thi thu c bao nhiu gam CO2 va bao nhiu gam H2O ? Cu 6. m gam hn hp gm C3H6, C2H4 v C2H2 chy hon ton thu c 4,48 lt kh CO2. Nu hiro ho hon ton m gam hn hp trn ri t chy ht hn hp thu c V lt CO2. Tm V Cu 7. Chia hn hp gm C3H6, C2H4, C2H2 thnh hai phn u nhau. Phn 1: t chy hon ton thu c 2,24 lt CO2. Phn 2: Hiro ho ri t chy ht sn phm th thu c m gam CO2. Tm m. Cu 8. t chy hon ton 20 ml hn hp X gm C3H6, CH4, CO (th tch CO gp hai ln th tch CH4), thu c 24 ml CO2 (o cng nhit v p sut). Bit d X/H 2 = a . Tm a. Cu 9. t chy hon ton a gam hn hp eten, propen, but-2-en cn dng va b lt oxi ( ktc) thu c 53,76 lit CO2 v 43,2 gam nc. Tm b. Cu 10. t chy ht V lt hn hp X gm CH4, C2H4 thu c 0,15 mol CO2 v 0,2 mol H2O. Gi tr ca V l? Cu 11. t chy hon ton 2,24 lit hn hp gm CH4, C4H10 v C2H4 thu c 0,14 mol CO2 v 0,23mol H2O. Tm th tch ca anken v ankan trong hn hp. Cu 12. t chy hon ton 6,11 lit ( 136,50C, 2,2 atm) hn hp kh X gm 1 ankan v 1 anken ri cho sn phm chy hp th ht qua bnh cha nc vi trong d thy khi lng bnh tng 61,2 g ng thi xut hin 90 g kt ta. Xc nh CTPT ca hidrocacbon? Cu 13. t chy ht 7 gam hn hp gm C2H4, C3H6, C4H8 cn V lit oxi. Tm V. Cu 14. t chy ht 1,12 lit hn hp A gm 1 anken v 1 ankan c t l mol tng ng l 3:2 thu c 3,36 lit CO2. Tm CTPT ca cc cht trong A. Cu 15. Hn hp X gm C2H4, C3H6, C4H8 v C5H10. t chy ht 3,2 gam X cn V lit Oxi thu c a gam CO2 v 4,5 gam H2O. Tm a v V.

Mucdong_Tay Bac


7

Cu 16. Hn hp X gm C2H4, C3H6, C4H8 v C5H10. Cho m gam X tc dng vi dung dch Brom th thy c 32 gam Brom tham gia phn ng. Nu hidro ha hon ton m gam hn hp X bng lng H2 va ri t chy ht lng sn phm th thu c 22 gam CO2 v a gam H2O. Tm m v a. Cu 17. Hn hp X gm 2 anken l ng ng k tip. Hirat ha hon ton 2,24 lit X ri t chy ton b lng sn phm hu c th thu c 7,65 gam H2O. Tm CTPT ca 2 anken. Cu 18. t chy hon ton 1 anken A bng Oxi thu c CO2 v H2O. Bit tng th tch ca CO2 v H2O ng bng tng th tch ca Oxi v anken. Tm CTPT ca A. Cu 19. t chy hon ton 6,72 lit hn hp gm 2 hirocacbon A, B thuc cng dy ng ng ankan hoc anken (c t l khi lng mol phn t l 22: 13) ri cho ton b sn phm chy vo bnh ng dung dch Ba(OH)2 d th thy khi lng bnh tng 46,5 gam v c 147,75 gam kt ta. a. Hai hirocacbon trn thuc dy ng ng no ?. b. Xc nh CTCT ca A, B v tnh % th tch tng cht trong hn hp. Cu 19. Mt hn hp X gm CO v mt hirocacbon A mch h. t chy hon ton 1,96g hn hp X c 4,84g CO2 v 1,44g H2O. 1. Tm dy ng ng ca A. 2. Tm cng thc phn t ca A, bit d X/H2 = 19,6.

II. PHN NG CNG H2 Cu 1. Cho 5,6 lit C2H4 tc dng vi 7,84 lit H2 (Ni, t0) thu c hn hp A. Cho A li qua bnh ng dung dch Brom n khi phn ng xy ra hon ton thy c 8 gam Brom tham gia phn ng. Tnh Hp hiro ha anken. Cu 2. Cho 0,1 mol anken A tc dng vi 0,08 mol H2 (Ni, t0) thu c hn hp A. Cho A li qua bnh ng dung dch Brom n khi phn ng xy ra hon ton thy c 8 gam Brom tham gia phn ng. Tnh H p hiro ha anken. Cu 3. Dn 2,24 lit H2 v 2,24 lit C2H4 qua bt Niken nung nng thu c hn hp kh X , d X/Y = 0,6 . Tnh hiu sut ca phn ng hiro ho anken Cu 4. Hn hp A gm C2H4 v H2 c d A/H 2 = 7,5 . em hn hp A qua Ni, to thu c hn hp B c d B/H 2 = 9 a. Gii thch ti sao t khi hi tng. b. Tnh % th tch mi kh trong hn hp A, B. c. Tnh hiu sut phn ng. Cu 5. Hn hp kh X gm H2 v C2H4 c t khi so vi He l 3,75. Dn X qua Ni nung nng, thu c hn hp kh Y c t khi so vi He l 5. Tnh hiu sut ca phn ng hiro ho anken.

Cu 6. Cho hn hp X gm anken A v H2 c t khi so vi heli bng 3,33. Cho X i qua bt niken nung nng n khi phn ng xy ra hon ton, thu c hn hp Y c t khi so vi heli l 4. Tm CTPT ca A.Cu 7. Cho hn hp X gm etilen v H 2 c t khi so vi H2 bng 4,25. Dn X qua bt niken nung nng thu c hn hp Y. Bit hiu sut phn ng l 75%. Tnh t khi ca Y so vi H2 ? Cu 8. Mt hn hp X gm anken A v H2 c d X/H 2 = 10 cho qua niken , un nng A b hiro ha hon ton thu c hn hp Y, d Y/H 2 = 15 . Tm CTPT ca A.

Mucdong_Tay Bac


8

Cu 9. Hn hp A gm 2 anken. Khi dn 3,696 lit A i qua bnh ng nc brom d thy bnh nng thm 7 g. Khi cho 7,392 lit A vi 3,696 lit H2 i qua Ni nung nng thu c hn hp kh B. Tnh t khi ca B so vi Oxi Cu 10. Cho H2 v 1 olefin A (t l mol 1:1) qua Niken un nng ta c hn hp X. Bit d X/H 2 = 23,2 . Hiu sut phn ng hiro ho l 75%. Tm CTPT ca A. Cu 11. Hn hp kh X gm H2 v mt anken A, d X/H 2 = 9,1 . un nng X c xc tc Ni, sau khi phn ng xy ra hon ton, thu c hn hp kh Y, d Y/H 2 = 13 . Tm CTPT ca A. Cu 12. Hn hp X gm 0,15 mol C2H4 v 0,25 mol H2. Dn X qua bt Ni nung nng thu c hn hp Y. Cho Y qua dung dch Brom d thy khi lng bnh Brom tng thm 1,82 gam . Tm Hiu sut phn ng hidro ha anken. Cu 13. Hn hp X gm 0,2 mol C2H4, 0,3 mol C3H6 v 0,5 mol H2. Dn X qua bt Niken nung nng. Sau 1 thi gian thu c hn hp kh Y. Cho Y tc dng vi dung dch Brom d thy khi lng bnh tng thm 3,64 gam v c 16 gam Brom tham gia phn ng.Coi hiu sut hidro ha ca 2 anken l nh nhau. Tm Hp ? Cu 14. un nng 20,16 lt hn hp kh X gm C2H4 v H2 dng Ni xc tc th thu c 13,44 lt hn hp kh Y. Cho Y li tht chm qua bnh ng dung dch Br2 d th thy khi lng bnh tng 2,8 gam. Tnh hiu sut phn ng hidro ha anken? Cu 15. Hn hp A gm 1 ankan , 1 anken v H2. em 5,6 lit hn hp A qua Ni, to, sau phn ng thu c 4,48 lit hn hp B, dn tip B qua dung dch brm d th thy khi lng bnh brm tng ln 3,15 gam v c 2,8 lit hn hp kh C gm 2 kh thot ra. Bit d C/H 2 = 17,8 cc th tch o ktc. a. Tnh th tch tng cht trong hn hp A. b. Xc nh CTPT ca ankan v anken Cu 16. Hn hp A gm 1 ankan , 1 anken v H2. em 100 ml hn hp A qua Ni, to, sau phn ng thu c 70 ml mt hrcacbon duy nht. Cn em t chy 100 ml hn hp A th thu c 210 ml kh CO 2 . cc th tch o ktc. a. Xc nh CTPT ca ankan v anken v % th tch tng cht trong hn hp A. b. Trnh by cch tch ring ankan ra khi hn hp A. Cu 17. Hn hp kh X gm c H2, ankan A v anken B. t chy 150 ml hn hp X thu c 315 ml CO 2. Mt khc khi nung nng 150 ml hn hp X vi Ni th sau phn ng thu c 105 ml mt ankan duy nht. a. Xc nh % th tch mi kh trong hn hp X ban u b. Xc inh CTPT ca A v B c. Tnh t khi hn hp X i vi khng kh. Cu 18. Hn hp kh A gm H2 v 2 olefin l ng ng k tip nhau. Cho 19,04 lt (kc) hn hp kh A qua bt Ni nung nng thu c hn hp kh B (gi s hiu sut phn ng l 100%) v tc phn ng cng H 2 ca 2 olefin bng nhau. Cho mt t hn hp kh B qua dung dch brom th brom b nht mu. Nu t chy 1/2 hn hp kh B th thu c 43,56 gam CO2 v 20,43 gam nc. a. Xc nh cng thc phn t, vit cng thc cu to 2 olefin b. Tnh % th tch cc kh trong hn hp A c. Tnh t khi hi ca B i vi khng kh.

III. PHN NG CNG Br2

Mucdong_Tay Bac


9

Cu 1. Mt hn hp X gm ankan A v anken B u th kh. Khi cho 6,72 lt kh X i qua nc brom d, khi lng bnh brom tng ln 2,8 gam; th tch kh cn li ch bng 2/3 th tch hn hp X ban u. Nu t chy ht 3,36 lit X th thu c 17,6 gam CO2. Tm CTPT ca A, B v khi lng ca hn hp X. Cu 2. Mt hn hp X gm ankan A v mt anken B c cng s nguyn t C v u th kh. Cho hn hp X i qua nc Br2 d th th tch kh Y cn li bng na th tch X, cn khi lng Y bng 15/29 khi lng X. Tm CTPT A, B . Cu 3. Cho 10,8 lt hn hp X gm metan v 1 olefin qua dung dch brom d thy c 1 cht kh bay ra, t chy hon ton kh ny thu c 5,544 gam CO2. Tnh % th tch metan v olefin trong hn hp X ? Cu 4. Cho 8,96 lit anken X qua dung dch brom d. Sau phn ng thy khi lng bnh brom tng 22,4 gam. Bit X c ng phn hnh hc. Tm CTCT ca X ? . Cu 5. Cho hirocacbon X phn ng vi Dung dch Brom theo t l mol 1:1, thu c cht hu c Y (cha 74,074% Br v khi lng). Khi X phn ng vi HBr thu c hai sn phm hu c khc nhau. Tm CTCT ca X. Cu 6. Hirocacbon X cng HCl theo ti l mol 1:1 tao san phm co ham lng clo la 55,04% v khi lng. Tm CTPT ca X. Cu 7. Mt hirocacbon X cng hp vi axit HCl theo t l mol 1:1 to sn phm c thnh phn khi lng clo l 45,223%. Tm CTPT ca X. Cu 8. 0,05 mol hirocacbon X lam mt mau va u dung dich cha 8 gam brom cho ra san phm co ham lng brom at 69,56%. Tm CTPT ca X. Cu 9. Dn 1,68 lt hn hp kh X gm 1 anken v 1 ankan vo bnh ng dung dch brom (d). Sau khi phn ng xy ra hon ton, c 4 gam Brom phn ng . Nu t chy hon ton 1,68 lt X th sinh ra 2,8 lt kh CO2. Tm CTPT ca hai hirocacbon ? Cu 10. Hn hp X gm 2 hirocacbon A v B (B c s cacbon ln hn A, A v B u phn ng vi dung dch Br2). Ly 8,96 lit hn hp X tc dng ht vi nc brom cn ti thiu 64g brom. Mt khc, t 8,96 lit hn hp X thu c l 48,4 gam CO2. Hiu s hi nc sinh ra ca B so vi A l 12,6g. Xc nh CTPT ca A, B. Cu 11. Hn hp X gm hai anken A v B ( MA< MB) , t khi hi ca X i vi H2 l 19,6. Trong X s mol B chim 40 % s mol hn hp X. a. Xc nh CTPT, CTCT ca A, B. Bit B c ng phn cis-trans. b. Nu cho lng X trn tc dng vi dung dch Brm d , thy c 80 gam Br 2 tham gia phn ng. Tnh phn trm khi lng ca A, B trong X. Cu 12. Hn hp A gm 1 ankan v 1 anken. em 22, 4 lt hh A li qua dd brm d th thy c 11,2 lt kh thot ra v khi lng bnh ng dung dch brm tng ln 28 gam. t chy hon ton kh thot ra ri dn qua dung dch NaOH th thu c 106 gam Na2CO3 v 84 gam NaHCO3. Xc nh CTPT ca ankan v anken

Cu 13. Hn hp X gm mt ankan A v mt anken B c cng s nguyn t C. T kh hi hn hp X i vi H 2bng 21,66. Nu cho 3,36 lt hn hp X qua bnh ng dung dch brom d thy c 8 gam brom phn ng. a. Xc nh cng thc phn t ca A, B. b. t chy hon ton 3,36 lt hn hp X ni trn cn bao nhiu lt khng kh.

Cu 14. Dn 3,36 lt hn hp gm CH4 v anken A qua dung dch brom d thy khi lng bnh brom tng 4,2gam. Kh thot ra khi bnh c th tch 1,12 lt . Xc nh CTPT ca A.

Mucdong_Tay Bac

L thuyt v bi tp anken 10 Cu 15. Cho 4,48 lt hn hp X gm 2 hirocacbon mch h li t t qua bnh cha 1,4 lt dung dch Br20,5M. Sau khi phn ng hon ton, s mol Br2 gim i mt na v khi lng bnh tng thm 6,7 gam. Cng thc phn t ca 2 hirocacbon l Cu 16. Hn hp X gm anken A v ankan B c s mol bng nhau,dX/H =18 . 2

Cho 6,72 lit hn hp X qua bnh

ng 100 gam dung dch Brom 16% n phn ng hon ton thu c hn hp kh c d B/H2 =20 . Tm CTPT ca A, B. Cu 20. Cho 1,12 lit anken A tc dng vi dung dch Brom va thu c 4,32 gam sn phm cng. Tm CTPT ca anken A.

IV. PHN NG OXI HA Cu 1. Thi 0,25 mol kh etilen qua 125 ml dung dch KMnO4 1M trong mi trng trung tnh (hiu sut 100%) Tnh khi lng etylen glicol thu c .

Cu 2. kh hon ton 200 ml dung dch KMnO4 0,2M to thnh cht rn mu nu en cn V lt kh C2H4 ( ktc). Gi tr ti thiu ca V l? .Cu 3. Cho 3,5g mt anken tc dng vi dung dch KMnO4 long th c 5,2g sn phm hu c. Tm CTPT ca anken. Cu 4. Mt hn hp hai olefin ng ng k tip nhau c th tch 17,92 lt (o 0 oC v 2,5 atm) dn qua bnh cha dung dch KMnO4 d, thy khi lng bnh cha dung dch KMnO4 tng 70g. 1. Xc nh cng thc phn t, vit cng thc cu to hai olefin. 2. Tnh % khi lng 2 olefin trong hn hp.

V. PHN NG TRNG HP Cu 1. Hin nay PVC c iu ch theo s sau: C2H4 CH2ClCH2Cl C2H3Cl PVC. Nu hiu sut ton b qu trnh t 80% th lng C2H4 cn dng sn xut 5000 kg PVC l?

VI. PHN NG CNG TCH Cu 1. Thc hin phn ng tch H2 6,72 lit hn hp X gm C2H6 v C3H8 thu c 11,2 lit hn hp Y gm cc anken, ankan v H2. Tnh th tch dung dch Brom 1M cn dng tc dng ht vi Y. Cu 2. Dn 2,24 lit kh propan qua bnh ng Niken nung nng thu c 3,92 lit hn hp kh Y. Dn ton b Y qua bnh ng dung dch Brom d th thy c m gam Brom tham gia phn ng. Tm m. Cu 3. Thc hin phn ng cracking 3,36 lit propan thu c hn hp A gm 2 hidrocacbon. Cho A qua bnh ng Vml dung dch Brom 0,5M thy dung dch Brom mt mu v kh Y thot ra khi bnh c d Y/H 2 = 9,5 . Tm V. Cu 4. Cracking 18 gam ankan A ri cho ton b sn phm thu c li qua bnh ng dung dich Brom d thy cn li 5,6 lit hn hp kh B gm cc ankan, d B/H 2 =13,6 . Tm CTPT ca A.

Mucdong_Tay Bac

Documents

anken hay