Bin son v ging dy : Gio vin Nguyn Minh Tun T Ha Trng THPT Chuyn Hng Vng Ph Th

MC LCTr ang

Li gii thiu Phn 1: Gii thiu cc chuyn ha hu c 11Chuyn 1 : i cng ha hu c Chuyn 2 : Hirocacbon no Chuyn 3 : Hirocacbon khng no Chuyn 4 : Hirocacbon thm Chuyn 5 : Dn xut halogen Phenol Ancol Chuyn 6 : Anehit Xeton Axit cacboxylic

2 9 9 49 87 147 173 239

Phn 2 : p n

315

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Trn bc ng thnh cng khng c du chn ca k li bing

1

Li gii thiuT gii thiuH v tn : Nguyn Minh Tun Gii tnh : Nam Ngy, thng, nm sinh : 31 05 1980 Trnh vn ha : 12/12 Trnh chuyn mn : C nhn Sinh Ha Tt nghip HSP H Ni 2 thng 06 2002 Hin l gio vin ging dy b mn ha hc Ngy vo ngnh : 31 12 2002 Ngy vo ng : 29 12 2009 Ngy vo ng chnh thc : 29 12 2010 i ch nh ring : S nh 16 T 9A Khu 5 Phng Gia Cm Vit Tr Ph Th S in thoi : 01689 186 513 a ch email : n gu yen m in h tu an ch v@yah oo.com .vn a ch facebook: nguyn minh tun (Vit Tr) h t t p : //www.fa ceb ook .com /n gu yen .m in h t u a n .1650? sk =wa ll Cc trng tng cng tc : Trng THPT Phng X (t thng 09 2002 n 04 2003) Trng THPT Xun ng (t thng 04 2003 n 08 2007) Trng THPT Chuyn Hng Vng (t thng 09 2007 n nay)

B ti liu n thi i hc, cao ng mn ha hcB ti liu trc nghim n thi i hc, cao ng mn ha hc do thy bin son gm 12 quyn :

Quyn 1 : Gii thiu 7 chuyn ha hc 10 Quyn 2 : Gii thiu 3 chuyn ha hc i cng v v c 11 Quyn 3 : Gii thiu 6 chuyn ha hc hu c 11 Quyn 4 : Gii thiu 4 chuyn ha hc hu c 12 Quyn 5 : Gii thiu 4 chuyn ha hc i cng v v c 12 Quyn 6 : Gii thiu cc chuyn phng php gii nhanh bi tp ha hc Quyn 7 : Gii thiu 40 luyn thi trc nghim mn ha hc Quyn 8 : Hng dn gii 7 chuyn ha hc 10 Quyn 9 : Hng dn gii 3 chuyn ha hc i cng v v c 11 Quyn 10 : Hng dn gii 6 chuyn ha hc hu c 11 Quyn 11 : Hng dn gii 4 chuyn ha hc hu c 12 Quyn 12 : Hng dn gii 4 chuyn ha hc i cng v v c 12

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Nhng iu thy mun ni :iu th nht thy mun ni vi cc em rng : la tui ca cc em, khng c vic g l quan trng hn vic hc tp. Hy c gng ln cc em nh, tng lai ca cc em ph thuc vo cc em y. iu th hai thy mun ni rng : Nu cc em c mt c m trong sng th ng v nhng kh khn trc mt m t b n. Thy tng cc em cu chuyn di y (do thy su tm), hi vng cc em s hiu c gi tr ca c m.

i bng v GNgy xa, c mt ngn ni ln, bn sn ni c mt t chim i bng. Trong t c bn qu trng ln. Mt trn ng t xy ra lm rung chuyn ngn ni, mt qu trng i bng ln xung v ri vo mt tri g di chn ni. Mt con g mi tnh nguyn p qu trng ln y. Mt ngy kia, trng n ra mt ch i bng con xinh p, nhng bun thay ch chim nh c nui ln nh mt con g. Chng bao lu sau, i bng cng tin n ch l mt con g khng hn khng km. i bng yu gia nh v ngi nh ang sng, nhng tm hn n vn khao kht mt iu g cao xa hn. Cho n mt ngy, trong khi ang chi a trong sn, i bng nhn ln tri v thy nhng ch chim i bng ang si cnh bay cao gia bu tri. " - i bng ku ln - c g ti c th bay nh nhng con chim ". By g ci m ln: "Anh khng th bay vi nhng con chim c. Anh l mt con g v g khng bit bay cao". i bng tip tc ngc nhn gia nh tht s ca n, m c c th bay cao cng h. Mi ln i bng ni ra m c ca mnh, by g li bo n iu khng th xy ra. l iu i bng cui cng tin l tht. Ri i bng khng m c na v tip tc sng nh mt con g. Cui cng, sau mt thi gian di sng lm g, i bng cht. Trong cuc sng cng vy: Nu bn tin rng bn l mt ngi tm thng, bn s sng mt cuc sng tm thng v v, ng nh nhng g mnh tin. Vy th, nu bn tng m c tr thnh i bng, bn hy eo ui c m ... v ng sng nh mt con g!

Chng tr nh n thi i hc cao ng mn ha hc Mn ha hc lp 10Chuyn s 01 02 03 04 05 06 07 Tn chuyn n tp ha hc 9 Nguyn t Bng tun hon cc nguyn t ha hc v nh lut tun hon Lin kt ha hc Phn ng ha hc Nhm halogen Nhm oxi Tc phn ng ha hc v cn bng ha hc S bui hc 05 06 05 05 10 07 07 05 50 bui

Mn ha hc lp 11Chuyn s 01 02 03 04 05 06 07 08 09 Tn chuyn S in li Nhm nit Nhm cacbon i cng ha hu c Hirocacbon no Hirocacbon khng no Hirocacbon thm Dn xut halogen. Ancol Phenol Anehit Xeton Axit cacboxylic S bui hc 06 06 03 06 05 10 04 10 10 60 bui

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Mn ha hc lp 12Chuyn s 1 2 3 axit Protein 4 polime 5 loi 6 Kim loi kim th Nhm 7 v mt s kim loi khc 8 Phn bit mt s cht v c. Ha hc v vn pht trin kinh t, x hi v mi trng Tn chuyn S bui hc Este Lipit 07 Cacbohirat 03 Amin Amino 07 Polime Vt liu 03 i cng v kim 07 Kim loi kim 10 Crom, st, ng 10 05 52 bui

Phng php gii nhanh bi tp ha hcChuyn s Tn chuyn 1 ng cho 2 chn lng cht 3 ton nguyn t 4 ton khi lng 5 gim khi lng, s mol, th tch kh 6 ton electron 7 i 8 dng phng trnh ion rt gn 9 ton in tch 10 dng cc gi tr trung bnh 6Trn bc ng thnh cng khng c du chn ca k li bing

S bui hc Phng php 02 Phng php t 02 Phng php bo 02 Phng php bo 02 Phng php tng 02 Phng php bo 02 Phng php quy 02 Phng php s 02 Phng php bo 02 Phng php s 02 20 bui

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luyn thi tr c nghim mn ha hcMi bui hc cha 02 , 40 cha trong 20 bui.

Hnh thc hc tpHc theo tng chuyn , mi chuyn ng vi mt chng trong sch gio khoa, quy trnh hc tp nh sau : + Tm tt l thuyt c bn ; ch trng, khc su kin thc trng tm m thi thng hay khai thc. + Phn dng bi tp c trng, c cc v d minh ha.

Trn bc ng thnh cng khng c du chn ca k li bing

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+ Cung cp h thng bi tp trc nghim theo cu trc : L thuyt trc, bi tp sau. Cc bi tp tnh ton c chia theo tng dng hc sinh d dng nhn dng bi tp v p dng phng php gii cc v d mu vo cc bi tp ny nhm rn luyn v nng cao k nng gii bi tp. + Cung cp h thng p n chnh xc hc sinh kim tra, nh gi kt qu hc tp ca mnh. + i vi nhng bi tp kh, hc sinh khng lm c, thy s hng dn gii bng nhiu cch sau cht li cch ngn gn nht (khong t 1 n 10 dng). + Sau mi chuyn s c mt bi kim tra trn lp, thng qua kt qu ca bi kim tra nh gi, xp loi, pht hin ra nhng im mnh v im yu ca tng hc sinh. Trn c s , s pht huy nhng im mnh v khc phc nhng im yu ca cc em nhm mc ch quan trng nht l nng cao thnh tch hc tp cho cc em.

Lu : i vi mt s em sinh d thi khi A, B v nhng l do no , n ht hc k 1 ca lp 12 m kin thc ha hc cn yu, khng p ng yu cu thi i hc, cao ng th c th n thy xin theo hc ly li kin thc. i vi cc em hc sinh nh vy thy s c mt chng trnh ring km cp cc em trong khong 40 bui : Ha i cng v v c hc 20 bui. Ha hu c hc 20 bui. Sau 40 bui hc cc em s ly li c nhng kin thc c bn nht v kt qu im thi i hc mn ha hc ca cc em s t c khong t 5 n 6 im hoc c th hn mt cht, tt nhin t c iu th cc em phi hc tp tht s nghim tc theo ng nhng hng dn ca thy. V tnh cht c bit nn nhng lp hc ny ch khong 1 n 5 hc sinh. T chc lp hc- a ch t chc lp hc : Tng 2 S nh 16 T 9A Khu 5 Phng Gia Cm Vit Tr (pha trong khu th Trm So). Phng hc sch s, c y nh sng, my iu ha, h thng cch m vi bn ngoi.

Cc em hc sinh khu vc Vit Tr, Lm Thao, Phong Chu, Ph Ninh c nhu cu hc thm nng cao kin thc v mua sch tham kho mn ha hc hy lin h vi thy gio Nguyn Minh Tun Gio vin trng THPT Chuyn Hng Vng theo s in thoi 01689186513 hoc email n gu yen m inh tu an ch v@yah oo.com .vn Cc em hc sinh tnh ngoi nu cn mua sch th ng k vi thy, thy s gi sch qua ng bu in cho cc em.

Cu chuyn ca hai v nhn

C mt cu hc sinh 18 tui ang gp kh khn trong vic tr tin hc. Cu ta l mt a tr m ci, v cu ta khng bit i ni u kim ra tin. Th l anh chng ny bn ny ra mt sng kin. Cu ta cng mt ngi bn khc quyt nh t chc mt bui nhc hi ngay trong khun vin trng gy qu cho vic hc. H tm n ngi ngh s dng cm i ti Ignacy J Paderewski. Ngi qun l ca Paderewski yu cu mt khon ph bo m $2000 cho ng y c biu din. Sau khi h tha thun xong, hai ngi sinh vin y bt tay ngay vo cng vic chun b cho bui trnh din c thnh cng. Ngy trng i y cui cng n. Paderewski cui cng cng biu din ti Stanford. Th nhng khng may l v vn cha c bn ht. Sau khi tng kt s tin bn v li, h ch c c $1600. Qu tht vng, h n ch ca ca Paderewski trnh by hon cnh ca mnh. Hai ngi sinh vin y a Paderewski ton b s tin bn v, cng vi 1 check n $400, v ha rng h s tr s n y sm nht c th. KHNG, Paderewski ni. Ci ny khng th no chp nhn c. ng ta x t check, tr li $1600 cho hai chng thanh nin v ni : y l 1600 , sau khi tr ht tt c cc chi ph cho bui biu din th cn bao nhiu cc cu c gi ly cho vic hc. Cn d bao nhiu th hy a cho ti. Hai cu sinh vin y v cng bt ng, xc ng cm n Paderewski.. y ch l mt lm nh, nhng chng minh c nhn cch tuyt vi ca Paderewski. Ti sao ng y c th gip hai ngi m ng y thm ch khng h quen bit. Chng ta tt c u bt gp nhng tnh hung nh vy trong cuc sng ca mnh. V hu ht chng ta u ngh : Nu chng ta gip h, chng ta s c g ?. Th nhng, nhng ngi v i h li ngh khc: Gi s chng ta khng gip h, iu g s xy ra vi nhng con ngi ang gp kh khn y?. H khng mong i s n p, H lm ch v h ngh l vic nn lm, vy thi. Ngi ngh s dng cm tt bng Paderewski hm no sau ny tr thnh Th Tng ca Ba Lan. ng y l mt v lnh o ti nng. Th nhng khng may chin tranh th gii n ra, v t nc ca ng b tn ph nng n. C hn mt triu ri ngi Ba Lan ang b cht i, v by gi chnh ph ca ng khng cn tin c th nui sng h c na. Paderewski khng bit i u tm s gip . ng ta bn n C Quan Cu Tr Lng Thc Hoa K nh s tr gip. Ngi ng u c quan chnh l Herbert Hoover, ngi sau ny tr thnh Tng Thng Hp Chng Quc Hoa K. ng Hoover ng gip v nhanh chng gi hng tn lng thc cu gip nhng ngi Ba Lan ang b i kht y. Thm ha cui cng cng c ngn chn. Th Tng Paderewski lc by gi mi cm thy nh nhm. ng bn quyt nh i sang M t mnh cm n ng Hoover v c ch cao qu ca Trn bc ng thnh cng khng c du chn ca k li bing 8

ng y gip ngi dn Ba Lan trong nhng lc kh khn. Th nhng khi Paderewski chun b ni cu cm n th ng Hoover vi ct ngang v ni : Ngi khng cn phi cm n ti u, tha ngi Th Tng. C l ngi khng cn nh, nhng nhiu nm v trc, ngi c gip hai cu sinh vin tr tui bn M c tip tc i hc, v ti l mt trong hai chng sinh vin y Th gii ny ng tht l tuyt vi, khi bn cho i th g, bn s nhn c nhng iu tng t.

Trn bc ng thnh cng khng c du chn ca k li bing

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PHN 1: GII THIU CC CHUYN HA HU C 11CHUYN 1 :L THUYTI. HA HC HU C V HP CHT HU C 1. Khi nim v hp cht hu c v ha hc hu c - Hp cht hu c l hp cht ca cacbon (tr CO, CO2, HCN, mui cacbonat, mui xianua, mui cacbua). - Ha hc hu c l nghnh ha hc nghin cu cc hp cht hu c. 2. c im chung ca hp cht hu c - c im cu to : Lin kt ha hc ch yu trong hp cht hu c l lin kt cng ha tr. - Tnh cht vt l : + Nhit nng chy, nhit si thp. + Phn ln khng tan trong nc, nhng tan nhiu trong cc dung mi hu c. - Tnh cht ha hc : + Cc hp cht hu c thng km bn vi nhit v d chy. + Phn ng ha hc ca cc hp cht hu c thng xy ra chm v theo nhiu hng khc nhau, nn to ra hn hp nhiu sn phm. II. PHN LOI V GI TN CC HP CHT HU C 1. Phn loi - Hp cht hu c thng chia thnh hai loi : + Hirocacbon : L nhng hp cht hu c trong phn t ch cha hai nguyn t C, H. Hirocacbon li c chia thnh cc loi : Hirocacbon no (CH4, C2H6) ; hirocacbon khng no (C2H4, C2H2) ; hirocacbon thm (C6H6, C7H8). + Dn xut ca hirocacbon : L nhng hp cht hu c m trong phn t ngoi cc nguyn t C, H th cn c nhng nguyn t khc nh O, N, Cl, S. Dn xut ca hidrocacbon li c chia thnh dn xut halogen nh CH3Cl, C6H5Br,; ancol nh CH3OH, C2H5OH,; anehit nh HCHO, CH3CHO. 2. Nhm chc - L nhng nhm nguyn t (-OH, -CHO, -COOH, -NH2) gy ra phn ng c trng ca phn t hp cht hu c. 3. Danh php hu c a. Tn thng thng Tn thng thng ca hp cht hu c thng hay c t theo ngun gc tm ra chng, i khi c th c phn ui ch r hp cht thuc loi no. V d : HCOOH : axit fomic ; CH3COOH : axit axetic ; C10H20O : mentol (formica : Kin) (acetus : Gim) (mentha piperita : Bc h)

I CNG HA HC HU C

10

Trn bc ng thnh cng khng c du chn ca k li bing

b. Tn h thng theo danh php IUPAC Tn gc - chc Tn gc - chc Tn phn gc CH3CH2 - Cl (etyl || clorua)

Tn phn nh chc CH3 CH2 - O - CH3 (etyl metyl || ete)

CH3CH2 -O-COCH3 (etyl || axetat )

Tn thay thH | Vi d : H C H | H H | Cl C H | H

H H | | H C C H | | H H

H H | | Cl C C H | | H H

Metan Clometan Etan Cloetan Tn thay th c vit lin (khng vit cch nh tn gc - chc), c th c phn lm ba phn nh sau : H3C -CH3 (et + an) etan 1 1 12 3 4

H3C -CH2Cl (clo + et + an) cloetan1 2 3 4

H2C =CH2 (et + en) eten1

OH2| 3

HC CH (et + in) etin4

CH2=CH-CH2-CH3

CH3-CH=CH-CH3

CH3 CH CH= CH 2

but-1-en but-2-en but-3-en-2-ol gi tn hp cht hu c, cn thuc tn cc s m v tn mch cacbon S m 1 m ono 2 i 3 t ri 4 t etra 5 p enta 6 h exa 7 h epta 8 o cta 9 n Mch cacbon chnh ona 10 eca

C met C-C et C-C-C prop C-C-C-C but C-C-C-C-C pent C-C-C-C-C-C hex C-C-C-C-C-C-C hep C-C-C-C-C-C-C-C oct C-C-C-C-C-C-C-C-C non C-C-C-C-C-C-C-C-C-C ec

Trn bc ng thnh cng khng c du chn ca k li bing

1 1

Khng xut pht t s m Xut pht t s m III. S LC V PHN TCH NGUYN T 1. Phn tch nh tnh - Mc ch : Xc nh nguyn t no c trong hp cht hu c. - Nguyn tc : Chuyn cc nguyn t trong hp cht hu c thnh cc cht v c n gin ri nhn bit chng bng cc phn ng c trng.

12

Trn bc ng thnh cng khng c du chn ca k li bing

2. Phn tch nh lng - Mc ch : Xc nh thnh phn % v khi lng cc nguyn t c trong phn t hp cht hu c. Nguyn tc : Cn chnh xc khi lng hp cht hu c, sau chuyn nguyn t C thnh CO2, H thnh H2O, N thnh N2, sau xc nh chnh xc khi lng hoc th tch ca cc cht to thnh, t tnh % khi lng cc nguyn t. Biu thc tnh ton : 12.mCO2 m = gam ; m = 2.m H2 O gam ; m = 28.VN2 gamC

44 m C .100

H

18 m H .100 ; %N =

N

22, 4

; %O = 100% - %C - %H - %N a a a IV. CNG THC PHN T HP CHT HU C 1. Cng thc tng qut (CTTQ) - Cho bit trong phn t hp cht hu c c cha nhng nguyn t no. V d ng vi cng thc CxHyOzNt ta bit hp cht hu c ny c cc nguyn t C, H, O, N. 1. Cng thc n gin nht (CTGN) a. nh ngha -Cng thc n gin nht l cng thc biu th t l ti gin v s nguyn t ca cc nguyn t trong phn t. b. Cch thit lp cng thc n gin nht - Thit lp cng thc n gin nht ca hp cht hu c CxHyOzNt l thit lp t l : m m m m x : y : z : t = n : n : n : n= C : H : O : N hoc x : y : z : t = %C : %H : %O : %NC H O N

- Tnh c : %C =

; %H =

m N .100

12

1

16

14

12

1

16

14

c. Cng thc thc nghim (CTTN): CTTN = (CTGN)n (n : s nguyn dng). 2. Cng thc phn t a. nh ngha - Cng thc phn t l cng thc biu th s lng nguyn t ca mi nguyn t trong phn t. b. Cch thit lp cng thc phn t - C ba cch thit lp cng thc phn t Cch 1 : Da vo thnh phn % khi lng cc nguyn t - Cho CTPT CxHyOz: ta c t l M 100 = 12.x %C = 1.y = 16.z

%H %O M.%C T ta c : x = ; M.%H M.%O y= ; z= 12.100 1.100 16.100

Cch 2 : Da vo cng thc n gin nht. Cch 3 : Tnh trc tip theo khi lng sn phm chy. V. CU TRC PHN T HP CHT HU C 1. Ni dung ca thuyt cu to ho hc a. Trong phn t hp cht hu c, cc nguyn t lin kt vi nhau theo ng ho tr v theo mt th t nht nh. Th t lin kt c gi l cu to ho hc. S thay i th t lin kt , tc l thay i cu to ho hc, s to ra hp cht khc.

V d : Cng thc phn t C2H6O c hai th t lin kt (2 cng thc cu to) ng vi 2 hp cht sau : H3C-O-CH3 : imetyl ete, cht kh, khng tc dng vi Na. H3C-CH2-O-H : ancol etylic, cht lng, tc dng vi Na gii phng hiro. b. Trong phn t hp cht hu c, cacbon c ho tr 4. Nguyn t cacbon khng nhng c th lin kt vi nguyn t ca cc nguyn t khc m cn lin kt vi nhau thnh mch cacbon. V d : CH3-CH2-CH2-CH3 ; CH3-CH-CH3 ; CH2-CH2 CH3(mch khng nhnh) (mch c nhnh)

CH2-CH2(mch vng)

CH2

c. Tnh cht ca cc cht ph thuc vo thnh phn phn t (bn cht, s lng cc nguyn t) v cu to ho hc (th t lin kt cc nguyn t). V d :- Ph thuc thnh phn phn t : CH4 l cht kh d chy, CCl4 l cht lng khng chy ;

CH3Cl l cht kh khng c tc dng gy m, cn CHCl3 l cht lng c tc dng gy m. cht ho hc. 2. Hin tng ng ng, ng phn a. ng ng- Ph thuc cu to ho hc : CH3CH2OH v CH3OCH3 khc nhau c v tnh cht vt l v tnh

Cc hirocacbon trong dy : CH4, C2H6, C3H8, C4H10, C5H12, ..., CnH2n+2, cht sau hn cht trc 1 nhm CH2 nhng u c tnh cht ho hc tng t nhau. Cc ancol trong dy : CH3OH, C2H5OH, C3H7OH, C4H9OH,... CnH2n+1OH cng c thnh phn hn km nhau mt hay nhiu nhm CH2 nhng c tnh cht ho hc tng t nhau. Khi nim : Nhng hp cht c thnh phn phn t hn km nhau mt hay nhiu nhm CH2 nhng c tnh cht ho hc tng t nhau l nhng cht ng ng, chng hp thnh dy ng ng. Gii thch : Mc d cc cht trong cng dy ng ng c cng thc phn t khc nhau nhng nhm CH2 nhng do chng c cu to ho hc tng t nhau nn c tnh cht ho hc tng t nhau. b. ng phn Etanol (C2H5OH) v imetyl ete (CH3OCH3) l 2 cht khc nhau (c tnh cht khc nhau) nhng li c cng cng thc phn t l C2H6O. Metyl axetat (CH3COOCH3), etyl fomiat (HCOOC2H5) v axit propionic (CH3CH2COOH) l 3 cht khc nhau nhng c cng cng thc phn t l C3H6O2. Khi nim : Nhng hp cht khc nhau nhng c cng cng thc phn t l nhng cht ng phn. Gii thch : Nhng cht ng phn tuy c cng cng thc phn t nhng c cu to ho hc khc nhau, chng hn etanol c cu to H3C-CH2-O-H, cn imetyl ete c cu to H3C-O-CH3, v vy chng l nhng cht khc nhau, c tnh cht khc nhau. 3. Lin kt tr ong phn t hp cht hu c a. Cc loi lin kt tr ong phn t hp cht hu c Theo Li-ut (Lewis), cc nguyn t c xu hng dng chung electron t c 8 electron

lp ngoi cng (Quy tc bt t), (i vi H ch cn t 2 electron). V d :

hoc

hoc

Lin kt to bi 1 cp electron dng chung l lin kt n. Lin kt n thuc loi lin kt . Lin kt n c biu din bi 2 du chm hay 1 gch ni gia 2 nguyn t. Lin kt to bi 2 cp electron dng chung l lin kt i. Lin kt i gm 1 lin kt v 1 lin kt , biu din bi 4 du chm hay 2 gch ni. Lin kt to bi 3 cp electron dng chung l lin kt ba. Lin kt ba gm 1 lin kt v 2 lin kt , biu din bi 6 du chm hay 3 gch ni. Lin kt i v lin kt ba gi chung l lin kt bi. Nguyn t C s dng obitan lai ho to lin kt theo kiu xen ph trc (hnh a, b) v dng obitan p to lin kt theo kiu xen ph bn (hnh c).

b. Cc loi cng thc cu to Cng thc cu to biu din th t v cch thc lin kt ca cc nguyn t trong phn t. C cch vit khai trin, thu gn v thu gn nht.

Cng thc cu to khai trin : Vit tt c cc nguyn t v cc lin kt gia chng. Cng thc cu to thu gn : Vit gp nguyn t cacbon v cc nguyn t khc lin kt vi n thnh tng nhm. Cng thc cu to thu gn nht : Ch vit cc lin kt v nhm chc, u mt ca cc lin kt chnh l cc nhm CHx vi x m bo ho tr 4 C.

4. ng phn cu to a. Khi nim ng phn cu to Nhng hp cht c cng cng thc phn t nhng c cu to ho hc khc nhau gi l nhng ng phn cu to. b. Phn loi ng phn cu to - ng phn cu to chia lm ba loi : ng phn mch cacbon ; ng phn nhm chc v ng phn v tr nhm chc. - Nhng ng phn khc nhau v bn cht nhm chc gi l ng phn nhm chc. Nhng ng phn khc nhau v s phn nhnh mch cacbon gi l ng phn mch cacbon. Nhng ng phn khc nhau v v tr ca nhm chc gi l ng phn v tr nhm chc. 5. ng phn lp th a. Khi nim v ng phn lp th V d : ng vi cng thc cu to CHCl = CHCl c hai cch sp xp khng gian khc nhau dn ti hai cht ng phn :

ng phn lp th ca CHCl = CHCl

Kt lun : ng phn lp th l nhng ng phn c cu to ho hc nh nhau (cng cng thc cu to) nhng khc nhau v s phn b khng gian ca cc nguyn t trong phn t (tc khc nhau v cu trc khng gian ca phn t). VI. PHN NG HU C 1. Phn loi phn ng hu c Da vo s bin i phn t hp cht hu c khi tham gia phn ng ngi ta phn phn ng hu c thnh cc loi sau y : a. Phn ng th Mt hoc mt nhm nguyn t phn t hu c b th bi mt hoc mt nhm H3C-H + Cl-Cl as H3C-Cl + HCl nguyn t khc. b. Phn ng cng Phn t hu c kt hp thm vi cc nguyn t hoc phn t khc. c. Phn ng tch Mt vi nguyn t hoc nhm nguyn H3C-OH + H-Br H3C-Br + HOHo

HC CH + 2H2 xt, H3C - CH3 t

t b tch ra khi phn t.

H2 C CH 2| |

H2C=

H ,t 0

+

CH2 + H2O

H OH

2. Cc kiu phn ct lin kt cng ho tr a. Phn ct Trong s phn ct ng li, i electron dng chung c ng li chia u cho hai nguyn t lin kt to ra cc tiu phn mang electron c thn gi l gc t do. Gc t do m electron c thn nguyn t cacbon gi l gc cacbo t do. Gc t do thng c hnh thnh nh nh sng hoc nhit v l nhng tiu phn c kh nng phn ng cao. b. Phn ct d li

Trong s phn ct d li, nguyn t c m in ln hn chim c cp electron dng chung tr thnh anion cn nguyn t c m in nh hn b mt mt electron tr thnh cation. Cation m in tch dng nguyn t cacbon c gi l cacbocation. Cacbocation thng c hnh thnh do tc dng ca dung mi phn cc. 3. c tnh chung ca gc cacbo t do v cacbocation Gc cacbo t do (k hiu l R ), cacbocation (k hiu l R ) u rt khng bn, thi gian tn ti rt ngn, kh nng phn ng cao. Chng c sinh ra trong hn hp phn ng v chuyn ho ngay thnh cc phn t bn hn, nn c gi l cc tiu phn trung gian. Ngi ta ch nhn ra chng nh cc phng php vt l nh cc phng php ph, m thng khng tch bit v c lp c chng. Quan h gia tiu phn trung gian vi cht u v sn phm phn ng c thy qua cc v d sau : Cht u Tiu phn trung gian Sn phmg

+

PHNG PHP LP CNG THC CA HP CHT HU C Lp cng thc phn t hp cht hu c khi bit cng thc n gin nht Phng php gii- Bc 1 : t cng thc phn t ca hp cht hu c l : (CTGN)n (vi n N ) - Bc 2 : Tnh bt bo ha ( ) ca phn t (ch p dng cho hp cht c cha lin kt cng ha tr, khng p dng cho hp cht c lin kt ion). + i vi mt phn t th 0 v N . + i vi cc hp cht c nhm chc cha lin kt nh nhm CHO, COOH, th s lin kt nhm chc (v gc hirocacbon cng c th cha lin kt ). - Bc 3 : Da vo biu thc chn gi tr n (n thng l 1 hoc 2), t suy ra CTPT ca hp cht hu c. Lu : Gi s mt hp cht hu c c cng thc phn t l CxHyOzNt th tng s lin kt v vng ca phn t c gi l bt bo ha ca phn t . Cng thc tnh bt bo ha :*

x ( 4 2) + y (1 2) + z (2 2) + t (3 2) + 2 2x ( 0 v N ) = = y+t+2 2 2

Cc v d minh ha V d 1: Hp cht X c CTGN l CH3O. CTPT no sau y ng vi X ? A. C3H9O3. B. C2H6O2. C. CH3O. D. Khng xc nh c. Hng dn gii t cng thc phn t (CTPT) ca X l (CH3O)n (n N ). 2n 3n + 2 2 n bt bo ha ca phn t = = 0. 2 2 V bt bo ha ca phn t N nn suy ra n = 2. Vy cng thc phn t ca A l C2H6O2. p n B. V d 2: Hp cht X c CTGN l C4H9ClO. CTPT no sau y ng vi X ? A. C4H9ClO. B. C8H18Cl2O2. C. C12H27Cl3O3. D. Khng xc nh c. Hng dn gii t cng thc phn t ca X l (C4H9OCl)n (n N ). 8n 10n + 2 2 2n bt bo ha ca phn t = = = 1 n 0 . 2 2 V bt bo ha ca phn t N nn suy ra n = 1. Vy cng thc phn t ca X l C4H9OCl. p n B. V d 3: Axit cacboxylic A c cng thc n gin nht l C3H4O3. A c cng thc phn t l : A. C3H4O3. B. C6H8O6. C. C18H24O18. D. C12H16O12.* *

Hng dn gii t cng thc phn t ca X l (C3H4O3)n (n N ).*

bt bo ha ca phn t =

6n 4n + 2

2 2 V bt bo ha ca phn t N nn suy ra n = 2.

=

2 + 2n

2

3n

2 n 0 n 2 . 2

Vy cng thc phn t ca X l C6H8O6. p n B. 3n : Mt chc axit COOH c 2 nguyn t O c mt lin kt . Vy Gii thch ti sao 2 phn t axit c 3n nguyn t O th c s lin kt l axit cng c th c cha lin kt . 3n 2 . Mt khc, gc hirocacbon ca phn t

Lp cng thc n gin nht, cng thc phn t hp cht hu c khi bit thnh phn phn tr m v khi lng ca cc nguyn t; khi lng ca cc nguyn t v khi lng phn t ca hp cht hu c Phng php gii- Bc 1 : Lp t l mol ca cc nguyn t trong hp cht hu c : nC : n H : nO : n N %C %H %O %N m C m H m O (1) : : : = = : : : mN 12 1 16 14 12 1 16 14

- Bc 2 : Bin i t l trn thnh t l ca cc s nguyn n gin nht (thng ta ly cc s trong dy (1) chia cho s b nht ca dy . Nu dy s thu c vn cha phi l dy s nguyn ti gin th ta bin i tip bng cch nhn vi 2 ; 3 ;), suy ra cng thc n gin nht. - Bc 3 : t CTPT = (CTGN)n n.MCTGN = M (M l KLPT ca hp cht hu c) n CTPT ca hp cht hu c.

Cc v d minh ha V d 1: Mt cht hu c A c 51,3% C ; 9,4% H ; 12% N ; 27,3% O. T khi hi ca A so vi khng kh l 4,034. a. Xc nh CTGN ca A. b. Xc nh CTPT ca A. Hng dn gii a. Xc nh CTGN ca 51, 3 9, 4 27, 3 12 A: Ta c : n C : n H : n O : n N = 16 Vy cng thc n gin nht ca A l C5H11O2N. b. Xc nh CTPT ca A : t cng thc phn t ca A l (C5H11O2N)n. Theo gi thit ta c : : : : = 4, 275 : 9, 4 :1, 706 : 0, 857 = 5 :11: 2 :1 14

12

1

(12.5 + 11 + 16.2 + 14).n = 4,034.29 n = 1 Vy cng thc phn t ca A l C5H11O2N.

V d 2: Cht hu c A cha 7,86% H ; 15,73% N v khi lng. t chy hon ton 2,225 gam A thu c CO2, hi nc v kh nit, trong th tch kh CO2 l 1,68 lt (ktc). CTPT ca A l (bit MA < 100) : A. C6H14O2N. B. C3H7O2N. C. C3H7ON. D. C3H7ON2. Hng dn gii 1, 68 Ta c : n = n 2 = 0, 9 gam 2 = 0, 075 mol C CO = C %C = m 22, 4 0, 9 .100 = 40, 45% . 2, 225

Do : %O = (100 40,45 15,73 7,86)% = 35,96%. 40, 45 7, 86 35, 96 15, 73 : : : = 3, 37 : 7, 86 : 2, 2475 :1,124 = 3 : 7 : 2 :1 nC : nH : nO : n N = 12 1 16 14 Cng thc n gin nht ca A l C3H7O2N. t cng thc phn t ca A l (C3H7O2N)n. Theo gi thit ta c : (12.3 + 7 + 16.2 + 14).n < 100 n < 1,12 n =1 Vy cng thc phn t ca A l C3H7O2N. p n B. V d 3: Mt hp cht hu c Z c % khi lng ca C, H, Cl ln lt l : 14,28% ; 1,19% ; 84,53%. CTPT ca Z l : A. CHCl2. B. C2H2Cl4. C. C2H4Cl2. D. mt kt qu khc. Ta c : n C : n H : n Cl = Hng dn gii 14, 28 1,19 84,53 : : = 1:1: 2 12 1 3, 35

cng thc n gin nht ca Z l CHCl2. t cng thc phn t ca A l (CHCl2)n (n N ). 2n 3n + 2 2 n bt bo ha ca phn t = = 0. 2 2 V bt bo ha ca phn t N nn suy ra n=2. Vy cng thc phn t ca Z l : C2H2Cl4. p n B. V d 4: Cht hu c X c M = 123 v khi lng C, H, O v N trong phn t theo th t t l vi 72 : 5 : 32 : 14. CTPT ca X l : A. C6H14O2N. B. C6H6ON2. C. C6H12ON. D. C6H5O2N. Hng dn gii 72 5 32 14 Ta c : n C : n H : n O : n N = : : : = 6 : 5 : 2 :1 . 12 1 16 14 Cn c vo cc phng n ta thy CTPT ca X l : C6H5O2N. p n D.*

Lp cng thc phn t ca hp cht hu c da vo kt qu ca qu tr nh phn tch nh lng.Cch 1 : T cc gi thit ca bi, ta tin hnh lp CTGN ri t suy ra CTPT.

Phng php gii- Bc 1 : T gi thit ta tnh c nC, nH, nN mC, mH, mN. p dng nh lut bo ton khi lng cho cc nguyn t trong hp cht hu c (hchc), suy ra mO (trong hchc)= mhchc - mC - mH mN nO (trong hchc) - Bc 2 : Lp t l mol ca cc nguyn t trong hp cht hu c : n C : n H : n O : n N (1) - Bc 3 : Bin i t l trn thnh t l ca cc s nguyn n gin nht (thng ta ly cc s trong dy (1) chia cho s b nht ca dy . Nu dy s thu c vn cha phi l dy s nguyn ti gin th ta bin i tip bng cch nhn vi 2 ; 3 ;), suy ra cng thc n gin nht. - Bc 4 : t CTPT = (CTGN)n n.MCTGN = M (M l KLPT ca hp cht hu c) n CTPT ca hp cht hu c.

Cc v d minh ha V d 1: Khi t chy hon ton mt amin n chc X, thu c 16,80 lt kh CO2 ; 2,80 lt N2 (cc th tch o ktc) v 20,25 gam H2O. CTPT ca X l : A. C4H9N. B. C3H7N. C. C2H7N. D. C3H9N. Hng dn gii Ta c : n C = n CO = 16,8 = 0, 75 mol; H = 2.n H O = 2. 20, 25 = 2, 25 2 2 n 22, 4 mol; 18 2, 8 n N = 2.n 2 = 2. = 0, 25 mol. 22, 4 N n C : n H : n N = 0, 75 : 2, 25 : 0, 25 = 3 : 9 :1 . Cn c vo cc phng n ta thy CTPT ca X l C3H9N. p n D. V d 2: Oxi ha hon ton 4,02 gam mt hp cht hu c X ch thu c 3,18 gam Na2CO3 v 0,672 lt kh CO2. CTGN ca X l : A. CO2Na. B. CO2Na2. C. C3O2Na. D. C2O2Na. Hng dn gii Ta c : nNa

3,18 6, 72 3,18 = 2.n = 0, 06 mol;C = n CO + n Na 2CO3 = + = 0, 06 2 3 2 2 Na CO = 2. n 106 mol 22, 4 106 4, 02 0, 06.23 0, 06.12 = 0,12 mol n : n : n := 0, 06 : 0, 06 : 0,12 = 1:1:C H O

n 2O (hchc)

=

16

Vy CTGN ca X l : CNaO2. p n A. Trn y l nhng v d n gin. Ngoi ra c nhng bi tp tm cng thc phn t ca hp cht hu c ta phi p dng mt s nh lut nh : nh lut bo ton nguyn t, nh lut bo ton khi lng. i vi nhng bi tp m lng cht phn ng v lng sn phm thu c l nhng i lng c cha tham s, khi ta s dng phng php t chn lng cht chuyn bi tp phc tp thnh bi tp n gin.

V d 3: t chy hon ton m gam mt amin X bng lng khng kh va thu c 17,6 gam CO2, 12,6 gam H2O v 69,44 lt N2 (ktc). Gi thit khng kh ch gm N2 v O2 trong oxi chim 20% th tch khng kh. X c cng thc l : A. C2H5NH2. B. C3H7NH2. C. CH3NH2. D. C4H9NH2. Hng dn gii 12, 6 = 1, 4 mol . Ta c : n = n 2 = 17, 6 = 0, 4 mol;H = 2.n H O = 2. 2 C CO 18 n 44 p dng nh lut bo ton nguyn t i vi oxi suy ra : 2.n CO + n H O 2 2 n O2 (kk ) = 0, 75 mol 2 (kk ) = 0, 75.4 = 3 mol. N = 2 n 69, 44 n = 2.( 3) = 0, 2 mol n : n = 0, 4 :1, 4 : 0, 2 = 2 : 7 :1 :n Do : N (hchc) C H N 22, 4 Cn c vo cc phng n ta thy cng thc ca X l C2H5NH2. p n A. V d 4: t chy hon ton 1,47 gam cht hu c X (ch cha C, H, O) bng 1,0976 lt kh O2 ( ktc) lng dng va th sau th nghim thu c H2O, 2,156 gam CO2. Tm CTPT ca X, bit t khi hi ca X so vi khng kh nm trong khong 3< dX < 4. A. C3H4O3. B. C3H6O3. C. C3H8O3. D. p n khc. Hng dn gii p dng nh lut bo ton khi lng ta c : m X + m O2 = m CO + m2H O m H O = 0,882 gam 2 2 n C = n CO = 2,156 = 0, 049 mol; H = 2.n H O = 2. 0,882 = 0, 098 2 2 2 n 44 mol 18 nO(hchc)

=

1, 47 0, 049.12 0, 098 = 0, 049 16 mol

n C : n H : n O = 0, 049 : 0, 098 : 0, 049 = 1: 2 :1 CTGN ca X l : CH2O t cng thc phn t ca X l (CH2O) n. Theo gi thit ta c : 3.29 < 30n < 4.29 2,9 < n < 3,87 n =3 Vy CTPT ca X l C3H6O3. p n B. V d 5: t chy hon ton 1,88 gam cht hu c A (cha C, H, O) cn 1,904 lt O2 (ktc) thu c CO2 v hi nc theo t l th tch 4 : 3. Hy xc nh cng thc phn t ca A. Bit t khi ca A so vi khng kh nh hn 7. A. C8H12O5. B. C4H8O2. C. C8H12O3. D. C6H12O6. Hng dn gii Theo gi thit: 1,88 gam A + 0,085 mol O2 4a mol CO2 + 3a mol H2O. p dng nh lut bo ton khi lng ta c :

m CO + m H2

2O

= 1,88 + 0, 085.32 = 46 gam

Ta c : 44.4a + 18.3a = 46 a = 0,02 mol Trong cht A c: nC = 4a = 0,08 mol ; nH = 3a.2 = 0,12 mol ; nO = 4a.2 + 3a 0,085.2 = 0,05 mol

nC : nH : nO = 0,08 : 0,12 : 0,05 = 8 : 12 : 5 Vy cng thc ca cht hu c A l C8H12O5 c MA < 203. p n A. V d 6: Phn tch x gam cht hu c X ch thu c a gam CO2 v b gam H2O. Bit 3a = 11b v 7x = 3(a + b). T khi hi ca X so vi khng kh nh hn 3. CTPT ca X l : A. C3H4O. B. C3H4O2. C. C3H6O. D. C3H6O2. Hng dn gii n gin cho vic tnh ton ta chn : b = 18 gam a = 66 gam, x = 36 gam. Ta c : n C = n CO = 66 = 1, 5 mol;H = 2.n H O = 2. 18 = 2 mol; 2 2 2 n 44 n 18 n C : n H : n O = 1, 5 : 2 :1 = 3 : 4 : 2 Cn c vo cc phng n ta thy CTPT ca X l C3H4O2. p n B. V d 7: t chy hon ton m gam ancol X, sn phm thu c cho i qua bnh ng dung dch nc vi trong d thy khi lng bnh tng thm p gam v c t gam kt ta. Cng thc ca X l m+p (Bit p = 0,71t ; t = ): 1, 02 A. C2H5OH. B. C3H5(OH)3. C. C2H4(OH)2. Hng dn gii D. C3H5OH.O (hchc)

=

36 1, 5.12 2 = 1 mol. 16

Chn t =

m + p = 100 gam 1, 02

p = 71 gam ; m = 31 gam Gi cng thc tng qut ca ancol R l CxHyOz Phng trnh phn ng : y z y (x + )O xCO + H O Cx HyOz + CO2 4 2 2 + Ca(OH)2 CaCO + H O 3 2 n = nC 2 2 2

(1)

(2)

Theo phng trnh (2)

CO 2

= n

CaCO3

= 1 mol

Khi lng bnh tng ln: p = m m V nH2O

CO2

+m

H2 O

= 71 44 = 27 gam n = 1, 5 mol H O2

>n

H2O

CO2

nn ancol X l ancol no

31 (12 + 1, 5.2 ) nO = = 1 mol 16 Vy ta c x : y : z = nC : nH : nO = 1 : 3 : 1 t cng thc phn t (CTPT) ca X l (CH3O)n (n N ). 2n 3n + 2 2 n bt bo ha ca phn t = = 0. 2 2 V bt bo ha ca phn t N nn suy ra n = 2.*

Vy cng thc phn t ca A l C2H6O2 hay CTCT l C2H4(OH)2. p n C. V d 8: Hn hp X gm 2 hirocacbon A v B c khi lng a gam. Nu em t chy hon ton 132a gam CO2 v 45a gam H2O. Nu thm vo X mt na lng A c trong X ri X th thu c 41 41 t chy hon ton th thu c 60, 75a gam H O. Tm cng thc phn t ca 165a gam CO2 2 v 41 41

A v B. Bit X khng lm mt mu dung dch nc brom v A, B thuc loi hirocacbon hc. Hng dn gii Gi s a = 41 gam Khi t chy X: n CO = 132 = 3 mol ; H O = 45 = 2, 5 2 2 n 44 mol 18 Khi t chy X + 1 2 1 2 165 60, 75 A: n CO 2 = = 3, 75 mol ; H2 O = = 3,375 n 44 mol 18 A ta thu c: n = 0, 75 mol ; n = 0, 875 mol

Vy khi t chy

CO 2

H2O

V nCO2

1.

MN 2

p n C. V d 9: t chy hon ton m gam hirocacbon A. Sn phm thu c hp th vo nc vi trong d th to ra 4 gam kt ta. Lc kt ta, cn li bnh thy khi lng bnh nc vi trong gim 1,376 gam. A c cng thc phn t l : A. CH4. B. C5H12. C. C3H8 . D. C4H10. Hng dn gii Do Ca(OH)2 d nn CO2 chuyn ht vo kt ta CaCO3. Ta c : n = = n = 0, 04 mol. n CO CaCO3 C 2 2 Cho sn phm chy gm CO2 v H2O vo bnh nc vi trong d. Lc kt ta cn li bnh thy khi lng bnh nc vi trong gim 1,376 gam iu c ngha l khi lng kt ta b tch ra khi dung dch ln hn khi lng H2O v CO2 hp th vo bnh. Suy ra : m CaCO3 m H2O m CO2 = 1,376 gam m H2O = 0,864 gam n H2 O = 0, 048 mol n H = 0, 096 mol n C : n H = 0, 04 : 0, 096 = 5 :12 Vy A c cng thc phn t l C5H12. p n B.

V d 10: t chy hon ton mt hp cht hu c X cn 7,84 lt O2 (ktc). Sn phm chy gm chy hp th ht vo bnh ng dung dch Ba(OH)2 thy c 19,7 gam kt ta xut hin v khi lng dung dch gim 5,5 gam. Lc b kt ta, un nng nc lc li thu c 9,85 gam kt ta na. CTPT ca X l : A. C2H6. B. C2H6O. C. C2H6O2. D. C3H8. Hng dn gii Cc phn ng xy ra khi cho sn phm chy vo bnh ng dung dch Ba(OH)2 : CO2 + Ba(OH)2 BaCO3 + H2O (1) 2CO2 + Ba(OH)2 Ba(HCO3)2 (2) Ba(HCO3)2 BaCO3 + CO2 + H2O (3) = Theo (1) : n = 0,1 molCO ( p )

n BaCO2 3

Theo (2), (3): n

CO2 ( p ) )

= 2.n = 2.n = 0,1 mol BaCO3 3 Ba(HCO 2

Tng s mol CO2 sinh ra t phn ng t chy hp cht hu c l 0,2 mol. Theo gi thit khi lng dung dnh gim 5,5 gam nn ta c : = = = = 19, 7 0, 2.44 m H 2O 5, 5 m H 2O 5, 4 gam n H 2.n H2 O 0, 6 mol. p dng inh lut bo ton nguyn t i vi oxi ta c : = + = + = . Nh vy trong X khng c oxi. n O( hchc) 2.n n H O 2.n O ( b ) 2.0, 2 0, 3 0, 35.2 0CO22 2

n C : n H = 0, 2 : 0, 6 = 2 : 6 Vy CTPT ca X l C2H6. p n A. V d 11: t chy hon ton mt hirocacbon A. Sn phm thu c hp th hon ton vo 200 ml dung dch Ca(OH)2 0,2M thy thu c 3 gam kt ta. Lc b kt ta, cn li phn dung dch thy khi lng tng ln so vi ban u l 0,28 gam. Hirocacbon trn c CTPT l : A. C5H12. B. C2H6. C. C3H8 . D. C4H10. Hng dn gii Theo gi thit ta c : nCa (OH) = 0, 04 mol; 2 nCaCO3

= 0, 03 mol. Do c hai trng hp xy ra :

Trng hp 1 : Ca(OH)2 d, ch xy ra phn ng to kt ta : CO2 + Ca(OH)2 CaCO3 + H2O (1) mol: 0,03 0,03 0,03 = n CO 0, 03 mol. 2 2 Lc b kt ta, cn li phn dung dch thy khi lng tng ln so vi ban u l 0,28 gam c ngha l khi lng CO2 v H2O hp th vo dung dch Ca(OH)2 ln hn khi lng kt ta CaCO3 b tch ra. Suy ra :

m H2 O + m CO2 m CaCO3 = 0, 28 gam m H2 O = 0, 28 + 3 0, 03.44 = 1, 96 gam n H2 O = 0,1088 mol n H = 0, 217 mol n C : n H = 0, 03 : 0, 217 = 1: 7, 3 (loai). Trng hp 2 : Ca(OH)2 phn ng ht : CO2 + Ca(OH)2 CaCO3 + H2O mol: 0,03 0,03 0,03 (1)

2CO2 + Ca(OH)2 Ca(HCO3)2 mol: 0,02 0,01 = n CO 0, 05 mol.2

(2)

Lp lun tng t nh trn ta c : m H2 O + m CO2 m CaCO3 = 0, 28 gam m H O = 0, 28 + 3 0, 05.44 = 1, 08 gam 2 n2H O = 0, 06 mol n H = 0,12 mol n C : n H = 0, 05 : 0,12 = 5 :12 . Vy CTPT ca ankan l C5H12. p n A. V d 12: t chy 1 lt hi hirocacbon vi mt th tch khng kh (lng d). Hn hp kh thu c sau khi hi H2O ngng t c th tch l 18,5 lt, cho qua dung dch KOH d cn 16,5 lt, cho hn hp kh i qua ng ng photpho d th cn li 16 lt. Xc nh CTPT ca hp cht trn bit cc th tch kh o cng iu kin nhit , p sut v O2 chim 1/5 khng kh, cn li l N2. A. C2H6. B. C2H4. C. C3H8. D. C2H2. Hng dn gii Theo gi thit, ta c : V = 2 lt ; V (d) = 0,5 lt ; = 16 lt V (ban u) = 4 lt. VCO 2 O2 N2 O2

S phn ng : CxHy + O2 CO2 + H2O + O2 d lt: 1 4 2 a 0,5 p dng nh lut bo ton nguyn t i vi cc nguyn t C, H, O ta c : 1.x = 2.1 x= 2 1.y y = a.2 4.2 = 2.2 + a = 6 + 0.5.2 a= 3 p n A. V d 13: Cho 0,5 lt hn hp gm hirocacbon v kh cacbonic vo 2,5 lt oxi (ly d) ri t. Th tch ca hn hp thu c sau khi t l 3,4 lt. Cho hn hp qua thit b lm lnh, th tch hn hp kh cn li 1,8 lt v cho li qua dung dch KOH ch cn 0,5 lt kh. Th tch cc kh c o trong cng iu kin. Tn gi ca hirocacbon l : A. propan. B. xiclobutan. C. propen. D. xiclopropan. Hng dn gii Theo gi thit, ta c : VH O = 1, lt ; V = 1, 3 lt V (d) = 0,5 lt. O O2 CO 2 6 ;2

Cng thc ca hirocacbon l C2H6.

S phn ng : lt: (CxHy + CO2) + O2 CO2 + H2O + O2 d a b 2,5 1,3 1,6 0,5

p dng nh lut bo ton nguyn t i vi cc nguyn t C, H, O ta c : a.x + b.1 = 1, 3 x= 3 a.y = 1, 6.2 y b.2 + 2, 5.2 = 1, 3.2 + 1, 6.1+ Cng thc ca hirocacbon l C3H8. 0, 5.2 = 8 a = 0, 4 a + b = 0, 5 p n A. b = 0,1

V d 14: Np mt hn hp kh c 20% th tch ankan A (CnH2n+2) v 80% th tch O2 (d) vo kh nhin k. Sau khi cho n ri cho hi nc ngng t nhit ban u th p sut trong kh nhin k gim i 2 ln. Cng thc phn t ca ankan A l : A. CH4. B. C2H6. C. C3H8 . D. C4H10. Hng dn gii n gin cho vic tnh ton ta chn s mol ca A l 1 mol v ca O2 l 4 mol (V ankan chim 20% v O2 chim 80% v th tch). Phng trnh phn ng : CH + ( 3n + 1ot )O

nCO

+ (n + 1)H O2 2

(1)

n

2 n+ 2

b: p:

1 1 (

2 4 3n + 1 2 )

2

: mol n (n+1) : mol

sp:

0

4-(

3n + 1 )2

n

(n+1)

: mol

V sau phn ng hi nc ngng t nn ch c O2 d v CO2 gy p sut nn bnh cha. Tng s mol kh trc phn ng : n1 = 1 + 4 = 5 mol 3n + 1 Tng s mol kh sau phn ng : n = 4 - ( ) + n = (3,5 0,5n) mol 2 2 Do nhit trc v sau phn ng khng i nn : n1 = n 2 p2 Vy A l C2H6. p n B. V d 15: Trn mt hirocacbon X vi lng O2 va t chy ht X, c hn hp A 0 C o v p sut P1. t chy hon ton X, thu c hn hp sn phm B 218,4 C c p sut P2 gp 2 ln p sut P1. Cng thc phn t ca X l : A. C4H10. B. C2H6. C. C3H6. D. C3H8. Hng dn gii n gin cho vic tnh ton ta chn s mol ca X (CxHy) l 1 mol th t gi thit v phng y trnh phn ng ta thy s mol O2 em phn ng l (x + ) . 4 Phng trnh phn ng : t oy (x +o ) O CH + o

p1

5

=

p1

= 2 n = 2

3, 5 0, 5n 0, 5p1

xCO b:

x

y

+ 4 y (x + ) 4 (x + y 42

y

H O2

(1) 22

1

: mol y 2 y

p:

1

)

x

: mol

sp:

2 218,4 C nc th hi v gy p sut ln bnh cha.o

0

0

x

: mol

Tng s mol kh trc phn ng : n1 = [1 + (x + y y) Tng s mol kh sau phn ng : n2 = (x + 2

z) 4

) ] mol

) mol

Do nhit trc v sau phn ng thay i i nn : y 1+x+ = n1 p 1 2 p (218, 4 + 273) T 4 = 0, 9 0, 2y 0,1x = 1 x 2 1 = = = 0, 9 n2 p2 T1 2p1 .273 x+ y 2 y = 6

Vy A l C2H6. p n B.

V d 16: Hn hp kh X gm 2 hirocacbon no, mch h A v B l ng ng k tip. t chy X vi 64 gam O2 (d) ri dn sn phm thu c qua bnh ng Ca(OH)2 d thu c 100 gam o kt ta. Kh ra khi bnh c th tch 11,2 lt 0 C v 0,4 atm. Cng thc phn t ca A v B l : A. CH4 v C2H6. B. C2H6 v C3H8. C. C3H8 v C4H10. D. C4H10 v C5H12. Hng dn gii T gi thit suy ra : 100 64 11, 2.0, = 1, 8 mol. = = 1 mol; n = n = n n = 4 nCO 2 CaCO 3

100

O 2 p

O 2 b

O 2 d

322 n+ 2

0, 082.273

t cng thc phn t trung bnh ca A v B l : C Hn

Phng trnh phn ng chy : CHn 2n + 2

+

3n + 1 2 3n + 1 .x 2

n O CO2 2

+ ( n +1) H O2

(1)

mol:

x

nx

nx = 1 n= Theo gi thit ta c : 3n + 1, .x 1, 8 667 x = 0, 6 = 2 V hai ankan l ng ng k tip v c s C trung bnh bng 1,667 nn cng thc ca hai ankan l CH4 v C2H6. p n A.

V d 17: X l hn hp 2 ankan A v B. t chy ht 10,2 gam X cn 25,76 lt O2 (ktc). Hp th ton b sn phm chy vo nc vi trong d c m gam kt ta. a. Gi tr m l : A. 30,8 gam. B. 70 gam. C. 55 gam. D. 15 gam b. Cng thc phn t ca A v B l : A. CH4 v C4H10. B. C2H6 v C4H10. C. C3H8 v C4H10. D. C A, B v C. Hng dn gii t cng thc phn t trung bnh ca hai ankan A v B l : C Hn 2 n+ 2

Phng trnh phn ng chy :

CHn 2n + 2

+

3n + 1 2 3n + 1 .x 2

n O CO2 2

+ ( n +1) H O2

(1)

mol:

x

nx

CO2 + Ca(OH)2 CaCO3 + H2O mol: nx nx

(2)

nx = 0, 7 (14n + 2)x = 10, 2 x = 0, 2 Theo gi thit ta c : 3n + 1 n = 3, 5 .x = 1,15 2 Vy : n CaCO = n = 0, 7 mol CaCO = 0, 7.100 = 70 gam. CO 2 3 3 m Vi s C trung bnh bng 3,5 nn phng n A hoc B hoc C u tha mn. p n BD. V d 18: t chy hon ton hn hp X gm hai hirocacbon thuc cng dy ng ng ri hp th ht sn phm chy vo bnh ng nc vi trong d thu c 25 gam kt ta v khi lng nc vi trong gim 7,7 gam. CTPT ca hai hirocacon trong X l : A. CH4 v C2H6. B. C2H6 v C3H8. C. C3H8 v C4H10. D. C4H10 v C5H12. Hng dn gii Theo gi thit ta c : n = n = 0, 25 mol.CO2 CaCO3

Khi lng dung dch gim 7,7 gam nn suy ra : = = = 25 0, 25.44 m H 2O 7, 7 m H 2O 6,3 gam n H 2O 0,35 mol. Hn hp X gm hai cht ng ng, t chy X cho s mol nc ln hn s mol CO2 chng t X gm hai ankan. t cng thc phn t trung bnh ca hai ankan trong X l : C H .n 2 n+ 2

Phng trnh phn ng chy : CHn 2n + 2

+

3n + 1

n O CO2 2

+ ( n +1) H O2

(1)

2 T phn ng ta suy ra : nH O2

=

n+1

=

0,35

n = 2, 5 hoc2

2

n CO 2 2

n=

n CO 2 2

= 2, 5 0, 25 n

n H O n CO

Vi s C trung bnh bng 2,5 v cn c vo cc phng n ta thy hai ankan l : C2H6 v C3H8. p n B.

V d 19: Nung m gam hn hp X gm 3 mui natri ca 3 axit hu c no, n chc vi NaOH d, thu c cht rn D v hn hp Y gm 3 ankan. T khi ca Y so vi H2 l 11,5. Cho D tc dng vi H2SO4 d thu c 17,92 lt CO2 (ktc). a. Gi tr ca m l : A. 42,0. B. 84,8. C. 42,4. D. 71,2. b. Tn gi ca 1 trong 3 ankan thu c l : A. metan. B. etan. C. propan. D. butan. Hng dn gii t CTPT trung bnh ca 3 mui natri ca 3 axit hu c no, n chc l : C H COONan 2 n+1

Phng trnh phn ng : C o aO C H COONa + NaOH C Hn 2 n+1

,t

+ Na COn 2 n+ 2 2 3

(1)

Na 2 CO 3 + H SO 2 4 Na 2SO4 Theo (1), (2) v gi thit ta c : n Na CO = n C H 2 3n

+ H 2 O + CO2

(2)

= n NaOH = 2n +2 n CO

2 2

=

17, 92 22, 4

= 0,8 mol.

p dng nh lut bo ton khi lng ta c : m +m = m +m = 0, 8.106 + 11, 5.2.0, 8 0,8.40 = 71, 2 gam. mX NaOH C Hn 2n +2

Na 2 CO 3

X

M Y = 14n + 2 = 23 n = 1, 5 . Vy trong Y chc chn phi c mt ankan l CH4. p n DA.

BI TP TRC NGHIMCu 1: Trong cc nhn xt di y, nhn xt no sai ? A. Tt c cc ankan u c cng thc phn t CnH2n+2. B. Tt c cc cht c cng thc phn t CnH2n+2 u l ankan. C. Tt c cc ankan u ch c lin kt n trong phn t. D. Tt c cc cht ch c lin kt n trong phn t u l ankan. Cu 2: C bao nhiu ng phn cu to c cng thc phn t l C5H12 ? A. 3 ng phn. B. 4 ng phn. C. 5 ng phn. D. 6 ng phn. Cu 3: C bao nhiu ng phn cu to c cng thc phn t l C6H14 ? A. 3 ng phn. B. 4 ng phn. C. 5 ng phn. D. 6 ng phn. Cu 4: C bao nhiu ng phn cu to c cng thc phn t l C4H9Cl ? A. 3 ng phn. B. 4 ng phn. C. 5 ng phn. D. 6 ng phn. Cu 5: C bao nhiu ng phn cu to c cng thc phn t l C5H11Cl ? A. 6 ng phn. B. 7 ng phn. C. 5 ng phn. D. 8 ng phn. Cu 6: Hp cht X c cng thc cu to thu gn nht l : Hy cho bit trong phn t X cc nguyn t C dng bao nhiu electron ho tr to lin kt CH. A. 10. B. 16. C. 14. D. 12. Cu 7: Phn trm khi lng cacbon trong phn t ankan Y bng 83,33%. Cng thc phn t ca Y l : A. C2H6. B. C3H8. C. C4H10. D. C5H12. Cu 8: Cng thc n gin nht ca hirocacbon M l CnH2n+1. M thuc dy ng ng no ? A. ankan. B. khng d kin xc nh. C. ankan hoc xicloankan. D. xicloankan. Cu 9: Cho cc ankan sau :CH 3 CH CH 2 CH 3 (1) | CH 3 CH CH 3 | C CH (2) | CH3

3

CH 3 CH CH 3 (3) | CH 3

3

CH CH

3 3

CH

2

CH 2

(4)

CH 3 | CH 3 C CH3

2

CH (5)

| CH

3

Tn thng thng ca cc ankan sau y c tn tng ng l : A. (1) : iso-pentan ; (2) : tert-butan ; (3) : iso-propan ; (4) : n-butan ; (5) : neo-hexan. B. (1) : iso-pentan ; (2) : neo-pentan ; (3) : iso-propan ; (4) : n-butan ; (5) : neo-hexan. C. (1) : iso-pentan ; (2) : neo-pentan ; (3) : sec-propan ; (4) : n-butan ; (5) : neo-hexan. D. (1) : iso-pentan ; (2) : neo-pentan ; (3) : iso-butan ; (4) : n-butan ; (5) : neo-hexan.

Cu 10: Cho cc cht : (X) (Y) (P) (Q) Tn thng thng ca cc ankan sau y c tn tng ng l : A. (X) : iso-butan ; (Y) : n-butan ; (P) : iso-butan ; (Q) : n-pentan. B. (X) : iso-pentan ; (Y) : n-butan ; (P) : iso-propan ; (Q) : n-pentan. C. (X) : iso-pentan ; (Y) : n-butan ; (P) : iso-butan ; (Q) : n-hexan. D. (X) : iso-pentan ; (Y) : n-butan ; (P) : iso-butan ; (Q) : n-pentan. Cu 11: Ankan CH3 CH CH 2 CH CH 2 CH 2 CH 3 c tn ca X l : | | CH3 CH3 A. 1,1,3-trimetylheptan. B. 2,4-imetylheptan. C. 2-metyl-4-propylpentan. D. 4,6-imetylheptan. Cu 12: Ankan CH3 CH CH CH3 c tn l : | CH3

| C 2 H5 B. 2,3-imetylpentan. D. 2-etyl-3-metylbutan.

A. 3,4-imetylpentan. C. 2-metyl-3-etylbutan.

Cu 13: Ankan CH3 CH 2 CH CH 2 CH3 c tn l : | CH CH3 | CH3 A. 3- isopropylpentan. C. 3-etyl-2-metylpentan. Cu 14: Ankan CH3

B. 2-metyl-3-etylpentan. D. 3-etyl-4-metylpentan.

C 2H5 | C CH CH CH c tn l : CH | CH2

3

| CH

2

3

3

B. 2,4-ietyl-2-metylhexan. D. 3-etyl-5,5-imetylheptan. C 2H 5 | Cu 15: Tn gi ca cht hu c X c CTCT : CH 3 CH CH CH3 l : | Cl A. 3-etyl-2-clobutan. C. 2-clo-3-etylpentan. Cu 16: Tn gi ca cht hu c X c CTCT : B. 2-clo-3-metylpetan. D. 3-metyl-2-clopentan. CH 3 CH | NO2

A. 2-metyl-2,4-ietylhexan. C. 3,3,5-trimetylheptan.

CH CH 2 | CH3 CH 3 A. 4-metyl-3-nitropentan. C. 2-metyl-3-nitropentan. B. 3-nitro-4-metylpetan. D. 2-nitro-3-metylpentan.

l :

Cu 17: Tn gi cu cht hu c X c CTCT : CH 3 CH CH CH 2 CH 3 | |

l :

Cu

Cu

Cu Cu Cu Cu Cu

NO2 Cl A. 3-clo-2-nitropentan. B. 2-nitro-3-clopetan. C. 3-clo-4-nitropentan. D. 4-nitro-3-clopentan. 18: Cho ankan c CTCT l: (CH3)2CHCH2C(CH3)3. Tn gi ca ankan l : A. 2,2,4-trimetylpentan. B. 2,4-trimetylpetan. C. 2,4,4-trimetylpentan. D. 2-imetyl-4-metylpentan. 19: Hp cht hu c X c tn gi l: 2-clo-3-metylpentan. Cng thc cu to ca X l: A. CH3CH2CH(Cl)CH(CH3)2. B. CH3CH(Cl)CH(CH3)CH2CH3. C. CH3CH2CH(CH3)CH2CH2Cl. D. CH3CH(Cl)CH3CH(CH3)CH3. 20: 2,2,3,3-tetrametylbutan c bao nhiu nguyn t C v H trong phn t ? A. 8C,16H. B. 8C,14H. C. 6C, 12H. D. 8C,18H. 21: Hp cht 2,2-imetylpropan c th to thnh bao nhiu gc ha tr I ? A. 1 gc. B. 4 gc. C. 2 gc. D. 3 gc. 22: Hp cht 2,3-imetylbutan c th to thnh bao nhiu gc ha tr I ? A. 6 gc. B. 4 gc. C. 2 gc. D. 5 gc. 23: S gc ankyl ha tr I to ra t isopentan l : A. 3. B. 4. C. 5. D. 6. 24: Cc gc ankyl sau y c tn tng ng l :CH 3 CH CH 2 (1) | CH 3 CH3

CH

3

CH 3 | C | CH 3

(2)

CH 3 CH (3) | CH 3 CH (5)

3

CH CH 2 CH (4) |

CH 3 CH 2 CH

2

2

A. (1) : iso-butyl ; (2) : tert-butyl ; (3) : sec-propyl ; (4) : sec-butyl ; (5) : n-butyl. B. (1) : iso-butyl ; (2) : neo-butyl ; (3) : iso-propyl ; (4) : sec-butyl ; (5) : n-butyl. C. (1) : sec-butyl ; (2) : tert-butyl ; (3) : iso-propyl ; (4) : iso-butyl ; (5) : n-butyl. D. (1) : iso-butyl ; (2) : tert-butyl ; (3) : iso-propyl ; (4) : sec-butyl ; (5) : n-butyl. Cu 25: Ankan ha tan tt trong dung mi no sau y ? A. Nc. B. Benzen. C. Dung dch axit HCl. D. Dung dch NaOH. Cu 26: Phn t metan khng tan trong nc v l do no sau y ? A. Metan l cht kh. B. Phn t metan khng phn cc. C. Metan khng c lin kt i. D. Phn t khi ca metan nh. Cu 27: iu kin thng hirocacbon no sau y th kh ? A. C4H10. B. CH4, C2H6. C. C3H8. D. C A, B, C.

Cu 28: Trong cc cht di y, cht no c nhit si thp nht ? A. Butan. B. Etan. C. Metan.

D. Propan.

Cu 29: Cho cc cht sau : C2H6 (I) C3H8 (II) n-C4H10 (III) i-C4H10 (IV) Nhit si tng dn theo dy l : A. (III) < (IV) < (II) < (I). B. (III) < (IV) < (II) < (I). C. (I) < (II) < (IV) < (III). D. (I) < (II) < (III) < (IV). Cu 30: Trong s cc ankan ng phn ca nhau, ng phn no c nhit si cao nht ? A. ng phn mch khng nhnh. B. ng phn mch phn nhnh nhiu nht. C. ng phn isoankan. D. ng phn tert-ankan. Cu 31: Cho cc cht sau : CH3 CH2CH2CH2CH3 (I) CH3 CH 2 CH CH 3 (II) | CH3 Th t tng dn nhit si ca cc cht l : A. I < II < III. B. II < I < III. Cu 32: Cho cc cht : CH3 CH2 CH CH2 CH3 (I) CH3 | CH3 CH3 CH2 CH (III) | CH3 CH3 Th t tng dn nhit si ca cc cht l : A. I < II < III. B. II < I < III. Cu 33: Cho cc cht sau : CH3CH2CH2CH3 (I) CH3 CH CH CH3 (III) | | CH3 CH3 CH 3 C CH3 (III)| CH3 CH3 |

C. III < II < I. CH3 | CH 3 C | CH3

D. II < III < I.

(II)

C. III < II < I.

D. II < III < I.

CH3CH2CH2CH2CH2CH3 (II) CH3 CH2 CH CH3 (IV)| CH3

Th t gim dn nhit si ca cc cht l : A. I > II > III > IV. B. II > III > IV > I. C. III > IV > II > I. D. IV > II > III > I. Cu 34: Phn ng c trng ca hirocacbon no l : A. Phn ng tch. B. Phn ng th. C. Phn ng cng. Cu 35: Cc ankan khng tham gia loi phn ng no ? A. Phn ng th. B. Phn ng cng. C. Phn ng tch. D. Phn ng chy.

D. C A, B v C.

Cu 36: Sn phm ca phn ng th clo (1:1, nh sng) vo 2,2-imetylpropan l : (1) CH3C(CH3)2CH2Cl (2) CH3C(CH2Cl)2CH3 (3) CH3ClC(CH3)3 A. (1) ; (2). B. (2) ; (3). C. (2). D. (1).

Cu 37: Khi cho 2-metylbutan tc dng vi Cl2 theo t l mol 1:1 th to ra sn phm chnh l : A. 1-clo-2-metylbutan. B. 2-clo-2-metylbutan. C. 2-clo-3-metylbutan. D. 1-clo-3-metylbutan. Cu 38: Cho iso-pentan tc dng vi Br2 theo t l 1 : 1 v s mol trong iu kin nh sng khuch tn thu c sn phm chnh monobrom c cng thc cu to l : A. CH3CHBrCH(CH3)2. B. (CH3)2CHCH2CH2Br. C. CH3CH2CBr(CH3)2. D. CH3CH(CH3)CH2Br. Cu 39: Cho hn hp iso-hexan v Cl2 theo t l mol 1 : 1 ngoi nh sng th thu c sn phm chnh monoclo c cng thc cu to l : A. CH3CH2CH2CCl(CH3)2. B. CH3CH2CHClCH(CH3)2. C. (CH3)2CHCH2CH2CH2Cl. D. CH3CH2CH2CH(CH3)CH2Cl. Cu 40: Cho neo-pentan tc dng vi Cl2 theo t l s mol 1 : 1, s sn phm monoclo ti a thu c l : A. 2. B. 3. C. 5. D. 1. Cu 41: Hp cht Y c cng thc cu to : CH 3 CH CH CH2 3

CH 3 Y c th to c bao nhiu dn xut monohalogen ng phn ca nhau ? A. 3. B. 4. C. 5. D. 6. Cu 42: Iso-hexan tc dng vi clo (c chiu sng) c th to ti a bao nhiu dn xut monoclo ? A. 3. B. 4. C. 5. D. 6. Cu 43: Khi clo ha C5H12 vi t l mol 1:1 thu c 3 sn phm th monoclo. Danh php IUPAC ca ankan l : A. 2,2-imetylpropan. B. 2-metylbutan. C. pentan. D. 2-imetylpropan. Cu 44: khi clo ha mt ankan c cng thc phn t C6H14, ngi ta ch thu c 2 sn phm th monoclo. Danh php IUPAC ca ankan l : A. 2,2-imetylbutan. B. 2-metylpentan. C. n-hexan. D. 2,3-imetylbutan. Cu 45: Hirocacbon mch h X trong phn t ch cha lin kt v c hai nguyn t cacbon bc ba trong mt phn t. t chy hon ton 1 th tch X sinh ra 6 th tch CO2 ( cng iu kin nhit , p sut). Khi cho X tc dng vi Cl2 (theo t l s mol 1 : 1), s dn xut monoclo ti a sinh ra l : A. 3. B. 4. C. 2. D. 5. Cu 46: Khi clo ha hn hp 2 ankan, ngi ta ch thu c 3 sn phm th monoclo. Tn gi ca 2 ankan l : A. etan v propan. B. propan v iso-butan. C. iso-butan v n-pentan. D. neo-pentan v etan. Cu 47: Ankan no sau y ch cho 1 sn phm th duy nht khi tc dng vi Cl2 (as) theo t l mol (1 : 1): CH3CH2CH3 (a), CH4 (b), CH3C(CH3)2CH3 (c), CH3CH3 (d), CH3CH(CH3)CH3 (e) A. (a), (e), (d). B. (b), (c), (d). C. (c), (d), (e). D. (a), (b), (c), (e), (d).

Cu 48: C bao nhiu ankan l cht kh iu kin thng khi phn ng vi clo (c nh sng, t l mol 1:1) to ra 2 dn xut monoclo ? A. 4. B. 2. C. 5. D. 3. Cu 49: Dy ankan no sau y tha mn iu kin : mi cng thc phn t c mt ng phn khi tc dng vi clo theo t l mol 1 : 1 to ra 1 dn xut monocloankan duy nht ? A. CH4, C3H8, C4H10, C6H14. B. CH4, C2H6, C5H12, C8H18. C. CH4, C4H10, C5H12, C6H14. D. CH4, C2H6, C5H12, C4H10. Cu 50: Khi clo ha mt ankan thu c hn hp 2 dn xut monoclo v 4 dn xut iclo. Cng thc cu to ca ankan l : A. CH3CH2CH3. B. (CH3)2CHCH2CH2CH3. C. (CH3)2CHCH2CH3. D. CH3CH2CH2CH3. Cu 51: Khi clo ha mt ankan thu c hn hp 3 dn xut monoclo v 7 dn xut iclo. Cng thc cu to ca ankan l : A. CH3CH2CH2CH2CH2CH3. B. (CH3)2CHCH2CH2CH3. C. (CH3)3CCH2CH3. D. (CH3)2CHCH(CH3)2. Cu 52: Khi thc hin phn ng hiro ha hp cht X c CTPT C5H12 thu c hn hp 3 anken ng phn cu to ca nhau. Vy tn ca X l : A. 2,2-imetylpentan. B. 2-metylbutan. C. 2,2-imetylpropan. D. pentan. Cu 53: t chy mt hn hp gm nhiu hirocacbon trong cng mt dy ng ng nu ta thu c s mol H2O > s mol CO2 th CTPT chung ca dy l : A. CnHn, n 2. B. CnH2n+2, n 1 (cc gi tr n u nguyn). C. CnH2n-2, n 2. D. Tt c u sai. Cu 54: t chy cc hirocacbon ca dy ng ng no di y th t l mol H2O : mol CO2 gim khi s cacbon tng. A. ankan. B. anken. C. ankin. D. aren Cu 55: Khi t chy ankan thu c H2O v CO2 vi t l tng ng bin i nh sau : A. tng t 2 n + . B. gim t 2 n 1. C. tng t 1 n 2. D. gim t 1 n 0. Cu 56: Khng th iu ch CH4 bng phn ng no ? A. Nung mui natri malonat vi vi ti xt. B. Canxicacbua tc dng vi nc. C. Nung natri axetat vi vi ti xt. D. Nhm cacbua tc dng vi nc. Cu 57: Trong phng th nghim c th iu ch metan bng cch no sau y ? A. Nhit phn natri axetat vi vi ti xt. B. Crackinh butan. C. T phn ng ca nhm cacbua vi nc. D. A, C. Cu 58: Thnh phn chnh ca kh thin nhin l : A. metan. B. etan. C. propan. D. n-butan.

Cu 59: Trong cc phng trnh ha hc : Al4C3 + 12H2O C4H10C rack

3CH4 + 4Al(OH)3 inh

(1) + (3) (4) CH4 + NaOH + (5) D. (5). CH4 (2)

C

CH3COONa + NaOH CH3COONa + H2 O

CH4 + Na2CO3 aO, to CH + 22Na CO 4 3 pdd

CaO, t o

C3H6

CH2(COONa)2 + 2NaOH

CO2 + H2 Cc phng trnh ha hc vit sai l : A. (2), (5), (4). B. (2), (3), (4). C. (2), (3), (5). Cu 60: Phn ng no sau y iu ch c CH4 tinh khit hn ? A. Al4C3 + 12H2O B. CH3COONa (rn) + NaOH (rn) C. C4 H10 Ni, to

CaO, to

3CH4 + 4Al(OH)3 CH4 + Na2CO3

C

rack

inh

C3 H 6 +

CH4

D. C + 2H2

CH4

Cu 61: Ankan Y phn ng vi brom to ra 2 dn xut monobrom c t khi hi so vi H2 bng 61,5. Tn ca Y l : A. butan. B. propan. C. Iso-butan. D. 2-metylbutan. Cu 62: Khi brom ha mt ankan ch thu c mt dn xut monobrom duy nht c t khi hi i vi hiro l 75,5. Tn ca ankan l : A. 3,3-imetylhecxan. C. isopentan. B. 2,2-imetylpropan. D. 2,2,3-trimetylpentan Cu 63: Khi cho ankan X (trong phn t c phn trm khi lng cacbon bng 83,72%) tc dng vi clo theo t l s mol 1:1 (trong iu kin chiu sng) ch thu c 2 dn xut monoclo ng phn ca nhau. Tn ca X l : A. 3-metylpentan. B. 2,3-imetylbutan. C. 2-metylpropan. D. butan. Cu 64: Khi clo ha metan thu c mt sn phm th cha 89,12% clo v khi lng. Cng thc ca sn phm l : A. CH3Cl. B. CH2Cl2. C. CHCl3. D. CCl4. Cu 65: Khi tin hnh phn ng th gia ankan X vi kh clo c chiu sng ngi ta thu c hn hp Y ch cha hai cht sn phm. T khi hi ca Y so vi hiro l 35,75. Tn ca X l : A. 2,2-imetylpropan. B. 2-metylbutan. C. pentan. D. etan. Cu 66: Khi crackinh hon ton mt th tch ankan X thu c ba th tch hn hp Y (cc th tch kh o cng iu kin nhit v p sut); t khi ca Y so vi H2 bng 12. Cng thc phn t ca X l : A. C6H14. B. C3H8. C. C4H10. D. C5H12. Cu 67: Khi crackinh hon ton mt ankan X thu c hn hp Y (cc th tch kh o cng iu

kin nhit v p sut); t khi ca Y so vi H2 bng 14,5. Cng thc phn t ca X l : A. C6H14. B. C3H8. C. C4H10. D. C5H12 Cu 68: Crakinh 8,8 gam propan thu c hn hp A gm H2, CH4, C2H4, C3H6 v mt phn propan cha b crakinh. Bit hiu sut phn ng l 90%. Khi lng phn t trung bnh ca A l : A. 39,6. B. 23,16. C. 2,315. D. 3,96.

Cu 69: Crakinh 40 lt n-butan thu c 56 lt hn hp A gm H2, CH4, C2H4, C2H6, C3H6, C4H8 v mt phn n-butan cha b crakinh (cc th tch kh o cng iu kin nhit v p sut). Gi s ch c cc phn ng to ra cc sn phm trn. Hiu sut phn ng to hn hp A l : A. 40%. B. 20%. C. 80%. D. 60%. Cu 70: Crakinh n-butan thu c 35 mol hn hp A gm H2, CH4, C2H4, C2H6, C3H6, C4H8 v mt phn butan cha b crakinh. Gi s ch c cc phn ng to ra cc sn phm trn. Cho A qua bnh nc brom d thy cn li 20 mol kh. Nu t chy hon ton A th thu c x mol CO2. a. Hiu sut phn ng to hn hp A l : A. 57,14%. B. 75,00%. C. 42,86%. D. 25,00%. b. Gi tr ca x l : A. 140. B. 70. C. 80. D. 40. Cu 71: Cho etan qua xc tc ( nhit cao) thu c mt hn hp X gm etan, etilen, axetilen v H2. T khi ca hn hp X i vi etan l 0,4. Hy cho bit nu cho 0,4 mol hn hp X qua dung dch Br2 d th s mol Br2 phn ng l bao nhiu ? A. 0,24 mol. B. 0,16 mol. C. 0,40 mol. D. 0,32 mol. Cu 72: Cho butan qua xc tc ( nhit cao) thu c hn hp X gm C4H10, C4H8, C4H6, H2. T khi ca X so vi butan l 0,4. Nu cho 0,6 mol X vo dung dch brom (d) th s mol brom ti a phn ng l : A. 0,48 mol. B. 0,36 mol. C. 0,60 mol. D. 0,24 mol. Cu 73: Khi t chy hon ton V lt hn hp kh gm CH4, C2H6, C3H8 (ktc) thu c 44 gam CO2 v 28,8 gam H2O. Gi tr ca V l : A. 8,96. B. 11,20. C. 13,44. D. 15,68. Cu 74: Khi t chy hon ton 7,84 lt hn hp kh gm CH4, C2H6, C3H8 (ktc) thu c 16,8 lt kh CO2 (ktc) v x gam H2O. Gi tr ca x l : A. 6,3. B. 13,5. C. 18,0. D. 19,8. Cu 75: t chy hon ton 2,24 lt hn hp A (ktc) gm CH4, C2H6 v C3H8 thu c V lt kh CO2 (ktc) v 7,2 gam H2O. Gi tr ca V l : A. 5,60. B. 6,72. C. 4,48. D. 2,24. Cu 76: Oxi ho hon ton 0,1 mol hn hp X gm 2 ankan. Sn phm thu c cho i qua bnh (1) ng H2SO4 c, bnh (2) ng dung dch Ba(OH)2 d th khi lng ca bnh (1) tng 6,3 gam v bnh (2) c m gam kt ta xut hin. Gi tr ca m l : A. 68,95 gam. B. 59,1 gam. C. 49,25 gam. D. Kt qu khc. Cu 77: t chy hon ton 6,72 lt hn hp A (ktc) gm CH4, C2H6, C3H8, C2H4 v C3H6, thu c 11,2 lt kh CO2 (ktc) v 12,6 gam H2O. Tng th tch ca C2H4 v C3H6 (ktc) trong hn hp A l : A. 5,60. B. 3,36. C. 4,48. D. 2,24. Cu 78: t chy hon ton hn hp A gm CH4, C2H2, C3H4, C4H6 thu c x mol CO2 v 18x gam H2O. Phn trm th tch ca CH4 trong A l : A. 30%. B. 40%. C. 50%. D. 60%. Cu 79: t chy hon ton m gam hn hp X gm hai hirocacbon thuc cng dy ng ng cn dng 6,16 lt O2 v thu c 3,36 lt CO2. Gi tr ca m l : A. 2,3 gam. B. 23 gam. C. 3,2 gam. D. 32 gam. Cu 80: t chy mt hn hp hirocacbon ta thu c 2,24 lt CO2 (ktc) v 2,7 gam H2O th th

tch O2 tham gia phn ng chy (ktc) l : A. 5,6 lt. B. 2,8 lt.

C. 4,48 lt.

D. 3,92 lt.

Cu 81: t chy hon ton mt th tch kh thin nhin gm metan, etan, propan bng oxi khng kh (trong khng kh, oxi chim 20% th tch), thu c 7,84 lt kh CO2 ( ktc) v 9,9 gam nc. Th tch khng kh ( ktc) nh nht cn dng t chy hon ton lng kh thin nhin trn l : A. 70,0 lt. B. 78,4 lt. C. 84,0 lt. D. 56,0 lt. Cu 82: Craking m gam n-butan thu c hp A gm H2, CH4, C2H4, C2H6, C3H6, C4H8 v mt phn butan cha b craking. t chy hon ton A thu c 9 gam H2O v 17,6 gam CO2. Gi tr ca m l : A. 5,8. B. 11,6. C. 2,6. D. 23,2. Cu 83: Khi tin hnh craking 22,4 lt kh C4H10 (ktc) thu c hn hp A gm CH4, C2H6, C2H4, C3H6, C4H8, H2 v C4H10 d. t chy hon ton A thu c x gam CO2 v y gam H2O. Gi tr ca x v y tng ng l : A. 176 v 180. B. 44 v 18. C. 44 v 72. D. 176 v 90. Cu 84: Cho 224,00 lt metan (ktc) qua h quang c V lt hn hp A (ktc) cha 12% C2 H 2 ;10% CH4 ; 78%H2 (v th tch). Gi s ch xy ra 2 phn ng : 2CH4 C2H2 + 3H2 (1) CH4 C + 2H2 (2) Gi tr ca V l : A. 407,27. B. 448,00. C. 520,18. D. 472,64. Cu 85: Trn 2 th tch bng nhau ca C3H8 v O2 ri bt tia la in t chy hn hp. Sau phn ng lm lnh hn hp ( hi nc ngng t) ri a v iu kin ban u. Th tch hn hp sn phm khi y (V2) so vi th tch hn hp ban u (V1) l : A. V2 = V1. B. V2 > V1. C. V2 = 0,5V1. D. V2 : V1 = 7 : 10. Cu 86: Hn hp kh A gm etan v propan. t chy hn hp A thu c kh CO2 v hi H2O theo t l th tch 11:15. Thnh phn % theo khi lng ca hn hp l : A. 18,52% ; 81,48%. B. 45% ; 55%. C. 28,13% ; 71,87%. D. 25% ; 75%. Cu 87: t chy 13,7 ml hn hp A gm metan, propan v cacbon (II) oxit, ta thu c 25,7 ml kh CO2 cng iu kin nhit v p sut. Thnh phn % th tch propan trong hn hp A v khi lng phn t trung bnh ca hn hp A so vi nit l : A. 43,8% ; bng 1. B. 43,8 % ; nh hn 1. C. 43,8 % ; ln hn 1. D. 87,6 % ; nh hn 1. Cu 88: n gin ta xem xng l hn hp cc ng phn ca hexan v khng kh gm 80% N2 v 20% O2 (theo th tch). T l th tch xng (hi) v khng kh cn ly l bao nhiu xng c chy hon ton trong cc ng c t trong ? A. 1 : 9,5. B. 1 : 47,5. C. 1 : 48. D. 1 : 50 Cu 89: t chy hon ton mt hirocacbon X thu c 6,72 lt CO2 (ktc) v 7,2 gam nc. Cng thc phn t ca X l : A. C2H6. B. C3H8. C. C4H10. D. CH4. Cu 90: oxi ha hon ton m gam mt hirocacbon X cn 17,92 lt O2 (ktc), thu c 11,2 lt CO2 (ktc). CTPT ca X l : A. C3H8. B. C4H10. C. C5H12. D. C2H6. Cu 91: Np mt hn hp kh c 20% th tch ankan A v 80% th tch O2 (d) vo kh nhin k. Sau khi cho n ri cho hi nc ngng t nhit ban u th p sut trong kh nhin k gim

i 2 ln. Cng thc phn t ca ankan A l : A. CH4. B. C2H6.

C. C3H8 .

D. C4H10.

Cu 92: Hirocacbon X chy cho th tch hi nc gp 1,2 ln th tch CO2 (o cng k). Khi tc dng vi clo to mt dn xut monoclo duy nht. X c tn l : A. isobutan. B. propan. C. etan. D. 2,2- imetylpropan. Cu 93: t chy hon ton mt hirocacbon X thu c 0,11 mol CO2 v 0,132 mol H2O. Khi X tc dng vi kh clo thu c 4 sn phm monoclo. Tn gi ca X l : A. 2-metylbutan. B. etan. C. 2,2-imetylpropan. D. 2-metylpropan. Cu 94: t chy hon ton 0,2 mol hirocacbon X. Hp th ton b sn phm chy vo nc vi trong c 20 gam kt ta. Lc b kt ta ri un nng phn nc lc li c 10 gam kt ta na. Vy X khng th l : A. C2H6. B. C2H4. C. CH4. D. C2H2. Cu 95: t chy hon ton mt hirocacbon A. Sn phm thu c hp th hon ton vo 200 ml dung dch Ca(OH)2 0,2M thy thu c 3 gam kt ta. Lc b kt ta cn li phn dung dch thy khi lng tng ln so vi ban u l 0,28 gam. Hirocacbon trn c CTPT l : A. C5H12. B. C2H6. C. C3H8 . D. C4H10. Cu 96: t chy hon ton m gam hp cht hu c A. Sn phm thu c hp th vo nc vi trong d th to ra 4 gam kt ta. Lc kt ta cn li bnh thy khi lng bnh nc vi trong gim 1,376 gam. A c cng thc phn t l : A. CH4. B. C5H12. C. C3H8 . D. C4H10. Cu 97: Cho hn hp 2 ankan A v B th kh, c t l s mol trong hn hp: nA : nB = 1 : 4. Khi lng phn t trung bnh l 52,4. Cng thc phn t ca hai ankan A v B ln lt l : A. C2H6 v C4H10. B. C5H12 v C6H14. C. C2H6 v C3H8. D. C4H10 v C3H8 Cu 98: Mt hn hp 2 ankan lin tip trong dy ng ng c t khi hi vi H2 l 24,8. a. Cng thc phn t ca 2 ankan l : A. C2H6 v C3H8. B. C4H10 v C5H12. C. C3H8 v C4H10. D. Kt qu khc. b. Thnh phn phn trm v th tch ca 2 ankan l : A. 30% v 70%. B. 35% v 65%. C. 60% v 40%. D. 50% v 50%. Cu 99: t chy hon ton hn hp X gm hai ankan k tip trong dy ng ng c 24,2 gam CO2 v 12,6 gam H2O. Cng thc phn t 2 ankan l : A. CH4 v C2H6. B. C2H6 v C3H8. C. C3H8 v C4H10. D. C4H10 v C5H12 Cu 100: Khi t chy hon ton hn hp 2 ankan l ng ng k tip thu c 7,84 lt kh CO2 (ktc) v 9,0 gam H2O. Cng thc phn t ca 2 ankan l : A. CH4 v C2H6. B. C2H6 v C3H8. C. C3H8 v C4H10. D. C4H10 v C5H12. Cu 101: t chy hon ton hn hp kh X gm 2 hirocacbon A v B l ng ng k tip cn dng 85,12 lt O2 (ktc), thu c 96,8 gam CO2 v m gam H2O. Cng thc phn t ca A v B l : A. CH4 v C2H6. B. C2H6 v C3H8. C. C3H8 v C4H10. D. C4H10 v C5H12. Cu 102: t chy hon ton hn hp X gm 2 hirocacbon l ng ng lin tip, sau phn ng thu c V H O V = 1 : 1,6 (o cng k). X gm : : CO2 2

A. CH4 v C2H6. B. C2H4 v C3H6. C. C2H2 v C3H6. D. C3H8 v C4H10. Cu 103: Hn hp kh X gm 2 hirocacbon A v B l ng ng k tip. t chy X vi 64 gam O2 (d) ri dn sn phm thu c qua bnh ng Ca(OH)2 d thu c 100 gam kt ta. Kh ra

khi bnh c th tch 11,2 lt 0 C v 0,4 atm. Cng thc phn t ca A v B l : A. CH4 v C2H6. B. C2H6 v C3H8. C. C3H8 v C4H10. D. C4H10 v C5H12

o

Cu 104: t chy hon ton hn hp gm hai hirocacbon c phn t lng km nhau 14 vC c m gam H2O v 2m gam CO2. Hai hirocacbon ny l : A. 2 anken. B. C4H10 v C5H12. C. C2H2 v C3H4. D. C6H6 v C7H8. Cu 105: t chy hon ton hn hp hai hirocacbon ng ng c khi lng phn t hn km nhau 28 vC, ta thu c 4,48 lt CO2 (ktc) v 5,4 gam H2O. CTPT ca 2 hirocacbon trn l : A. C2H4 v C4H8. B. C2H2 v C4H6. C. C3H4 v C5H8. D. CH4 v C3H8. Cu 106: Hn hp kh gm 2 hirocacbon no A v B thuc cng dy ng ng, c t khi i vi H2 l 12. a. Khi lng CO2 v hi H2O sinh ra khi t chy 15,68 lt hn hp ( ktc) l : A. 24,2 gam v 16,2 gam. B. 48,4 gam v 32,4 gam. C. 40 gam v 30 gam. D. Kt qu khc. b. Cng thc phn t ca A v B l : A. CH4 v C2H6. B. CH4 v C3H8. C. CH4 v C4H10. D. C A, B v C. Cu 107: X l hn hp 2 ankan. t chy ht 10,2 gam X cn 25,76 lt O2 (ktc). Hp th ton b sn phm chy vo nc vi trong d c m gam kt ta. a. Gi tr m l : A. 30,8 gam. B. 70 gam. C. 55 gam. D. 15 gam b. Cng thc phn t ca A v B l : A. CH4 v C4H10. B. C2H6 v C4H10. C. C3H8 v C4H10. D. C A, B v C. Cu 108: t chy hon ton hn hp X gm hai hirocacbon thuc cng dy ng ng ri hp th ht sn phm chy vo bnh ng nc vi trong d thu c 25 gam kt ta v khi lng nc vi trong gim 7,7 gam. CTPT ca hai hirocacon trong X l : A. CH4 v C2H6. B. C2H6 v C3H8. C. C3H8 v C4H10. D. C4H10 v C5H12. Cu 109: Hn hp X gm hai hirocacbon thuc cng dy ng ng. t chy hon ton hn hp X, sn phm chy thu c cho li qua bnh (1) ng H2SO4 c, sau qua bnh (2) ng 250 ml dung dch Ca(OH)2 1M. Khi kt thc phn ng, khi lng bnh (1) tng 8,1 gam v bnh (2) c 15 gam kt ta xut hin. CTPT ca hai hirocacbon trong X l : A. CH4 v C4H10. B. C2H6 v C4H10. C. C3H8 v C4H10. D. A hoc B hoc C. Cu 110: t chy hon ton hn hp 2 hirocacbon c khi lng phn t hn km nhau 28 vC. Sn phm c hp th ton b vo nc vi trong d thu c 65 gam kt ta, lc kt ta thy khi lng dung dch gim so vi ban u 22 gam. Hai hirocacbon thuc h : A. Xicloankan. B. Anken. C. Ankin. D. Ankan. Cu 111: t chy mt s mol nh nhau ca 3 hirocacbon K, L, M ta thu c lng CO2 nh nhau v t l s mol nc v CO2 i vi K, L, M tng ng l 0,5 : 1 : 1,5. Xc nh CT K, L, M (vit theo th t tng ng) : A. C2H4, C2H6, C3H4. B. C3H8, C3H4, C2H4. C. C3H4, C3H6, C3H8. D. C2H2, C2H4, C2H6. Cu 112: Nung m gam hn hp X gm 3 mui natri ca 3 axit hu c no, n chc vi NaOH d, thu c cht rn D v hn hp Y gm 3 ankan. T khi ca Y so vi H2 l 11,5. Cho D tc dng vi H2SO4 d thu c 17,92 lt CO2 (ktc).

a. Gi tr ca m l : A. 42,0. b. Tn gi ca 1 trong 3 ankan thu c l : A. metan.

B. 84,8. B. etan.

C. 42,4. C. propan.

D. 71,2. D. butan.

BI 2 : XICLOANKANL THUYTI. KHI NIM DANH PHP 1. Khi nim - Xicloankan l mt loi hirocacbon no m trong phn t ch gm lin kt n v c mt vng khp kn. C CTTQ l CnH2n (n 3). - V d: (xiclopropan) 2. Danh php (xiclobutan)

Tn xicloankan = S ch v tr nhnh + tn nhnh + xiclo + tn mch chnh (vng) + an

- V d:

-CH3

(metylxiclopropan)

xiclohexan

metylxiclopentan

1,2-imetylxiclobutan

1,1,2-trimetylxiclopropan

II. TNH CHT VT L Xicloankan t nc, C t s, C Khi lng ring 3 g/cm (nhit ) Mu sc Tnh tan 127 33 0,689 ( 40C) Khng mu. Khng tan trong nc, tan trong dung mi hu c. 90 13 0,703 (0C) 94 49 0,755 (20C) 7 81 0,778 (20C)

II. TNH CHT HA HC a. Phn ng cng m vng - Cc xicloankan c vng ba cnh c th tham gia phn ng cng m vng vi H2, dung dch Br2 v dung dch HCl, HBr. - Cc xicloankan c vng bn cnh c th tham gia phn ng cng m vng vi H2.

V d :

+ H2 + Br2 + HBr + H2

N

i, 80

o

C

CH3CH 3 2

CH

(propan)

BrCH2CH2CH2Br Ni, 120o C

(1,3-ibrompropan) (1-brompropan)

CH3CH2CH2Br

CH3CH2CH2CH3 (butan)

- Xicloankan vng 5, 6 cnh tr ln khng c phn ng cng m vng trong nhng iu kin trn. b. Phn ng th : Phn ng th xicloankan tng t nh ankan. V d :

as, t

o

c. Phn ng oxi ho 3n o t CnH2n + O2 nCO2 + nH2O 2t C6H12 + 9O2 o 6CO + 6H O 2 2 2

H < 0 H = -3947,5 kJ

Xicloankan khng lm mt mu dung dch KMnO4. III. IU CH V NG DNG 1. iu ch Ngoi vic tch trc tip t qu trnh chng ct du m, xicloankan cn c iu ch t ankan, v d : CH3[CH2]4CH3 t , xto

+ H2

+ 3H2

t , xt

o

2. ng dng Ngoi vic dng lm nhin liu nh ankan, xicloankan cn c dng lm dung mi, lm nguyn liu iu ch cc cht khc, v d : t , xto

+

3H2

BI TP TRC NGHIMCu 113: Hp cht X c cng thc cu to thu gn nht l :

Hy cho bit hp cht X c bao nhiu nguyn t cacbon bc 2 ? A. 4. B. 5. C. 3. Cu 114: Cho cc cht sau : CH3 (I) (II) (III) (IV) Nhng cht no l ng ng ca nhau ? A. I, III, V. B. I, II, V. C. III, IV, V. Cu 115: Hp cht X c cng thc cu to thu gn nht l :

D. 6. CH2 CH3 (V) D. II, III, V.

Hy cho bit cn bao nhiu ng phn cu to mch vng c cng thc phn t ging nh X ? A. 2. B. 5. C. 3. D. 4. Cu 116: Cho cc hp cht vng no sau : Xiclopropan (I) xiclobutan (II) xiclopentan (III) xiclohexan (IV) bn ca cc vng tng dn theo th t no ? A. I < II < III < IV. B. III < II < I < IV. C. II < I < III < IV. D. IV < I < III < II. Cu 117: Hirocacbon X c CTPT C6H12 khng lm mt mu dung dch brom, khi tc dng vi brom to c mt dn xut monobrom duy nht. Tn ca X l : A. metylpentan. B. 1,2-imetylxiclobutan. C. 1,3-imetylxiclobutan. D. xiclohexan. Cu 118: Xicloankan (ch c mt vng) A c t khi so vi nit bng 3. A tc dng vi clo c chiu sng ch cho mt dn xut monoclo duy nht, xc nh cng thc cu to ca A ?CH3 CH3

A.

CH3

. Cu 119: Hai xicloankan M v N u c t khi hi so vi metan bng 5,25. Khi tham gia phn ng th clo (as, t l mol 1:1) M cho 4 sn phm th cn N cho 1 sn phm th. Tn gi ca cc xicloankan N v M l : A. metylxiclopentan v imetylxiclobutan. B. Xiclohexan v metylxiclopentan. C. Xiclohexan v n-propylxiclopropan. D. C A, B, C u ng. Cu 120: C bao nhiu ng phn cu to ca xicloankan c thc phn t l C5H10 phn ng c o vi H2 (t , Ni) ? A. 0. B. 2. C. 3. D. 4. Cu 121: C bao nhiu ng phn cu to ca xicloankan c thc phn t l C6H12 phn ng c o vi H2 (t , Ni) ?

.

B.

.

C.

H3C

.

D.

H3 C

CH3

A. 8.

B. 9.

C. 7.

D. 10.

Cu 122: C bao nhiu ng phn cu to ca xicloankan c thc phn t l C5H10 lm mt mu dung dch brom ? A. 0. B. 2. C. 3. D. 4. Cu 123: C bao nhiu ng phn cu to ca xicloankan c thc phn t l C6H12 lm mt mu dung dch brom ? A. 6. B. 5. C. 3. D. 4. o Cu 124: Cho cc cht : H2 (t , Ni), Cl2 (as), dung dch HCl, dung dch Br2, dung dch KMnO4. Cho xiclopropan v xiclobutan ln lt phn ng vi cc cht trn th s xy ra bao nhiu phn ng? A. 8. B. 6. C. 7. D. 9. o Cu 125: Hp cht X l 1-etyl-2-metylxiclopropan. Cho X tc dng vi H2 (Ni, t ). S sn phm cng ti a c th to ra l : A. 2. B. 3. C. 4. D. 1. Cu 126: Cho cc cht : (X) (Y) (P) (Q) o Hy cho bit cht no trn c th l sn phm ca phn ng gia metylxiclopropan vi H2 (Ni, t ). A. X, Y. B. P, Q. C. X, Q. D. Y, P. Cu 127*: Hp cht X l dn xut ca monoxiclopropan (c cha vng 3 cnh). Cho X cng o H2 (Ni, t ) th thu c hn hp cc sn cng phm trong c hp cht Y. Cng thc cu to thu gn nht ca Y l : Hy cho bit c my ng phn cu to ca X tha mn tnh cht trn ? A. 2. B. 4. C. 5. D. 3. Cu 128: Dn hn hp kh A gm propan v xiclopropan i vo dung dch brom s quan st c hin tng no sau y : A. Mu ca dung dch nht dn, khng c kh thot ra. B. Mu ca dung dch nht dn, v c kh thot ra. C. Mu ca dung dch mt hn, khng cn kh thot ra. D. Mu ca dung dch khng i. Cu 129: Cho phn ng : A + Br2 BrCH2CH2CH2 Br A l cht no trong phn ng sau y ? A. propan. B. 1-brompropan. C. xiclopopan. D. A v B u ng. Cu 130: Hp cht X c CTPT C3H6, X tc dng vi dung dch HBr thu c mt sn phm hu c duy nht. Vy X l : A. propen. B. propan. C. ispropen. D. xicloropan. Cu 131: Xicloankan vng khng bn c phn ng cng m vng. Hp cht X l xicloankan, khi cho X tc dng vi dung dch Br2 th sn phm thu c c cng thc cu to l : CH3CHBr CH2CHBrCH3. X s l cht no sau y ? A. metyl xiclobutan. B. etylxiclopropan. C. 1,2-imetylxiclopropan. D. 1,1-imetylxiclopropan.

Cu 132: Xicloankan vng khng bn c phn ng cng m vng. Hp cht X l xicloankan, khi cho X tc dng vi dung dch Br2 th sn phm thu c c cng thc cu to l : CH3CHBr CH2CHBrCH2CH3. X s l cht no sau y ? A. 1,2-imetylxiclobutan. B. 1-etyl-2-metylxiclopropan. C. 1,3-imetylxiclobutan. D. etylxiclobutan. Cu 133: Metylxiclopropan phn ng vi dung dch Br2 to ra hai sn phm, cng thc ca hai sn phm l : A. CH3CHBrCHBrCH3 v CH2BrCH2CHBrCH3. B. CH2BrCH(CH3)CH2Br v CH2BrCH2CHBrCH3. C. CH2BrCH(CH3)CH2Br v CH3CHBrCHBrCH3. D. CH3CHBrCHBrCH3 v CH2BrCHBrCH2CH3. Cu 134: Cht X c cng thc phn t l C5H10. X tc dng vi dung dch Br2 thu c 2 dn xut ibrom. Vy X l cht no sau y ? A. 1,1,2-trimetyl xiclopropan. B. 1,2-imetylxiclopropan. C. 2-metylbut-2- en. D. 2-metylbut-1- en. Cu 135: Cht X c cng thc phn t l C6H12. X khng tc dng vi dung dch KMnO4, X tc dng vi dung dch Br2 thu c 1 dn xut ibrom duy nht. Vy X l cht no sau y ? A. 1,2,3-trimetyl xiclopropan. B. 1,1,2-trimetylxiclopropan. C. 2-metylpent-2-en. D. 2-metylpent-1-en. Cu 136: Xiclohexan c th c iu ch theo s :

X + YCng thc cu to ca X v Y ln lt l

+H2 Ni, t0

A. CH2=CHCH=CH2 v CH CH. B. CH2=CHCH=CH2 v CH2=CH2. C. CH3CH=CHCH3 v CH3CH3. D. CH3CH=CHCH3 v CH2=CH2. Cu 137: t chy ht a gam hn hp X gm 2 monoxicloankan th thu c 3,36 lt CO2 (ktc). Gi tr ca a l : A. 2,1. B. 2,4. C. 2,6. D. 3,0. Cu 138: t chy ht hn hp X gm 2 monoxicloankan th cn a lt O2 v thu c 3,36 lt CO2. Cc th tch kh u o ktc. Gi tr ca a l : A. 2,24. B. 4,48. C. 5,04. D. 5,16. Cu 139: Hp cht X l monoxicloankan vng bn v phn t c 2 nguyn t cacbon bc 1. t chy ht 0,1 mol hp cht X th khi lng CO2 thu c ln hn khi lng H2O l 18,2 gam. S ng phn cu to tha mn X l : A. 2. B. 3. C. 4. D. 1. Cu 140: t chy 2,14 gam hn hp A gm hp cht ankan X v xicloankan Y (t l mol tng ng l 2 : 3) th thu c 3,36 lt CO2 (ktc). S nguyn t cacbon c trong phn t ca X v Y tng ng l : A. 3 v 4. B. 3 v 3. C. 2 v 4. D. 4 v 3.

Cu 141: Hn hp A gm hp cht ankan X v xicloankan Y (t l mol tng ng l 2 : 3) c t khi so vi H2 bng 21,4. t chy 3,36 lt hn hp A th thu c a lt CO2 (ktc). Gi tr ca a l : A. 9,86. B. 8,96. C. 10,08. D. 4,48. Cu 142: t chy ht hn hp X gm butan, xiclobutan, xiclopentan v xiclohexan th thu c 0,375 mol CO2 v 0,40 mol H2O. Phn trm khi lng ca butan c trong hn hp X l : A. 27,358. B. 27,38. C. 31,243. D. 26,13. Cu 143: Hp cht X l hirocacbon no phn t c 5 nguyn t cacbon. Khi cho X th clo iu kin nh sng, t l mol 1:1 th ch to ra 1 sn phm th. Hn hp A gm 0,02 mol X v 1 lng hirocacbon Y. t chy ht hn hp A thu c 0,11 mol CO2 v 0,12 mol H2O. Tn gi ca X, Y tng ng l : A. neopentan v metan. B. metylxiclobutan v etan. C. xiclopentan v etan. D. xiclopentan v metan.

Cu chuyn v nhng ht muiMt chng trai tr n xin hc mt ng gio gi. Anh ta lc no cng bi quan v phn nn v mi kh khn. i vi anh, cuc sng ch c nhng ni bun, v th hc tp cng chng hng th g hn. Mt ln, khi chng trai than phin v vic mnh hc mi m khng tin b, ngi thy im lng lng nghe ri a cho anh mt tha mui tht y v mt cc nc nh. Con cho tha mui ny vo cc nc v ung th i. Lp tc, chng trai lm theo. Cc nc mn cht. Chng trai tr li. Ngi thy li dn anh ra mt h nc gn v mt tha mui y xung nc: By gi con hy nm th nc trong h i. Nc trong h vn vy thi, tha thy. N chng h mn ln cht no. Chng trai ni khi mc mt t nc di h v nm th. Ngi thy chm ri ni: Con ca ta, ai cng c lc gp kh khn trong cuc sng. V nhng kh khn ging nh tha mui ny y, nhng mi ngi ha tan n theo mt cch khc nhau. Nhng ngi c tm hn rng m ging nh mt h nc th ni bun khng lm h mt i nim vui v s yu i. Nhng vi nhng ngi tm hn ch nh nh mt cc nc, h s t bin cuc sng ca mnh tr thnh ng cht v chng bao gi hc c iu g c ch.

CHUYN 3 :

HIROCACBON KHNG NO BI 1 : ANKEN (OLEFIN)

L THUYT NG NG-C2H4 v cc ng ng ca n to thnh dy ng ng , gi chung l anken hay olefin. -Anken l cc hirocacbon khng no, mch h, trong phn t c 1 lin kt i C = C. -Cc anken c cng thc chung l CnH2n (n 2).

NG PHNa. 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: - Bc 1 : Vit mch cacbon khng phn nhnh. t lin kt lin kt i vo cc v tr khc nhau trn mch chnh. - Bc 2 : Vit mch cacbon phn nhnh. + B 1 cacbon lm nhnh, t nhnh vo cc v tr khc nhau trong mch. Sau ng vi mi mch cacbon 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 nguyn t cacbon khc cho n khi khng b c na th dng li. b. ng phn hnh hc - L ng phn v v tr khng gian ca anken. - Gm 2 loi : ng phn cis (cc nhm th c khi lng ln nm cng pha) v trans (cc nhm th c khi lng ln nm khc pha). iu kin c ng phn hnh hc : - Cho anken c CTCT : abC=Ccd. iu kin xut hin ng phn hnh hc l : a b v c d. a c C=C b d - V d but2en c mt cp ng phn hnh hc l :

II. DANH PHP1. 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 V d : CH2 = CH2 CH2 = CH -H

Eten

Vinyl (Etenyl)

CH2 = CH CH3 CH2 = CH CH2 -H H Propen anlyl (allyl) (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. - V d:4 3 2 1

C H3 - C H = C H - C H31 2 3

(C4H8) But-2-en

C H 2 = C(CH 3 ) - C H3

(C4H8) 2-Metylprop-1-en

Lu : Gia s v s c du phy, gia s v ch c du gch -

TNH CHT VT L- 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 t.

o + ng phn cis-anken c t nc thp hn nhng c ts o cao hn so vi ng phn trans-anken. n

o + Khi cu trc phn t cng gn th tnc cng cao cn tso 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.

III. TNH CHT HAHC 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 hir o to ankan CnH2n + H2 CnH2n+2 b. Cng halogen X2 (Cl2, Br 2) CnH2n + X2 CnH2nX2 Br2 (dd) CH2BrCH2Br (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, HOH) CnH2n + HX CnH2n+1X CnH2n + HOH CnH2n+1OH CH2=CH2 + CH2=CH2 + HOH CH3CH2OH H+ t o , Ni

CH2=CH2 + HBr CH3CH2Br - Cc anken c cu to phn t khng i xng khi cng HX c th cho hn hp hai sn phm. CH3-CH=CH2 + HBr CH3CH2CH2Br(spp) 1-brompropan CH3CHBrCH3(spc) 2-brompropan Quy tc Maccopnhicop : Trong phn ng cng HX vo lin kt i, nguyn t H (phn mang in dng) ch yu cng vo nguyn t C bc thp hn (c nhiu H hn), cn nguyn hay nhm nguyn t X (phn mang in m) cng vo nguyn t C bc cao hn (t H hn). 2. Phn ng tr ng 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). nA A n - n gi l h s trng hp. - Phn trong ngoc gi l mt xch ca polime. o nCH2 = CH2 100at mt 0 , xtt , p, xt0

Peoxit,100 300 C

( CH2 2 ) CH

n

(polietilen, n = 300 40000)

nCH 2 = CH

|

CH2 CH |

(polipropilen)

CH3

CH3

n

Phn ng trng hp l qu trnh kt hp lin tip nhiu phn t nh ging nhau hoc tng t nhau to thnh nhng phn t rt ln gi l polime. 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 CnH2n +3n 2

O2

t

0

nCO 2

2+

nH O

- Trong phn ng chy lun c : n CO 2 = n H O 2 b. Phn ng oxi ha khng hon ton - Dn kh C2H4 vo dung dch KMnO4 (mu tm) thy dung dch mt mu tm : 3C2H4 + 2KMnO4 +4H2O 3HOCH2 2OH + 2MnO2 + 2KOH CH (etylen glicol) - Phn ng tng qut : 3CnH2n + 2KMnO4 +4H2O 3CnH2n(OH)2 + 2MnO2 + 2KOH Phn ng lm mt mu tm ca dung dch kali pemanganat c dng nhn ra s c mt ca lin kt i anken.

IV. IU CH V NG DNG1. iu ch a. hir o ha ankan CnH2n+2 CnH2n + H2 b. Phng php cr acking c rack inh CnH2n+2 CaH2a+2 + CbH2b c. T ankin (l hp cht c ni ba C C), ankaien (c 2 ni i) CnH2n-2 + H2 CnH2n d. T dn xut halogena nco l CnH2nX + KOH CnH2n + KX + H2O t o , Pd t o , xt

e. T dn xut ihalogen CnH2nX2 + Zn CnH2n + ZnX2 f. Tch nc ca ancol no n chc 1o 70 C ,H S O CnH2n+1OH CnH2n + H2O 2. ng dngo 2 4

to

Trong cc ho cht hu c do con ngi sn xut ra th etilen ng hng u v sn lng. S d nh vy v etilen cng nh cc anken thp khc l nguyn liu quan trng ca cng nghip tng hp polime v cc ho cht hu c khc. a. Tng hp polime

Trng hp etilen, propilen, butilen ngi ta thu c cc polime ch to mng mng, bnh cha ng dn nc... dng cho nhiu mc ch khc nhau. Chuyn ho etilen thnh cc monome khc tng hp ra hng lot polime p ng nhu cu phong ph ca i sng v k thut.

CH2=CH2 CH | C l2

2 - CH 2 |

500 0 C

= CH

x

t,

CH HCl

2 2

to , p|

CH 2 CH|

Cl

Cl

Cl

Cl n

vinyl clorua b. Tng hp cc ho cht khc T etilen tng hp ra nhng ho cht hu c thit yu nh etanol, etilen oxit, etylen glicol, anehit axetic, 1 2

poli(vinyl clorua) (PVC)

CH2 = CH2 + O2

A , to

g CH 2 CH 2

Oetilen oxit

Cha khng b r i conVo nm 1989 ti Ar menia c mt trn ng t ln 8,2 Richter san bng ton b t nc v lm thit mng hn 30.000 ngi tr ong vng cha y bn pht. Gia khung cnh hn lon , mt ngi cha chy vi n trng hc m con ng ang theo hc. Ta nh trc kia l trng hc nay ch cn l ng gch vn nt. Sau cn sc, ng nh li li ha vi con mnh rng: Cho d chuyn g xy ra i na, cha s lun bn con! v nc mt ng li tro ra. Nhn vo ng nt m trc kia l trng hc ng khng cn hy vng. Nhng trong u ng lun nh li li ha ca mnh vi cu con trai. Sau ng c nh li ca hnh lang m ng vn a a con i hc qua mi ngy. ng nh li rng phng hc ca con trai mnh pha ng sau bn tay phi ca trng. ng vi chy n v bt u o bi gia ng gch v. Nhng ngi cha, ngi m khc cng chy n v t khp ni vang ln nhng ting ku than: i, con trai ti!, i, con gi ti!. Mt s ngi khc vi lng tt c ko ng ra khi ng nt v ni i ni li: mun qu ri!, Bn nh cht ri!, ng i i, khng cn lm c g na u!, ng ch lm cho mi vic kh khn thm thi!. Vi mi ngi, ng ch t mt cu hi Anh c gip ti khng? v sau ng li d tng ming gch, tip tc o bi tm a con mnh. Vin ch huy cu ha cng c sc khuyn ng ra khi ng nt: Xung quanh y u ang chy v cc ta nh ang sp . ng ang trong vng nguy him. Chng ti s lo cho mi vic. ng hy v nh i?. ng tip tc chu ng mt mnh, v ng phi t mnh tm ra cu tr li cho iu day dt: Con trai ng cn sng hay cht?. ng o tip... 12 gi... 24 gi... sau ng lt nga mt mng tng ln v cht nghe ting con trai ng. ng ku ln tn con v ng nghe ting tr li vng li: Cha i! Con y, cha! Con ni vi cc bn ng s v nu cha cn sng cha s cu con v khi cha cu con th cc bn cng s c cu. Cha ha vi con l d trng hp no cha cng bn con. - Con c sao khng? - ng hi. - Bn con cn li 14 trn tng s 33, cha ! Bn con s lm. i, kht... Nhng by gi bn con c cha y. Khi ta nh , y to ra mt khong khng nh v th l bn con cn sng. - Ra y i con! - ng kh gi trong nh nhm.

- Khoan cha! cc bn ra trc, con bit rng cha khng b con. C chuyn g xy ra con bit l cha chc chn s khng b ri con.

PHNG PHP GII BI TP V ANKEN Phn ng cng X2, HX, H 2O, H 2 Phng php gii1. Bi tp tm cng thc ca hirocacbon khng no trong phn ng cng HX, X 2 (X l Cl, Br, I) Nu bi cho bit s mol ca hirocacbon v s mol ca HX hoc X2 tham gia phn ng th ta n tnh t l T = n HX hoa c T = X 2 t suy ra cng thc phn t tng qut ca hirocacbon.

nC H H T = 1 suy ra cng thc phn t tng qut ca hirocacbon l CnH2n. Bit c cng thc tng qut ca hirocacbon s bit c cng thc tng qut ca sn phm cng. Cn c vo cc gi thit khc m cho tm s nguyn t C ca hirocacbon. 2. Bi tp lin quan n phn ng cng H 2 vo hirocacbon khng no Khi lm bi tp lin quan n phn ng cng H2 vo anken cn ch nhng iu sau : + Trong phn ng khi lng c bo ton, t suy ra : n h on h p c phan .M h on h ptr c phan n g = nh on h p .M h on h p sau phan n g tr sauxHy x y

nC

ng

ng

phan

+ Trong phn ng cng hiro s mol kh gim sau phn ng bng s mol hiro phn ng. + Sau phn ng cng hiro vo hirocacbon khng no m khi lng mol trung bnh ca hn hp thu c nh hn 28 th trong hn hp sau phn ng c hiro d.

Cc v d minh ha V d 1: 0,05 mol hirocacbon X lm mt mu va dung dch cha 8 gam brom cho ra sn phm c hm lng brom t 69,56%. Cng thc phn t ca X l : A. C3H6. B. C4H8. C. C5H10. D. C5H8. Hng dn gii 8 1 n n = = 0, 05 mol; = 0, 05 mol 2 = X la C H . Br nBr2

160

X

nX

1

n

2n

Phng trnh phn ng : 80.2 Theo gi thit ta c : = 69, 56 CnH2n + Br2 14n p n C. 100 69, 56 CnH2nBr2 (1) n = 5 X l C H .5 10

V d 2: Cho 8960 ml (ktc) 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. CTCT ca X l : A. CH2=CHCH2CH3. B. CH3CH=CHCH3. C. CH3CH=CHCH2CH3. D. (CH3)2C=CH2. Hng dn gii Phng trnh phn ng : CnH2n + Br2 CnH2nBr2 (1) Theo gi thit ta c :

n = mX

8, 96

= 0, 4 mol;

= 22, 4 gam = MX X

22, 4

= 56 gam / mol X : C H

22, 4

0, 4

4

8

V X c ng phn hnh hc nn X l : CH3CH=CHCH3. p n C. V d 3: Cho hirocacbon X phn ng vi brom (trong dung dch) theo t l mol 1 : 1, thu c cht hu c Y (cha 74,08% Br v khi lng). Khi X phn ng vi HBr th thu c hai sn phm hu c khc nhau. Tn gi ca X l : A. but-1-en. B. but-2-en. C. Propilen. D. Xiclopropan. Hng dn gii X phn ng vi Br2 theo t l mol 1:1 nn X c cng thc l CnH2n. Phng trnh phn ng : 80.2 Theo gi thit ta c : = 74, 08 CnH2n + Br2 CnH2nBr2 (1)

n = 4 X l C H .4 8

14n 100 74, 08 Khi X phn ng vi HBr th thu c hai sn phm hu c khc nhau nn X l but-1-en. CH3CH2CH2CH2Br CH2=CHCH2CH3 + HBrphm ph) (sn

CH3CH2CHBrCH3 (sn phm chnh)

p n B. V d 4 : Dn 3,36 lt (ktc) 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. a. CTPT ca 2 anken l : A. C2H4 v C3H6. B. C3H6 v C4H8. C. C4H8 v C5H10. D. C5H10 v C6H12. b. Thnh phn phn % v th tch ca hai anken l : A. 25% v 75%. B. 33,33% v 66,67%. C. 40% v 60%. D. 35% v 65%. Hng dn gii a. Xc nh cng thc phn t ca hai anken : t CTPT trung bnh ca hai anken trong X l : C H .n 2n

Theo gi thit ta c : 3,36 n = = 0,15 mol; mC H

= 7, 7 gam C H H MC Hn 2n n 2n

7, 7 = 11 = 0,15

154

14n =

154

n =

n

2n

22, 4

3

3

3

V hai anken l ng ng k tip v c s nguyn t C trung bnh l

11 3

= 3, 667 nn suy ra

cng thc phn t ca hai anken l C3H6 v C4H8. b. Tnh thnh phn phn trm v th tch ca cc anken : p dng s ng cho cho s nguyn t C trung bnh ca hn hp C3H6 v C4H8 ta c :

nC4 H8

4 11 3

11 2 3 3= 3 4 11 1 3 = 3

nC

4

H8

H nC H3 6

3

= 2 nC H 1 3 6

Vy thnh phn phn trm v th tch cc kh l : 1 %C H = .100 = 33, 33%; %C = (100 33,33)% = 66, 67%. H3 6

p n BB.

3

4

8

V d 5: Hirocacbon X cng HCl theo t l mol 1:1 to sn phm c hm lng clo l 55,04%. X c cng thc phn t l : A. C4H8. B. C2H4. C. C5H10. D. C3H6. Hng dn gii X phn ng vi HCl theo t l mol 1:1 nn X c cng thc l CnH2n. Phng trnh phn ng : HCl CnH2n+1Cl (1) 35, 5 Theo gi thit ta c : = 55, 04 n = 2 X l C H . CnH2n + 14n + 1 100 55, 04 p n B. V d 6: Hn hp X gm hai anken k tip nhau trong dy ng ng. t chy hon ton 5 lt X cn va 18 lt kh oxi (cc th tch kh o cng iu kin nhit v p sut). a. Cng thc phn t ca hai anken l : A. C2H4 v C3H6. B. C3H6 v C4H8. C. C4H8 v C5H10. D. A hoc B. b. Hirat ha mt th tch X trong iu kin thch hp thu c hn hp ancol Y, trong t l v khi lng ca cc ancol bc 1 so vi ancol bc 2 l 28 : 15. Thnh phn phn trm khi lng ca mi ancol trong hn hp Y l : A. C2H5OH : 53,49% ; iso C3H7OH : 34,88% ; n C3H7OH : 11,63%. B. C2H5OH : 53,49% ; iso C3H7OH : 11,63% ; n C3H7OH : 34,88%. C. C2H5OH : 11,63% ; iso C3H7OH : 34,88% ; n C3H7OH : 53,49%. D. C2H5OH : 34,88% ; iso C3H7OH : 53,49% ; n C3H7OH : 11,63%. Hng dn gii a. Xc nh cng thc phn t ca hai anken : t cng thc phn t trung bnh ca hai anken trong X l : C Hn

2

4

2n

Phng trnh phn ng chy : 3n CH + O

to

n CO

+n 2n

nH O2

(1)

2 3n 2 .5

2

2

lt:

5

Theo gi thit v (1) ta c :

3n 2

.5 = 18 n = 2, 4 .

Do hai anken l ng ng k tip v c s cacbon trung bnh l 2,4 nn cng thc ca hai anken l : C2H4 v C3H6. p n A. b. Xc nh thnh phn phn trm khi lng ca mi ancol trong hn hp Y : p dng s ng cho cho s nguyn t C trung bnh ca hai anken ta c : n C H2 2 2,4 nC H3 6

3 2,4 = 0,6

4

nC 2

H4

0, 6 = 0, 4 =

3 2

nC H2,4 2= 0,43

6

3

Vy chn s mol ca C2H4 l 3 th s mol ca C3H6 l 2. Phn ng ca hn hp hai anken vi nc : + C2 H 4 mol: 3 CH2=CHCH3 mol: x+y + H2 O + H2 Ot ,Ho +

3 CH3CH2CH2OH x

to , H

C2H5OH (2)

(3)

(4) CH3CHOHCH3 y 3.46 + x.60 28 = x = 0, 5 Theo (2), (3), (4) v gi thit ta c : y.60 15 = y 5 1, x+y = 2 Thnh phn phn trm khi lng ca mi ancol trong hn hp Y l : %C 2 H 5OH = 3.46 1, 5.60 .100 = 53, 49%; %i C 3 H 7 OH .100 = 34, 88% 3.46 + 2.60 3.46 + 2.60 =

%n C 3H 7OH = 100% 53, 49% 34, 88% = 11, 63%. p n A. V d 7: Cho H2 v 1 olefin c th tch bng nhau qua niken un nng ta c hn hp A. Bit t khi hi ca A i vi H2 l 23,2. Hiu sut phn ng hiro ho l 75%. Cng thc phn t olefin l : A. C2H4. B. C3H6. C. C4H8. D. C5H10. Hng dn gii Theo gi thit ta chn : nH = n C H2 n

H2 n

= 1 mol.

Phng trnh phn ng :t , CnH2n + H2 Ni CnH2n+2o

(1)

Theo (1) ta thy, sau phn ng s mol kh gim mt lng ng bng s mol H2 phn ng. Hiu sut phn ng l 75% nn s mol H2 phn ng l 0,75 mol. Nh vy sau phn ng tng s

mol kh l 1+1 0,75 = 1,25 mol. p dng nh lut bo ton khi lng ta c : khi lng ca H2 v CnH2n ban u bng khi lng ca hn hp A. 1.2 + 1.14n = 23, 2.2 n = 4 . MA = 1, 25

Vy cng thc phn t olefin l C4H8. p n C. V d 8: Cho hn hp X gm anken v hiro 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. CTPT ca X l : A. C2H4. B. C3H6. C. C4H8. D. C5H10. Hng dn gii V M Y = 4.4 = 16 nn suy ra sau phn ng H2 cn d, CnH2n phn ng ht. p dng nh lut bo ton khi lng ta c : n X M Y = 4.4 1, mX = mY nX. = nY. M Y = = 1 n Y M X 3, 33.4 MX Chn nX = 1,2 mol v nY =1 mol H2 ( p ) = n H = n X n Y = 0,2 mol. Cn 2 n n Ban u trong X c 0,2 mol CnH2n v 1 mol H2 0, 2.14n + 1.2 = 3, 33.4 n = 5 Cng thc phn t olefin l C5H10. Ta c : M X = 1, 2 p n D. V d 9: 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. Hiu sut ca phn ng hiro ho l : A. 20%. B. 40%. C. 50%. D. 25%. Hng dn gii p dng s ng cho ta c : n 2 28 15 1 H = C th tnh hiu sut phn ng theo hoc theo C H H = nC 2H 4

H

15 2

1

2

2

4

Phng trnh phn ng : H2 + C2 H 4 C2H6Ni,t o

p dng nh lut bo ton khi lng ta c : nX M Y mX = mY nX. = nY. M Y MX nY Chn nX = 4 mol n = nH2 C 2 H4

5.4 = =

4

= MX

3, 75.4 3 = n n = 1 mol.

= 2 mol ; nH2 ( p )

X

Y

1 Hiu sut phn ng : H = .100% = 50% . 2 p n C.

Phn ng oxi ha3. Phn ng oxi ha khng hon ton 3C2H4 + 2KMnO4 +4H2O 3HOCH2 2OH + 2MnO2 + CH 2KOH (etylen glicol) 3CnH2n + 2KMnO4 +4H2O 3CnH2n(OH)2 + 2MnO2 + 2KOH 4. Phn ng oxi ha hon ton CnH2n +3n 2

O2

to

nCO + nH2O 2 2

Nhn xt : Trong phn ng chy anken ta lun c : n CO 2 = n H O 2

Phng php giiKhi gii bi tp lin quan n phn ng t chy hn hp cc hirocacbon ta nn s dng phng php trung bnh chuyn bi ton hn hp nhiu cht v mt cht; mt s bi tp m lng cht cho di dng tng qut th ta s dng phng php t chn lng cht nhm bin cc i lng tng qut thnh i lng c th cho vic tnh ton tr nn n gin hn. Ngoi ra cn phi ch n vic s dng cc nh lut nh bo ton nguyn t, bo ton khi lng, phng php ng cho gii nhanh bi tp trc nghim.

Cc v d minh ha V d 1: 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 : A. 2,240. B. 2,688. C. 4,480. D. 1,344. Hng dn gii Cch 1 : p dng nh lut bo ton electron : 3 3 = .n = .0, 2.0, 2 = 0, 06 mol = 2.n 3.n n VKMnO 4 C 2 H4 C2 H4

= 0, 06.22, 4 = 1,344 lt.

2 Cch 2 : Tnh ton theo phng trnh phn ng : 3C2H4 + 2KMnO4 +4H2O 3HOCH2 2OH + 2MnO2 + CH 2KOH mol: 0,06 0,04 p n D. Nhn xt : Cch 1 nhanh hn cch 2 do ch cn xc nh s thay i s oxi ha ca cc cht, ri