Luận văn HydroCacBon

8/3/2019 Lun vn HydroCacBon

1/115

Lunvnttnghip GVHD:CV Th Th

SVTH:PhanTh Thy 1

PHN MUI. L do chn ti

Ha hc hydrocacbon l phn mu ca chng trnh ha hc hu cph thng.Tt c nhng khi nim cbn, nhng l thuyt cho ca chng trnh ha hc hu c

ph thng u c trnh by trong phn hydrocacbon. Nu cc em hc sinh hiu r phnny th vic hc ha hc ph thng s thun li hn.

Nhng lm sao cc em c th hiu, nhv vn dng bi mt cch tt nht c ?Ki

n th

c v

hydrocacbon

ph

thng r

t nhi

u trong khi

s

gi

hc

trn lp l

i

khng gio vin c th trnh by ht kin thc phn ny, cc em hc sinh cng cmthy lng tng, i khi l khng hiu kp bi, khng lm c bi tp.

Thc tin chng minh cch tt nht c th hiu v vn dng kin thc hc lgii bi tp. Nhng vn t ra l bi tp nhiu lm sao gii ht c. Thc t cho thy,thng cc em hc sinh ch lm c nhng bi tp quen thuc v lng tng khi gp bi tpmi mc d khng kh do cc em khng nhn ra c dng ton, cha bit vn dng cc

phng php gii ton hoc do cc em khng hc bi.Nu c th h thng ha l thuyt v a ra phng php gii bi tp th hc sinh s

d dng tip thu bi hn, hiu r bi hn, thm yu thch mn hc hn v gio vin cng t

tin hn trc hc sinh.Vi suy ngh ti quyt nh chn ti : Phn loi v phng php gii mt sbi tp v hydrocacbon trong chng trnh THPT.

II. Mc ch ca ti- Nhm gip cho hc sinh, sinh vin c ci nhn h thng v l thuyt v bi tp ha

hu cTHPT c bit l phn hydrocacbon chng trnh hc k 2 lp 11, t to iukin thun li cho qu trnh tip thu bi ging v cc kin thc ha hc, gp phn nng caohiu qu ging dy trng ph thng.

III. Nhim v ca ti-Nghin cu csl lun v bi tp ha hc- Tm tt l thuyt, phn loi, h thng v xut phng php gii cc dng bi tp

v hydrocacbon.- Tm hiu thc trng dy v lm bi tp trng THPT.

IV. Khch thv i tng nghin cu- Khch th nghin cu: qu trnh dy v hc mn ha trng THPT.- i tng nghin cu: bi tp v hydrocacbon.

V. Gi thuyt khoa hc


2/115


SVTH:PhanTh Thy 2

- Nu hiu r l thuyt, nm vng phng php gii bi tp hydrocacbon chng trnhTHPT s gip gio vin v hc sinh h thng ha v hiu su sc bi tp ny cng nh cnn tng vng chc trong b mn ha hu ctrng THPT.

VI. Phm vi nghin cu- Chng trnh ha hc THPT : chng trnh ha hu c11

VII. Phng tin v phng php nghin cu- Nghin cu, tham kho cc ti liu c lin quan- Tng hp, phn tch, xut phng php gii- a ra cc dng bi tp tiu biu minh ha sau c bi tp tng t- Trao i, iu tra thc t


3/115


SVTH:PhanTh Thy 3

CHNGI:

C S LLUNCA TI


4/115


SVTH:PhanTh Thy 4

I.1 CSL LUN V BI TPHA HC

I.1.1 KHI NIM BI TP HA HC :Bi tp ha hc l mt dng bi lm gm nhng bi ton, nhng cu hi hay ng

thi c bi ton v cu hi thuc v ha hc m trong khi hon thnh chng, hc sinh nmc mt tri thc hay knng nht nh.

Cu hi l nhng bi lm m trong qu trnh hon thnh chng, hc sinh phi tinhnh mt hot ng ti hin. Trong cc cu hi, gio vin thng yu cu hc sinh phinh li ni dung ca cc nh lut, quy tc, khi nim, trnh by li mt mc trong schgio khoa,cn bi ton l nhng bi lm m khi hon thnh chng, hc sinh phi tin

hnh mt hot ng sng to gm nhiu thao tc v nhiu bc.V d : Thno l phn ng th? Nhng loi hydrocacbon no hc tham gia cphn ng th? Mi loi cho mt v d?

lm c bi ny, hc sinh phi nhli c nh ngha phn ng th tc ti toli kin thc. Ngoi ra cc em cn h thng ha li c CTTQ, nh ngha cchydrocacbon, tnh cht ha hc c trng ca mi hydrocacbon .

Nh vy, chnh cc bi tp Ha hc gm bi ton hay cu hi, l phng tin cc kquan trng pht trin t duy hc sinh. Ngi ta thng la chn nhng bi ton v cuhi a vo mt bi tp l c tnh ton n mt mc ch dy hc nht nh, l nm hayhon thin mt dng tri thc hay k nng no . Vic hon thnh v pht trin k nng

gii cc bi ton Ha hc cho php thc hin nhng mi lin h qua li mi gia cc trithc thuc cng mt trnh ca cng mt nm hc v thuc nhng trnh khc nhau canhng nm hc khc nhau cng nh gia tri thc v k nng.

I.1.2 TM QUAN TRNG CA BI TP HA HC :Bi tp Ha hc gi vai tr rt quan trng trong vic thc hin mc tiu o to chung

v mc tiu ring ca mn Ha hc.Bi tp Ha hc va l mc ch, va l ni dung, li va l phng php dy hc

hiu nghim. L lun dy hc coi bi tp l mt phng php dy hc c th, uc pdng ph bin v thng xuyn cc cp hc v cc loi trng khc nhau, c s dng

tt c cc khu ca qu trnh dy hc : nghin cu ti liu mi, cng c, vn dng, khiqut ha h thng ha v kim tra, nh gi kin thc, k nng, k xo ca hc sinh. Ncung c p cho hc sinh c kin thc, c con ng dnh ly kin thc, m cn mang linim vui sng ca s pht hin, ca vic tm ra p s.

Bi tp Ha hc c nhiu ng dng trong dy hc vi t cch l mt phng phpdy hc ph bin, quan trng v hiu nghim. Nh vy, bi tp Ha hc c cng dng rngri, c hiu qu su sc trong vic thc hin mc tiu o to, trong vic hnh thnh phng

php chung ca vic t hc hp l, trong vic rn luyn k nng t lc, sng to.Bi tp Ha hc l phng tin cbn dy hc sinh tp vn dng cc kin thc

hc vo thc ti sng, sn xut v tp nghin cu khoa hc. Kin thc hc sinh tip thu

c ch c ch khi s dng n. Phng php luyn tp thng qua vic s dng bi tp l


5/115


SVTH:PhanTh Thy 5

mt trong cc phng php quan trng nht nng cao cht lng dy hc b mn. ivi hc sinh, vic gii bi tp l mt phng php dy hc tch cc.

I.1.3 TC DNG CA BI TP HA HC :1)Bi tp Ha hc c tc dng lm cho hc sinh hiu su hn v lm chnh xc

ha cc khi nim hc.Hc sinh c th hc thuc lng cc nh ngha ca cc khi nim, hc thuc lng cc

nh lut, nhng nu khng qua vic gii bi tp, hc sinh cha th no nm vng nhngca m hc sinh thuc lng. Bi tp Ha hc s rn luyn cho hc sinh k nng vn dngc cc kin thc hc, bin nhng kin thc tip thu c qua cc bi ging ca thythnh kin thc ca chnh mnh. Khi vn dng c mt kin thc no , kin thc sc nhlu.

V d : Cc hp cht sau, cht no l ru?CH3 CH2 OH, C6H5 OH, NaOH, C6H5 CH2 OH, HO CH2 CH2 OH

Khi lm c bi tp ny, hc sinh nhc nh ngha ru, CTPT ca ru vcch phn bit cc hp cht c cha nhm -OH tc cc em chnh xc ha cc khi nimv khng b ln ln gia cc cht gn ging nhau v hnh thc.

2) Bi tp Ha hc o su mrng shiu bit mt cch sinh ng phong phkhng lm nng n khi lng kin thc ca hc sinh

V d : Trong tinh du chanh c cht limonen.a)Hy vit phng trnh phn ng khi hidro ha limonen c metan v CTCT

metan.

b)Limonen thuc dy ng ng no trong chng trnh ha hc hc bit

limonen:

CH3

CH3

CH2

Khi cho hc sinh lm bi ny, cc em rt thch th v bit c mt cht trong

chanh. Vic vit phng trnh phn ng khng phi l kh i vi cc em. Tuy nhin, quav d ny hc sinh bit ankadien c nhiu loi mch khc nhau. Nhvy m kin thc hohc gn lin vi thc t cuc sng c thi vo tr nhca cc em mt cch d dng, .

Hoc mt v d khc l cc phn bi tp v ru, cc bi tp tnh hiu sut, iuch cng rt gn gi vi cuc sng. Nhng bi tp ny cng gp phn ng k trong vic

gn kin thc ha hc vi cuc sng lm cho cc em thm yu thch mn ha, khng lmnng n kin thc ca hc sinh, t cc em cm thy ha hc khng phi l nhng khinim kh nh, kh hiu m rt thit thc, gn gi i vi cc em

3) Bi tp Ha hc cng c kin thc c mt cch thng xuyn v h thng hacc kin thc hc :

Kin thc c nu chn thun l nhc li s lm cho hc sinh chn v khng c gmi v hp dn. Bi tp Ha hc s n tp, cng c v h thng ha kin thc mt cchthun li nht. Mt sng k bi tp i hi hc sinh phi vn dng tng hp kin thcca nhiu ni dung, nhiu chng, nhiu bi khc nhau. Qua vic gii cc bi tp Ha hc


6/115


SVTH:PhanTh Thy 6

ny, hc sinh s tm ra mi lin h gia cc ni dung ca nhiu bi, chng khc nhau t s h thng ha kin thc hc.V d : Cht A c CTPT l C5H12, khi tc dng vi Cl2 (c chiu sng) th to ra mt sn

phm duy nht tm CTCT ca A? A c my ng phn?c tn cc ng phn?Ch

v

i m

t v d

nh

nh

th, h

c sinh

c n v

thuyt c

u t

o ha h

c, cch

vit cc ng phn, phn ng th v cch xc nh cht tha bi, c n v danh php.Nh vy cc em c cng c kin thc c, h thng ha cc kin thc hc. Ccdng bi tp v phn bit, tch cht, iu ch hoc bi ton ha hc cng c ngha ln ivi tc dng ny.

4)Bi tp Ha hc thc y thng xuyn srn luyn cc k nng k xo vha hc :

Cc knng, kxo v ha hc nh knng s dng ngn ng ha hc, lp cng thc,cn bng phng trnh ha hc; cc tnh ton i s: qui tc tam sut, gii phng trnh v

h phng trnh; knng nhn bit cc ha cht, V d : Mt hn hp gm 2 chtngng ankan ktip c khi lng 24,8g. Th

tch tngng l 11,2lt.a) Hy xc nh CTPT ca ankanb) Tnh % thtch ca 2 ankan.

lm bi tp ny hc sinh phi hiu cc khi nim ng ng, ankan, ankan k tip,CTTQ, vit c h phng trnh v khi lng v s mol, bit quy i th tch ra s mol.Bit cng thc tnh % theo th tch 2 cht .

Qua vic thng xuyn gii cc bi tp hn hp, lu dn hc sinh s thuc cc k hiu

ha hc, nhha tr, s oxi ha ca cc nguyn t, 5) Bi tp ha hc to iu kin tduy hc sinh pht trin:

Bi tp ha hc pht trin nng lc nhn thc, rn luyn tr thng minh cho hc sinh.Khi gii mt bi tp, hc sinh c rn luyn cc thao tc t duy nh phn tch, tng hp,so snh, din dch, qui np. Mt bi ton c th c nhiu cch gii khc nhau: c cch giithng thng, theo cc bc quen thuc, nhng cng c cch gii ngn gn m li chnhxc. Qua vic gii nhiu cch khc nhau, hc sinh s tm ra c cch gii ngn m hay,iu s rn luyn c tr thng minh cho cc em.

Vd :bi v d trn:

Mt hn hp gm 2 chtngng ankan k tip c khi lng 24,8g. Th tch tngng l 11,2lt.

a)Hy xc nh CTPT ca ankanb) Tnh % thtch ca 2 ankan.

Vi bi ny c 2 cch gii:- Cch 1: Da vo khi lng v th tch bi cho a v phng trnh 2 n s (gia

s C ca mt ankan (ln hoc b) vi s mol ca hn hp) v bin lun.- Cch 2: dng phng php trung bnh tm c s C trung bnh( n ) ta s suy c 2

gi tr (n, m) ng vi 2 ankan ng ng k tip. cch 2 gii nhanh, chnh xc hn cch 1v t tnh ton hn cch 1.


7/115


SVTH:PhanTh Thy 7

Cch gii 2 : t CTPT trung bnh ca 2 ankan :2n2n

HC+

, t phng trnh tnh khi

lng ca hn hp [(14 n +2).11,2/22,4=24,8] n =3,4 2 ankan l C3H8 v C4H10T nhiu cch gii nh vy hc sinh s chn ra cho mnh mt phng php gii

thch hp nht nhvy m t duy cc em pht trin.

6)Tc dng gio dc ttng:

Khi gii bi tp ha hc, hc sinh c rn luyn v tnh kin nhn, tnh trung thctrong lao ng hc tp, tnh c lp, sng to khi x tr cc vn xy ra. Mt khc, vic tmnh gii cc bi tp ha hc cn gip cho hc sinh rn luyn tinh thn k lut, bit t kimch, c cch suy nghv trnh by chnh xc, khoa hc, nng cao lng yu thch b mnha hc.

Tc dng ny c th hin r trong tt c cc bi tp ha hc. Bi ton ha hc gmnhiu bc i n p s cui cng. Nu cc em sai bt k mt khu no s lm cho hthng bi ton b sai.

Vd: C4H10O c bao nhiu ng phn ?y l mt bi tp rt n gin, di vi hc sinh nhng khng phi hc sinh no

cng lm ng hon ton v cc em khng cn thn, ch quan khi lm bi.Tuy nhin, tc dng gio dc t tng ca bi tp c c pht huy hay khng, iu

ny cn ph thuc vo cch dy ca gio vin.Bi tp ha hc c ni dung thc nghim cn c tc dng rn luyn tnh cn thn,

tun th trit qui nh khoa hc, chng tc phong lum thum da vo kinh nghim ltvt cha khi qut vi phm nhng nguyn tc ca khoa hc.

Vd : Trong phng th nghim ha hc no u c ni qui phng th nghim, cc chailu c nhn v nhng v tr cnh

7) Gio dc kthut tng hp:

B mn ha hc c nhim v gio dc kthut tng hp cho hc sinh, bi tp ha hcto iu kin tt cho gio vin lm nhim v ny.

Nhng vn ca kthut ca nn sn xut yu cu c bin thnh ni dung ca ccbi tp ha hc, li cun hc sinh suy nghv cc vn ca kthut.

Bi tp ha hc cn cung cp cho hc sinh nhng s liu l th ca kthut, nhng sliu mi v pht minh, v nng sut lao ng, v sn lng ngnh sn xut hn hp tc gip hc sinh ha nhp vi s pht trin ca khoa hc, k thut thi i mnh ang

sng. Vd1: Tnh lng Crm c thiu chc t1 t crmit cnh (FeCr2O4) ThanhHa.

Vd2: Cho bit thnh phn chnh ca kh thin nhin, kh cracking, kh than v khl cao (kh ming l). Mun iu chmi chtdi y ta c thi t loi kh no nitrn: CCl4, C2H5OH, CH3NH2?

I.1.4 PHN LOI BI TP HA HC:Hin nay c nhiu cch phn loi bi tp khc nhau trong cc ti liu gio khoa. V

vy, cn c cch nhn tng qut v cc dng bi tp da vo vic nm chc cc csphn

loi.1. Phn loi da vo ni dung ton hc ca bi tp:


8/115


SVTH:PhanTh Thy 8

Bi tp nh tnh (khng c tnh ton) Bi tp nh lng (c tnh ton)

2. Phn loi da vo hot ng ca hc sinh khi gii bi tp: Bi tp l thuyt (khng c tin hnh th nghim)

Bi tp thc nghim (c tin hnh th nghim)3. Phn loi da vo ni dung ha hc ca bi tp: Bi tp ha i cng

-Bi tp v cht kh-Bi tp v dung dch-Bi tp vin phn

Bi tp ha v c-Bi tp v cc kim loi-Bi tp v cc phi kim-Bi tp v cc loi hp cht oxit, axit, baz, mui,

Bi tp ha hu c-Bi tp v hydrocacbon-Bi tp v ru, phenol, amin-Bi tp v andehyt, axit cacboxylic, este,

4. Da vo nhim v v yu cu ca bi tp: Bi tp cn bng phng trnh phn ng Bi tp vit chui phn ng Bi tp iu ch Bi tp nhn bit Bi tp tch cc cht ra khi hn hp Bi tp xc nh thnh phn hn hp Bi tp lp CTPT. Bi tp tm nguyn t cha bit

5. Da vo khi lng kin thc, mc n gin hay phc tp ca bi tp: Bi tp dng cbn Bi tp tng hp

6. Da vo cch thc tin hnh kim tra: Bi tp trc nghim Bi tp t lun

7. Da vo phng php gii bi tp: Bi tp tnh theo cng thc v phng trnh. Bi tp bin lun Bi tp dng cc gi tr trung bnh

8. Da vo mc ch s dng: Bi tp dng kim tra u gi Bi tp dng cng c kin thc Bi tp dng n tp, n luyn, tng kt Bi tp dng bi dng hc sinh gii Bi tp dng pho hc sinh yu,


9/115


SVTH:PhanTh Thy 9

Mi cch phn loi c nhng u v nhc im ring ca n, ty mi trng hp cth m gio vin s dng h thng phn loi ny hay h thng phn loi khc hay kt hpcc cch phn loi nhm pht huy ht u im ca n.

Thng gio vin sdng bi tp theo hng phn loi sau:Bi tp gio khoa:Thng di dng cu hi v khng tnh ton nhm lm chnh xc khi nim; cng

c, h thng ha kin thc; vn dng kin thc vo thc tin.Cc dng hay g p: vit phng trnh phn ng, hon thnh chui phn ng, nhn

bit, iu ch, tch cht, gii thch hin tng, bi tp v tnh cht ha hc cc cht, C th phn thnh 2 loi :+ Bi tp l thuyt (cng c l thuyt hc)+ Bi tp thc nghim : va cng c l thuyt va rn luyn cc knng, kxo thc

hnh, c ngha ln trong vic gn lin l thuyt vi thc hnh.

Bi tp ton:L nhng bi tp gn lin vi tnh ton, thao tc trn cc s liu tm c s liu

khc, bao hm 2 tnh cht ton hc v ha hc trong bi.Tnh cht ha hc: dng ngn ng ha hc & kin thc ha hc mi gii c (nh

va , hon ton, khan, hidrocacbon no, khng no, ) v cc phng trnh phn ng xyra.

Tnh cht ton hc: dng php tnh i s , qui tc tam sut, gii h phng trnh, Ha hc l mt mn khoa hc t nhin, tt yu khng trnh khi vic lin mi vi

ton, l, c im ny cng gp phn pht trin t duy logic cho hc sinh. Hin nay, hu

ht cc bi tp ta ha nh nhn vic rn luyn t duy ha hc cho hc sinh, gim dnthut ton.

I.1.5 MT S PHNG PHP GII BI TP:1- Tnh theo cng thc v phng trnh phn ng2- Phng php bo ton khi lng3- Phng php tng gim khi lng4- Phng php bo ton electron5- Phng php dng cc gi tr trung bnh

Khi lng mol trung bnh

Ha tr trung bnh S nguyn t C, H, trung bnh S lin kt trung bnh G hydrocacbon trung bnh S nhm chc trung bnh,

6- Phng php ghp n s7- Phng php t chn lng cht8- Phng php bin lun

I.1.6IU KINHOC SINH GII BI TPC TT:

1. Nm chc l thuyt: cc nh lut, qui tc, cc qu trnh ha hc, tnh cht l ha hcca cc cht.


10/115


SVTH:PhanTh Thy 10

2. Nm c cc dng bi tp cbn, nhanh chng xc nh bi tp cn gii thucdng bi tp no.

3. Nm c mt s phng php gii thch hp vi tng dng bi tp4. Nm c cc bc gii mt bi ton hn hp ni chung v vi tng dng bi ni

ring5. Bit c mt s th thut v php bin i ton hc, cch gii phng trnh v h

phng trnh bc 1,2,

I.1.7 CC BC GII BI TP TRN LP:1. Tm tt u bi mt cch ngn gn trn bng. Bi tp v cc qu trnh ha hc c

th dng s.2. X l cc s liu dng th thnh dng cn bn (c th bc ny trc khi tm tt

u bi)3. Vit cc phng trnh phn ng xy ra (nu c)4. Gi v hng dn hc sinh suy nghtm li gii:

- Phn tch d kin ca bi xem t cho ta bit c nhng g- Lin h vi cc dng bi tp cbn gii- Suy lun ngc t yu cu ca bi ton

5. Trnh by li gii6. Tm tt, h thng nhng vn cn thit, quan trng rt ra t bi tp (v kin thc,

knng, phng php)

I.1.8 CSTHC TIN:Thc t nhiu trng ph thng, s tit ha trong tun t, phn ln dng vo vic

ging bi mi v cng c cc bi tp cbn trong sch gio khoa. Bi tp gio khoa mrng v cc bi tp ton chc cp mc thp. Khi c bi tp ha nhiu hc sinh

b lng tng khng nh hng c cch gii, ngha l cha hiu r bi hay cha xc nhc mi lin h gia gi thit v ci cn tm.

Cc nguyn nhn lm hc sinh lng tng v sai lm khi gii bi tp ha hc: Cha hiu mt cch chnh xc cc khi nim, ngn ng ha hc (v d nh :

nng mol, dd long, c, va , ) Cha thuc hay hiu c th vit ng cc phng trnh phn ng, cha nm

c cc nh lut cbn ca ha hc Cha thnh tho nhng knng cbn v ha hc, ton hc (cn bng phn ng,

i s mol, V, nng , lp t l, ) Khng nhn ra c mi tng quan gia cc gi thit, gi thit vi kt lun

c th la chn v s dng phng php thch hp i vi tng bi c th.


11/115


SVTH:PhanTh Thy 11

I.2 C S LTHUYTCHUYN

NGNH

I.2.1 THUYT CU TO HA HC :Ni dung :1. Trong phn t cht hu c, cc nguyn t lin kt vi nhau theo ng ha tr v theo mtth t nht nh. Th t lin kt gi l cu to ha hc. S thay i th t s to nncht mi.

2. Trong phn t hp cht hu c, cacbon c ha tr IV. Nhng nguyn t cacbon c thkt hp khng nhng vi nhng nguyn t ca cc nguyn t khc m cn kt hp trc tipvi nhau thnh nhng mch cacbon khc nhau (mch khng nhnh, c nhnh, mch vng).3. Tnh cht ca cc hp cht ph thuc vo thnh phn phn t (bn cht v s lng ccnguyn t) v cu to ha hc (th t lin kt cc nguyn t)

I.2.2 NG NG NG PHN :1. ng ng :- ng ng l hin tng cc cht c cu to v tnh cht tng t nhau, nhng v thnh

phn phn t khc nhau mt hay nhiu nhm CH2. Nhng cht c gi l nhng cht

ng ng vi nhau, chng hp thnh mt dy ng ng.2. ng phn :- ng phn l hin tng cc cht c cng CTPT, nhng c cu to khc nhau nn c tnhcht ha hc khc nhau. Cc cht c gi l nhng cht ng phn.

I.2.3 CC LOI CNG THC HA HU CVic nm vng ngha ca mi loi cng thc ha hu cc vai tr rt quan trng.

iu ny cho php nhanh chng nh hng phng php gii bi ton l p CTPT, dngton cbn v ph bin nht ca bi tp hu c. Cc bi ton lp CTPT cht hu cnhnchung ch c 2 dng :

- Dng 1 : Lp CTPT ca mt cht- Dng 2 : Lp CTPT ca nhiu cht.Vi kiu 1, c nhiu phng php khc nhau gii nh : tm qua CTG, tm trc

ti p CTPTKiu 2 ch yu dng phng php tr s trung bnh (xem phn tr s trungbnh). Nhng d dng phng php no chng na th cng vic u tin l t cng thctng qut ca cht , hoc cng thc tng ng cho hn hp mt cch thch hp nht,vic t cng thc ng chim 50% yu t thnh cng.1. Cng thc thc nghim : cho bit thnh phn nh tnh, t l v s lng cc nguyn ttrong phn t.V d : (CH2O)n (n 1, nguyn dng nhng cha xc nh )

2. Cng thc n gin : c ngha nh cng thc thc nghim nhng gi tr n = 1


12/115


SVTH:PhanTh Thy 12

3. Cng thc phn t : cho bit s lng nguyn t ca mi nguyn t trong phn t, tc lcho bit gi tr n4. Cng thc cu to : ngoi vic cho bit s lng nguyn t ca mi nguyn t trong

phn t cn cho bit trt t lin kt gia cc nguyn t trong phn t.

C nhiu loi CTCT khc nhau, chng hn CTCT y , CTCT vn tt, CTCT bnkhai trinNguyn tc chung vit CTCT bn khai trin l c th bt cc lin kt ngia cc nguyn t cc nguyn t, cc lin kt bi trong nhm chc (nu thy khng cnthit) nhng nht thit khng c b lin kt bi gia cc C-C.

Cc loi cng thc CTTN, CTG, CTPT trng hau khi gi tr n = 1. Cng thc tng qut : cho bit thnh phn nh tnh cht c cu to nn t nhng

nguyn t no, i vi CTTQ ca mt dy ng ng c th th cn cho bit thm t lnguyn t ti gin hoc mi lin h gia cc thnh phn cu to .V d : CTTQ ca hydrocacbon l CxHy hoc CnH2n+2-2k nhng vi hydrocacbon c th lankan th CTTQ l : CnH2n+2, anken l : CnH2n ,

I.2.4 TM TT HA TNH CC HYDROCACBON

ANKAN :- Hydrocacbon no, mch h, trong phn t ch c lin kt n gia C-C v C-H- CTTQ : CnH2n +2 , n1, nguyn

a) Tnh cht ho hc :1. Phn ng oxiha :+ Phn ng oxy ha hon ton :CnH2n +2 + (3n +1)/2 O2

Cto n CO2 + (n+1)H2ON

u thi

u oxi :

CnH2n +2 + (n +1)/2 O2 Cto n C + (n+1)H2O

+ Phn ng oxy ha khng hon ton : nu c xc tc th ankan s b oxi ha to nhiu snphm : andehyt, axitCH4 + O2

COV o300,52 HCHO + H2O(andehyt fomic)

n-C4H10 + 5/2 O2 2CH3COOH + H2O2. Phn ng phn hy+ Bi nhit :CnH2n +2

Co1000 n C + (n+1)H2+ Bi Clo :CnH2n +2 + (n +1)Cl2

', cuctimsto n C + 2(n+1)HCl3. Phn ng th vi cc halogen :CnH2n +2 + mX2

kt',sto CnH2n+2-m Xm + mHX4. Phn ng hidro ha (tch hydro) : to sn phm c th c mt hay nhiu ni i hockhp vng.CnH2n +2

CNiFe o600,, CnH2n + H2 (n 2)V d :CH3CH3 Ctxt

o, CH2CH2 + H2

n-hexan Ctxto, xiclohexan + H2


13/115


SVTH:PhanTh Thy 13

(C6H14) (C6H12)5. Phn ng cracking (b gy mch cacbon)CnH2n +2

cracking CmH2m + CxH2x+2iu kin : n 3, m 2, nguyn

x 1n=m+xTng qut :Ankan (3C)

otxt, Ankan + ankenC3H8

otxt, CH4 + C2H4

XICLOANKAN- L hydrocacbon no, mch vng, trong phn t ch tn ti lin kt n.- CTTQ : CnH2n , n3 nguynXicloankan c y tnh cht ca mt hydrocacbon no (vng C5 trln ), ngoi ra cn ctnh cht ca vng:cc vng nh c sc cng ln, km bn, d tham gia phn ng cng mvng (vng C3, C4 ) :

CH2CH

2CH

2

Br Br

+ Br2

ANKEN :- L nhng hydrocacbon mch hc mt ni i trong phn t- CTTQ : CnH2n ,n 2, nguyn1. Phn ng cng

CnH2n + H2 o

txt, CnH2n+2CnH2n + Br2

otxt, CnH2nBr2CnH2n + HA

otxt, CnH2n+1A(vi HA l cc axit nh HCl, HBr, H2SO4)CnH2n + H2O

otxt, CnH2n +1OH- Phn ng cng ca anken tun th quy tc Maccopnhicop : nguyn t H (hay phn mangin tch dng) cng vo nguyn t Cacbon c nhiu H hn, cn phn m ca tc nhn(nguyn t X)gn vo C ca ni i mang in dng (C t H hn).

2. Phn ng oxiha :

+ oxiha hon ton :CnH2n + 3n/2 O2 nCO2 + nH2O+ oxi ha khng hon ton bi ddKMnO4 :CnH2n + [O] + H2O 4

ddKMnO CnH2n(OH)2CH2CH2 + [O] + H2O 4ddKMnO HOCH2CH2 OH3. Phn ng trng hpCH2CH2 P,txt,

o

[CH2CH2]n(Poly etilen) (nha PE)


14/115


SVTH:PhanTh Thy 14

CH3 CH=CH2xt,t

o,p CH CH2

CH3

nn

Polypropylen(nha PP)

Tng qut :nRCH CHR' CH

R

CH

R'

xt,to,P

n

ANKADIEN :- L hydrocacbon khng no, mch h, trong phn t c 2 ni i C=C.- CTTQ : CnH2n-2, n 2 nguyn

1. Phn ng cng :

CH2=CH CH=CH2 + Br21:1

cong 1,2

CH2=CH CHBr CH2Br

cong 1,4CH2 CH=CH CH2BrBr

1CH2CHCHCH2 + 2Br2 BrCH2CHBrCHBrCH2Br

cong 1,4

CH2=CH CHCl CH3cong 1,2

1:1CH2=CH CH=CH2 + HClCH2=CH CH2 CH2Cl

(spc)

(spp)

CH3 CH=CH CH2Cl (sp duy nhat)

2. Phn ng trng hp :nCH2CHCHCH2 PC,tNa,o

[CH2CHCHCH2]nnCH2CCHCH2 xtP,,thop,trng

o

[CH2CCHCH2]n CH3 CH3

3. Phn ng oxi ha :+ Oxi ha hon ton :CnH2n-2 +(3n-1)/2O2

Cto nCO2 + (n-1)H2O+ Oxi ha khng hon ton :3CH2CHCHCH2 + 4KMnO4 + 8H2O

CH2OHCHOHCHOHCH2OH + 4MnO2 + 4KOH

ANKIN

- L nhng hydrocacbon khng no, mch hc mt ni ba trong phn t- CTPTTQ : CnH2n-2, n 2 , nguyn

1. Phn ng cng :CnH2n-2 + H2

COtPd, CnH2nCnH2n-2 + 2H2

COtNi, CnH2n+2

CnH2n-2 + X2 CnH2nX2 + 2X

CnH2nX4Vi X : halogen


15/115


SVTH:PhanTh Thy 15

HC CH + X2 XHC=CHX + 2X X2HC-CHX2

CnH2n-2 + HA CtoXt, CnH2n-1A

Vi HA : cc axit nh : HCl, HCN, H2SO4HC CH + H2O

CHgSO O80,4 CH3CHO

Lu : trong phn ng cng gia ankin bt i v tc nhn bt i, sn phm chnh cxc nh theo quy tc Maccopnhicop.2. Phn ng oxi ha :CnH2n-2 +(3n-1)/2O2

Cto nCO2 + (n-1)H2O3C2H2 + 8KMnO4

OH 3K2C2O4 +8MnO2 + 2KOH + 2H2OC2H2 + 2KMnO4 + 3H2SO4 2CO2 + 2MnSO4 + K2SO4 + 4 H2O5CH3C CH +8KMnO4 + 12H2SO4 5CH3COOH + 5CO2 + 8MnSO4 + 4K2SO4 +12H2O (Hin tng mu tm dung dch nht dn hoc mt hn)3. Phn ng trng hp

2HC CH CClNHCuCl o100,/ 4

CH2CHC CH (Trng hp)(Vinylaxetyl hay vinylaxetilen)3HC CH C

o600hoat tinh,C C6H6 (benzen) (Tam hp)nHC CH CCu

o280, (CHCH) (Cupren)4. Phn ng bi kim loi ca Ankin-1 :HC CH + 2AgNO3 + 2NH3 AgC CAg + 2NH4NO3RC CH + AgNO3 + NH3 RC CAg + NH4NO3Vit tt :HC CH + Ag2O 33/NHddAgNO AgC CAg + H2O

2RC

CH + Ag2O 33/NHddAgNO 2

RC

CAg + H2OTrong dy ngng ankin, chc axetilen c ththhai ln vi ion kim loiHC CH + 2Na NaC CNa + H2

HYDROCACBON THM :Aren hay hydrocacbon thm l loi hydrocacbon c c trng trong phn t bi s

c mt mt hay nhiu vng benzen1. Phn ng thVi Halogen :

H

Br2

Br

+ Fe + HBr

(Brombenzen)Vi axit nitric (xc tc H2SO4, t

oC) (phn ng nitro ha)H

HO NO2

H2SO

4NO2

+d, toC

+ H2O

Nitrobenzen


16/115


SVTH:PhanTh Thy 16

NO2

HO NO2

H2SO

4NO2

NO2

+d, toC

+ H2Odu

(1,3-dinitrobenzen)Vi axit H2SO4, bo ha SO3 (phn ng sunfo ha)

H

HO SO3H

SO3HSO3

+ + H2O

axit benzensunfonic ng ng ca benzen cng cho phn ng thC mch nhnh vi Halogen trong

iu kin chiu sng :CH

3

Br2

CH2Br

+ + HBr

a's'kt

Brombenzyl 2. Phn ng cng :

+ 3 H2

Ni,tOC

xiclohexan

Cl

Cl Cl

Cl

Cl

Cl

a's'

hexacloxiclohenxan

+ 3 Cl2

(C6H

6Cl

6) (666)

3. Phn ng oxi ha :CnH2n-6 + (3n-3)/2O2 nCO2 + (n-3)H2OC6H5CH3 + 2KMnO4 Ct

o

C6H5COOK + 2MnO2 + KOH +H2OToluen Kalibenzoat

* Benzen bn, khng b oxiha bi ddKMnO4, ch c mch nhnh ca vng benzen mi boxiha phn ng dng phn bit benzen v cc ng ng ca n.

II.5 IU CH CC HYDROCACBON1. iu ch ankan :

Nguyn liu ly t thin thin nh kh than , kh du m Tng hp t cc dn xut halogen hoc cc mui ca cc axit hu c

RX + 2Na + XR RR + 2NaXC2H5Cl + 2Na + ClCH3 C2H5CH3 + 2NaClR1(COONa)m + mNaOH(r)

COtCaO, R1Hm + mNa2CO3iu ch Metan :C + 2H

2 COtNi, CH

4CO + 3H2 COt CH4+ H2O


17/115


SVTH:PhanTh Thy 17

CH3COONa + NaOHr COtCaO, CH4 + Na2CO3

Al4C3 + 12 H2O 4Al(OH)3 + 3CH42. iu ch anken :+ Phn ng cracking v phn ng hydro ha :CH3CH2OH C

o180,SOH 42 CH2CH2 + H2ORCHXCH2R ruouKOH,Dd RCHCHR + HXRCHXCHXR + Zn RCHCHR + ZnX2RCHOHCH2R CO

O350,AlMgO, 32 RCHCHR + H2ORC CR + H2 C

OtPd, RCHCHRCnH2n+2

Cto CmH2m + CxH2x+2CnH2n+2

Cto CmH2m + (n + 1 - m)H23. iu ch Ankin, Ankadien :

RCHXCHXR ruouKOH,2

RC CR + 2HXRCHX2CHX2R +2Zn RC CRRC CH + Na RC CNa +1/2H2RC CNa + X-R RC CR (Phn ng tng mch C)CaC2 + 2H2O Ca(OH)2 + C2H22CH4

llnC,1500o C2H2 + 3H22C + H2

CO3000 C2H24. iu ch ankadien

2CH3CH2OH Co500400OAl 32 CH2CHCHCH2 + 2H2O

2HC CH CClNHCuClo100,/ 4 CH CCHCH2

CH CCHCH2 + H2 COtPd, CH2CHCHCH2

CH2CHOHCHOHCH3 Co180,SOH 42 CH2CHCHCH2 +2H2O

CH2CHCH2CH3 hidrhaDe CH2CCHCH2 CH3 CH3 (Isopren)

CH CCHCH2 + HCl CH2CCHCH2Cl (Cloropren)

5. iu ch hydrocacbon thm v cc hydrocacbon khc :3C2H2

hoat tinhCC,600o C6H6C6H12(xicloankan)

COtPd, C6H6 + 3H2C6H14

COtPd, C6H6 + 4H2C6H5COOH + 2NaOH

COt C6H6 + Na2CO3 + H2OC6H5X +2Na +X-CH3

C6H5CH3 + 2NaXC6H6 + CH3X 3

AlCl C6H5CH3 + HX

Nhn xt :


18/115


SVTH:PhanTh Thy 18

Hydrocacbon no (ankan), phn ng c trng l phn ng th, khng c phn ng cngv kh b oxiha bi dd KMnO4 Hydrocacbon khng no (anken, ankadien, ankin) phn ng c trng l phn ng cng.(anken c phn ng thnhit cao, th

H )

Phn ng cng Hidro :+ xt Ni/toC th xicloankan (C3, C4), anken v ankin, ankadien cng H2 c ankan; arencng H2c xicloankan

+ xt Pd/toC th ankin, ankadien cng H2c anken Phn ng cng HX vo anken, ankadien, ankin phi ch sn phm chnh ph v slng sn phm.

t chy CxHy: t2

2

COmolSo

OHmolSoT = th :

T>1 => CxHy l ankan, CTTQ : CnH2n+2T = 2 CxHy l CH4T=1 => CxHy l anken, xicloankan CTTQ : CnH2nT CxHy l ankadien, ankin, CTTQ : CnH2n-2 hoc l aren, CTTQ : CnH2n-6T = 0,5 CxHy l C2H2 hoc C6H6.


19/115


SVTH:PhanTh Thy 19

CHNG2:

PHNLOIVPHNGPHP

GIIMTS BITPVHYDROCACBONTRONGCHNGTRNHTHPT


20/115


SVTH:PhanTh Thy 20

II.1BITPGIOKHOA

I.1.1 BI TP V CNG THC CU TO NG NG NG PHN DANH PHP

I.1.1.1 Bi tp vng ng

Phng php :C 2 cch xc nh dy ng ng ca cc hydrocacbon :- Da vo nh ngha ng ng- Da vo electron ha tr xc nh

Lu :C lun c ha tr IV tc l c 4e ha tr

nC s c 4ne ha trH lun c ha tr I tc l c 1e ha tr

- Parafin chnh l ankan, dy ng ng parafin chnh l dy ng ng ca CH4.- Olefin chnh l anken, dy ng ng olefin chnh l dy ng ng ca C2H4- Ankadien cn c gi l ivinyl- Aren : dy ng ng ca benzen.- Hydrocacbon : CxHy : y chn, y 2x + 2

Bi tp v d :

V d 1: Vit CTPT mt vi ng ng ca CH4. Chng minh cng thc chungca dy ng ng ca CH4 l CnH2n+2.

GII :

Da vo nh ngha ng ng, CTPT cc ng ng ca CH4 l C2H6,

C3H8, C4H10,, C1+kH4+2k

Chng minh CTTQ dy ng ng metan CH4 l CnH2n+2 :

Cch 1: Da vo nh ngha ng ng th dy ng ng ca metan phi l:CH4 + kCH2 = C1+kH4+2k

Tm mi lin h gia s nguyn t C v s nguyn t Ht nC = 1 + k = nnH = 4 + 2k = 2(k + 1) + 2 = 2n + 2

Vy dy ng ng farafin l CnH2n+2 (n 1)


21/115


SVTH:PhanTh Thy 21

Cch 2: Da vo s electron ha tr :- S e ha tr ca nC l 4n- S e ha tr ca 1C dng lin kt vi cc C khc l 2 S e ha tr ca nC dng lin kt vi cc C khc l [2(n-2)+2] =

2n2 (v trong phn t ch tn ti lin kt n)(Sd+2 v 1C u mch ch lin kt vi 1C nn dng 1e ha tr, 2Cu mch dng 2e ha tr.- S e ha tr dng lin kt vi H: 4n2n-2 = 2n + 2- V mi nguyn t H ch c 1 e ha tr nn s e ha tr ca (2n+2)nguyn t H trong phn t l 2n + 2. Cng thc chung ca ankan l CnH2n+2 (n 1)

Cch 3: Metan c CTPT CH4 dng CnH2n+2 dy ng ng ca ankan lCnH2n+2

V d 2: CT n gin nht ca 1 ankan l (C2H5)n. Hy bin lun tm CTPTca cht trn.

GII :

CT n gin ca ankan l (C2H5)n. Bin lun tm CTPT ankan :

Cch 1: Nhn xt: CT n gin trn l 1 gc ankan ha tr 1 tc c kh nngkt hp thm vi 1 gc nh vy na n = 2 CTPT ankan C4H10

Cch 2:

Cch 3:

CTPT ca ankan trn : (C2H5)n = CxH2x+2 2n = x v 5n = 2x + 2 5n = 2.2n + 2 n = 2. CTPT ankan : C4H10

Ankan trn phi tha iu kin s H 2.s C + 2 5n 2.2n + 2n 2n =1 th s H l loin= 2 CTPT ankan l C4H10 (nhn)Vy CTPT ankan l C4H10

V d 3 :

Phn bit ng phn vi ng ng. Trong s nhng CTCT thu gn di y, nhngcht no l ng ng ca nhau? Nhng cht no l ng phn ca nhau.?


22/115


SVTH:PhanTh Thy 22

CH3CH2CH3 CH3CH2CH2Cl CH3CH2CH2CH3

CH3CHClCH3 (CH3)2CHCH3 CH3CH2CH=CH2

CH3CH=CH2 CH2 CH2

CH2 CH2

CH3C=CH2

CH3

(1) (2) (3)

(4) (5) (6)

(7)

(8) (9)

GII :

Phn bit ng phn vi ng ng : xem I.2.2/12 Nhng cht l ng ng ca nhau l : 1 v 5 hoc 1 v 3(ankan); 6 v 7 hoc 6 v 9(anken). Nhng cht l ng phn ca nhau : 2 v 4; 3 v 5; 6 v 9 v 8.

Bi tp tng t:

1) Vit CTPT mt vi ng ng ca C2H4. Chng minh CTTQ ca dy ng ng caetilen l CnH2n , n 2 nguyn

2) Vit CTPT mt vi ng ng ca C2H2 . Chng minh CTTQ ca dy ng ng caaxetilen l CnH2n-2, n 2 nguyn

3) Vit CTPT mt vi ng ng ca C6H6. Chng minh CTTQ ca cc aren l CnH2n-6,n 6 nguyn

II.1.1.2 Bi tp vng phn danh php : Phng php vit ng phn :

Bc 1: - T CTPT suy ra cht thuc loi hydrocacbon hc no.- Vit cc khung cacbon

Bc 2 :- ng vi mi khung cacbon, di chuyn v tr lin kt bi (nu c), di chuyn v trcc nhm th (nu c).

- Nu c ni i hoc vng trong CTCT ca cht th xt xem c ng phn hnhhc khng.

Bc 3 : - in Hidro.

Lu : lm xong phi kim tra li xem cc nguyn t ng ha tr cha.

Bi tp v d :

V d 1 :

a) Nu iu kin mt phn t c ng phn hnh hc?b) Vit tt c cc CTCT cc ng phn ca C5H10; Trong cc ng phn , ng phn noc ng phn hnh hc? c tn cc ng phn .

GII :

a) iu kin mt phn t c ng phn hnh hc (ng phn cis-trans) :


23/115


SVTH:PhanTh Thy 23

Xt ng phn :

C=C

a b

fd

iu kin : a d v b f- Nu a > d v b>f (v kch thc phn t trong khng gian hoc v phn t lng M)* tac ng phn cis.- Nu a > d v b


24/115


SVTH:PhanTh Thy 24

Bc 1 : Vit tt c cc khung mch C ng vi CTPT bi cho (nhp)Bc 2 : Thc hin cc phn ng theo bi v xc nh s sn phm. CTCT no tha mns sn phm bi th ta chn (nhp)

Bc 3 : Xc nh li CTCT va tm c, vit ptp chng minh. (v)

ng v

i pentan C

5H

12c cc d

ng khung C sau :

C C C C C C C C C

C

C C C

C

C

(1) (2) (3)

1 2 3 4 5 1 2 3 4 1 2 3

a) Khi thc hin phn ng th :(1) c 3 v tr th (C1, C2, C3) to 3 sn phm (loi)(2) c 4 v tr th (C1, C2, C3, C4) to 4 sn phm (nhn)(3) c 1 v tr th (C1 hoc C3) to mt sn phm (loi)

Vy CTCT ca pentan l (2) : 2-metylbutan (isopentan)

Ptp :

CH3 CHCH2 CH3

CH3 (2)

1 2 3 4+ Cl2

askt

CH2Cl CH CH2 CH3

CH3CH3 CCl CH2 CH3

CH3CH3 CH CHCl CH3

CH3CH3 CH CH2 CH2Cl

CH3 b) Tng t :CTCT ca pentan l (3): 2,2-dimetylpropan (neopentan), khi cracking ch cho 2 sn phm :

CH3 C CH3

CH3

CH3

(3)

1 2 3 cracking,to

CH2 C

CH3

CH3 + CH4

V d 3 :Vit CTCT ca cc cht c tn sau v gi tn li cho ng nu cn :a) 3,3-Diclo-2-etyl propan

b) 4-Clo-2-isopropyl-4-metylbuten-2c) 2-isopropyl penten-1

GII :

i vi loi bi ny th nguyn tc l t tn gi vit CTCT ca cht . Sau xtxem ngi ta gi tn ng cha bng cch chn mch chnh, nh s ch v trnhnhnu sai th gi tn li.a) 3,3-Diclo-2-etylpropan

CH3 CH CH Cl

Cl

CH2 CH3

12

3 4 1,1-Diclo-2-metylbuta


25/115


SVTH:PhanTh Thy 25

(tn sai do chn mch chnh 3C cha phi l mch di nht)b) 4-Clo-2-isopropyl-4-metylbuten-2

CH3 C

CH

CH CH

CH3

Cl

CH3

CH312

3 4 5

6

c) 2-isopropylpenten-1

CH2 C CH2 CH2 CH3

CH

CH3

CH3 CH

CH3

CH3

CH2 C CH2 CH2 CH31 2 3 4 5

1

2

3 4 5 6

(i) (ii)

Cch c tn trn l ng. Nu chn mch chnh l 6C (ii) l sai v mch ny khngcha ni i.

Bi tp tng t:

1) Vit CTCT ca cht X c CTPT C5H8. Bit rng khi hydro ha cht X, ta thu cisopren. Mt khc, cht X c kh nng trng hp cho ra cao su tng hp. c tn danh

php IUPAC cc ng phn mch hca X2) Cho aren c CTPT C8H10. vit CTCT v gi tn cc ng phn ca A.3) Vit CTCT v gi tn li cho ng nu cn. Xt xem ng phn no c ng phn hnhhc.

a) 1,2- Diclo-1-metyl hexanb) 2,3,3-Tri metyl butanc) 1,4-Dimetyl xiclobutan.c) Diallyld) 3-allyl-3-metylbuten-1e) 2,2,5,5- tetrametylhexin-3f) 3-metylpentin-1

Cch gi tn trn sai v chnmch chnh sai.Tn ng l :5-Clo-2,3-Dimetylhexen-3


26/115


SVTH:PhanTh Thy 26

to

II.1.2 CHUI PHN NG IU CH

Phng php :

1) Mun lm bi tp chui phn ng cn lu :

-Mi mi tn ch vit mt phng trnh phn ng.-Bt u t phn ng trong c CTCT ca mt cht ta bit chnh xc (phn ng khngc sai CTCT ca cht) da vo cc iu kin phn ng suy lun tm ra cc cht cn li.

-Xem trong chui c phn ng no ct bt mch hay tng mch cacbon khng.2) Cc phn ng ct bt mch hoc ct t mch cacbon th dng cc phn ng :

-Ct bt mch th dng cch nhit phn mui :R COONa + NaOH(r)

caotCaO, o RH + Na2CO3-Cch t th dng phng php cracking

C4H10 to C2H4

+

C2H6

C3H6 CH4

+

3) Ni di thm (tng mch) cacbon : dng mt trong hai cch n gin ca chng trnhha hc ph thng :a) Trng hp :

2HC CH Co

4Cl,100NHCuCl, CH2=CH-C CHb) Ni hai gc ankyl :

RCl + 2Na + RCl RR + 2NaCl3) Bi tp iu ch l mt dng khc ca chui phn ng, y bi ch cho bit nguynliu ban u v yu cu iu ch mt cht no . lm c bi ny, hc sinh phi nh

v vit cc ptp trung gian c ghi km y iu kin phn ng. C nhiu cch iu chkhc nhau vi cng mt bi iu ch.Lu : nu bi yu cu vit siu ch (hoc s tng hp) th ta ch cn vitdi dng mt chui phn ng t nguyn liu n sn phm, trn cc mi tn c ghi kmiu kin phn ng.* Thnh phn ch yu ca :

- Kh thin nhin : ch yu l Metan (90%), cn li l etan, propan, butan v mt sng ng cao hn.

- Kh cracking : Hydrocacbon cha no (C2H4, C3H6, C4H8), ankan (CH4, C2H6, C4H10)v H2.

- Kh than : ch yu l H2(60%), CH4 (25%) cn li l CO, CO2, N2- Kh l cao : CO2, CO, O2, N2,

Bi tp v d :

V d 1 : Chui phn ng cho bit CTPT cc cht :Hon thnh chui phn ng sau :C2H5COONa

)1( C2H6 )2( C2H5Cl

)3( C4H10 )4( CH4

)5( CO2

GII :

Nhn xt : bi cho bit CTPT cc cht, ta ch cn nhv vit phn ng c y

iu kin hon thnh phn ng khng cn suy lun nhiu. Loi bi ny thng cdng tr bi hoc lm bi tp cbn trong tit bi tp.


27/115


SVTH:PhanTh Thy 27

(1) ct bt mch nhit phn mui.(3) tng mch cacbon ni hai gc ankyl.Ptp :(1) C2H5COONa + NaOH (r)

caotCaO, o C2H6 + Na2CO3

(2) C2H6 + Cl2 kts'a'

C2H5Cl + HCl(3) C2H5Cl + 2Na + C2H5Cl

ot C4H10(4) C4H10

Cracking CH4 + C3H6(5) CH4 + 2O2 CO2 + 2H2O

V d 2 : bi khng cho bit CTPT ca cc cht nhng cho bit iu kin phn ng.+ Vit phng trnh phn ng, xc nh CTCT cc cht :AlC3 + L E + X (1)E lln,1500

O C Y + Z (2)

CH3COOH + Y xt,to

A (3)nA trunghop B (4)

GII :

Phn tch : iu kin phn ng chnh l du hiu suy lun tm CTCT cc cht.- Da vo (2) E : CH4- Y hoc l C2H2 hoc H2(4)A c phn ng trng hp trong phn t A c C=C;(3) CH3COOH + Y A Y l C2H2 v Z : H2.

(1) L : H2O; X : Al(OH)3(3) CH3COOH + C2H2(Y) CH3COOCH =CH2 (A)(4) (B) :

CH CH2

OCOCH3 n Ptp :Al4C3 + 12H2O 3CH4 + 4Al(OH)32CH4

lln,1500O C C2H2 + 3H2

CH3COOH + HC CH + ot,Hg2

CH2=CHOCOCH3

nCH CH2

OCOCH3

n CH2=CHOCOCH3xt,t

o,p

V d 3: bi khng cho iu kin phn ng, ch cho bit duy nht CTPT ca mt cht

+ B tc chui phn ng sau :A D + FD F + C

F + Br2 GG + KOH J + +


28/115


SVTH:PhanTh Thy 28

Ru

600oC, cacbon hoat tnh

J B (tam hp)B + Cl2 C6H6Cl6J + C D2J X

X + C EGII :

Nhn xt : gia cc phn ng u c mi lin h vi nhau, mi ch ci ng vi mtcht nht nh v cc cht khng trng nhau. bi ny, t phn ng to 666, ta tm c B, da vo cc du hiu khc, suy lun tm racc cht cn li

Phn tch :B + Cl2 C6H6Cl6 B : C6H6J B J : C2H2

F + Br2 G G c hai nguyn t Brom trong phn t.M G + KOH C2H2 (J) G : C2H4Br2 F : C2H4C2H2(J) + CD C : H2 v D : C2H6 (D khng th l C2H4c v trng F)DC2H4(F) + CA D(C2H6) + F(C2H4) A : C4H10Vy A : C4H10; C:H2 ; D:C2H6 ; F : C2H4; G:C2H4Br2 ; J:C2H2

Ptp :C4H10

otCracking, C2H6 + C2H4

C2H6

ot

C2H4 + H2C2H4 + Br2 C2H4Br2C2H4Br2 + 2KOH C2H2 + 2KBr + 2 H2O

3C2H2 C6H6C6H6 + 3Cl2 C6H6Cl6C2H2 + 2H2

CtNi, o C2H6

V d 4 :

Vit s phn ng tng hp PVC t vi v than .

GII :

S :

CaCO3 CaO CaC2 C2H2 C2H3Cl CH2 CH

Cln

1000oC + C

lo ien

+ H2O + HCl trung hp

Bi tp tng t:

Hon thnh cc chui phn ng sau. Ghi y iu kin phn ng :1)


29/115


SVTH:PhanTh Thy 29

Ru butylic(1)

Butilen Butan Metan axetilen PE

Etilen glicol

Etilen(2) (3) (4) (5) (6)

(7)

2)

C2H4 C2H5OH C2H4

Etyl Clorua

Etilen glicol

PE(1) (2) (3) (4)

(5)

3)

CH3COONa

Al4C3C3H8C

CH4

CO2

CH3Cl

(1)(2)

(3)

(4)

(5)

(6)

C2H2

4)

Ankan A

B

xt,to

D

E

PP

cao su Isopren

C CH2

CH3

CH3

n

5) vivi sngcanxicacbuaaxetilenvinyl axetilenDivinylcaosu Buna

6*)

CxHy(A)

A1

B1

A2 A3 TNT

B2 B3 Etylen

p n : A: CH4; A1:C2H2 ; A2 :C6H6 ; A3: C6H5CH3;B1:C2H6 ; B2: C2H5Cl; B3: C2H5OH

7*)

CxHy(X)

X1 X2 cao su Buna

X4

(1) (2) + X3

(3)

(4)(5)

C2H5OH X4(6)

(7)

p n : X:C2H2; X1:C4H4 (vinyl axetilen); X2 : C4H6 (Butadien-1,3) ; X3: C6H5CH=CH2; X4:C2H4; X5: C2H5OH8*)

+H2

H2Oxt

Buna SBunaN

A6

A5A4A3A2A1Ato


30/115


SVTH:PhanTh Thy 30

Bit A v A3 c cng s C.p n : A:C4H10; A1:C2H4; A2: C2H5OH; A3 :C4H6 (Divinyl); A5:C4H8; A6:CH3-CH(OH)-CH39*)

T

kh thin nhin vit ph

ng trnh ph

n

ng

iu ch

caosu Isopren, cao su

Cloropren, Caosu Buna N, CCl4. Cho cc cht v cv iu kin th nghim coi nh.10) Vit phng trnh phn ng tng hp tng hp caosu t cht u l isopentan. Cciu kin phn ng v cc cht v ccoi nh.

11) Vit phng trnh phn ng iu ch C2H5OH t kh cracking.


31/115


SVTH:PhanTh Thy 31

II.1.3 TCH TINH CH

II.1.3.1 Tch cc hydrocacbon :

Nguyn tc :

Tch ri l tch ring tt c nguyn cht ra khi hn hp bng cch tch dn tng chtmt. Th nghim ny kh, i hi phi chn ho cht thch hp tch v hon nguyn licht .S :

A,B

A (nguyen chat)

BX

+ X

+ YB (nguyen chat)

XY (loai bo)

Phng php:

* Phng php vt l :- Phng php chng ct tch ri cc cht lng ha ln vo nhau, c th dng

phng php chng ct ri ngng t thu hi ha cht.- Phng php chit (dng phu chit) tch ring nhng cht hu ctan c trong

nc vi cc cht hu ckhng tan trong nc (do cht lng s phn thnh 2 lp)- Phng php lc (dng phu lc) tch cc cht khng tan ra khi dd.

* Phng php ha hc :- Chn nhng phn ng ha hc thch hp cho tng cht ln lt tch ring cc

cht ra khi hn hp, ng thi ch dng nhng phn ng ha hc m sau phn ng ddng ti to li cc cht ban u.

- Mt s phn ng tch v ti to:Hidrocacbon Phn ng tch Phn ng ti to Phng php thu hiAnken

c c

R-CH=CH2 + Br2R-CHBr-CH2Br

R-CHBr-CH2Br C

otZn, R-CH=CH2

Thu ly kh ankenbay ra (hoc chitly anken lng phnlp)

Etilen

CH2=CH2

CH2=CH2 + H2SO4

CH3CH2OSO3HCH3CH2OSO3H

CotZn, CH2=CH2+H2SO4

Ankin-1 vaxetilenR-C CH

2R-C CH +Ag2O

Co3 t,NH 2R-C CAg + 2H2O

RCCAg + HClRCCH + AgCl

Lc b kt ta thuhi ankin lng hocthu ly ankin kh.

Benzen vcc ngng ca

benzen

Khng tan trongnc v trong cc ddkhc nn dng

phng php chit tch.


32/115


SVTH:PhanTh Thy 32

- Nu c anken v ankin th tch ankin trc bng dd AgNO3/NH3 v ankin cng cho

phn ng cng vi dd Br2 nh anken.

Bi tp v d :

Tch ring tng kh ra khi hn hp kh gm CH4, C2H4, C2H2 v CO2.

GII :

Nhn xt: CO2 tan trong dd nc vi trong, CH4, C2H4, C2H2 th khng, nn dng cc phnng bng trn tch:S tch

CH4C2H2C2H4CO2

Dd Ca(OH)2

CaCO3

CH4

C2H4C2H2

Dd AgNO3/NH

3

AgC CAg

(vang)

CH4C2H4

Dd Brom

C2H4Br2Zn

C2H4to

CH4

to

CO2

HClC2H2

Li gii v phng trnh phn ng:

Dn hn hp kh qua dd Ca(OH)2 d, thu c CaCO3CO2 + Ca(OH)2 CaCO3 + H2O

Thot ra ngoi l hn hp kh CH4, C2H4, C2H2 c dn qua dd AgNO3/NH3 th

C2H2 b gi li trong C2Ag2, cc kh CH4, C2H4 thot raC2H2 + 2AgNO3 (dd) + 2NH3 C2Ag2 + 2NH4NO3 Tip tc dn hn hp kh CH4, C2H4 qua dd nc Br th C2H4b gi li, CH4 thot ra

ta thu c CH4.C2H4 + Br2 C2H4Br2

Ti to CO2 bng cch nhit phn kt ta CaCO3 Ti to C2H2 bng cch cho kt ta C2Ag2 tc dng vi dd HCl

C2Ag2 + 2HCl C2H2 + 2AgCl Ti to C2H4 bng cch cho cht lng C2H4Br2 tc dng vi Zn/ru:

C2H4Br2 ru C2H4 + ZnBr2+ Zn

Bi tp tng t:Tch ri cc kh sau ra khi hn hp gm :a) Benzen, styren, phenol

b) NH3, butin-1, butadien v butanc) Kh HCl, butin-1 v butan

II.1.3.2 Tinh ch :


33/115


SVTH:PhanTh Thy 33

Nguyn tc : Tinh ch l lm sch ha cht nguyn cht no bng cch loibi tp cht ra khi hn hp (nguyn cht v tp cht).

Phng php : Dng ha cht tc dng vi tp cht m khng phn ng vinguyn cht to ra cht tan hoc to ra cht kt ta lc bi.

S

tinh ch

:

A,B

A(nguyen chat)

BX (loai bo)

+X

Trong X l ha cht ta phi chn tc dng vi B loi B ra khi hn hp.

Bi tp v d :

Cc phng trnh phn ng u l nhng phng trnh phn ng quen thuc gp trn.Do phn hng dn gii cha ra cc s tinh ch.

V d 1 :

Tinh ch (lm sch) Propilen c ln propin, propan v kh sunfur

GII :

Lu : SO2 v C3H6u lm cho phn ng vi dd Brom nn phi tch SO2 trc ri midng dd Brom tch ly C3H6 ra khi hn hp ri tinh ch.S tinh ch

ddAgNO3/NH3

C3H6C3H4C3H8SO2

C3H6,C3H8SO2

CH3C2Ag

dd Ca(OH)2

CaSO3

C3H6

C3H8

ddBr2C3H8

C3H6Br2Zn C3H6

V d 2:Tinh ch C6H6 c ln C6H12, C6H5CH3

GII :

S :

ddBr2C6H6C6H12C6H5CH3 C6H12Br2

C6H6C6H5CH3

DdKMnO4 C6H6

C6H5COOH

V d 3:Tinh ch Styren c ln benzen, toluen, hexin-1.

GII :

S :


34/115


SVTH:PhanTh Thy 34

C8H8ZnC8H8Br2

C6H6C6H5CH3ddBr2

C6H

5C CAg

C8H8,C6H6C6H5CH3

ddAgNO3/NH3

C8H8C6H6C6H5CH3

C5H11C CH

Bi tp tng t:1) Tinh ch C3H8 ln NO2 v H2S, hi nc2) Tinh ch C2H6 ln NO, NH3, CO23) Lm sch etan c ln etilen v lm sch etilen c ln etan.4) Lm sch etan c ln axetilen v ngc li5) Lm sch etilen c ln axetilen v ngc li.


35/115


SVTH:PhanTh Thy 35

II.1.4 NHN BIT PHN BIT

Phng php:Tng qut:

- Lm th nghim vi cc mu th+ Ch dng nhng phn ng c trng ca hidrocacbon nhn bit+ Cc phn ng dng nhn bit phi n gin, d thc hin v du hiu phn ngquan st c (mu sc, , si bt kh, )

- Khi c c cht hu cv v cnn phn bit cht v ctrc, nu c.Cch nhn bit vi cht kh v cquen thuc:

CO2, SO2 : lm c nc vi trong nhng SO2 to kt ta vng khi sc vo dd H2Shoc lm mt mu nu ca dd nc Brom.

2H2S + SO2 3S(vng) + H2OSO2 + Br2 + H2O 2HBr + H2SO4

H2O (hi) : i mu trng ca CuSO4 khan thnh xanh N2, kh tr: khng chy NH3 : lm xanh mu qu tm m hoc to khi trng (NH4Cl) vi kh HCl HCl (kh) : lm qu tm m ha hoc to khi trng vi NH3(kh) HCl (dd) : lm qu tm , si bt CO2 vi CaCO3. NO : chuyn thnh nu khi gp khng kh (NO + O2 NO2)

nu NO2 : kh mu nu H2 : cho qua CuO nung nng, CuO chuyn t mu en sang mu .

CuO + H2 Cu + H2O(en) () CO : cho li qua dd PdCl2, sn phm kh thu c cho sc vo dd nc vi trong

d th nc vi trong bc.CO + PdCl2 + H2O CO2 + Pd + 2HClCO2 + Ca(OH)2 CaCO3 + H2O

Thttnginhn bit cc hydrocacbon

Hidrocacbon Thuc th Du hiu Phng trnh phn

ngAnkin u mch dd AgNO3/NH3 vng nht CHCH + 2AgNO3

+ 2NH3AgCCAg +2NH4NO3

Dd Br2 mu nu Mu nu cadd Br2 b nhthay mt mu

CnH2n+2-2k+ kBr2CnH2n+2Br2k

CxHy cha no(anken, akin,ankadien, )

Dd KMnO4l (tm) Mu tm ca ddKMnO

4b

nht

hay mt mu


36/115


SVTH:PhanTh Thy 36

Benzen & ankan Cl2, askt Ch benzen tom trng

Toluen Dd KMnO4l Mt mu tm C6H5CH3 +3[O] 4ddKMnO

C6H5COOH + H2O Nhng im cn lu thm khi nhn bit cc hydrocacbon :

1) Phn bit anken vi cc hydrocacbon mch hkhc c s lin kt nhiu hnBng cch ly cng th tch nh nhau ca cc hydrocacbon ri nh tng lng dd Br2 (cngnng ) vo. Mu no c th tch Br2 b mt mu nhiu hn ng vi hydrocacbon c slin kt nhiu hn.

2) Phn bit axetilen vi cc ankin-1 khc

- Bng cch cho nhng th tch bng nhau ca cc cht th tc dng vi lng d ddAgNO3 trong NH3 ri nh lng kt ta kt lun.CH CH + 2AgNO3 + 2NH3 AgC CAg + 2NH4NO3R C CH + AgNO3 + NH3 R C CAg + NH4NO3

3) Phn bit ankin-1 vi cc ankin khcAnkin-1 to kt ta vng nht vi dd AgNO3 trong NH3

4) Phn bit benzen v ng ng khc ca benzenBenzen khng lm mt mu dd thuc tm (KMnO4) trong khi cc ng ng ca benzen

lm mt mu hoc nht mu dd thuc tm.

*Nu hn hp phc tp nn lp bng nhn bit* Lu : t hin tng suy ra chtVd:Khi lm c nc vi trong v to vng vi dd H2S l SO2 ()Kh SO2 lm c nc vi trong v to vng vi dd H2S l SO2 (ng v mt khoa hcnhng khi nhn bit nh vy l sai qui tc)

Bi tp v d :

Nhn bit cc l kh mt nhn :Bi 1:

a)N2, H2, CH4, C2H4, C2H2b) C3H8, C2H2, SO2, CO2.

GII :

a) N2, H2, CH4, C2H4, C2H2C 3 cch gii :Cch 1 :

Nhn xt:- N2 : khng cho phn ng chy


37/115


SVTH:PhanTh Thy 37

- H2 : phn ng chy, sn phm chy khng lm c nc vi trong- CH4 : phn ng chy, sn phm chy lm c nc vi trong- Cc kh cn li dng cc phn ng c trng nhn bit.

Tm tt cch gii:- Ly mi kh mt t lm mu th.- Dn ln lt cc kh i qua dd AgNO3/NH3. Kh no to c kt ta vng l C2H2

ddAgNO3/NH3+ H2OAgC CAgAg2O+C2H2

(vang) - Dn cc kh cn li qua dd nc Brm (mu nu ). Kh no lm nht mu nc brom lC2H4

H2C=CH2 + Br2 BrH2CCH2Br- Ln lt t chy 3 kh cn li. Kh khng chy l N2. Sn phm chy ca hai kh kiac dn qua dd nc vi trong. Sn phm chy no lm c nc vi trong l CH4. Mu

cn li l H2.CH4 + 2O2 CO2 + 2H2OCO2 + Ca(OH)2 CaCO3 + H2O

H2 + O2 H2OCch 2 :- Dn 5 kh trn ln lt qua dd Brom, c 2 kh lm mt mu dd nc Brom (nhm 1) gmC2H4 v C2H2. 3 kh cn li khng c hin tng g thot ra ngoi (nhm 2) gm CH4 vCO2, H2.- Sau nhn bit cc kh trong mi nhm trn tng t cch 1.Cch 1 ti u hn cch 2.

b) C3H8, C2H2, SO2, CO2.Nhn xt:C 3 cch :Cch 1 :

- Dn bn kh trn ln lt qua dd nc vi trong d. C 2 kh lm c nc vi trong(nhm 1) v 2 kh kia khng lm c nc vi trong (nhm 2).

- Cho 2 kh mi nhm ln lt qua dd nc Brom. Kh nhm 1 lm mt mu nu ca dd Brom l SO2 v kh nhm 2 cng c hin tng nh vy l C2H2. Hai kh cnli l CO2 v C3H8.

Cch 2 :- Dng phn ng c trng nhn bit.- Th t nhn bit C2H2, SO2, CO2, C3H8

Cch 3 :- Dn 4 kh trn ln lt vo dd Brom, c 2 kh lm mt mu nu ca dd Brom

(nhm 1) v 2 kh kia khng c hin tng g (nhm 2).- Dn ln lt 2 kh nhm 1 qua dd AgNO3/NH3. Kh no to kt ta vng nht l

C2H2, kh cn li l SO2.- Dn ln lt 2 kh nhm 2 qua dd nc vi trong. Kh no lm c nc vi trong

l CO2, cn li l C3H8.

Vy c nhiu cch gii bi ny nhng cch 2 l ti u hn c.


38/115


SVTH:PhanTh Thy 38

Bi tp tng t:

1) Ch dng 1 thuc th nhn bit 3 cht lng: benzen, toluen, styren2) Pentan, penten-1, pentin-1, dd AgNO3, nc, dd NH4OH, nc Br, dd HCl, dd HI (chs dng qu tm)

3) Ch dng 1 ha cht nhn bit : n-butan, buten-2, butadien-1,3 , vinylacetylen.4) Nhn bit : n-hexan, hexen-2, hexen-1, n-heptan, toluen, styren v benzen5*)Nhn bit cc l mt nhn sau :a) Kh etan, etylen, acetylen (bng 2 cch)b) Kh metan, etylen, SO2, NO2 v CO2.


39/115


SVTH:PhanTh Thy 39

II.1.5BITPVITPHNGTRNHPHNNGGIACCCHT

Nhng ch khi lm loi bi tp ny :- Phi nm vng cc phn ng ha hc ca cc hydrocacbon.- Nhcc im c bit trong cc phn ng, v d : Ankan :- Phn ng th : t C3 trln nu th vi Cl2 (askt, 1:1) s thu c hn hp sn phm lng phn ca nhau.- Phn ng cracking : ch c ankan t C3 trln.- Phn ng hidro ha i khi cng c gi l phn ng cracking nhng xc tc l Ni,to- Lu : phn ng cng H2 v H2u c xc tc l Ni,t

o .

Xicloankan :- Vng C3, C4 ch c phn ng cng mvng khng c phn ng th. Vng C5 trlnkhng c phn ng cng ch c phn ng th. Aken, ankadien, ankin :- Phn ng cng : nu tc nhn bt i cng vi anken bt i th sn phm chnh c xcnh theo quy tc Macopnhicop. Ch n s sn phm.- i vi ankin th cn ch n xc tc bit 1 hoc 2 lin kt s bt.- Phn ng trng hp : cn ch cc phn ng trng hp 1,4 thng to thnh cao su. Aren :- Cn ch n quy tc th vo vng benzen.

Bi tp p dng :Bi 1:a) Vit phng trnh phn ng khi cho propen, propin, divinyl tc dng vi Br2 theo t lmol 1: 1.

b) Hi khi cho 3 cht trn tc dng vi HCl (c xt) theo t l 1: 1 th thu c nhng snphm g? Gi tn chngc) Hy cho bit CTCT v tn gi ca sn phm khi cho isopren v pentadien-1,4 tc dngvi dung dch Br2, HCl theo t l mol 1: 1. Vit CTCT ca polime thu c khi trng hp 2ankadien cho trn

GII :a) Phn ng cng gia hydrocacbon khng no vi tc nhn i xng th tng i ngin. Ty vo t l s mol m 1 hoc 2 kin kt s bt.Ptp : xem phn tm tt ha tnh (I.2.4/14)

b) Tc dng vi HCl (1:1)p dng quy tc Maccopnhicop* Propen cng HCl cho 2 sn phm


40/115


SVTH:PhanTh Thy 40

CH3 CH CH2 + HCl

CH3 CH

Cl

CH3

CH3 CH2 CH2Cl

2-Clopropan (spc)

1-Clopropan (spp) * Propin cng HCl to 2 sn phm

CH3 C CH + HCl

CH3 C

Cl

CH2

CH3 CH CHCl

2-Clopropen (spc)

1-Clopropen-1(spp)* Divinyl th c 2 hng cng- Cng 1,2 (hay 3,4) th to 2 sn phm

CH2 CH CH CH2 + HCl

CH2 CH CHCl

CH3

CH2 CH CH2 CH2Cl

3-Clobuten-1

4-Clobuten-1 - Cng 1,4 to 1 sn phm duy nhtCH2 CH CH CH2 + HCl CH3 CH CH CH2Cl

1-Clobuten-2

c) CTCT v tn gi ca sn phm khi cho* Isopren tc dng vi HCl (1:1)- Cng 1,2 to 2 sn phm

+ HClCH2 C CH

CH3

CH22 3 4

CH3 CCl

CH3

CH CH2

CH2Cl CH

CH3

CH CH2

3-Clo-3-metylbuten-1

4-Clo-3-metylbuten-1 - Cng 3,4 to 2 sn phm

+ HClCH2 C CH

CH3

CH22 3 4

CH2 C CHCl

CH3

CH3

CH2 C CH2

CH3

CH2Cl


4-Clo-2-metylbuten-1 - Cng 1,4 to 2 sn phm


41/115


SVTH:PhanTh Thy 41

+ HClCH2 C CH

CH3

CH24

CH2Cl C CH

CH3

CH3

CH3 C CH

CH3

CH2Cl


1-Clo-3-metylbuten-2 * Pentadien-1,4 tc dng vi HCl (1:1)- Cng 1,2 hay 3, 4 : tng t nh Divinyl- Khng c phn ng cng 1,4 do hai lin kt khng lin hp.- CTCT cc polime thu c khi trng hp 2 ankadien trn :

CH2 C CH

CH3

CH2nTH1,2 CH2 C

CH3

CH=CH2

n

CH2 C CH

CH3

CH2nTH3,4 CH2 CH

CCH3 CH2

n

CH2 C CH

CH3

CH2TH1,4

nCH2 C CH

CH3

CH2

CH2 CH CH2 CH CH2n

TH1,2 CH2 CH

CH2 CH CH2

n

- Pentadien-1,4 khng c sn phm trng hp 1,4 do khng c 2 lin kt lin hp.

Bi 2 :a) Pht biu quy tc thvng benzen.

b) T benzen vit phng trnh phn ng iu ch ortho-bromnitrobenzen v meta-Bromnitrobenzen (ghi r iu kin phn ng).

GII :

a) Quy tc thvng benzen :

- Khi vng benzen c nhm thy electron(gc ankyl hoc OH, NH2, Cl, Br) phnng th xy ra d hn so vi benzen v u tin th vo v tr ortho hoc para.- Khi vng benzen c nhm th rt electron (nhm th c lin kt nh NO2, - COOH, -CHO, -SO3H,) phn ng th kh hn (so vi benzen) v u tin th vo v tr meta.

b) Cc phng trnh phn ng :* iu ch ortho bromnitrobenzen :

+ Br2Fe

Br

+ HBr


42/115


SVTH:PhanTh Thy 42

Br

+ HNO3

NO2

Br

+ H2OH2SO4

* iu ch meta bromnitrobenzen :

H2SO4 + H2O

NO2

+ HNO3

NO2 NO2

Br

+ Br2Fe + HBr

Bi tp tng t:

1) Vit phng trnh phn ng ca butin-1, butadien-1,3 vi H2, Br2, HCl, H2O. Gi tn snphm.2) Khi trng hp butadien-1,3 vi xc tc Na ta thu c cao su Buna c ln 2 sn phm

ph A v B. A l mt cht do khng c tnh n hi, mi mt xch c mt mch nhnh lnhm vinyl. B l hp cht vng c tn l 1-vinyl xiclohexan-3 c phn t bng 108. Vitcc phng trnh phn ng xy ra di dng CTCT.3) Phn ng cracking l g? Vit cc phng trnh phn ng dng tng qut khi cracking

mt ankan.- Khi cracking butan thu c mt hn hp gm 7 cht, trong c H2 v C4H8. Hi CTCTca butan l n hay iso? Vit cc phng trnh phn ng xy ra?4) Olefin l g? Vi CTPT CnH2n c th c cc cht thuc dy ng dng no? Nu tnhcht ha hc cbn ca n?Vit phng trnh phn ng khi cho propylen tc dng vi O2; dd Br2; HCl; dd KMnO4;

phn ng trng hp.Hp cht C6H12 khi cng hp HBr ch thu c mt sn phm duy nht, nh CTCT c thc ca olefin ny v vit phng trnh phn ng.5) Vit phng trnh phn ng (nu c) ca cc hp cht sau vi dung dch AgNO3/NH3

a) Axetylenb) Butin-1c) Butin-26) Vit phng trnh phn ng (nu c) gia cc cht sau vi Brom, ghi r iu kin: dd, to,kh(nu c):a) Isopren (1:1)

b) Toluenc) Benzend) Styren7) Vit phng trnh phn ng (nu c) gia cc cht sau:

a) Toluen + dd KMnO4b) Propylen + AgNO3/NH3 d


43/115


SVTH:PhanTh Thy 43

c) Styren + dd KMnO4 + Ba(OH)2d) Axetylen + dd KMnO4_+ H2SO4e) Propin +dd KMnO4_+ H2SO48) Mun iu ch n-pentan, ta c th hidro ha nhng anken no? Vit CTCT ca chng.9) Vi

t ph

ng trnh ph

n

ngiu ch

cc h

p ch

t sau

y t

nh

ng anken thch h

p :

a) CH3CHBr CHBrCH3b) CH3CHBr CBr(CH3)2c) CH3CHBr CH(CH3)2


44/115


SVTH:PhanTh Thy 44

II.1.6 BI TP SO SNH GII THCH CU TO, TNH CHTHA HC CA CC HYDROCACBON

Nguyn tc : Da vo s so snh vc im cu to cc cht ri suy ra tnhcht ha hc ca cc cht . Bi tp v d :

Bi 1 :So snh v mt CT v ha tnh ca cc hp cht sau, vit phng trnh phn ng minh

ha.a) Etan, etylen, axetylen

b) hexan, hexen, benzenc) butin-1, butin-2 v butadien-1,3

GII :a) Etan, Etilen, Axetilen :* Ging nhau :

- Thnh phn cu to ch gm C v H* Khc nhau :Phn t C2H6 C2H4 C2H2Cu to Trong phn t ch

tn ti cc lin ktn ( ) bn gia

C v C, gia C vH

Trong phn t cmt lin kt igm mt kin kt

(

) bn v mtlin kt () linhng km bn.

Trong phn t cmt lin kt bagm mt lin kt

(

) bn v hai linkt () linh ngkm bn.

c im lin kt Lin kt n ( )rt bn vng rtkh bt khi thamgia phn ng hahc

Lin kt () linh ng km bn rt d bt khi tham gia phn ng ha hc.

Tnh cht hahc

Tnh cht ha hcc trng l phn

ng th, kh b oxiha.Ngoi ra cn cphn ng hydroha nhit thch hp v xctc thch hp.

Tnh cht ha hc c trng l phn ngcng. Ring vi axetilen th khi tham gia

phn ng ha hc ty iu kin xc tcm mt hay c hai lin kt () s bt.Ngoi ra cn c phn ng trng hp,oxiha.

Phng trnhphn ng

Xem I.2.4/14 Xem I.2.4/15

Ring axetilen c hai nguyn t H linh

ng nn n cn c kh nng tham giaphn ng th vi ion kim loi. iu ny


45/115


SVTH:PhanTh Thy 45

c gii thch nh sau : do lin kt bart ngn nn hai nhn C rt gn nhau,in tch tp trung nhiu v 2 C ny nncc H gn trc tip vi C ca ni ba tr

nn rt linh ng.b) n-hexan, n-hexen, benzen.* Ging nhau :

- Thnh phn cu to gm C v H* Khc nhau :

- n-hexan v n- hexen so snh cu to v tnh cht ha hc tng t cu trn. Ringn-hexan cn c phn ng b gy mch C khi c xc tc nhit cao.Ankan caotxt,

o

ankan + ankenCnH2n+2

caotxt, o CmH2m + 2 + CxH2x

m 1, x 2, n = m + x.- Benzen :c im cu to : trong phn t c mt vng kn v 3 lin kt benzen c phn

ng c trng l phn ng cng. nhng 3 lin kt ny li lin hp vi nhau to thnh mth thm bn vng lm cho kh nng t lin kt tham gia phn ng ha hc b hnch benzen kh tham gia phn ng cng, ch c cng vi H2, d tham gia phn ng thv bn vi tc nhn oxiha.

Phng trnh phn ng : xem phn ha tnh (I.2.4/14).

c) So snh c im cu to ca butin-1, butin-2, divinylTng t cu a.

Nhng kh nng tham gia phn ng cng ca lin kt i (Divinyl) hi d hn so vilin kt ba v : lin kt 3 ngn, hai nhn C gn nhau nn lin kt 3 hi bn hn so vi linkt i.

Bi 2 : C7H8 l ng ng ca benzen. Khi cho C6H6 v C7H8 tc dng vi Brom khan (cbt Fe lm xc tc) th phn ng no xy ra d hn? Gii thch (vit phng trnh phn ngtheo t l 1:1 v s mol)

GII :

C7H8 tham gia phn ng thnhn d hn so vi benzen v nhm CH3y electronv nhn lm nhn giu electron hn.

* C6H6 cho 1 sn phm :

+ Br2Fe

Br

+ HBr

* C6H5CH3 cho hn hp hai sn phm.


46/115


SVTH:PhanTh Thy 46

+ HBr

CH3

Fe+ Br2

CH3

BrCH3Fe + HBr

CH3

+ Br2

Bi 3 : So snh phn ng trng hp v phn ng cng.

GII :

* Ging nhau : u l phn ng cng hp cc phn t nh thnh mt phn t mi.* Khc nhau :Phn ng cng : Chn thun cng 2 phn t nh (monome) thnh mt phn t mi cngl monome. Ch cn mt trong hai monome ban u c t nht mt lin kt trong phn t.Phn ng trng hp : khng ch cng hai m cng nhiu phn t ging nhau hoc tng tnhau thnh mt phn t mi c khi lng v kch thc rt ln gi l nhng polime.Cc monome tham gia phn ng trng hp nht thit phi c t nht mt lin kt trong

phn t.

Bi 4 : So snh di lin kt dC-C trong ankan (C C), anken (C=C), ankin (C C). Gii

thch kh nng tham gia phn ng ca ankan km anken nhng ankin km anken.? Mc dphn t ankin c nhiu lin kt hn anken?

GII :

Thc nghim cho bit dC-C trong etan (C-C) l : 1,54Ao

Etilen(C=C) : 1,34AoAxetilen (C C):1,2 Ao

* C th gii thch nh sau :Khi hnh thnh lin kt C-C trong phn t ankan th 2C xy ra s xen ph trc lin

nhn lm cho khong cch 2 nhn xa nhau nn dC-C ln.

Khi hnh thnh lin kt C=C trong phn t anken th lin kt c hnh thnh nhcch trn, cn lin kt c hnh thnh do s xen ph bn lm cho khong cch gia 2nhn C gn nhau hn.

Tng t vi ankin c 2 lin kt nn xy ra 2 s xen ph bn lm cho khong cchgia hai nhn cng gn nhau hn.

Do dC-C C C > C=C > C C.

* Gii thch v kh nng tham gia phn ng :- S xen ph trc xy ra vi mt ln lm cho lin kt bn vng.- S xen ph bn xy ra vi mt nh nn lin kt km bn vng d bt khi c tc

nhn tn cng kh nng tham gia phn ng ca ankan< anken, ankin.


47/115


SVTH:PhanTh Thy 47

- y do lin kt 3 lm cho khong cch 2 nhn C rt gn nhau nn lin kt 3 hi bn hnlin kt i nn kh nng tham gia phn ng ca ankin hi km hn anken.- V cng do khong cch gia hai nhn C b m mt in tch tp trung hu ht nhnnn cc ankin-1 c H linh ng tham gia c phn ng th vi ion kim loi

Bi tp tng t:1) Gii thch quy tc cng Maccopnhicop? Minh ha bng v d c th.2) Gii thch ti sao di lin kt n C-C trong butadien-1,3 ch bng 1,46Ao ngn hnlin kt n C-C bnh thng?3) Ti sao khi nhit phn mui axetat vi xt iu ch ankan tng ng li phi dngxc tc CaO,to?4) So snh nhit si ca cc hydrocacbona) Khi khi lng phn t tng dn?

b) C cng CTPT nhng khc nhau dng khung Cacbon?5) Khi thc hin phn ng phn hy ankan bi nhit li c tin hnh nhit trn1000oC ti sao li nhn mnh trong iu kin khng c khng kh?6) So snh kh nng tham gia phn ng th ca cc halogen Flo, Clo, Brom, Iod vi ccankan?7) Ti sao cao su khi chy li c nhiu khi en? Lm th no khi en t li?8) Trong phn ng iu ch axetilen t metan c tin hnh nhit 1500oC cn ghikm iu kin lm lnh nhanh?9) So snh cao su thng v cao su lu ha v thnh phn, bn, ng dng?10) Gii thch v sao cao su tng hp c tnh n hi km cao su thin nhin?11)Phn bit cc khi nim:a) CTN, CTG, CTPT v CTCT

b) Lin kt , . Ly propen lm v dc) ng ng, ng phn l g? Nu cc loi ng phn, cho v d?d) C th coi nguyn t Br trong phn t CnH2n+1Br l mt nhm chc c khng? Tisao?


48/115


SVTH:PhanTh Thy 48

II.2 BI TON LP CTPT

HYDROCACBONII.2.1 CC PHNG PHP LP CNG THC PHN TCA

HYDROCACNON

II.2.1.1 Phng php khi lng hay % khi lng.

1) Phng php gii :Bc 1 : Tm MA : ty theo gi thit bi cho m s dng cc cch tnh sau tm MATm MA da trn cc khi nim cbn, cc nh lut cbn. C nhiu cch tmkhi lng phn t, ty tng gi thit bi cho m dng cch tnh thch hp.

1. Da vo khi lng ring DA (ktc)MA = 22,4 . DA vi DAn v g/l

2. Da vo t khi hi ca cht hu cAMA = MB . dA/BMA = 29 . dA/KK

3. Da vo khi lng (mA ) ca mt th tch VA kh A ktcMA = (22,4 . mA)/ VAmA: khi lng kh A chim th tch VAktc

4. Da vo biu thc phng trnh Mendeleep Claperon:Cho mA (g) cht hu cA ha hi chim th tch VA (l) nhit T(oK) v p sut P(atm)

PV = nRT pV

mRTM = (R = 0,082 atm/ oKmol)

5. Da vo nh lut Avogadro:nh lut: cngiu kin nhit v p sut, mi thtch kh bngnhau u cha cng mt sphn tkh.

VA = VB => nA = nBB

B

A

A

M

m

M

m=

=> MA = mAB

B

m

M

Bc 2 : t CTPT cht A: CxHy

Xc nh thnh phn cc nguyn t trong hydrocacbon.


49/115


SVTH:PhanTh Thy 49

Cch 1 :Dng khi bi-Khng cho khi lng hydrocacbon em t chy-Tnh c mC, mH t mCO2, mH2O

Tnh khi lng cc nguyn t c trong A v mA (g) cht A.

- Xc nh C:22,4

V12.n.12

44

m12.)CO(trongmCA)(trongmC CO2CO2

CO22 ====

- Xc nh H

OH

O

OH nn 22

2.2

18

m22.1H2O)(trongmHA)mH(trong H ====

- Xc nh mA mA = mH + mA

* Xc nh CTPT cht hu cA: CxHyDa trn CTTQ cht hu cA: CxHy

;12.m

m.MxmM

my

m12

A

CA

A

A

HC

x ==>== A

HA

mm.My =

Cch 2 : Khi bi cho bit thnh phn % cc nguyn t trong hn hp* Dng cng thc sau:

12.100

C.%Mx

100%

M

%H

y

C%

12 AAx ==>== ;100

H.%My A=

CTPT A.

Cch 3 : * Tm CTG nht => CTN => CTPT A:

1

m:

12

my:x HC == hoc :

1

%H:

12

%Cy:x ==

- CTG nht : CH=> CTTN : (CH)n

- Xc nh n: bin lun t CTTN suy ra CTPT ng ca A :y 2x + 2; y chn, nguyn dng ; x 1, nguyn dng.

T xc nh c CTPT ng ca cht hu cA.Lu : Khi bi tan yu cu xc nh CTG nht ca cht hu cA (hay CTN ca A) hockhi khng cho d kin tm MA th ta nn lm theo cch trn.

2) Cc v d :V d 1 :

Mt hydrocacbon A c thnh phn nguyn t: % C = 84,21; %H = 15,79; T khi hii vi khng kh bng dA/KK= 3,93. Xc nh CTPT ca A

GII

Bc 1: Tnh MA:Bit dA/KK=> MA = MKK. dA/KK= 29.3,93 = 114

Bc 2 : t A : CxHy


50/115


SVTH:PhanTh Thy 50

100

M

%H

y

%C

12x A==

812.100

114.84,21

12.100

.%CMx A ===

181.100

114.15,791.100

.%HMy A ===

Suy ra CTPT A: C8H18

V d 2 :Mt hydrocacbon A th kh c th tch gp 4 ln th tch ca lu hunh ioxit c

khi lng tng ng trong cng iu kin. Sn phm chy ca A dn qua bnh ngnc vi trong d th c 1g kt ta ng thi khi lng bnh tng 0,8g. Tm CTPT A.

GII

* Tm MA :1VA = 4VSO2(cng iu kin )nA = 4nSO2

22

2 414SOASO

SO

A

A

MMM

m

M

m== (A v SO2 c khi lng tng ng nhau)

164

64

4

MM 2SOA ===

Cch 1 : gii theo phng php khi lng hay % khi lng :t A : CxHy

Bnh ng Ca(OH)2 hp th CO2 v H2OCa(OH)2 + CO2 CaCO3 + H2O

m = mCaCO3 = 1gnCO2 = nCaCO3 = 1/100= 0,01molnC = nCO2 = 0,01mol mC = 12.0,01=0,12g

mCO2 = 0,01.44 = 0,44gmbnh = mCO2 + mH2O

mH2O= 0,8-0,44 = 0,36g

gm

mOH

H 04,0

18

36,02

18

2 2 ===

LBT khi lng (A) :mA = mC + mH = 0,12 +0,04 = 0,16

Ta c 116,0.12

12,0.16

12.m

m.Mx

m

M

m

y

m

12

A

CA

A

A

HC

x ====>==

416,0

04,0.16

m

m.My

A

HA ===

Vy CTPT A : CH4Cch 2 : Bin lun da vo iu kin y 2x + 2; y chn, nguyn dng ; x 1, nguyn x =1 v y = 4CTPT A.

V d 3:


51/115


SVTH:PhanTh Thy 51

t chy hon ton 2,64g mt hydrocacbon A thu c 4,032 lt CO2 (ktc). TmCTPT A?

GII

* Tm thnh phn cc nguyn t

:

mC (trong A) = mC (trong CO2) = (4,032/ 22,4)*12 = 2,16gmH = mA mC = 2,64 2,16 = 0,48g

C Hm m 2,16 0,48x:y= : = : =3:812 1 12 1

CTN : C3H8 CTTN : (C3H8)nBin lun :S H 2 s C +2 8n 6n + 2 n 1 m n nguyn dng n = 1CTPT A : C3H8

II.2.1.2) Phng php da vo phn ng chy:Du hiu nhn bit bi ton dng ny : bi t chy mt cht hu cc cp n

khi lng cht em t hoc khi lng cc cht sn phm (CO2, H2O) mt cch trc tiphoc gin tip (tc tm c khi lng CO2, H2O sau mt s phn ng trung gian).

1) Phng php gii:Bc 1 : Tnh MA (phn II.2.1.1)Bc 2 : t A : CxHy

* Vit phng trnh phn ng chy.

OH2yxCOO4yxHC 22t

2yx0 +

++

MA(g) 44x 9ymA(g) mCO2 mH2O

* Lp t l tnh x,y

OHCOA

A

22m

9y

m

44x

m

M== hoc

2puA O CO2 H2O

y yx+1 x4 2= = =

n n n n

A

OHA

A

COA

9m

.mMy,

44m

.mMx 22 ==

* T suy ra CTPT AMt slu :1) Nu bi cho: oxi ha han tan mt cht hu cA th c ngha l t chy han tancht hu cA thnh CO2 v H2O2) Oxi ha cht hu cA bng CuO th khi lng oxy tham gia phn ng ng bng gim khi lng a(g)ca bnh ng CuO sau phn ng oxi ha. Thng thng trong biton cho lng oxi tham gia phn ng chy, tm khi lng cht hu cA nn ch nnh lut bo ton khi lng

mA + a = mCO2 + mH2O

3) Sn phm chy (CO2, H2O) thng c cho qua cc bnh cc cht hp th chng.


52/115


SVTH:PhanTh Thy 52

4) Bnh ng CaCl2 (khan), CuSO4 (khan), H2SO4c, P2O5, dung dch kim, hp thnc.

Bnh ng cc dung dch kimhp th CO2.Bnh ng P trng hp th O2.

5) tng khi lng cc bnh chnh l khi lng cc cht m bnh hp th.6) Nu bi ton cho CO2phn ng vi dung dch kim th nn ch n mui to thnh xc nh chnh xc lng CO2.7) Vit phng trnh phn ng chy ca hp cht hu cvi oxy nn oxy li cn bngsau t v sau n v trc. Cc nguyn t cn li nn cn bng trc, t v trc ra v sau

phng trnh phn ng.

2) Bi tp v d :V d 1 :

t hon ton 0,58g mt hydrocacbon A c 1,76g CO2 v 0,9g H2O. Bit A c

khi lng ring DA 2,59g/l. Tm CTPT ATm tt :0,58g X + O2 (1,76g CO2; 0,9 g H2O)DA 2,59g/l. Tm CTPT A?

GII :

* Tm MA :Bit DA => MA = 22,4.2,59 58

* Vit phng trnh phn ng chy, lp t l tm x,y

OH2yxCOO

4yxHC 22

t2yx

0 +

++

MA(g) 44x 9ymA(g) mCO2 mH2O

OHCOA

A

22m

9y

m

44x

m

M== =

0,9

9y

1,76

44x

0,58

58==

x = 4y =10

Vy CTPT A : C4H10

V d 2 : Khi t chy han tan 0,42 g mt Hydrocacbon X thu tan b sn phm quabnh 1 ng H2SO4c, bnh 2 ng KOH d. Kt qu, bnh 1 tng 0,54 g; bnh 2 tng 1,32g. Bit rng khi ha hi 0,42 g X chim th tch bng th tch ca 1,192 g O2cng iukin. Tm CTPT ca XTm tt :

0,42g X (CxHy)+O2 CO2

H2O

Bnh 1ng ddH2SO4

-H2O, m1=0,54gCO2

Bnh 2 ng KOHd

-CO2, m2=1,32g Tm CTPT X?

GII


53/115


SVTH:PhanTh Thy 53

* Tnh MX :0,42g X c VX = VO2 ca 0,192g O2 (cng iu kin)

=> nX = nO2 =>2O

O2

X

X

M

m

M

m=

=> 700,192

0,42.32m

.MmM2

2

O

OXX ===

* Gi X : CxHy

OH2

yxCOO

4

yxHC 22

t2yx

0

+

++

MX 44x 9y (g)0,42 mCO2 mH2O (g)

Ta c :

H2OCO2X

X

m

9y

m

44x

m

M== (1)

bi cho khi lng CO2, H2O gin tip qua cc phn ng trung gian ta phi tm khilng CO2, H2O

* Tm mCO2, mH2O :- Bnh 1 ng dd H2SO4 s hp th H2O do tng khi lng bnh 1 chnh l

khi lng ca H2O :m1 = mH2O=0,54g (2)

- Bnh 2 ng dd KOH d s hp th CO2 do tng khi lng bnh 2 chnh lkhi lng ca CO2 :

m2 = mCO2 =1,32g (3)

(1), (2), (3) 0,549y

1,3244x

0,4270 ==

x = 5y = 10

Vy CTPT X : C5H10 (M = 70vC)

II.2.1.3 Phng php th tch (phng php kh nhin k):

Phm vi ng dng : Dng xc nh CTPT ca cc cht hu cth kh hay th lng d bay hi. Cskhoa hc ca phng php : Trong mt phng trnh phn ng c cc cht

kh tham gia v to thnh (cng iu kin nhit , p sut) h st trc cng thc cacc cht khng nhng cho bit t l s mol m cn cho bit t l th tch ca chng.

1) Phng php giiBc 1 : Tnh th tch cc kh VA, VO2, VCO2, VH2O (hi)Bc 2 : Vit v cn bng cc phng trnh phn ng chy ca hydrocacbon A di dngCTTQ CxHyBc 3 : Lp cc t l th tch tnh x,y

OH2

y

xCOO4

y

xHC 22t

2yx

0

+

++


54/115


SVTH:PhanTh Thy 54

1(l)

+4

yx (l) x(l)

2

y(l)

VA(l) VO2 (l) VCO2 (l) VH2O(hi)(l)

OHCOA 222V2

y

VxV 4

yx

V1 ==

+

= hayOHCOOA n

y

nx

n

yx

n22

2412

==

+

=

;n

n

V

Vx

A

CO

A

CO 22 ==

A

OH

n

n22

2

V

2Vy

A

OH ==

Cch khc : Sau khi thc hin bc 1 c th lm theo cch khc:- Lp t l th tch VA: VB : VCO2 : VH2O ri a v t l s nguyn ti gin m:n:p:q.- Vit phng trnh phn ng chy ca hp cht hu cA di dng:

mCxHy + nO2 ot pCO2 + qH2O

- Dng nh lut bo ton nguyn t cn bng phng trnh phn ng chy s tm c xv y =>CTPT A* Mt slu :- Nu VCO2 : VH2O = 1:1 => C : H = nC : nH = 1: 2- Nu tan cho oxy ban u d th sau khi bt tia la in v lm lnh (ngng t hinc) th trong kh nhin k c CO2 v O2 cn d. Bi tan l lun theo CxHy- Nu tan cho VCxHy = VO2 th sau khi bt tia la in v lm lnh th trong kh nhin kc CO2 v CxHy d. Bi tan l lun theo oxy.- Khi t chy hay oxi ha han ton mt hydrocacbon m gi thit khng xc nh r sn

phm, th cc nguyn t trong hydrocacbon s chuyn thnh oxit bn tng ng tr:N2 kh N2Halogen kh X2 hay HX (ty bi)

2. Bi tp v d

V d 1:Trn 0,5 l hn hp C gm hydrocacbon A v CO2 vi 2,5 l O2 ri cho vo kh nhin

kt chy th thu c 3,4 l kh, lm lnh ch cn 1,8 l. Cho hn hp qua tip dung dchKOH (c) ch cn 0,5 l kh. Cc V kh o cng iu kin. Tm CTPT ca hydrocacbon A.

Tm tt :CxHy : a (l)

Gi 0,5 l hn hpCO2 : b (l)

0,5l hn hp + 2,5l O2 ot CO2 ,O2 d,H2O ll(- H2O) CO2,O2dKOH(- CO2) O2 d

GII :

* O2 d , bi tan l lun theo Hydrocacbon A


55/115


SVTH:PhanTh Thy 55

OH2

yxCOO

4

yxHC 22

t2yx

0

+

++

a a

+4

yx ax a

2

y(lt)

CO2 CO2 b b (lt)

Ta c Vhh = a + b = 0,5 (1)VCO2 = ax + b = 1,8 0,5 = 1,3 (2)

VH2O = a2

y= 3,4 1,8 = 1,6 (3)

VO2 d = 2,5 - a

+4

yx = 0,5

ax + a

4

y= 2 (4)

ax + 3,2/4 = 2 ax = 1,2 (5)(2), (3) VCO2 = b = 0,1

Vhh = a + b = 0,5 a = 0,4 x = ax /a = 3 y = ay/a = 8Vy CTPT ca A l C3H8

V d 2 :Trn 12 cm3 mt hydrocacbon A th kh vi 60 cm3 oxi (ly d)

ri t chy. Sau khi lm lnh nc ngng t ri a viu kin ban u th th tchkh cn li l 48 cm3, trong c 24cm3 b hp th bi KOH, phn cn li b hp th bi P.Tm CTPT ca A (cc th tch kh o trong cng iu kin nhit v p sut)Tm tt :

12cm3

CxHy

60cm3

O2 (d)

otCO2H2O

O2d

lam lanh

-H2O

CO2O2 d

(V=48cm3)

24cm3

kh b hap thu bi KOH

kh con lai b hap thu bi P

(- CO2)

(-O2) GII :

* Tnh cc V:VCO2 = 24cm

3VO2 d = 48 24 = 24cm

3 VO2 p = 60 24 = 36 cm3

* Tm CTPT :Cch 1: Tnh trc tip t phng trnh phn ng t chy:

OHy

xCOOy

xHt

y 222x 24C

0

+

++

12

+4

yx 12 12x (cm3)

VCO2 =12x = 24 => x = 2


56/115


SVTH:PhanTh Thy 56

VO2 d = 60 12

+4

yx = 24 => y = 4

CTPT ca A: C2H4

Cch 2: Lp t l th tchOHCOOA 222

V2

y

VxV 4

yx

V1 ==

+

=

OH2

yxCOO

4

yxHC 22

t2yx

0

+

++

1

+4

yx x

2

y(cm3)

12 36 24 (cm3)

22 COOA V

x

V

4

yx

V

1=

+=

24

x

36

4

yx

12

1=

+= => x = 2 v y = 4

CTPT ca A: C2H4Cch 3:

Nhn xt: t 12 cm3 A dng 36 cm3 oxy v to ra 24 cm3 CO2Suy ra O?H24CO36OH12C 22

t2yx

0

++

LBT (O): => O24H24CO36OH12C 22t

2yx

0

++

LBT (C): 12x = 24 => x = 2LBT (H) :12y = 48 => y = 4

Vy CTPT ca A l C2H4

V d 3 :Trong mt bnh kn th tch 1dm3 c mt hn hp ng th tch gm hydrocacbon A

v O2133,5oC, 1 atm. Sau khi bt tia la in v a v nhit ban u (133,5 oC) th

p sut trong bnh tng ln 10% so vi ban u v khi lng nc to ra l 0,216 g. TmCTPT ATm tt :

V = 1dm3

CxHy(A)

O2

t=133,5oC,P1=1atm

ot sp chay

V=1dm3

t=133,5oC, P2 tang 10%

(lng H2O tao ra la 0,216g)

GII :

Tm CTPT A?

0,03(mol)133,5)0,082.(273

1.1

RT

PVn1 =+

==

V hn hp ng th tch nn nA = nO2 = 0,03/2 = 0,015 mol=> C

xH

yd

, bin lu

n theo O

2


57/115


SVTH:PhanTh Thy 57

Sau khi a v nhit ban u, cc kh to p sut c trong bnh gm H2O, CO2,CxHy d c s mol l :

n2 = n1 . P2/P1 = 0,03.110/100 = 0,033 molnH2O = 0,216/18 = 0,012 mol

LBT kh

i l

ng (O) : nO2

= nCO

2+ 1/2n

H

2O

=> nCO2 = nO2 1/2nH2O = 0,015-0,012/2 = 0,009molnCxHyd = n2 - nCO2 - nH2O = 0,033-0,012-0,009 =0,012mol

=>nCxHyphn ng = 0,015-0,012 = 0,003 mol

OH2

yxCOO

4

yxHC 22

t2yx

0

+

++

1

+4

yx x

2

y(mol)

0,003 0,015 0,009 0,012 (mol)Ta c :

012,02

009,0015,04

003,0

1y

xyx

==+

=

=> x = 3y = 8

Vy CTPT A : C3H8

II.2.1.4 Phng php gi tr trung bnh (xc nh CTPT ca hai hay nhiu chthu ctrong hn hp):

L ph

ng php chuyn h

n h

p nhi

u gi tr

v

m

t gi tr

t

ng

ng, nhiu ch

t

v mt cht tng ng c imPhng php gi tr trung bnh c dng nhiu trong ha hu ckhi gii bi tan v

cc cht cng dy ng ng. Mt phn bn cht ca gi tr trung bnh c cp n vic tnh phn trm n v v khi lng hn hp kh trong bi tan t khi hi chngu lp 10. Do , hc sinh d dng lnh hi phng php ny xc nh CTPT ca haihay nhiu cht hu ctrong hn hp.

II.1.4.1 Phng php khi lng phn ttrung bnh ca hn hp ( hhM )

Cht tng ng c khi lng mol phn t hhM l khi lng mol phn t trungbnh ca hn hp. Cc bc gii :Bc cbn : Xc nh CTTQ ca hai cht hu cA,BBc 1 : Xc nh CTTB ca hai cht hu cA, B trong hn hpBc 2 : Tm hhM qua cc cng thc sau :

( )100

M%A100%A.M

100

%B.M%A.M

nn

.Mn.Mn

n

mM BABA

BA

BBAA

hh

hhhh

+=

+=

+

+==

Hoc( )100

M%A100%A.M

V

.MV.MV

VV

.MV.MV.MdM

BABBAA

BA

BBAAXhh/Xhh

+=

+=+

+==


58/115


SVTH:PhanTh Thy 58

Gi s MA< MB => MA< hhM < MBBc 3 : Bin lun tm MA, MB hp l => CTPT ng ca A v B

Phm vi ng dng: s dng c li nhiu i vi hn hp cc cht cng dy ng ng

1) Phng php CTPT trung bnh ca hn hp: Phm vi p dng: Khi c hn hp gm nhiu cht, cng tc dng vi mt cht

khc m phng trnh phn ng tng t nhau (sn phm, t l mol gia nguyn liu v snphm, hiu sut, phn ng tng t nhau), c th thay th hn hp bng mt cht tngng, c s mol bng tng s mol ca hn hp. Cng thc ca cht tng ng gi lCTPT trung bnh. Phng php gii :

Bc 1 : t CTPT ca hai cht hu ccn tm ri suy ra CTPT trung bnh ca chng :t A : CxHy ; B : CxHy CTPTTB : x yC H

Bc 2 : Vit phng trnh phn ng tng qut v d liu bi cho tnh y,x Bc 3 : bin lun

Nu x


59/115


SVTH:PhanTh Thy 59

OHnCOnOn

HCnn 22222

)1(2

13++

++

+

X

X

M

m

X

X

M

m. n

Cch 1: phng php s C trung bnh ( n )S mol hn hp

X

XX

M

mn =

S mol CO2 : nCO2 =X

X

M

m. n = 1,3

2,6n3,1n2n14

19,2==

+

Hn hp gm 2 ankan lin tip CnH2n+2

CmH2m+2 ; n M = mhh / nhh = 19,2/0,5 = 38,4MA < 38,4 < MB = MA + 14

A CH4 C2H6 C3H8 C4H10

MA 16 30 44 58 M 38,4 38,4 38,4 38,4 MB 30 44 58 72

Vy A : C2H6

B : C3H8


60/115


SVTH:PhanTh Thy 60

II.2.1.5 - Phng php bin lun

1. Da vo gii hn xc nh CTPT ca mt hydrocacbon:- Khi s phng trnh i s thit lp c t hn sn cn tm, c th bin lun da vogii hn :A : CxHy th : y 2x + 2; y chn, nguyn dng ; x 1, nguyn.- Nu khng bin lun c hay bin lun kh khn c th dng bng tr s tm kt qu.- iu kin bin lun ch yu ca loi ton ny l : ha tr cc nguyn t. Phng php

bin lun trnh by trn ch c th p dng xc nh CTPT ca mt cht hoc nu nmtrong 1 hn hp th phi bit CTPT ca cht kia.

2. Bin lun theo phng php ghp n s xc nh CTPT ca mthydrocacbon :

a) Cc bc cbn :Bc 1 : t s mol cc cht trong hn hp l n s.Bc 2 : ng vi mi d kin ca bi ton ta lp mt phng trnh ton hc.Bc 3 : Sau ghp cc n s li rt ra h phng trnh ton hc. Chng hn : a + b = P(vi a, b l s mol 2 cht thnh phn)

an + bm = Q (vi n, m l s C ca 2 hydrocacbon thnh phn)Bc 4 : c th xc nh m, n ri suy ra CTPT cc cht hu cthnh phn, c th pdng tnh cht bt ng thc :Gi s : n < m th n(x + y) < nx + my < m(x + y)

nx+myn< nCO2 (A) thuc dy ng ng ankan

ptp : n 2n+2 2 2 23n+1C H + O nCO + (n+1)H O2

* nH2O = nCO2 (A) thuc dy ng ng anken hay olefinhoc (A) l xicloankan

ptp : n 2n 2 2 23n

C H + O nCO + nH O2

* nH2O < nCO2 (A) thuc dy ng ng ankadien, ankin hoc benzen

ptp : n 2n-2 2 2 23n-1

C H + O nCO + (n-1)H O2

( ng ng ankin hoc ankadien)

n 2n-6 2 2 23n-3C H + O nCO + (n-3)H O2 ( ng ng benzen)


61/115


SVTH:PhanTh Thy 61

Cch 2 : Da vo CTTQ ca hydrocacbon A :

* Bc 1 : t CTTQ ca hydrocacbon l :CnH2n+2-2k(y k l s lin kt hoc dng mch vng hoc c 2 trong CTCT A)iu kin k 0, nguyn. Nu xc nh c k th xc nh c dy ng ng ca A.

- k = 0 A thuc dy ng ng ankan- k = 1 A thuc dy ng ng anken- k = 2 A thuc dy ng ng ankin hay ankadien- k = 4 A thuc dy ng ng benzen.

chng minh hai ankan A, B thuc cng dy ng ng, ta t A : CnH2n+2-2k; B :CmH2m+2-2k. Nu tm c k = k th A,B cng dy ng ng.* Bc 2 : Sau khi bit c A,B thuc cng dy ng ng, ta t CTTQ ca A l CxHy.V B l ng ng ca A, B hn A n nhm CH2- th CTTQ ca B :CxHy (CH2)n hayCx+nHy+2n.* Bc 3 : Da vo phng trnh phn ng chy ca A, B, da vo lng CO2, H2O, O2hoc s mol hn hp thit lp h phng trnh ton hc, ri gii suy ra x, y, n Xc nhc CTPT A, B.

Cch 3 : da vo khi nim dy ng ng rt ra nhn xt :

- Cc cht ng ng k tip nhau c khi lng phn t lp thnh mt cp s cngcng sai d = 14.

- C mt dy n s hng M1, M2, ,Mn lp thnh mt cp s cng cng sai d th ta c :+ S hng cui Mn = M1 + (n-1)d

+ Tng s hng S =2

M1 nM+ .n

+ Tm M1, , Mn suy ra cc chtTrong mt bi ton thng phi kt hp nhiu phng php.

V d :t chy mt hn hp gm 2 hydrocacbon A, B (c M hn km nhau 28g) th thu

c 0,3mol CO2 v 0,5 mol H2O. Tm CTPT & tn A, B

GII :

Hydrocacbon A, B c M hn km nhau 28g A, B thuc cng dy ng ng.Cch 1 :A, B + O2 CO2 + H2O

67,13,0

5,0

n

n

2

2

CO

OH == >1 A, B thuc dy ng ng ankan.

t CTTB A, B :2n2n

HC+

: a mol

O1)Hn(COnO2

1n3HC 2222n2n ++

++

+

a a n a( n +1) (mol)

Ta cn

1n

3,0

5,0

n

n

2

2

CO

OH +== n = 1,5


62/115


SVTH:PhanTh Thy 62

t CTTQ A, B : CnH2n+2 v CmH2m+2Gi s n< m n< 1,5 n = 1 CTPT A : CH4 (M = 16) MB = 16 + 28 = 44 CTPT B : C3H8.

Cch 2 : t CTTQ A, B : CnH2n+2 : a mol v CmH2m+2 : b molCc ptp chy :

Ok)H-1(nnCOO2

k-13nHC 2222k-22nn ++

+++

a an a(n+1-k) (mol)

Ok)H-1(mmCOO2

k-13mHC 2222k-22mm ++

+++

b bm b(m+1-k) (mol)Ta c :

=+++

=+

0,5k)b-1(mk)a-1(n

0,3bman

(a+b)(1-k) = 0,2 k = 0 v ch c k = 0 th phng trnh mi c ngha. a + b = 0,2 v an + bm = 0,3Gi s n < m n(a+b) < m (a+b)

n nCO2 hai hydrocacbon trn thuc dy ng ng ankan.CTPT trung bnh 2 ankan l :

22 +nnHC

OHnCOnOn

HCnn 22222

)1(2

13++

++

+

x (3 n +1)/2x x n x ( n +1) (mol)nCO2 = x n = 1nH2O = x(n +1) = 1,6x = 0,6n = 1,671 < n =1,67 < m= n + 1 n= 1 v m = 2 CTPT 2 ankan l CH4 v C2H6

Bi 5 :t chy 560cm3 hn hp kh (ktc) gm 2 hydrocacbon c cng s nguyn t

cacbon ta thu c 4,4g CO2 v 1,9125g hi nc.a) Xc nh CTPT cc cht hu c.

b) Tnh %khi lng cc cht.c) Nu cho lng CO2 trn vo 100 ml dd KOH 1,3M; Tnh CM mui to thnh.


67/115


SVTH:PhanTh Thy 67

GII

bi ny, ta dng phng php s nguyn t H trung bnh kt hp vi phng phpbin lun gii.a) Xc nh CTPT cc hydrocacbon :

t CTPT 2 hydrocacbon trn :

y'x

yx

HC:BHC:A

CTPT trung bnh 2 hydrocacbon trn :yx

HC

Gi s y < y y < y < y

S mol hn hp kh nhh = 025,04,22

56,0= mol

nCO2 = 4,4/44 = 0,1 (mol)nH2O = 1,9125/18 = 0,10625 (mol)

OH2yCOxO4yxHC 22t

2yx

0

+

++

0,025 0,025x 0,025 y /2

=

=

==

==

8,5y

4x

0,106252

y0,025n

0,10,025n

H2O

CO2

CTPT A, B c dng : A : C4Hy v B : C4HyTa c y < y < y hay y < 8,5


68/115


SVTH:PhanTh Thy 68

b b b (mol)

Ta c :

==+

==+

13,0nb2a

1,0nba

KOH

CO2

=

=

0,07b

0,03a(mol)

CM(K2CO3 ) = 3,00,1

0,03

= (M) CM(KHCO3) = 7,00,1

0,07

= (M)

Bi 6 :t chy hon ton hn hp gm ankin (A) v ankan (B) c V = 5,6 lt (kc) c

30,8g CO2 v 11,7g H2OXc nh CTPT A,B. Tnh % A,B. Bit B nhiu hn A mt C

GII

bi ny, t chy hn hp 2 hydrocacbon khng phi l ng ng ca nhau nnkhng dng ph

ng php trung bnh

c m s

dng ph

ng php ghp

n s

v bi

n

lun gii.

Gi 5,6 l hh :

+ b

a

:HC:B

:HC:A

22mm

2-2nn (mol) (n 2; m 1)

O1)H(nnCOO2

3nHC 2222-2nn ++

a an a(n-1) (mol)

O1)Hm(mCOO2

13mHC 22222mm ++

+++

b bm bm (mol)

n hn hp = a+ b = 25,04,22

6,5 = (mol) (1)

nCO2 = an + bm = 7,044

8,30= (mol) (2)

nH2O = a(n-1) + bm = 65,018

7,11= (mol) (3)

(2), (3) an - a + bm = 0,65 0,7 - a = 0,65a = 0,05 mol

(1) b = 0,25 a = 0,25-0,05 = 0,2 mol(2) an + bm = 0,05n +0,2m = 0,7n + 4m = 14

m 3,5n = 14 4mm = n +1 v B nhiu hn A mt C

Bin lun :m 1 2 3n 10 6 2

Vy m = 3 n =2


69/115


SVTH:PhanTh Thy 69

Vy CTPT A, B:

63

22

:

:

HCB

HCA

Bi 7 :Mt hn hp kh (X) gm 1 ankan, 1 anken v 1 ankin c V =1,792 lt (ktc) c

chia thnh 2 phn bng nhau:- Phn 1: Cho qua dung dch AgNO3/NH3 d to 0,735 g kt ta v th tch hn hp gim12,5%- Phn 2 : t chy hon ton ri hp th sn phm chy vo 0,2 lt dung dch Ca(OH)20,0125M thy c 11g kt taXc nh CTPT ca cc hydrocacbon.

GII :

bi ny, c nhiu th nghim vi nhiu d kin, ta nn dng phng php ghp ns gii.

nhh = 08,04,22

792,1 = mol

gi a, b, c ln lt l s mol ca ankan, anken, ankin trong mi phn a + b + c = 0,04 mol (1)

Phn 1 + dd AgNO3/NH3 d 0,735g Vhh gim 12,5% Vankin = 12,5%(1/2Vhh) nankin = c = 12,5%*0,04 = 0,005 mol (2)

M= 147005,0

735,0=

Trong phn t kt ta ch c mt nguyn t bc. Vy ankin ban u l ankin-1t CTPT kt ta CnH2n-3AgM = 14n +105 = 107 n= 3Vy CTPT ankin l C3H4.Ta c a + b = 0,04- c =0,04 -0,005 = 0,035 mol (3)

Phn 2 :C3H4 + 4O2 3CO2 + 2H2O0,005 0,015 (mol)

CmH2m + 3m/2O2 mCO2 + mH2Ob mb (mol)CnH2n+2 + (3n+1)/2O2 nCO2 + (n+1)H2O

a na (mol)nCO2 = 0,015 + mb + na (mol) (4)nCa(OH)2 = 0,2 * 0,0125 = 0,115 molnCaCO3 = 11/100 = 0,11(mol)Khi cho CO2 vo dd Ca(OH)2 c th xy ra cc phn ng sau :Ca(OH)2 + CO2 CaCO3 + H2O (5)Ca(OH)2 + 2CO2 Ca(HCO3 )2 (6)


70/115


SVTH:PhanTh Thy 70

TH 1 : S mol CO2 thiu so vi dd Ca(OH)2, ch xy ra phng trnh phn ngs (5)

nCO2 = 0,015 + mb + na = nCaCO3 = 0,11 mol mb + na = 0,095 mol (7)

Cch 1 : Dng phng php bin lun da vo gii hnGi s n < m na + nb < na + mb < ma + mbn(a + b) < na + mb < m(a + b)

n

Documents

Luận văn HydroCacBon