kiẾn thỨc trỌng tÂm vÀ cÁc phƯƠng phÁp giẢi nhanh bÀi tẬp hÓa hỮu cƠ - quÁch...

8/10/2019 KIN THC TRNG TM V CC PHNG PHP GII NHANH BI TP HA HU C - QUCH VN LONG - HONG T

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ThS. QUCH VN LONG ThS. HONG TH THU HNG

KIN THC TRNG TMV CC PHNG PHP GII NHANH

BI TPHOHU C

Sch tham kho dng ch:

GIO VIN HO HC

HC SINH 10, 11,12 V CHUYN HO HC

HC SINH N THI TT NGHIP, I HC V CAO NG

NH XUT BN I HC s PHM

WWW.FACEBOOK.COM/DAYKEM.QU

WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU

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W.DAYKEMQUYNHON.UCOZ.COM

g gp PDF bi GV. Nguyn Thanh T


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NH XUT BN ! HC s PHM

136 ng Xun Thy - Qun Cu Giy - H Ni

in thoi: (04)7.547.735 - Fax: (04)6.547.911

* *

Chu trch nhim xu t bn:

Gim c INH NGC BO

Tng bin tp INH VN VANG

Chu trch nhim ni dung v bn quyn:

NH SCH HNG N

Bin t p ni dung:

PHM HNG BAC

K thut vi tnh:

NH SCH HNG N

Trnh by ba:TIU VN ANH - V TH THA

M s:02.02.1214/1503.PT2011-268

KIN THC TRNG TM V PHNG PHP GI! NHANH BI TP HA HU c

In 2.000 cun, kh 16 X 24cm ti Cng ti In Vn Lang - TP.HGM.S ng k KHXB: 64-2011/CXB/1214-01/HSP k ngy 11/01/2011.QXB s: 059/Q-HSP ngy 10/11/2011In xong v Tp lu chiu qu I nm 2012.



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tck

Nhm gip cho cc em hc sinh nm vng kin thc v t kt qu caotrong cc k thi, c bit l k thi tt nghip Trung hc ph thng v tuynsinh i hc, Cao ng, chng ti bin son b sch:

"Kin thc trng tm v phong php gii nhanh bi tp ho hc".

B sch c chia lm ba tp vi 22 chng.TP I: HO HC I CNG

1. Nguyn t

2. Bng tun hon cc nguyn t ho hc v nh lut tun hon

3. Lin kt ho hc

4. Phn ng oxi ho - kh

5. Tc phn ng v cn bng ho hc

6 . S in li

TP II : HO HC v cd7. Nhm halogen

8 . Nhm oxi

9. Nhm nit

10. Nhm cacbon

11. i cng v kim loi

12. Kim loi kim - Kim th - Nhm

13. Crom - St - ng

14. Phn bit mt s cht v c - Chun dung dch - Ho hc vvn pht trin kinh t, x hi, mi trng.

TP III: HO HC HU C15. i cng v ho hc hu c

16. Hirocacbon

17. Dan xut halogen - Ancol - Phenol

18. Anehit - Xeton - Axit cacboxylic

19. Este - Lipit

2 0 . Cacbohirat

21. Amin - Aminoaxit - Protein

22. Polime - Vt liu polime

Chng ti hi vng b sch ny l ti liu qu bu ca bn c, nht lcc em hc sinh v cc hn ng n hip.

CC TC GI

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o CHNG 15------------- ------------------

I C N G V H f l H C H l c

. KIN THC TRNG TM

I. XC NH CNG THC PHN T DA VO % KHI

LNG CC NGUYN T V VO PHN NG CHYDa vo khi lng C2 , H2 O, N2(hay NH3) sinh ra khi phn tch cht huc nh CTPT CxHyOzNt bng cc cch:

+ Cch 1.Tnh trc tip

mc = [ 2.nCo 2 ;n iH = 2 riH2o ; mN =2 nN 2 ;m 0 = m x -( m c + mH + mN)

p dng cng thc:

1 2 x

m m.

16z

m,

14t M

m, m,

hay

c 11XH m o *"N

12x _ y _ 16z _ 14t _ M x% c ~ % H - % ~ % N ~ I 0

Suy ra: z = [Mx - ( 1 2 x + y + 14t)]16

+ Cch 2.Tnh gin tip

S dng cng thc:

x : y : z : t = m H m 0 . m N12 ' 1 * 1 6 14

= nco2-: 2nH 2o : no : 2nN2= a : p :'y : 5 (a, p, Y, e N.)

Cng thc thc nghim (CTTN) ca X: (CaHpOyNs)n

n = 1 - Cng thc n gin nht (CTG)

Mv

= % c _ %H . %0_ %N

1 2 1 : 16 ' 14

n12a + 3 + 16y + 146

-> CTPT ca X

5



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5/334

+ Cch 3.Da vo phn ng chy

CxHyOzNt + (x + ) 2 -> xC0 2 + - H 2O + - N 24 2 2 2

y ta ax a a

2 2

1 2 2Suy ra: x = - n co, ; y = - n Ho ; t = - n N

a a a

-> z = [Mx - ( I 2 x + y + 14t)]16

F/ d 1. t chy hon ton 0,295 gam cht hu c X cha c, H, o, N thuc 0,44 gam CO2 , 0,225 gam H2 O. Trong mt th nghim khc, phn tchmt khi lng cht X nh trn cho 55,8 cm3N2(o ktc); T khi hi caX i vi hiro l 29,5. Lp cng thc thc nghim v CTPT ca X.

GiiCct 1. Mx - 29,5.2 =59 gam/mol

mc = 1 2 .nC0 2 =0,12 gam; mH - 2 n H0 = 0,025 gam;

ntiN =28nN =28. = 0,07 gam2 22,4 5

-> mo = 0,295 - (0,12 + 0,025 + 0,07) = 0,08 gam

_ , 1 2 xTa c: == J L

_ 16z _ 14t _ M xm c m H m 0 m N m x

1 2 x y _ 16z. 14t 59

0,12 0,025 0,08 0,07 0,295

>x = 2 ; y = : 5 ; z = l ; t = i -* Cng thc phn t ca X l C2 H5ON.- > 1 2 0,025 0,08 0,07 _ , , ,Cch 2. X :y : z : t= =2 : 5 : 1 : 1

1 2 1 16 14- Cng thc thc nghim X: (C2 H5N),,

-> Mx = (12.2 + 5 + 16 +14)n = 59-> n = 1 -> CTPT ca X l C2H5ON.

0,295Cch 3. nx = = 0,005 mol

59

CxHyOzNt +' (x+ ~ - - ) 0 2 -> xC0 2 + - H 20 + - N 24 2 2 2

.0,005 ------ > 0,005x 0,0025y -> 0,0025t

> nco 2 = 0,005x = 0,01 -> X = 2

hh 2 0 = 0,0025y = 0,0125 -> y = 5

6



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6/334

nN, = 0,0025t = Q --0 5 ~ 8 = 0,0025 -> t = 12 22,4

-> z = (59 - 12.2 5 - 14) = 1 CTPT ca X l C2H5 ON.

Ch :

Trng hp X c CTTQ l CxHy ; CxHyOz hoc CxHyNt th ta vn xc nhCTPT da vo 3 cch trn nhung trong biu thc trn ta b z, t hoc c hai.

Trng hp X c cha Na -* CTTQ: CxHyOzNat th tng t nh trn tacng c cc biu thc:

Cch 1. Tnh trc tip12x _ y 16z _ 23t _ M x 12x _ y _ 16z _ 23t _ Mx

mc mH m0 mNamx ^ %c %H %0 %Na 100

Cch 2 . Tnh gin tip

_ m c m H m 0 m Na _ %c %H % 0 %Nax :y :z : - J : :

= a : p : y : (a , p, y, e N)

Cch 3.Da vo phn ng chy

CjiHyOzNat + (x + - - ) O2 -> ( x - - ) C 0 j + ^ H 20 + - N a 2COj

II. NHNG IM CN LU KHI GII TON- Nu ton cho oxi ha hon ton cht hu c tc l t chy hon ton

cht hu c.

- Nu oxi ha cht hu c bi CuOth khi lng ca CuO gim i l khi

lng ca oxi tham gia phn ng, lc tm khi lng ca chthu cem t cn lu nh lut bo ton khi lng:

mx + nigim = mCOj + mH j 0

- Sn phm chy thng c hp th bi bnh ng H2SO4c (hay P2O5)v bnh ng dung dch kim. Lu rng N2v O2d khng b hp th.

- Nhng cht hp th nc: CaC2 (khan), H2 SO4 c, P2O5 , CaO v dungdch baz kim (NaOH, KOH, Ba(OH)2 , Ca(OH)2 , ...) Khi lng ca bnhtng ln l khi lng H2O hp th.

- Nhng cht hp th CO2 : dung dch kim (NaOH, KOH, ..) v kim th(Ca(OH)2, Ba(OH)2). Khi lng ca bnh tng ln l khi lng ca CO 2hpth. Ty theo t l mol gia baz v CO2m mui to thnh l mui oxit haymui trung ha.

+ Trng hp CO2tc dng kim (NaOH, KOH)

C th xy ra 2 phn ng :

C0 2 + 2NaOH Na2C0 3 + H20 (1)

7



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C 0 2 + NaOH -> NaHC03 (2)

* Nu bi ton cho dung dch NaOH d hoc tnh c riNaOH ^ 2nco2 thc 2 trng hp ny mui to thnh l mui trung ha (ch c ( 1 )).

* Neu bi ton cho C2 d hoc tnh c nNa0H < nco2 th c 2 trng

hp ny mui to thnh l mui axit (ch c (2 )).

* Nu tnh c 1 < n~ to ra 2 mui (c (1) v (2))

nco,+ Trng hp CO2tc dng vi dung dch kim th (Ca(OH)2, Ba(OH)2 )

* Neu bi ton cho dung dch NaOH d hoc tnh c nCa(OH) 2^ nco2 th

c 2 trng hp ny mui to thnh l mui trung ha (ch c ( 1 )).

CO2 + Ca(OH) 2 - CaCOs'i' + H20

* Nu bi ton cho CO2 d hoc tnh c nca(OH), ^ tico, th c 21 2

trng hp ny mui to thnh l mui axit (ch c (2 )).

2

CO2

+ Ca(OH) 2

- Ca(HC03 ) 2

Cng c th nhn ra s c mt ca mui axit trong dung dch thu c thngqua hai d kin sau:

*Cho dung dch thu c tc dng vi dung dch baz thy c kt ta xut hin

Ca2+ + HCO3" + OH~ -> CaC03 +.H 20

* un nng dung dch thu c thy c kt ta xut hin v si bt kh thot ra

Ca(HC03 ) 2 1 > CaC0 3 + C0 2t + H20

* Nu tnh c < - Ca(0H)2- < 1_ to ra 2 mui

2 nco2

CO2 + Ca(OH) 2 >CaCCbi- + H20

2C 0 2 + Ca(OH) 2 ->Ca(HC0 3 ) 2Trng hp ny, nu lc tch kt ta, cho nc lc tc dng vi dung dch

OH - th li c kt ta xut hin.

Ca2* + HCO3 + OH~ - CaC 03>l + H20

- Cn phn bit khi lng bnh tng v khi lng dung dch tng

lbinh tng ( m co 2 m^cOhpth

m


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- Nu t chy cht hu c ch cha c, H hoc c, H, o ri cho sn phnchy i qua bnh I ng dung dch PdCh, bnh II ng nc vi d, iu C(ngha sn phm chy gm c o , C2 v H2 O. Trong c o b hp th bdung dch PdCh theo phn ng:

CO + PdCh + H20 -> P d + C02t + 2HCBinh nc vi hp th CO2c trong sn phm chy v CO 2 sinh ra do phi

ng trn v

ITC = m c(CO ) + m c(C 02)V d 2. t chy hon ton mt lng cht hu c X cn 6,72 lt O2 (ktc)

Khi cho ton b sn phm chy (ch gm CO2v H2 O) vo mt lng n'vi trong thu c 1 0 gam kt ta v 2 0 0 mi dung dch mui c nng 2 - > 5 = 1 ( m o 1 )

C 0 2 + Ca(OH) 2 -> CaCOsi + H200,1 Ca(HC03 ) 2

0,2 .*-------------- 0,1

-> L nCOi = 0,l+0,2 = 0,3 (mol)

rHdung dch tng ~ ( ^ c o 2 m H,0 )hp th ITkt ta

^ ^HjO idling dch tng rrikt ta tt ico , ~ ^ 4 4.0 ,3 5 ,4 (g am )

5,4n H , 0 = T T = 0 3 ( m o 1 )

lo

CxHyOz + (X + 2- ~ - ) 0 2 -> xC0 2 + - H 20

y za (x + - T- )a ax 0,5 ay

4 2

-> nGOj .= ax 5= 0,3 ; n H0 = 0,5ay = 0,3 -> ay 0,6

- nG = (x + )a ='= 0,3->a z =0,30 4 2 22,4

X: y : z = 0,3 :0,6 :0,3 = 1 : 2 : 1 - Cng thc n gin nht ca X l CH20.

V d 3. t chy hon ton 0,01 mol cht hu c X cha c , H, o cn 0,784 12 (ktc). Ton b sn phm chy c cho qua bnh I ng dung dcPdCh d, bnh II ng dung dch Ca(OH) 2 d. Su th nghim bnh I tn



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9/334

0,38 gam v xut hin 2,12 gam kt ta, cn bnh II c 3 gam kt ta. Xcnh cng thc phn t ca X.

Gii

npd = 1 = 0,01 (mol); nCaco 3 = = 0,03 (mol)

CO + PdCl2 + H20 -> P d + C 02t + 2HC1 (1)

0 , 0 1 *------ 0 , 0 1 -> 0 , 0 1

mbnh I tng = m H 2 0 + mco - m COz(i)

m H , 0 = m b in h I t n g m c o + m c o ( 1 ) = 0 , 3 8 + 1 6 , 0 , 0 1 = 0 , 5 4 ( g a m )

T> n H,0 = = 0 ,0 3 (m o l)

C 0 2 + Ca(OH) 2 CaC 03^ + H20

0,03 y = 6 ; n-co =0,011 = 0,01

>t ~ l \ nco (x l ).0 ,01= 0,02 >X = 3

n0 = (3 + - - - - 1 ) 0 , 0 1 =0,035 1 4 2 2

-> z = 1 -> CTPT ca X l C3 H6 0 .

. III. BIN LUN TM CNG THC PHN T1. Trung hp 1: s n ln hn s phng trinh ha hc => thiu 1phng trnhGi s c p n (s nguyn t c v s mol) m ch c p - 1 phng trnh (thiu

1 phng trnh) trong trurig hp ny gia 2 n s (thng l gia hai s nguynt cacbon n, m c A, B) lin h vi nhau bng biu thc:

na + mb = nCO;

Trong : a, b, n co bit.

Ta chn n hoc m nhng gi tr nguyn dng 1, 2, 3, ... ri tnh cc gitr tng ng ca n cn li. Ch gi li cc cp n, m sao cho c hai unguyn dng.

10



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V d L t chy mt hn hp X gm 2 hirocacbon ankan A(CnH2n+2) vanken B (CmH2m) thu c 15,68 lt C02 (ktc) v 14,4 gam H2 0 . Bit rnghn hp X chim th tch 6,72 lt ktc. Xc nh cng thc phn t ca A, B.

Gii

_ 15,68 _ A _ , .V _ 14,4_ n o , _ 6,72n co 2 = - ~ = >7 (mo1); n HjO = ^ = 8 (m oi); nx = ^ = 0,3 (moi)

C A , +2 + O2 -> nC02 + ( n + l ) H 20

a na (n + l)a

CmH2m + 0 2 - mC02 + mH202

b mb mb

- n HQ- nco = a = 0,8 - 0,7 = 0,1 mol

-> b = nx - a = 0,3 - 0,1= 0,2 mol

- nco = 0,1 n + 0 ,2 m = 0,7 -> n + 2m =7 (2 < m < 3)

m 1 3n 3 1

C 2 cp nghip ph hp

m = 2(C2H J Jm = 3(C3H6)< ' hoc thiu 2 phng trnh

Gi s c p n nhng ch c p - 2 phng trinh (thiu 2phng trnh).Trong trng hp ny ngi ta thng p dng tnh cht trung bnh (n < m):

n < n < m hoc Ma < M < Mb xc nh n, m.

Cng thc tnh n v M (xem thm chuyn 2).

- na + mbn = -----

a + b .

M A.a + M R.bM = -----

a + b

Ch : Ch s dng cng thc trung bnh trong trng hp cc cht trong

hn hp tham gia phn ng vi cng hiu sut.V d 2. Hn hp X gm hai hirocacbon thuc cng dy ng ng lin tip.t chy hon ton hn hp X, sn phm chy thu c cho hp th ht vobnh I ng H2SO4 c, bnh II ng dung dch Ca(OH) 2 d. Sau khi ktthc phn ng, khi lng bnh I tng 6,3 gam, bnh II c 25 gam kt taxut hin. Xc nh cng thc ca hai hirocacbon trong X.



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11/334

Gii

nwo = - =0,35 (mol); n co = ^ - = 0,25 (mo) < nHn o" J 18 2 100 2

- Hai hirocacbon trong X thuc dy ng ng ankan

- >n x = n H 0 - nco =0,35-0 ,25 = 0,1 (moi)

n x 0,1

n = 2 < n < m = n + l = 3

>CTPT ca hai hirocacbon trong X l C2 H6 v C3 H8-> Chn B.V d 3. Cho mt hn hp X gm mt anken A v mt ankin B. t chy m gam

hn hp X ri hp th ton b sn phm chy vo bnh ng dung dch ncvi trong c 25 gam kt ta v mt dung dch c khi lng gim 4,56 gamso vi ban u. Khi thm vo lng KOH d li thu c 5 gam kt ta na.Bit 50ml hn hp X phn ng ti a vi 80ml H2 (cc th tch kh o cngiu kin). Tm cng thc phn t ca A, B.

Gii

X + H2: CiaH2n + H2 -p r.. CnH2n +2

X X .

CmH2 m - 2 + 2H2 CmH2m * 2

y 2yX = 20 (mi)X + y 50

Ta c h: =>[x + 2y = 80 [y = 30 (ml)

-> A : ne = VA : Vb - 2 : 3

X + 0 2:C 0 2 + Ca(OH) 2 CaC03 + H20

0,25 0,25

2 CO2 + Ca(OH) 2 Ca(HC03 ) 2

0,1 CaCOs^ + K2 CO 3 + 2H20

0,05 S n c c , 2 = 0,25 + 0,1 = 0,35 (mo)

mdung dch gim m kt ta ( rn CQ + m HiQ)hp th

-> m Hj0 = m ktta- m C02 - m dUng dch gim= 2 5 -4 4 .0 ,3 5 - 4 ,5 6 = 5 ,04 (gam )

- nH0 = 0,28 (mo.l)

12



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12/334

> nB = nco2 - nHO =0,35 -0 ,2 8 = 0,07 (mol)

_ 2 2 0 ,1 4-> nA = ne = .0,07 = (mol)

3 3 3

0,14 _ 0,35 -> nx = 0,07 + - 2 = (mol)

3 3

0jl4n.-2 + m.0,07 -n .

- n - 31 ----------- = 3 - 2n + 3m = 5

W l + 0,07 n*3 3

> n = m = 3 >CTPT ca A l CjHe v B l C3H4 .

3. Trng hp 3: Thiu 3 phng trnh tr ln

Trong trng hp ny vn c th s dng tnh cht trung bnh: n < n < m

hoc Ma < M < Mb v trong mt s trng hp c bit vn c th xc nhc CTPT v thnh phn ca hn hp. Ta cng c th s dng cng thc tnh

s nguyn t H trung bnh: = ay. + by2

a + b

Nuyi yi < y < y2

Vi d 4.t chy hon ton 6,72 lt (ktc) hn hp X gm 2 hirocacbon A, Bthu c 8,96 lt (ktc) CO2v 9 gam H2O. Xc nh cng thc phn t caA, B.

Gii

nx = ~ = 0,3 (moi); nC0: = ^ = 0,4 (mol); nH j 0 = ^ = 0,5 (mol)

C- H - + (X + ^ ) 0 2 - ; xC 0 2 + - H 2O

0,3 0 ,3 X 0,3. 2 .

_ ^ 0 4

-* n co 2 = 0 ,3 X= 0 , 4 > X = - 1- = 1,33 -> X] = l < X < X2

>Trong X phi c 1 cht l CH4 (Gi s l A) yi =4

Hh, oV - - 05

0,3 - 0,5 -> y = 2 . - ^ =3,332 0,3

y2= 2 < y < y i = 4 >CTPT ca B l CnH2

13



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13/334

0,2.1-T n.o ,]0,4 >n = 2 (C ,H ,)

Ta c h:

J a + b = 0,3 Ja = 0 , 2

|2a + b = 0,5 7 |b = 0 , 1

V d 5.Hn hp kh X gom mt ankan v mt anken. T khi ca X so vi H2bng 11,25. t chy hon ton 4,48 lt X, thu c 6,72 lt C2 (cc thtch kh o ktc). Cng thc ca ankan v anken ln 'lt lA CH4 v C2 H4 . B. C2 H6 v C2H4 .

c . CH4 vC 3 H6. D.CH 4 vC 4 H8 .Gii

M x = 11,25.2 = 22,5 gam/mol - X c cha 1 cht l CH4+0,CH4 -

C nH2n

y

+0 ;

-> C 0 2

X

> nC2ny

16x + 14ny = 22,5.0,2

X + y = 0,2

X + n y = 0,3

X = 0,15

y = 0,05

n = 3

-> Cng thc ca anken l C3 H6 p n c

IV. NG NG - NG PHAN1 . ng ng- ng ng l nhng hp cht c thnh phn phn t hn km nhau mt

hay nhu nhm CH2nhng c tnh cht ho hc tng t nhau.V d.Dy ng ng ankan: CH4 , C2 H6 , C3 H8, C4 H 10 , C5H 1 2, C nH2 n +2- Gii thch: Mc d cc cht trong cng dy ng ng c cng thc phn

t khc nhau nhng nhm CH2nhng do c cu to ha hc tng t nhau nnc tnh cht ho hc tng t nhau.

- Khi lng moi cc cht trong cng dy ng ng lp thnh cp s cngc cng sai d = 14. '

2. ng phn

- ng phn l hin tng cc cht c cng thc phn t nh nhau nhngkhc nhau v cu to ha hc, do tnh cht ha hc khc nhau.

ng phn cu to:

+ ng phn mch cacbon (thng, nhnh, vng)Phn loi, + ng phn cch chia ct mch cacbon

+ ng phn v tr (ni i, ni ba, nhm th, nhm chc)+ ng phn lin kt

+ ng phn nhm chc

ng phn hnh hc: cis - tran

14



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14/334

+ ng phn mch cacbon: xut hin do s sp xp mch cacbon khc nhau.V d. CH3 - c h = C H - C H 3; CH3- C =CH 2 H2C CH2

CH3 H2C c h 2+ ng phn cch chia mch cacbon: xut hin do s chia ct mch cacbon

khc nhau.

V d. CH 3 COOCH3v HCOOC2 Hs

+ ng phn v tr: xut hin do s khc nhau v tr ca ni i, ni ba,nhm th hoc nhm chc trong phn t.

V d. CH 3- CH2 - CH 2- OH vCH3-CH - C H 3I

OH

CH2= CH - CH2- CH3v CH3- CH = CH - CHS+ ng phn nhm chc: xut hin do s thay i cu to nhm chc trong

phn t.

Vi d. CH3- CO - CH3v CH3 CH2 CHO

+ ng phn lin kt: xut hin do s thay i lin kt gia cc nguyn tcacbon v nhau.

Vi d. CH3- CH2- c = CH v CH2 = CH CH = CH2+ ng phn hnh hc (cis- trans):l loi ng phn khng gian (hay ng phn

lp th) gy nn bi s phn b khc nhau ca cc nguyn t hoc nhm nguyn t hai bn mt b phn "cng nhc" nh ni i, ni ba, vng no ,. ..

V d: H \ / H H3 C \ / Hr = r r = r

H s C ^ ^ C H 3 H / . ^ C H 3

cis - But - 2 - en trans- But - 2 - en

'CH 3

cis - \ l - imetyl xiclopropan

+ iu kin xut hin ng phn hnh hc:

* u kin cn: Phn t phi c lin kt i (mt lin kt i hay mt sliri kt i) hoc vng no (thng l vng nh) trong phn t. Coi l b

phn "cng nhc" cn tr s quay t do ca nguyn t (hay nhm nguyn t) b phn .

* iu kin : mi nguyn t cacbon ca lin kt i v t nht hai nguyntcacbon ca vng no phi c hai nguyn t hoc nhm nguyn t khc nhau.

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15/334

a &biu kin

c * d

Gi s xt hn cp tng i ta thy a > b v c > d, khi :

Nu a, c cng pha ni i (hoc mt phang vng no) ta c ng phn 670-(hay Z - ting c Zusammem ngha l "cng").

Nu a, c khc pha ta c ng phn trans- (hay E - ting c Entgegenngha l "i").

- Cch vit ng phn:

Trong chng trnh ho hc ph thng ch xt ng phn cu to v ngphn hnh hc. vit c y cc ng phn ta thc hin cc bc sau:

V d.Vit ng phn cho CTTQ: OxHyOzNtXv (X: halogen)

Bc 1. Xc nh bt bo ha (s lin kt 7t hoc s vng) ca phn ttheo cng thc:

_ 2 x + 2 - (y + v) + 1_ _ I

Bc 2. Xc nh cc ng phn cn vit theo yu cu bi ton:

Hp cht hu c thuc dy ng ng no?

Mch h hay mch vng?- Da vo gi tr ca A v s lng nguyn t c mt trong phn t

phn loi ng phn c th c.

Bc 3.Vit sn mch cacbon c th c, t mch di nht (mch thng) nmch ngn nht, nu l mch vng th t vng rng nht n vng nh nht.

Bc 4. Thm ni i, ni ba, nhm chc vo cc v tr thch hp trn tngmch cacbon. Cui cng bo ha ho tr ca cacbon bng s nguyn t H cho 4.

V d. Vit cc ng phn ca CHO.

2.3+ 2 - 6

A = ------

------ = 12

V A = 1 v ch c 1 nguyn t o trong phn t nn s c cc loi ng phn

sau y:

+ ng phn ancol khng no n chc

CH2 = C H - C H 2 - 0 H

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16/334

A 2H2C CH2 -OH+ Ete khng no CH2= CH - o - CH3

+ Et vng no

/ C H , C H3 - C H - C H 2

CH2 \?H2

+ Anehit no n chc

CH3 - CH2 - CHO

+ Xeton no n chc

CH3 - C O - C H 3Trong cc hp cht trn khng c ng phn hnh hc.

m a m

I BI TP T LUN1. t ch hon ton m gam cht hu c A (cha G, H) thu c 3,3rh gam

CO2 . Bit Ma < 80 vC. Xc nh CTPT, vit CTCT c th c ca A. BitA c cu to mch h.

2. t chy hon ton 0,175 mol hn hp X gm 1 anken A (CH2) v 1ankinB (CmH2m- 2) thu c 13,44 lt C02 (ktc) v 9,45 gam H20 . Xc nh cngthc phn t, cng thc cu to ca A, B.

3. t chy hon ton m gam hn hp X gm hai hirocacbon X|, X2cng dyng ng (th kh iu kin thng, X2c s nguyn t c nhiu hn Xi)cn dng 32 gam O2 thu c 26,4 gam CO2 . Xc nh cng thc phn tca X|, X2 .

4. t chy ht hn hp X gm hai hirocacbon A, B cng dy ng ng lintip thu c 37,4 gam CO2v 19,8 gam H2O.

a) Xe nh cng thc phn t, cng thc cu to ca A, B.

b) Tnh % khi lngrca A, B trong hn hp X.

5. t chy hon 5,6 lt (ktc) hn hp X gm hai hirocacbon A, B (Ma < Mb). trong VA : VB= 4 : 1 th thu c 7,84 lt CO2(ktc). Xc nh cng thc

phn t ca A, B.

6 . a) Xc nh cng thc phn t ca hirocacbon X, bit t khi ca X so vi

H2 l 28.'b) Hp cht hu c A (cha c, H, O), c t khi hi so vi H2 l 37. Xc

nh cng thc phn t ca A.

7. t chy hon ton m gam hn hp X gm hai hirocacbon A, B thuc cngdy ng ng lin tip, hp th ton b sn phm chy vo dung dch ncvi trong d thy c 60 gam kt ta v khi lng dung dch gim 18,3 gam.

+ Ancol vng no n chc

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17/334

a) Xc nh cng thc phn t ca A, B.

b) Tnh phn trm khi lng ca mi cht trong X.

8 . t chy hon ton 9 gam cht hu c X sinh ra 19,8 gam C2 v 10,8 gamH2 O. Xc nh cng thc phn t ca X.

9. Hy thit lp cng thc n gin ca cc cht t cc s iiu phn tch sau y:

a) 32%c, 6,67%H,18,67%N, cn li l oxi.

b) 4 0% c, 6,67%H, cn li l oxi.

10. t chy hon ton 1,04 gam m hp cht hu c D cn va 2,24 lt kh

O2 (ktc), ch thu c CO2 , hi H2O, theo t l th tch Vco2: Vh 2o = 2 : 1

cng iu kin nhit , p sut.

Xc nh cng thc phn t, cng thc cu to ca D, bit t khi hi ca Dso vi H2 l 52, D cha vng benzen v tc dng c vi dung dch brom.Vit cc phng trnh ha hc xy r.

11. Mt hp cht hu c A ch cha 3 nguyn t l c, H, N. t chy 100 cm3hn hp A v khng kh ly d sau phn ng thu c 105 Ch3hn hp kh

v hi. Lm lnh cn 91 cm3, tip tc cho li qu dung dch KOH cn li 83cm3. Tm cng thc phn t ca A (bit cc th tch o cng iu kin nhit v p sut. Khng kh ch cha oxi v nit theo t l th tch l 1 : 4).

II. BI TP TRC NGHIM1. t chy hon ton a moi mt hp cht hu c X thu c 3,36 lt CO2

(ktc) v 4,5 gam H20 . Gi tr ca a l

A. 0,05 B. 0,10 c . 0,15 D. 0,202. t chy hon ton mt hirocacbon X thu c 4,48 lt CO2 (ktc) v 5,4

gam H2 O. CTPT ca X l

A. C-4 B. C2 H6 C. C2H2 D. C2 H4

3. t chy hon ton 2,3 gam mt hp cht hu c X ch thu c CO2 vH2 O. Sn phm chy cho hp th ht vo bnh ng dng dch nc vitrong d, thy c 1 0 gam kt ta xut hin v khi lng bnh ng dungdch nc vi trong lng 7,1 gam. CTPT ca X l

A . C2H4O2 B. C 2H 60 c . C 2H 6 D . C4H10O

4. t chy hon ton mt hp cht hu c X cn dng 6,72 lt O2 (ktc). Snphm chy gm CO2 v H2 0, cho hp th ht vo bnh ng dng dch

Ba(OH ) 2 thy c 19,7 gam kt ta xut hin v khi lng dung dch gim5,5 gam. Lc kt ta, un nng nc lc li thu c 9,85 gam kt ta na.CTPT ca X l

A. C2H6 B .C 2 H60 C .C 2 H4 O2 D .C 2H6 0 2

5. t chy hon ton rnt hirocacbon X mch h, sn phm chy cho hp thht vo 2 0 0 ml dung dch Ca(OH) 2 IM thy c 10 gam kt ta xut hin v

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18/334

khi lng bnh ng dung dch Ca(0 H)2 tng 16,8 gam. Lc b kt ta chonc lc tc dng vi dung dch Ba(OH ) 2 d li thu c kt ta, tng khilng hai ln kt ta l 39,7 gam. Tn gi ca X l

A. propan B. propen c . propin D. buta - 1,3 - ien

6 . t chy hon ton m gam cht hu c X (cha c, H, N) cn dng 15,12 lt2 (ktc). Sn phm chy cho li chm qua bnh ng nc vi trong, dthy c 40 gam kt ta xut hin v c 1120 ml (ktc) kh bay ra. CTPT ca

X lA. C3 H9N B. C3H 10N2 C .C 4 H 11N D .C 4 H9N

7. Oxi ha hon ton 4,6 gam hp cht hu c X cn 9,6 gam O2 thu c 4,48lt CO2(ktc). Cng thc phn t ca X l

A.C 2H60 B.C2H6 c . C2H60 2 D. CH2O28 . t chy hon ton 2,2 gam axit X cn dng V lt O2 (ktc), sn phm chy

cho hp th hl vo dung dch Ba(OH ) 2 d thu c 19,7 gam kt t vkhi lng gim 13,5 gam. Gi tr ca V l

A.2,8 B. 5,6 c.3 ,36 D. 6,729. Mt hp cht X cha 3 nguyn t c , H, o c t l khi lng me : rriH : nio =2 1 : 2 : 4 . Hp cht X c cng thc n gin trng vi cng thc phn t. sng phn cu to thuc loi hp cht thm ng vi cng thc phn t caX l

A. 3 B. 6 c . 4 D. 5

1 0 . Mt hirocacbon thm X c cng thc n gin nht l C 4 H5v khng tcdng vi nc brom. s ng phn cu to ca X !

A. 1 B. 12 C.4 D. 5

11. Mt hp cht hu c n chc (cha c, H, o ) c khi lng phn t l60u. S cng thc cu to tha mn ca hp cht l

A. 4 B. 5 c . 6 D. I

12. Hp cht Y c thnh phn khi lng 35,03% C; 6,57% H v 58,4?^ Br. slng cc ng phn ph hp ca Y l

A. 5 B. 6 c . 4 D. 3

13. t chy hon ton mt amin no, n chc X (CnH 2n+3N) th sinh ra 17,6gam CO2v 9,9 gam H2 O. s lng amin ng phn cu to ng vi cngthc phn t ca X l

A. 7 B. 4 c . 8 D. 5

14. Ha hi 24,62 mg mt hirocacbon X 100C, 1 atm th thu c 7,53 m-th. Cng thc phn t ca X l

a CsH ^ B .C 4 H 10 C.CtH d . c 6h ,415. Tng s ng phn bn ng vi cng thc phn t CxHyOz (M = 60 vC) l

A. 4 B. 6 C. 3 D .5

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19/334

16. t chy hon ton 3,24 gam hn hp X gm hai cht hu c A v B, trong A hn B mt nguvn t cacbon, ngi ta ch thu c H?0 v 9,24 gamC2 - Bit t khi hi ca X so vi H2l 13,5. Cng thc phn t ca A v Bln lt l

A. C 2 H2v HCHO B. CH3CHO v CH4

c . C2H4 vCH4 D. C2H5OH v CH3OH

17. Tng s cht hu c mch h, c cng cng thc phn t C 2H4 O2l

A. 3. B. 1. C.2. D.4.18. Vi cng thc phn t C5H10 c bao nhiu ng phn cu to?

A. 8 . B . 6 . C.10. D.9 .

19. Hn hp X gm hai hirocacbon mch h X v X2 c cng s nguyn tcacbon. T khi ca X so vi H2 l 21. Cng thc phn t ca X[ v X2 lnlt l

A. C3 H4v C3 H6 B. C2H4v C2H6

c . C3H4 v C3H8 D. C2H4 v C2H220. t chy hon ton 8,96 lt (ktc) hn hp C2H4 , C2H6 , C3H6 , C4 H8 ri cho

ton b sn phm chy vo dung dch nc vi trong d thu c 1 1 0 gamkt ta v khi lng dung dch gim 40 gam. Phn trm th tch ca C2-6trong hn hp l

A. 37,5%. . B. 75%. c.25%. D. 50%.21. t chy hon ton 0,1 mol hn hp C2H4 v C4 H4 th th tch kh CO2

(ktc) v khi lng hi nc thu c ln lt l

A. 3,36 lt v 3,6 gam, B. 5,6 lt v 2,7gam.

c . 6,72 lt v 3,6 gam. D. 8,96 lt v 3,6 gam.

22. Cho cc cht c ng phn cu to CH3 -CH=CH2 , CH3-CH=CHC1,

CH3-CH=C(CH3)2 , CH5CH=CH-CH3 . s cht c ng phn hnh hc lA 3 . B.4. C.2. D. 1.

23. Cho cc pht biu sau:(a) Khi t chy hon ton mt hirocacbon X bt k, nu thu c s mol

CO2 bng so mol H2 O th X l anlcen.

(b) Trong thnh phn hp cht hu c nht thit phi c cacbon.

(c) Lin kt ho hc ch yu trong hp cht hu c l lin kt cng ho tr.

(d) Nhng hp cht hu c khc nhau c cng phn t khi l ng phnca nhau.

(e) Phn ng hu c thng xy ra nhanh v khng theo mt hng nht nh,(g) Hp cht CHnBrCI c vng benzen trong phn t. s pht biu ng l

A. 2. B. 5. c . 4. D. 3.24. Cht no sau y c ng phn hnh hc?

A. c h 2= c h - c h - c h 2 b. c h 3- c h = c h - c h = c h 2

c . CH3-CH=C(CH3) 2 d . CH2CH-CH 2-CH3

20



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20/334

B8S 3B 31B 2B 3B 4B 5C 6 C 7A 8 A 9D I0C IB 12C

13C 14C 5B 16A 17A 18C 19C 20C 21C 22C 23A 24B

D. HNG DN GIII. BI TP T LUN

1 . CxHy + (x + ) 2 > xC 2 + H20

m mx

1 2 x + y

mx

1 2 x + y

.44 = 3,3m = 12x + y y 4

- Cng thc ca A vit li thnh (C3H4)n >MA= 4n < 8 0 > n-< 2 -> n = 1 Cng thc phn t ca A l C3 H4

Cng thc cu to:2 .3 + 2 - 4

bt bo ha: A = 2 - Trong phn t ca A cha 2 lin kt 71

CH2 = c = CH2hoc CH =C-CH32. nCO = 0 ,6 (mol); nH 0 = 0,525 (mol) >ne = 0,6 - 0,525 = 0,075 (moi)

nA= 0,175-0,075 = 0,1 (mol)

> nco = n.o,! + m.0,075 = 0,6 -> 4n + 3m = 24 (2< n < 4V

n2

3 416 8m 3 4 3

(loi) {nhn) (//)Vy cp nghim ph hp l n = 3 v m - 4

- CTPT ca A: C3 H6 v B: C4 H6Cng thc cu to:

(A): CH3-CH=CH2

(B): CH3-C H 2-C =CH; CH3- O C - C H 3

3. nco = 0 ,6 (mol); n0i = 1 (mol) ->lco.

1

0,6> 1,5 - X, X2l ankan

nco, _ 0 , 6

2(n Uj - l,5 nCO) = 2(1 - 0,9)= 3

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21/334

-> 1 < ni < n = 3 < ri2< 4 (th kh k thng nn c Bi ton c hai cp nghim:

n = l(C H4) . - n = 2(C 2H 6)hoc n CO < n H j 0 '

A, B u l ankan (C -H ,- ) v 11 2 n+2 '

nco-> n x = 1,1 - 0,85 = 0,25 mol -> n = = 3,4n x 0,25

>nA 3 (C3 H8) < n < riB= 4 (C4 H10) Cng thc cu to:

(A): CH3-C H 2-CH3 ; (B): CH3-CH2-CH2-CH3; CH3-CH(CH3)-CH3

b) t nCjHg = a (mol); nc H(0= b (mol).

a + b = 0,25 fa = 0,15Ta c h: . _ , .

[3a + 4b = 0,85 [b = 0,l

^ %C 3H8 = 4 ^ 1^ = 5 3 ,2 2 %;0,15.44 + 0,1.58

%C4 H , 0 = 1 0 0 % - 53,22% = 46,78%

5. nx 0,25 (mol); nCOj = 0,35 (mo)

~ n co, 0,35 > n = = = 1,4 > I < n* < n = 1,4 < tin

n x 0,25

i/\ = 1 A l CH41-4 + m.l . .

Mt khc: n = - = 1,4 - m = 31 + 4

> B c th l C3H2 hoc C3H4 hoc C3H8 hoc C3H66 . a) t cng thc tng qut ca X l CxHy.

Ta c: 12x + y = 56 (1 < X< 4)

- y = 56 - 12x - 56 - 12x < 2x + 2 - X> 3,85 -> x = 4 v y = 8

Cng thc phn t ca X l C4 H8

b) t cng thc tng qut c A l CxHyOz.Ta c: 12x + y + I6 z =74 (1 Cng thc phn t ca A l C4H0O

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22/334

z = 2 -> 12x + y = 42 (1 < x< 3)

- y = 42 - I2x < 2x + 2 X> 2 ,8 57 - X = 3 v y = 6

Cng thc phn t ca A l C3H6 O2

z = 3 - I2x + y = 26 ( 1< x< 2)

-> y = 2 6 - 1 2 x < 2 x + 2 - > x > l 7 1 -> x = 2 vy = 2

-> Cng thc phn t ca A ! C 2 H2 O3

Bi ton c 3 nghim : CHO; CHOv CHO

7. a) C0 2 + Ca(OH ) 2 -> CaC0 3 >l' + H200,6


23/334

10. p s : CTPT ca D : CsHs11. Gi th tch ca A l a (cm3)

Th tch ca khng kh l 100 - a (cm3),Th tch oxi trong khng kh = (100 - a) : 5 = 20 - 0,5a (cm3)

Th tch nit trong khng kh: 100 - a - (20 - 0,2a) = 80 - 0,8a (cm3)Th tch H2 O to thnh: 105 - 91 = 14 (cm3)

Th tch CO2to thnh: 91 - 83 = 8 (cm3)

Th tch oxy d: 20 - 0,2a -15 = 5,- 0,2a(cm3)Th tch nit sinh ra do t chy: 83 - (5- 0,2a + 80 - 0,8a) = a - 2 (cm3)Xt phn ng chy:

CxHyNt + (x+ - ) 0 2 - >xC0 2 + - H 2 O + - N 24 2 2

_3 _3 ya _3 ta _3a cm xa cm cm - - cm

2 2ta 4

Ta c: = a - 2 - ta = 2a - 4 - a = 2

2

- tV a > 0; t nguyn dng nn ch c gi tr t = 1v a = 4.

xa = 8 X= 2 ; = 14 - a = 7 2

Vy, cng thc phn t ca A l : C2H7N

II. BI TP TRC NGHIM1. Chn B

n c o 2 = = 1 5 ( m o ) ; n H j O = = 2 5 ( m o 1 )

- > n H0 > n C 0 -> X l hp cht hu c no

-> nx = n H,o n c o , = 025 -0 ,1 5 = 0,1 (moi)

Ch :Hai hp cht hu c CxHy v CxHyOz u c cng cng thc tnh

bt bo (A) V! A khng ph thuc vo z ( A = + ^).

Khi t chy nu thu c nH, 0 > nco, th chng l hp cht hu c no v ta

lun c: nCH (hoc nCH;0z) = nH: - nC0:

2. Chn B

4 4 8 5 4nco2 = = 0,2 (mol); nH0 = ~ = 0,3 (mol)

-> n H:Q > nCO; X l ankan (CnH2ll +2)

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24/334

- nx = n H 0 - nco =0,3 -0 ,2 =0,1 mol

nco 0.2- >n = = 7 = 2 ( C 2H6)

nx 0,1

3. Chn B

ncaco3 = = 0,1 (mol)

C 0 2 + Ca(OH) 2 CaC034- + H200 , 1 n H2Q= 0,15 moi -> me = 12nco =1,2 gam; rriH = 2 n HQ= 0,3 (gam)

-> me + mH = 1,5 gam < 2,3 gam -> X c cha c , H, o.

Vi nH20 > nco: nx = nH20 ~ nco2 = 0 5 (mo1)

2 3>Mx = - 46 gam/mol

o;o5CxHyOz - 3 + > xC 0 2+ - H 20

2

0,05 -----> 0,05x 0,052

> nCOn = 0,05x =0,1 -> X =2; nH;t)= 0,05 = 0,15 > y = 6

z .= J_ ( 4 6 _ 12.2 - 6 ) = 1 CTPT ca X l C2H60

4. Chn B

n 2 = 2 2 4 = 0 , 3 nBaC0- = x~ m = 1(mol)

C 0 2 + Ba(OH) 2 -> BaC034 + H20

0 , 1 4 - 0 , 1

2C 0 2 + Ba(OH) 2 -> Ba(HC03 ) 2

0,1 2 5 BaC03i + C 02f + H2 0

9 850,05 n CO; = 0 , 1 + 0 , 1 = 0 , 2 (mol)

25



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niddgim rttktta (r o co , + ^ H ,o )

^ ^HiO * kt tarcidd gim ^co , 19,7 44.0,2 5,5 = 5,4 (gam)

> n H0 = 0,3 (mol).

V nn2o > nco; -> nx = nH20 - nC 0 3 = 0 , 1 (moi)

CNHyOz + (x+ ^ _ - ) 0 2 - xC02 + - H 20

0 , 1 (x+ I - - ) 0 , 1 0 ,lx 0 , l . 4 2 2

> nC0; = 0 ,l x = 0 ,2 - x = 2 ; n H;0= 0,1 . = 0,3 -> y = 6

n 0 = (x + - )0 ,1 = 0,3 -> z = 1 -> CTPT ca X i C2H60

5. Chn c

^Ca(OH), ~ 0,2

C 0 2 + Ca(OH) 2 -> CaC 03 + H20X X X

2C2 + Ca(OH)2 ->Ca(HC03)2

2y y y

Ca(HC03)2 + Ba(OH)2 CaC03i + BaCs^ + 2H20

y y yn Ca(OH)j = X + y = 0 ,2

-> mkt (a= 100(x + y )+ 1 9 7 y = 39,7 -> y = 0,1 m o l- x =0 ,1 mol- nCO; = X + 2y =0,3 mol

nibinh tng ITlco ^

- m H>0 = mbillhtllg- mCOj = 1 6,8-44 .0,3 = 3 ,6 (g am )- > n HO = 0,2mo

- riCOi > n H i 0 -> X l ankin hoc ankaien (CnH2n- 2)

0 3 . ___> nx - n COj - n H2Q= 0,1 (mol)>n = = 3 CTPT ca X l C3H4(propin).

0,1

6. Chn c

"N = = 0,05(m0l); n2 = ^224 = 0,675(m0l)

C 0 2 + Ca(OH) 2 -> C aC034- + H20

0,4


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CxHvNt + ( X+ )2 -> xC02 + - H 20 + - N 24 2 2y t

a (x + )a . ax a4 2

-> nCO; = ax =0 ,4; nN - .a = 0,05 - ta =0,1;

n0=- (x + )a = 0,675 - ay = 1,1

> x : y : z = 4 : II: 1 >cng thc thc nghim ca X l (C4 H|iN)n.

iu kin: 11 n < 2.4n + 3 - n < 1 - n = t -> CTPT ca X l C4 H 11N

7. Chn A

4 48 _ 9 6n co2 = = 2 (m o1) n 0, = = 3(mo1)

Theo LBTKL ta c:

5,4

m H0= 4,6 + 9,6 - 44.0,2 = 5,4 gam -> nH0= - z= 0,3 mol > n co2 2 18 2

' 4 6- nx = 0,3 - 0 , 2 = 0 , 1 mol -> Mx = - 2 - 46 gam/mol

Mt khc: n = r ~ = = 2 -> CTPT ca X l C2H60/I, 0,1

8 . Chon A

19 7- 13 5 - 0 144

nco, = ncaco- = >1 (mo1); n H,0 = ------------- = 1 ( m o1)1 0

11co, = n H,0 X i axit no, n chc.

CH2n0 2 + ^ 0 2 - nC0 2 + nH20

0,1 4- 0,1n

2 2 01- =2= = - n = 4 (C4H80 2)

14n + 32 n

-> n0j = M z l . M =0,125 (moi) -> v 0 2 = 2 , 8 lt

12. Chn c

t cng thc tng qut ca Y l CxHyBrz

27



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_ _ 3 5 ,0 3 6 ,5 7 5 8 ,4 n ,X : y : z = - : - r : = 4 : 9 : 1

1 2 1 80

>Cng thc nguyn ca Y C^Br => CTPT l: (C4 H9 Br)n>9n < 2.4n + 2 - n > n = 1 -> CTPT ca X l C4 H9 Br

CH3- CH2 - CH2 - CH2Br; CH3- CH2- CHBr - CH3;CH 3- CH(CH3) - CH2 Br; (CH3)3C - B r

14. Chn c

1 7 1O-3nx = - ....... = 0,24 62.10"3(mol)0 , 0 8 2 ( 2 7 3 + 1 0 0)

_> Mx = -~-3t =100 (gam/mo)0,2462.10"3

-> C T P T ca X i C 7H 1615. Chn B

Ta c: 12x + y + 16z = 60 - z < 2

z = 1 - I2 x + y = 44 ( 1 < X < 3)* -> y < 44 - 12x < 2x + 2

> X > 3 X = 3 ~ > y = 8 C3H8O c 3 ng phn:CH 3- CH2- CH2OH (1); GH3- CH(OH) - CH3(2); CH3- o - C2 H5(3)

z = 2 - 12 x + y = 2 8 ( 1 < X < 2) - > y = 2 8 - 1 2 x < 2 x + 2

-> X > 1,85 - X = 2 -> y = 4 - C2H4 O2c 3 ng phn bn:

CH3 COOH (4); HCOOCH3(5); HOCH2 - CHO (6 )

16. Chn A

Mx - . 3,5 =27 gam/moi

- Phi c mt cht c M < 27 > ch c th l CH4hoc C2 H2

3 24 9 24nx = =0 , 1 2 (mol); nCO: = = 0 , 2 1 (mol)

-> n = = 1,75 -> ne = < n


28/334

18. Chn cCH2=CH-GH2 -CH2 -CH3 ; CH2=CH-CH(CH3)-CH3 ;CH2 =C(CH3 )-CH2 -CH3; CH3-CH2 =CH-CH2-CH3; CH3-CH 2=C(CH3)-CH3

19.Chn cMx =42 - loi B, DT A, c -> X, l C3H4 v x 2 l C3Hy

Mi = 4 0 < Mx = 42 < M2= 36 + y - > y > 6 - > y = : 8 -> C3 H8

20.Chn c, , , 110 40 44.1,1

nco, = ncaco. = w (m o); nH,0 = ---------- TT----------- = 12 ( m 0I O

t chy C2 H4 , C3 H6, C4 H8to ra s mol CO2v H2 O bng nhau

-> n c H = 1 , 2 - 1 , 1 = 0 , 1 moi

21.Chn c

n H20 =2(nCiH4 +iCjHj) = 0,2moi-> I1 H 2 0 = 3,6 gam

4 n hh= 0 , 4 > n co > n Hy =0,2 - 4,48 lt < Vco2 < 8,96 lt

T p n VCo 2 = 6,72 lt

22. Chn cDa vo iu kin xut hin ng phn hnh hc ta thy ch c

CH3-CHCH CI , C6H5CH=CH-CH3l c ng phn hnh hc:

C]

Cis(Z) -1 -cloprop-1-en Tram(E) -1 -cloprop-1-en

C6 H5

H

c is-1-phenylprop-1-en Trarn-1-phenylprop-l-en

29



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23. Chn A

(b) Trong thnh phn hp cht hu c nht thit phi c cacbon.

(c) Lin kt ho hc ch yu trong hp cht hu c l lin kt cng ho tr.

24. Chn B

Da vo iu kin tn ti ng phn hnh hc ta d thy ch

CH3-CH=CH-CH=CH2l c ng phn hnh hc

CHj. r .H. CHK H

' c = c ;\ / \

H H H C2H3

. C/A-penta- 1,3-ien 7Vam-penta-l,3-ien

o CHNG 1 6 ..................... ... ....... ..... -_______ __

H I f } R O C f C B O N

. KIN THC TRNG TM

a) ng ngAnkan: metan (CH4), etan (C2 H6), propan (C3H8), cc butan (C4 H10), ....c

cng thc chung CaI2n + 2 (n > 1). Chng hp thnh dy ng ng gi l dyng ng metan.

b) ng phnAnkan t 4 ccbon tr ln bt u c ng phn cu to c th gm ng

phn mch cacbon, v tr nhnh. V d :CHi4c 5 ng phn cu to:

ch 3 ch 2ch 2ch 2ch 2ch 3 ch 3- ~ c h 2ch CH2CH3

CH3 CH CH2CH2CH3 CH3

CH3 CH3

CH3 CH CH CH3 CH3 CH2C - CH3I I Ik ch , c h 3

c) Bc ca cacbnBc ca mt nguyn t cacbon phn t ankan bng s nguyn t c lin k

trc tip vi n.



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30/334

I 1' H H CH3 CH3 H

Ii III I III IIV 11H H

! H H H CH3 H

I

2. Danh php

a) Ankan mch khng phn nhnh- Mun c ng tn ca hp cht hu c trc tin phi nh t gc (l t

dng biu th s nguyn t cacbon)

S c trongmch

1 . 2 3 4 5 6 7 8 9 10

T gc met et Prop but pent hex Hept oct non , dec

Cch nh M em phi bn phn ha hc ngoi ng

- Tn gi ca ankan mch khng phn nhnh theo danh php IUPAC

T gc + an

Ch : Tn ca nhm ankyl khng phn nhnh :

Tgc + yl

V d:Tn ca mi ankan v nhm ankyl khng phn nhnh u tin :

Ankan khng phn nhnh Ankyl khng phn nhnh

Cng thc Tn Cng thc Tn

c h 4 Metan c h 3- Metyl

CHCH Etn CIICH- Etyl

CHCHCH Propan CHCHCH- Propyl

CH3[CH]CH3 Butan CH3[CH2]2CH2- Butyl

CH3 [CH2]3CH3 Pentan CH3[CH2]3CH2- Pentyl

CH[CH]4CH3 Hexan CH[CH]4CH2- Hexyl

CH3[CH]CH3 Heptan CH3[CH2]5CH2- Meptyl

CH3[CH]CH3 Octan CH3[GH2]6CH2- Octyl

CH3[CH]CH Nonan CH3[CH2]7CH2- Nnyl

CH[CH]8CH3 ecan CH3 [CH2]8CH2- ecylb) Ankan mct phn nhnh

Bc 1.Chn mch chnh (mch di nht, c nhiu nhnh nht)Bc 2.nh s trn mch chnh, sao cho v tr nhnh l nh nht

Bc 3.Gi tn theo th t: ^S ch v tr nhnh - Tn nhnh (a, b, c , ...) + Tn ankan mch chnh

31



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31/334

V d: 1 ) C H3- C H2- C H2 - C H - C H - CH3i i 12

C2H5 c 2 h 5

4 - etyl - 3 - metylhptan

2 ) CH3 - CH - CH2 - C H - C H 2 - CH2 - CH-3

CH3 C H = c h 2

5 - metyl - 3 - propylhex - 1- en

3) CH 2 =


32/334

Trong phn t ch c lin kt nn nguyn t c quay c d dng quanhtrc lin kt on c - c - phn t ankan tn ti nhiu ng phn cu dng.

iu kin thng: Cl >C4l cht kh

C5>Cislcht lngC | 8 tr ln l cht rn.

Khi s nguyn t c tng - khi lng phn t M tng - t Vtng. Mch

cacbon cng phn nhnh -* b mt tip xc cng gim -> lc Van der Waals

gim - ts gim.

V khng c lin kt hiro gia ankan vi nc tt c hirocacbon ukhng tan trong nc (k nc) nhng tan nhiu trong dung mi hu c.

4. Tnh cht ha hca) Phn ng th halogen

C nH2n + 2 + ZX2 C nH 2n + 2 -z X z + Z H X

Cc ng phn ca metan cng tham gia phn ng th tng t meta. V d:

C H 3 - C H 2 - C H 3 ..c l^ Uc-as- > C H 3 - C H C I -C H 3 + C H 3 - C H 2 - C H 2 C I + H C 1

2 - clopropan, 57% 1- clopropan, 43%CH3CH2CH3 CH3CH BrCH 3+ CH3C H 2C H 2Br + HBr

, 2 - brompropan, 97% 1- bromprpan, 3%

(chnh) (ph)

Clo th cho H cacbon cc bc khc nhau. Brom hu nh ch th cho H cacbon bc cao. Flo phn ng mnh lit nn phn hy ankan thnh c v HF. lotqu yu nn khng phn ng vi ankan.

b) Phn ng tch (gy lin kt C -C v C-H)Di tc dng ca nhit v xc tc (Cr23, Fe, Pt, cc nkan khng

nhng b tch hiro to thnh hirocacbon khng no m cn b gy cc lin ktc - c to ra cc phn t nh hn.

CnI'2n+ 2 CmI 2m "* Cn_mH2(n-m) +2 (m > 2 , n m ^ 0 )

CH3CH = CHCHs + H2500 c

C H 3 C H 2 C H 2 C H 3 -------------------> C H 3C H = c h 2 + c h 4xt

> c h 2 = c h 2 + C H 3 C H 3

c) Phn ng ox ha

Phn ng chy (phn ng oxi ha hon ton)Khi t, cc ankan b chy to ra C2 , H2O v ta nhiu nhit.

CnH2 n + 2 + ^ - ^ - 0 2 -> nC02 + (n + 1)H20

3n + l .aa -> na -> (n + l)a

33



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33/334

nco ncoSuy ra: a= n H ; 0 - n q02; 11= =

lH2o COj

Phn ng ha khng hon tocm- Nu khng oxi, ankan b chy khng hon ton, khi ngoi C2 v

H2 O cn to ra cc sn phm nh c o , mui than, khng nhng lm gim nnsut ta nhit m cn gy c hi ch mi trng.

- Khi c cht xc tc, nhit thch hp, ankan b 0 X1ha khng hon to

to thnh dn xut cha oxi.V d: . i

CH4 + 0 2; ... ... >HCHQ + H20200atm,300c

- Nu mch cacbon di, khi oxi ha c th b b gy:

- t, p 'CH3 CH2GH2 CH 3 + 0 2 -;T3 > 2 CH3 COQH + H20

Mn

d) Phn ng phn hy

Phn hy bi nhit:

IOOOCCnH2 n + 2 > nC + (n + 1)H2

khng c kkc bit:

1500c2 CH4 ------- > C2H2 + 3H2

lm lnh nhanh

Phn hy bi clo: 0

CnH2n + 2 + (n + 1 )C 12 ------ ------> nC -t- 2 (n + 1)HC1 , ~ . as cc tm5. iu che

a) Khai thc t du m, kh thin ntin qua cc con ng cracknh vchng cat phn on

b) Phng php tng mch cacbon

Tng hp Wurtz (Php, 1855)

2CH2+ iX t 2Na (CH2n+ 1 )2 + 2NaXkhan

Ch : Nu dng hai loi dn xut halogen c gc nkyl khc nhau s th

c hn hp 3 ankan song kh tch khi nhau vl chng c nhit si xp xbng nhau:

r R - R t 2NaX

RX + R'X + -2Na -> s R - R' + 2NaX

R' -R ' + 2NaX

34



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34/334

Tng hp Kolbe (c, 1849)

2 RCOONa + 2H20 ...> R - R + 2C 02t + 2NaOH + H2fV-----, V------------------------ ,Anot Catot

c) Phng php gim mch cacbonPhng php Dumas:

RCOONa + NaOH Ca0;t- > RH + Na2 C 0 3V d

CH3COONa + NaOH Ca -,: -> CH4 + Na2 C 0 3

CH2(COONa) 2 + 2NaOH Ca-o / > CH4 + 2Na2 C 0 3 Phng php crackinh:

CnH2 n+ 2 : CmH2 m C(n _ m)H2(n - m) +2 (m ^ 2 , n > m)d) Phng php g i nguyn mch cacbon

Hiro ha anken, ankin, ankaien tng ng:

CH2 + H2 ) CnH2 n +2

CH2i,_2 + 2H2 nm > CH2+2 i t anco no, n chc:

CH2n + iOH + HI 200c > CnH2n +2 + H20e) MI s phng php khc T nhm cacbua ;

AI4 C3 + 12H20 -> 4AI(H)34 + 3CH4t

(iu ch AI4 C3 : 2 AI2O3 + 9C 2000c > AI4 C3 + 6 C O t ) TCvt

c + 2 H2 c -> CH4

II. XICLOANKAN (HIROCACBON NO MCH VNG,

35



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35/334

Xicloankan l nhng hirocacbon no mch vng.Xicloankan c I vng (n vng) gi l monoxicloankan. Xicloankan c

nhiu vng (a vng) gi l polixicloankan.

Monoxicloankan c cng thc chung l C nH 2n (n > 3).

b) ng phn v cch gi tn monoxicloankan

Quy tc:

S ch v tr - Tn nhnh + Xiclo + Tn mch chnh + anMch chnh l mch vng. nh s sao cho tng s ch v tr cc mchnhnh l nh nht,

b)V dn g vi cng thc CHi2c cc ng phn xicohkan l:

V -7

xiclohexan metylxiclopentan1

,2

-imetylxiclotutn etyxiclobutan

1 , 1 -imetylxiclobutan 2 -etyl-l -metylxiclopropan

1,2,3-trimety lxiclopropan isopropy Ixiclopropan

1 , 1 ,2 -trimetylxiclopropan propyxiclopropan

1-etyl- 1-metylxicopropan

Cch vit ng phn nich vng V vrig ln nht ri thu nh dn, ch t Hp cc loi nhnh (nu c).

Thm cc ngun t H vo mch m bo ng ha tr 4 ca cacbon.

II. TNH CHTI. Tnh cht vt l- nhit thng xiclpropan, xiclobutan, metylxiclopropan l cht kh

Cc xicloankan khc l cht lng hoc cht rn.

36



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36/334

- Khng mu, khng tan trong nc nhng tan c trong cc dung mi hu c.

2. Tnh cht ha hca) Phn ng cng m vng ca xclopropan v xidobutan

x \ + H2 CH3-CH 2-C H 3ou S' propan

+ Br2 - CH2 Br-CH 2-CH2Br

1,3 - ibrompropan/ X + HBr - CH3-CH 2-CH2Br

1 brompropanXiclobutan ch cng vi hiro:

+ h 2 N i -120 c > CH3- CH2- C 2- CH3

Xicloankan vng 5, 6 cnh tr ln khng c phn ng cng m vng trongnhng iu kin trn.

Ch : Trong cc xicloankan ch c xicloankan vng 3 cnh mi lm mtmu dung dch brom.

b) Phn ng thPhn ng th xicoankan tng t nh ankan. V d:

+ Cl2 -> C 7 + HC1

xiclopentan ; : ;c) Phn ng oxi ha hon ton

CnH2n + 0 2 ^ nC0 2 + nH2 02

C | 2 + 9 2 > 6 CO2 + 6 H2 O3. iu ch v ng dng

a) iu chNgoi vic tch trc tip t qu trnh chng ct du m, xicloankan cn

urc iu ch t ankan. V d :

CH 3 [CH2 ]4CH3 - xu:"> l J + H2b) ng dng

Ngoi vic ng lm nhin liu nh ankan, xicloankan c c dng lmdung mi, im nguyn liu iu ch cc cht khc.

xt,t > 1 ^ 1 + 3H2xicohexan benzen

37



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37/334

III. ANKEN (OLEFIN, G1H211,2)1. ng ng v danh phpa) Dy ng ng v tn thng thng ca ankenEtilen (C2H4), propilen (C3H6), butilen (C4 H8), .... u c mt lin kt i

c= c , c cng thc chung l CnH2n (n > 2 ). Chng hp thnh dy ng ng gil dy ng ng ca etilen.

Tn ca mt s anken n gin ly t tn ankan tng ng nhng i uian thnh ui ilen. V d :CHC H - C H CH2=CI I-CH-CH CH-CHCH-CH CH=C(CH)-CH

propilen a-butilen p-butilen isobutlenNhm CH2=CH- c gi l nhm vinyl; nhm CH2=CH-CH2- c gi l

nhn1 an!y!b) Tn thy th

Quy tcS ch v tr - Tn nhnh+ Tn mch chnh-S ch v tr - en

Mch chnh l mch di nht, cha ln kt i v c nhiu nhnh nht.nh s c mch chnh bt i t pha gn lin kt i hn.S ch v tr lin kt i ghi ngay trc ui en (khi mch chnh ch c hai

hoc ba nguyn t c th khng cn ghi)

V d:CH 2C H 2 CH2 CH-CH 3 CH2=CH-CH2-CH3

eten propen but-1 -en

C H 3 - C H C H - C H 3 CH2=C(CH3)-CH3but-2 -en 2 -metylpropen

2. Gu trc v ng phna) Cu trc

Hai nguyn t c mang ni i trng thi lai ho sp2 (lai ho tam gic).Lin kt i c=c phn t anken gm mt lin kt v mt lin kt 7t. Linkt c, to thnh do s xen ph trc (ca hai obitan lai ho sp2) nn tngi bn vng. Lin kt nc, hnh thnh do s xen ph bn (ca 2 obitan p)nn km bn hn so vi lin kt . Hai nhm nguyn t lin kt vi nhau bilin kt i c=c khng quay t do c quanh trc lin kt (do b cn tr bilin kt 7).

38



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39/334

n -*CH3- C H+ . I I

CH3 -> CH = CH2 + H Cl"___ Clt _ _____ I

*-r ' u__r u

CH3 (sp chnh)

^ CH3- CH2 - CH2 C1 (sp ph)

- Gii thch quy tc: Do gc ankyl y electron lm cho lin kt 71 bcc, nn ion dng gn vo cacbon mang in m v ngc li.

- Kh nng phn ng cng: HI > HBr > HC1 > HF

- C ch phn ng cng axit vo anker:

H

Cng 2O >ancol

CnH2n + H20 CnH2 n+lOH

CI-I2 = CH2 + H20 *> CH3 CH2 OHancol etylic

2 CH3-CH = CH2+ 2H20 h* > C H 3 - C H - C H 3 + C H 3 - C H 2 - C H 2 O H

; , OHancol isopropylic ancol propylic

(sp chnh) (sp ph)b) Phn ng trng hp

1 1n ) c = c ^ - ^ - *

V d:nCH 2 = CH 2 4 C H 2 - CH; -),r

polietilen (PE)

11CH2 = (pH --^ p~ > * C H 2 C H ^ r

C H 3 C H s

polipropilen (PP)

c) Phn ung ox ha

Phn ng chy

ChH2i, + 0 2 -> nC0 2 + nH20 AH < 02

40



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40/334

Oxi ha bi dung dch KMnOj- Dung dch KMn0 4 long nhit thng oxi ha ni i ca ant

thnh 1,2 3CnH2 + 2KMn04+ 4H20 -> 3CllH2 ll(OH) 2 + 2M n024 + 2 KOH

V d:

3CH= CH+ 2KMn0+ 4H20 - 3CH- CH + M n0 2 + KOH

(mu tm) OH H (mu eri).

Trong cng nghip, ngi ta oxi ha nh etilen sn xut CH3CHO2CHZ = CHg + 0 2 PdCK/CuCI ;t _ > 2CH3CHO

d) Phn ng th

nhit thp, cc anken d tham gia phn ng cng clo, nhng nh cao (500 - 600C), mt s anken u dy ng ng c th tham gia phng th bi clo. V d:

CH2 =CH - CH3 + Cl2 50Q- -> CH2= CH - CH2C1 + HCIAniyl clorua

5. iu ch ankena) hro ho ankan tng ng

CH2n + 2 CnH2n + H2

b) hirat ha (Uicol tng ng:CnH2n+iOH H2so4ac,t>i7o0c_ > CnF2n + h 2q

c) Cng H2 vo ankin (xt: Pd) hoc ankaen (xt: Ni) tng ng:

CnH2n- 2 + H2 CnH2ll

V d: CH = CH + I I2 ....Ptl--l0- > GH2= GH2

CH= CH - CH = CH + H 1- ,tP- > CH- CH- CH = Cl

d) Crackinh ankan

CH+ ^ .CmHm C_ mH(n - m)

e) Loi HX ra khi dn xut mnohalogen ca ankan tng ngr ' u V KOH/ancol v o _L LTV'-'*+A ------- -------- > L.nrln + HA

g) Lo iX2 ra khi dn xut a, p~ halogen ca ankan tng ng

R - CHX - CHX - R' + Zn > R - C H = C H - R ' + ZnX

IV. A N K A IEN : CnH2n2 (n > 3)

1. Phn loi v tnh cht vt l

Cn c vo v tr ca hai lin kt i m ngi ta chia lm 3 loi:

Lin kt i lin: Hai lin kt i lin nhau.

Vd: CH = c = CH : propain (anlen)

4



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41/334

Lin kt i lin hp: Hai lin kt i cch nhau mt lin kt n. V d:1 2 3 4 I 2. 3 4CH2 = CH - CH = CH2 ; CH2 = C(CH3) - CH = CH2

buta - 1,3 - ien (butaien) 2 - metylbuta - 1,3 ~ ien (isopren)

Lin kt i khng lin hp: Hai lin kt i cch nhau nhiu lin kt n.

vid*: , 2 3 4 5CH2 =CH - CH2- CH = CH2 : penta- 1 ,4 -ien

Hai ankaien quan trng l:- Butaien hay ivinyl l cht khng mu, c mi c trung, ts = - 4c.

- Isopren hay 2 - metylbuta - 1,3 - ien l cht lng, khng mu, si 34c.

u l nhng cht khng an trong nc nhng tan c trong cc dung mihu c.

2. Tnh cht ha hc

a) Phn ng cng

Cng hro

CH2 = CH - CH = CH2 + 2H2 - CH3- CH2- CH2- CHsCH2= c - CH =CH 2 + 2 H2 CH3 - C H - CH2- CH3

Ic h 3 c h 3

Cng halogen (X2) v hirhalogenua (HX)IButaien cng nh isopren c th tham gia phn ng cng X 2, HX v thfng

to ra hn hp cc sn phm cng - 1,2 v -1,4. nhit thp u tin to ra rasn phm cng -1,2 ; nhit ca u tin t ra sn phm cng-1 ,4 /

V d: (sn phm cng - 1,2) (sn phm cng - 1,4)

CH= CH-CH = CI- - ^ CHr-CH-CH = CH+ CH-CH = CH-CHI I , 1Br Br Br Br

-80C: 80% 20%

40C: 20% 80%

CH2= CH -C H =CH 2

(sn phm cng - 1,2) (sn phm cng - 1,4)

- - * 1 . > ' C H 2 - C H - C H = c h 2 + C H 2 - C H = C H - C H 2

H Br H Br

80C: 80% 20%

40C: 20% 80%

b) Phn ng trng hp

Trng hp - 1,4 cho polime c tnh n hi cao (ca su)

42



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42/334

nCH2 = CH - CH = CH2 ....f CH2 - c h = c h - CH2-^

polibutaien (thnh phn chnh c cao su buna)

nCH2 = c - CH = CH2 ...fC H2- c = CH - CH2 V

IC H j C H 3

poliisopren

nCH2= c - CH = CH2 (- CH2- c = CH - CH2 -frI

C 1 C 1

copren policlopren

Mt phn c th trng hp -1,2 to thnh polime khng n hi. iuny gii thch tnh n hi km ca mt s loi cao su tng hp.

1CH2= CH - CH = CH2 f C H a - C H V

I j .

c h = c h 23. iu ch ankaiena) iu ch buta -1 ,3 - ien

hir ha butan hoc bten nhit cao (6Q0C) c mt cht xc tc(Cr20 3): ' - ' T.-V

CH3 CH2 CH2CH3 xMp- > CH2= CH - CH = CH2 .+ 2 H2

C4Hs xtt0-P- > CH2 = CH - CH = CH2 + H2

hiro ha v hirat ha ancol eyic nh xc tc (ZnO hoc; MgO v A2 O3)

2CH3CH2OHMg0 /AI20 1,4S09C ^ CH2= CH- CH = CH2 + H2+ 2H20

c) i t axetilen

2CH = CH Cud/ NHjCij,150*c_ > CH2= C H -C = CH

C H 2 = C H - C S C H +H2^ t:Pd> > C H 2 = C H - C H = = c h 2

Tong hp Konbe (c):

2CH2= CH - CONa+ 2H20 - pJd- -> CH2= CH -C H = CH2+ 2C02T

+ 2NaOH + H2t

b) iu ch sopren

hiro ha isopentn v isopenten (sn phm cracknh du m)

CH3- CH - CHS - CH3 x0p > CH2 = c - CH = CH2 + 2H2I I

c h 3 c h 3

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2. Tnh cht ha hc

a) Phn ng cng

Cng hir:Phn ng xy ra qua 2 giai on (giai on I to sn phm dng cis)

Ankin * ^ ( 3 Anken xtt^ :"" > Ankan

Mun dng phn ng giai on th nht (anken) ngoi t l 1 : 1 cn ch

CH s CH + H2 -Pd/PbCO > CH2= CH2

CH = CH + 2H2 CH3-C H 3* Cng X (X= Cl, Br)Phn ng xy ra qua hai giai on (giai on I cng trans). Mun dng

phn ng giai on th nht th cn thc hin nhit thp

CH2ll-2 +X > CH2-2X2 > CnH2n_2X4 ' (I)V d: Br

Ic 2h 5 - c C - C2H5 (1) * C 2H s - c = c - C 2K 5 (2)

I Br

Br Br

C2H5 - C - C - C2H5

v:i- g r g r .*

(3)

Nhn xt:- Ni chung ankin lm mt mu nc brom chm hn anken.- Khi lng ca bnh ng dung dch brom tng ln l khi lng ca

ankin hp th.

. * Cng t x (X= C, Br, I)- Phn ng xy ra qua hai giai on v giai on sau kh hon giai on trc

CH 3 CH + HC1 HgCl2 > CH2 = CHC1 ** t0 > -fCH2 - CH^-150 - 200c p I

. Cl(vinyl clorua) (poiivinylclorua) (PVC)

CH2 = CHC1 + HC1 -> CII3-CHCI 2 (1,1-icloetan)- Phn ng cng vo ng ng axetilen tun theo quy tc Mac-cop-nhi-cop:

CH3- c s CH + HCI xt> 10 >CI-3-C = GH2 > CH3-CCI2- CH3p I xt,t

. Cl

45



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45/334

Cng nc (hirat ha)- Axetilen + H20 - Anehit axetic

C H s C H + H 20 - [CI-I2 = C H O H ] - C H 3 C H 0

(khng bn)- Cc ng ng + H2 O ->Xeton

Rl _ c = c - R2+ HOH - [Rj - CH = CH2= CH - c = c hvinylaxetilen

Ba phn t axetilen c th cng hp vi nhau to thnh benzen:

3CH CH 600 c. than hot tinh _c) Phn ng th bng ton kim loi ha tr I (Ag+y Cu+)

Nguyn t H nh vo cacbon mang ni ba linh ng hn rt nhiu so vi Hnh vo cacbon mang ni i, do n c th b thay th bng nguyn t kimloi. Nh vy, ch c axetilen v cc ank - 1 - in mi tc dng c VI dungdch AgN3/NH3 cho kt ta mu vng, tc ng v dng dch C11CI/NH3 chokt ta mu .

Cc phng trnh phn ng:CH = CH + 2AgN 0 3 + 2NH3 -> CAg = C A g + 2NH4NO3

bc axetilua(mu vng nht)

R - C = CH + AgN0 3+ NH3 -> R - e = CAg + NH4NO 3(mu vng nht)

CH = CH + 2CuCI + 2NH3 -> CCu = CCu + 2NH4 CIng (I) axetilua

(mu )

46



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46/334

R - c = CH + CuCI .+ NH3 R - c 3 CCu + NH4CI(mu )

d) Phn ng ox ha Phn ng chy

CH2 _ 2 + !-02 - - * nCOa + (n - l) H 202

Oxi ha bi dung dch KMn4

Phn ng oxi ha bi dung dch KMn4 : tng t anken, ankin d b oxi

ha bi KMn4sinh ra cc sn phm nh CO2 , HOOC - COOH,,..

3C2H2 + 8KMn04 -> 3KOOC -C 0 0 K + 8Mn02 + 2KOH + 2H20

C2H2 + 2KMn04 + 3 H2SO4 -> 2C02t + 2MnS04 + K2SO4 + 4H20

5CH3 - c = c h +. 8KMn04 + I2H2S 0 4 - 5GK3COOH + 5C 02f

+ 8Mn02 + 4 K2 SO4 + 12H20

Nhn xt:C th dng phn ng lm mt mu dung dch thuc tm nhnbit cc ankin. So vi anken th te phn ng din ra chm hn.

3. iu ch hkn

a) iu ch axetilen

T metan:Phng php chnh iu ch axetlen trong cng nghip hinnay l nhit phn metan 1500c, phn ng thu nhit mnh:

2 CH4 l500- > C2H2 + 3H2

T than , vi- CaC2(t n) 2 > C2H2

Than > than cc (C)

CaC03 - l000--c > CaO +C02t

CaO + 3C ....... 200Qllc > CaC2+c o

CaC2 + 2H20 ->Ca(OH)2 + C2H2t

Nhn xt: nhng ni cng nghip du kh cha pht trin, ngi ta thng iu ch C2 H2theo phng php trn.

TCvH2: 2C + H2 C2H2

T mui axetilenua:CAg 3CAg + 2HCI - CH = CHf + 2AgClb) iu ch ng ng cua axetn

hiro ha ankan tng ng:

cH2n+2: xl[; 1'' ) ch2_2 + 2H2

47



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47/334

*i t dn xut ihalogenH XI I

Ri - C - C - R2 + 2K 0H - anc- - > R) - c 3 c - R2 + 2KX + 2H20

X H

*i ( dan xut cha Ag ca ankin v dan xut cha halogen ca ankan:

- Phng php tng mch cacbon:Ag C - R i R'C1 ->! 6 .

- Khi coi vng benzen l mch chnh th cc nhm ankyl nh vi n lmch nhnh (cn gi l nhm th). Ankylbenzen c ng phrumch cacbon.

e gi tn chng, phi ch r v tr cc

nguyn t c ca vng bng cc ch s hoccc ch ci o(ortho), m (meta), p (para) nhhnh su:

48 4(p)



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48/334

V d :

Qmetylbenzen

Toluen

l,3-imetylbenzen

/w-imetylbenzen

m-metyltoluen0-xilen)

''3

1,4-imety lbenzen/?-imetylbenzen

/>-metyltoluer>(p-xilen)

l,2-imetylbenzen

-imetylbenzen

o-metyltoluen(o-xilen)

2. Tnh cht vt i

Benzen v cc ankylbenzen l nhng cht khng mu, hu nh khng tantrong nc nhng tan nhiu trong cc dung mi hu c nh ancol, te, xeton vng thi chnh chng cng l dung mi ha tan nhiu cht khc. Chng hn,

benzen ha tan brom, iot, lu hunh, cao su, cht bo,... Cc aren u nhngcht c mi nh benzen v toluen c mi thm nh, nhng c hi cho sc khenht l benzen. .

3. Tnh cht ha hc

r . a) Phn ng th Phn ng halogen ha- Benzen v cc ng ng ca benzen khng phn ng vi dung ch brom

nhung phn ng d dng vi brom khan khi c mt ca bt st lm xc tc

+ Br2

- Toluen tham gia phn ng brom ha d dng hn v to ra sn phm gmhai ng phn:

Ch :Nu khngmch nhnh.

I ' r.Br ' :

/7-bromtoluen -bromtoluen

Fe m chiu sng (as) th Br s th nguyn t H

49



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49/334

Phn ng nitro ha- Benzen tc dng vi hn hp HNO3 c v H2SO4 m c to t

nitrobenzen :

nitrobenzen

Nitrobenzen e dng vi hn hp axit HNO3 bc khi v H2SO4 m ng thi un nng th to thnh m-initrobenzen. '

w-initrobenzen

- Tluen tham gia phn ng nitro ha d dng hn benzen v to thnhphm th v tr orthovpara

uv tc th trn vng bem enKhi trn vng benzen c sn nhm ankyl (hay cc nhm -OH, -N H 2

-OCH 3 , ...), phn ng th vo vng benzen s d dng hn v u tin thvo v tr ortho hoc para. Ngc li, nu vng benzen c sn nhm NO(hoc cc nhm -COOH, -SO 3H, -COOR, ...), phn ng th vo vng s khhn v u tn th vo v tr meta. Ring nhm halgcnua (F , cr, Br, I"

lm cho kh nng phn ng th ca vng km hn so vi benzen nhng li nhhng cho nhm th m vo v tr ortho hoc para.

C ch phn ns th vng bemen

Phn t halogen hoc phn t axit nitric khng trc tip tn cng. Cc tiuphn mang in tch dng do, tc dng ca chng vi xc tc mi l tc nhn

H

+ HO-N 2

NO2

Q) + H2

n o 2

+ HO-N 2

N0 2

l_HjS04dac_NO2

+ h 20

HNO3, H2S04(0

,n o 2+ h 20

n 2 ........... ;; .p-nitrotoluen o-nitrotouen



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50/334

OqN - O H + H.+

$-* OqNH

+ H'.+

Phn ng sunfo haun nng benzen vi H2SO4 m c ng thi chng ct nc ra khi h

hp phn ng s thu c axit benzensunfonio:

Phn ng cinkyl h- Benzen tc dng vi ankylclorua c mt ca ACI

3khan s cho* ankylbenze

(phng php tng mch cacbori iu ch cc ng ng c benzen).

C 6H 6 + C nH 2 + 1 C 1 C 6H 5CH2rt + , + H C 1

- C th thay th dn xut halogen bng ancol hoc anken ankyl ha benzen

c 6 h 6 + CH 3CH 2OH > c 6h 5c h 2c h 3 + H 20

C6H6 + c h 2 = c h 2 C6 H5 CH2 CH3

"Nu anken i propilen ta s thu c sn phm chnh l cumri

C6 H6 + CH2= CH- CH 3 C6 H5CH(CH3) 2

b) Phn ng cng CngChBenzen v cc ankylbenzen khng lm mt mu dung ch brom nh c

hirocacbon khng no. Khi chiu sng, benzen cng vi o to thnh CHClt

* Oxi ha khng hon ton- Khc v'i etilen v axetiln, benzen khng phn ng vi dung dch KM11O4.

C6 H6 + HOSO3H 170c > c 6 h 5s o 3h + h 2o

(Cumen)

C 6H 6 + 3 H 2^ - C 6H 12

c) Phn ng oxi ha Oxi ha hoi ton

CH2 - 6 + 0 2 nC02 + (n- 3) H 20

5-



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51/334

- Toluen v cc ng ng khi un nng vi dung dch KMnCXt (hoK2Cr2 7) s b oxi ha mch nhnh (nhm anky) to ra mui v axit hu c.Phn ng ny dng nhn bit toluen:

C6H5CH3 +2KM n04 C6H5COOK + 2Mn024'+ KOH + H20 Nhn xt:Nu nhm ankyl vng benzen di hn nhm CH3 th phn ng

oxi ha mch nhnh vn u tin xy ra v tr a i vi vng.

C6H5CH2a-C H 2R Ky < - > C6H5COOK H3+--> C6H5COOH

4. iu ch benzen v arena) Chng c nha than v rifominh du m thu c mt lng ln

benzen, toluen v naphtaen (dng trong cng nghip)b) hiro ha ng vng texan v heptan thu c benzen v ioiuen:

CH3(CH 2)4CH3 > C6H6 + 4H2' ' 500 c,40atm

CH3(CH2)5CH3 xt;t0 p> C6HsCH3 + 4H2c) hiro ha xicloankan hoc metybcloankan

C6h 12 xt> tP > C6H6 + 3H2d) Trime ha axetilen

3C 2H 2 X P > C 6H 6

e) An ky l ha benzen iu ch ng ng ca benzen

C.U.. + CH2 + ,C1 - > C6H5CH2m + HC1

g) Cng H2vo mch nhnh khng n

C6H5CH = CH2 + H2 > C6H5CH2CH3

h) Ttrbenzen v ten iu ch etylbenzen

C 6H 6 + C H 2 = C H 2 > C 6H 5C H 2C H 3

VII. STIREN V NAPHTALEN

1. Stiren

a) cu to v tnh cht vt l

Stiren (vinylbenzen, phenyletilen) c cng thc cu to:

Stiren l cht lng khng mu, nh hn nc v khng tan trong nc,

t = 31c, t" = I45c.

b) Tnh cht ha hc Phn ng cngTong t nh anken, stiren c phn ng cng halogen (CI2, Br2), hirohalogenua

(HCl, HBr) vo nhm vinyl.

52



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52/334

C6H5CH = CH 2 + Br2 -> C6H5C B r- C H 2Br

CHCH -C H + HC1 CeHsCHCl - CH

Phn ng trng hp v ng trng hp

nCH =

Ic 6 h 5 c 6 h 5

nCH2= CH-CH = CH2 + nCH = CH2 ...xt' - p >

C* Hpolibutaienstiren) 6 5

f C H 2 -C H = C H -C H 2 - C H - C H 2 ^, I

CHs

(cao subuna-S)

Phn ng oxi ha

Ging nh etilen, stiren lm mt mu dung dch KMn4 v b oxi ha

nhm vinyl, cn vng benzen vn gi nguyn.3C6 H5CH = CH2+'2KMn04+ 4H20 -> 3C6 H5CH(OH) - CH2OH +

+ 2Mn02+ 2KOI

Phn ng trn dng nhn bit stiren.

2. Naphtaln

a) Cu to phn t v tnh cht vt

Naphtalen c cng thc phn t CioHg, c cu to bi hai nhn benzeic chung mt cnh.

Naphtalen l cht rn mu .trng, tnc = 80c, t s = 210c, thng hoa ngay f CH - CH2-)- polistiren (PS)

53



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53/334

Phn ng cng hro (hiro ha)

2H ,

Ni, 150 c

3H ,

Ni, 200c, 35 atm

*0 ecain'Phn ng oxi ha

Naphtalen khng b oxi ha bi dung dch KMn0 4 . Khi c V2O5 lmxc tc, nhit cao n b oxi ha bi oxi khng kh.

0 2 (kk)

V -A , 350 - 450c

C O \

^CO/

0

( 1)

L Bi ton v phn ng th halogen (CI2toc B2)

CnH2n +2. + ZX2 - CHn zX + zHX (1 < z< 2 n + 2)'-'ll**2n + 2 ' ^^2 r 11*12n + 2-Z-'VZ ' V* I 'V d:Khi a ng nghim cha y kh metan v clo ra ngoi nh sng th

mu vng ca kh clo b nht dn do cc phn ng:

CH + CI

CH + 2C1

CH + 3C1

CH + 4CI

-> CI3CI + HCiMonoclometan (hay metyl corua)

- > CH 2C I2 + 2 H C 1

iclometan (hay metylen clorua)

CHCI + 3HC1

Triclometan (hay clorofom)

CCI + 2HC1

Tetreclometan (hay cacbon tetraclorua)

Phn ng trn thng c dng xc nh cng thc cu to ng caankan khi bit tng s sn phm th hoc ngc li. Ch rng khi cng thc

cu to ca ankan c tm i xng th clo hoc brom vo cacbon no, tacng ch thu c mt sn phm th mono brom duy nht. V d:CH



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




54/334

V d 1. t chy hon ton mt hirocacbon X thu c 22 gam C2 v10,8 gam H2 O. Khi cho X tc dng vi CI2 theo t l moi 1 : 1 thu c sn

phm th. Tn gi ca A l

A. metan B. pentan c . neopentan D. 2,3-imetylbutan

Gii

nco 2- 0,5 mol; nn 2o = 0 , 6 mol -> X l ankan (CnH2n+2)

Hq a

________

nx = nH2o - nco = 0, mol - n = - = 5 ->CTPT ca X l C5 H 12n x

A + CI2 - Cho 1 sn phm th duy nht -> A c tm i xng >CTCT ph hp ca A l:

CH3

CH3 c CH3: neopentan > p n c

C H 3

V d 2.Hirocacbon mch h X trong phn t ch cha lin kt v c hainguyn t cacbon bc ba trong mt phn t. t chy hon ton 1 th tch Xsinh ra 6 th tch CO2( cng iu kin nhit , p sut). Khi cho X tc dngvi C2(theo t l s mol 1:1), s dn xut monoclo ti a sinh ra l

A.3. B.4. c . 2 . D.5

( th tuyn sinh i hc nm 2008 - Khoi B)Gii

Xl a nk an (C H2 *2) ^ n = - ^ - = - = 6 (C6H)n x 1

X c hai nguyn t cacbon bc b trong mt phn t

-> Cng thc ca X: (GH3)2CH-CH(GH3)2

CH3 - C H - C H - C H 3 + Ci2 CHj CHi

CH3 - C H - C C I - C H 3 + HC1CH3 CH3

CH3 - C H - C H - C H 3 + HC1c h 3 c h 212C1

>p n c2. Bi ton v phn ng tch (phn ng cracknh)

CnH2 n+ 2 ~ CmH2 m C(n-m)H2(n-m) +2 (n - m 0 m ^ 2 )b: a

p: X XX

d: a - X ( a - X > 0)

55



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




55/334

Gi X l hn hp cc ankan ban u, Y l hn hp thu c sau phn ng,ta c:

riankan p = nY- nx ; v ankan = VY- Vx

mx = m y (nh lut bo ton khi lng)

Hiu sut phn ng crackinh:

H = ibL ~ nx 1 0 0 %-= (ill-_ 1). 1 00% =f t - -1). 100% = ( -1). 100%1X nx px My

= [d(X/Y) - 1] .1 0 0 % = ( ^ - 1).1 0 0 %V\'

Do hm lng c, H X v Y l nh nhau nn t chy X hay Y ta cn smol 2nh nhau v s mol C 0 2 v H2O to ra cng nh nhau. V vy, trongkhi gii ton n gin, ta nn t chy Y thay v t chy X.

Vi d 1. Nhit phn C4H10 c hn hp Y gm CH4, C3H6, C2H6, C2H4, H2C4H8v C4Hiod. Bit My = 32,22 gam/mol, hiu sut phn ng crackinh l

A. 40% B. 80% c . 20% D. 60%

Gii Cch 1: C4 H 10 ...> C3 H6 + CH4

; X X X '

c 4 h 10 C2H4 + c 2 h 6

y y; yC4H10 - *-* - - > c 4h 8 + h 2

z z z

tnc4

H10

b = a mol - nc4

H10

d = a - (x + y + z);/ m Y m x 58a

nY = a + (x + y + z) = -ZZT- = -==A- = ^ M y M y 3 6 , 2 5

(v theo DLB TK L th ix my 58a)^ _ 21,75a

> (x + y + z) - 36,25

.. H - 3 - & y + z) .100% = 100% - (x + y + z) .100%a : : a

= 1 0 0 % - ^ - . 1 0 0 % = 60%32,22a

p n c

M c 4h 10 . 5 8 Cch 2: H = ( - I 410 - 1 ).1 00 % = - 1 ).1 0 0% = 6 0 %

M y 3 6 ,2 5

56



B

I

D

N

GT

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