bi mat de thi quoc gia tap 2 huu co

7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co

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Cu n 2)


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Cun sch ny gm 7 phn : cha tt c cc chiu hng ra thi , cc cch x !" #gii $uy%t nhanh m&t 'i t(n , )* h+c # )* hiu -

.hn /: hir(cac'(n 0 an1an 2 an13n 2 an1in 2 an1ai3n 2 '3n43n3 5

.hn 6: )n xut ha!(43n 2 r8u 2 h8p cht ph3n(!

.hn 9: an3hit 2 x3t(n

.hn : axit 2 3st3 2 !ipit

.hn ;: amin 2 amin( axit 2 p3ptit 2 pr(t3in

.hn < : cac'(nhirat

.hn 7: p(!im3


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.DRM /: DS=TUCLCFUM

CVC CDSWX DYOMZ TL =W HDS DS[TUCLCFUM

LM\LM E LM\]M E LM\SMQLM\L[S]M J F]M^]MChiu hng 1: l thuyt p

Chiu hng 2: bi tp vpnhit phn ankan ( tch loi H2, crckinh)

Chiu hng 3: bi tp vpt chyChiu hng 4: bi tp vpcng ( H2, X2, HX, H2O )

Chiu hng 5: bi tp vpthion kim loi ha tr1 (AgNO3/NH3 CuCl/NH3) ca ankin

NgNgNgNgy thy thy thy thnhnhnhnh ::::tttt HHHH_Y BY BY BY BTTTT =RUUUU

Ci g ko lm bn khut phc ci sto nn con ngi bn !

CDSWX DYOMZ /: Na HDXbH .DdM eMZ

Mguyfn tc h+c !" thuy%t1).BN KO THGHI NHHT L THUYT M!T LC "#C ? chnh v vy khi hc l thuyt chcn cc

bn c hiu v tm tt li l thuyt vi ln c ci tng quan trong u m tduy, cn vic ghi nhth cc

bn clm nhiu bi tp kin thc sttkhc su vo u.!n bn nu kh"ng nhc kin thc vo mt

l#c v n qu nhiu th cgisch ra $%m thoi mi. & phi lm nhiu n!n phi $%m nhiu .'%m nhiu th s

nhth"i.

( )t nht l vstduy snhnhanh c l thuyt trong thi gian ngn * chmt + ngy cc bn c th

nhc ht. &nhthno * gi in t"i hng dn.

( goi ra cn mt cch ghi nhc l thuyt na cng chmt + ngy, cc bn ginhng trang cui ca

cun sch ra v lm th%o hng dn .

2).NGUYN L CON CHIM:

C mt con chim bnht trong mt ci l$ng ,trong ci l$ng c c 100 ci l%,nhng ch&c mt ci

l%cha thc n v nc ung. Khi con chim i ,v sinh t$n b't buc phi con chim phi thtrc m


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vo 100 ci l% xem c thc n ha !h"ng. #au nhiu ln trc mvo th, cui c$ng n c%ng tm

ra (c l%cha thc n. & !)t*ln sau m%i !hi con chim i n stm n 'ng ci l%cha thc n

lu"n m !h"ng cn phi thnghim tm !im nhln trc.

(m tr'c nghim c%ng v !hi ) tim ra !t *uth s!h"ng bao gi*u+n p n. l -, ,C ha

l / .C iu cc bn c%ng phi ging nhcon chim ttm ti ra th mi nh(c.

LM\LM1)phn ng xi ho

1.1)hon ton (t chy )

CnH2n+2 + O2 CO2 + H2O

VD: C3H8 + O2CO2 + H2O

1.2)ko hon ton

CH4+ O2 HCHO + H2O

(anhit focmic)

CH3CH2CH2CH3 + 5 2 O2 CH3COOH + H2O

(axit axetic)

2)phn ng thhalogen

* theo cchtdo

* u tin thvo nguyn t!cacbon bc cao

CnH2n+2 + (2n+2)F2 nC + (2n+2)HF

CnH2n+2 + kX2(:)/ CnH 2n+2-k X k + k HX

( X2= Cl2, Br2, I2 - nguyn cht )

VD: C3H8 + Cl2(:) C3H7Cl + HCl

CH3-CH-CH3 + HCl (sp chnh )

CH3-CH2-CH3+ Cl2(:) Cl

CH3-CH2-CH2-Cl+ HCl (sp ph")

3)phn ng nhit

3.1)phn ng phn hu+

CnH2n+2

nC + (n+1)H2

VD: CH4

C + 2H2

2CH4

C2H2 + 3H2

(axetilen)

3.2)phn ng tch loi hir

CnH2n+2

CnH2n + H2. VD : C3H8

C3H6 + H2

CnH2n+2

CnH2n-2 + 2H2. VD : C3H8

C3H4 + 2H2

3.3)phn ng cr'ckinh

Ankan (c,)

Ankan(mi)+ Anken

VD: C4H10

CH4 + C3H6

VD: C4H10

C2H6+ C2H4

CU H-I L THUYT LIN QUAN N P.TH

Cu 1-A-2013:Khi c chiu sng hidrocacbon no sau y tham gia phn ng

thvi clo theo tlmol 1:1, thu c ba dn xut monoclo l ng phn cu to

ca nhau.

A.neo pentan B.pentan C.butan D.isopentan

Suy lun :

Hi)u thnht: bi ny ta cn khai trin tn gi ra cng thc cu to ri vit

phng trnh pvi Cl2theo tl1:1 , ci no cho 3 sn ph#m th ly.

Hi)u thhai :nhthno l pth?pthl pthay thnguyn t! ny b$ng

nguyn t!kia . Thhalozen (Cl2) theo tl1: 1 c ngh%a l 1 ng t!Cl sthay th

1 ng t!H trong ankan - u c H th Cl u c ththay thvo

Hi)u th3:c nhng v&tr thging nhau th n chcho cng mt sn ph#m (ging nhv&tr i xng thbn ny th thi bn ka)

Xt p n A: CH3

CH3 C CH3 + Cl2: c 4 v&tr thnhng chcho cng 1 sp

CH3

Xt p n B:

CH3-CH2-CH2-CH2-CH3+ Cl2:

Xt p n C:

CH3-CH2-CH2-CH3 + Cl2: c 4 v&tr thnhng chcho 2 sp

Xt p n D:

CH3-CH-CH2-CH3 + Cl2:

CH3

Cc bn xem cch vit ph/ng trnh pth0phn l thuyt ankan

Cu 2 :Ankan X c CTPT C6H14. Khi cho X tc d"ng vi clo trong k chiu

sng thu c ti a 3 dn xut monoclo . Hy cho bit X l cht no?

A.neo hexan B.so hexanC.3 metylpentan D.2,3-imetyl butan

Suy lun :bi ny lm tng tbi trn , vi ca tc gil C6H14 c rt nhiu

cng thc cu to , cc bn phi chn cng thc cu to no khi pvi Cl2

theo tl1:1 cho 3 sn ph#m monoclo

Ti sao bit n xy ra pc theo kiu tl1:1 ? l'may 1:2; 1:3 th sao ?

V bi cho thu c sn ph#m mono clo tc l sp cha 1 nguyn t!Cl nn

n phi tham gia ptheo tl1:1 mi cho ra sp cha 1 ngt!Cl ; nu l 1:2 th s

cho ra sp cha 2 ng!Cl ( xem li l thuyt ankan)

Cch lm :

Xtp n A: CH

3

CH3- C - CH CH3 + Cl2: 3 sn ph#m

CH3

Xt p n B:

CH3-CH-CH2-CH2-CH3 + Cl2: 5 sn ph#m

CH3

C 5 v&tr thnhng v&tr s2 v 4i xng nhau nn chtnh 1 sp, v&tr1 v 5 i xng nhau nn c(ng chtnh1 sp v v&tr s3 tnh 1 sp . Vy tngssp do cng thc ny to ra l 3 sp

C 5 v&tr thnhng c 2 v&tr CH3i xng nhau nn chtnh mt snph#m , cn cc v&tr khc m)i v&trcho 1 sp na tng l 4 sp


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VD: C3H8

CH4 + C2H4

I1U CH

1) i thircacbon ko no th cng vi H2

CnH2n + H2 CnH2n+2 VD: C3H6+ H2 C3H8

CnH2n-2+ 2H2 CnH2n+2 VD : C3H4+ 2H2 C3H8

2) i tdn xut monohalogen

2CnH2n+1Cl + 2Na (CnH2n+1)2+ 2NaCl

VD : 2CH3Cl + 2Na C2H6 + NaCl

3) i tmui ca axit cacboxylic

R(COONa)X+ NaOH!

RHx + Na2CO3

VD: CH3COONa+ NaOH!

CH4 + Na2CO3

HCOONa + NaOH!

H2 + Na2CO3

2 C34 5 K36 32C H3U C4 TH H3U C4

t phn ng h/n v c/C I1U khi bn hc l thuyth5u c/th c1 pbn phi n'm (c 3 dng vit pt ca

n. Mt l CCH VIT 0dng t6ng qut )lm bi tp

lin quan n xc nh CTPT, hai l 0dng phn t)

lm bi tp lin quan n tnh ton khi l(ng s mol,

th) tch., ba l CCH VIT 0dng CTCT )lm bi

tp lin quan n l thuyt p

Xt p n C :

CH3-CH2-CH-CH2-CH3 + Cl2: 4 sn ph#m

CH3

Xt p n D:

CH3-CH CH CH3 + Cl2:

CH3 CH3

Cu 3 :Ankan X l 1 cht kh nhit thng . Khi cho X tc d"ng vi clo

(as) th thu c 1 dn xut monoclo v 2 dn xut iclo. Hy cho bit X l chtno sau y:

A.metan B.etan C.propan D. butan

Suy lun: tc gicho bi ny l mun cc bn tm cht no trong cc p n

m khi tc d"ng tc d"ng vi Cl2theo tl1:1 cho ra 1 sn ph#m monoclo

V theo tl1:2 cho ra 2 sn ph#m i clo

Cch lm :

Xt p n A:

CH4 + Cl2

:

CH3Cl + HClCH4+ 2Cl2

:" CH2-Cl + 2HCl

Cl

Xt p n B: th*a mn

CH3-CH3 + Cl2: CH3-CH2Cl

CH3-CH-Cl

CH3-CH3+ Cl2:" Cl

CH2-CH2

Cl Cl

Xt p n C th propan: CH3-CH2-CH3ptheo tl1:1 cho 2 sp cn tl1:2 cho

4 sp

Xt p n D th butan: CH3-CH2-CH2-CH3ptheo tl1:2 cho 2 sp cn theo t

l1:2 cho 6sp

Cu 4 :X c CTPT l C6H14. Khi cho X tc d"ng vi clo trong k chiu sng

theo tl1 : 1 thu c 4 dn xut monoclo . Hy cho bit X l cht no

A.neo hexan B.iso hexan

C.3 metypentan D.2,3 imetylbutan

Cch lm : cu 4 lm ging cu 2

CU H-I L THUYT LIN QUAN N P.1HIRO HA

Cu 5 :Khi thc hin ptch 1 phn t!H2tisopentan th thu c bao nhiu

anken?

A.2 B.3 C.4 D.1

Suy lun: CH3 - CH-CH2-CH3 anken + H2

CH3

ptch loi H2th khng lm thay i cu trc mch ccbon vy cu trc mch

cacbon ca anken phi l C C C C

CH3

Suy lun tip : l anken l phi c ni i nn cc bn in ni i vo mch

ccbon ta sbit c c bao nhiu anken c thto ra

CH2= C CH2 CH3 CH3 C = CH CH3 CH3 CH CH = CH2

CH3 CH3 CH3

Cu 6 :hiro ho ankan A thu c isopren . Hy cho bit A l cht no:

4 v&tr CH3cho cng 1 sp; 2 v&trCH cho cng 1 sp na tng l 2san ph#m


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A.2 metylpentan B.2 metylbutan

C.2 metylpropan D.iso butan

Suy lun: Ankan CH2= C CH= CH2 + H2

CH3

P hiro ha hay cn gi l ptch loi H2th khng lm thay i cu trc

mch cc bon nn cu trc mch cacbon ca ankan phi l

CH3 CH CH2 CH3

CH3

NgNgNgNgy thy thy thy thhaihaihaihai:::: khkhkhkhngngngng i thi thi thi thkhkhkhkhng bao ging bao ging bao ging bao gi !!!!nnnn

LM\]MCnH2n ( n72 ; c 18t/ng ng vi 1 ni = )

1). PH9N.NG OXI HA

1.1)hon ton (t chy )

CnH2n + O2 CO2 + H2O

VD : C3H6 + O2 CO2 + H2O

1.2)ko hon ton ( lm mt mu dung dch KMnO4)

CnH2n + KMnO4 + H2O : CnH2n(OH)2 + KOH + MnO2;(en)

VD : C3H6 + KMnO4 + H2O C3H6(OH)2 + KOH + MnO2+

CH2=CH-CH3 + KMnO4 + H2O CH2 CH - CH3+ KOH +MnO2 +

OH OH

2) PH9N .NG C!NG

a) cng H2 (xc tc : Ni. Pt, Pd )

CnH2n + H2 CnH2n+2.

VD : C3H6+ H2 C3H8

CH2=CH-CH3+ H2 CH2 CH CH3 ( hay CH3 CH2 CH3)

H H

CU H-I L THUYT LIN QUAN N P.C!NG

Nhthno (c gi l pcng ? pcng l ptn cng vo ni = hoc ni


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b )cng X2( Br2, Cl2, I2)

CnH2n + X2 : CnH2nX2

VD : C3H6 + Br2(dd) C3H6Br2

CH2=CH-CH3 + Br2(dd) CH2 CH CH3

Br Br

Ch : pti to li anken t*dn xut i halozen

CnH2nX2 + Zn : CnH2n + ZnX2

VD : C3H6Cl2 + Zn C3H6 + ZnCl2

VD: C2H4Br2 + Zn C2H4 + ZnBr2

c ) cng HX (tun theo quy tc mocopcnhicop)

CnH2n + HX : CnH2n+1X

VD : C3H6 + HBr C3H7Br

CH2 CH CH3 (hay CH3 CH CH3 )

CH2=CH-CH3+ HBr H Br Br

CH2 CH CH3 (hay CH2 CH2 CH3 )

Br H Br

Ch : pti to anken

CnH2n+1X + KOH CnH2n+ KX + H2O

VD: C3H7Cl + KOH C3H6 + KCl + H2O

d )cng H2O (x/t axit) - tun theo quy tc maccopnhicop

CnH2n + H2O CnH2n + 1OH (tn r(u no /n chc)

VD : C3H6 + H2O C3H7OH

CH2 CH CH3 (hay CH3- CH CH3) chnh

CH2=CH-CH3+ H2O H OH OH

CH2 - CH CH3 (hay CH2 CH2 - CH3) ph"

OH H OH

Ch : pti to li anken t*r(u no /n chc

CnH2n+1OH#$%&'! *

*

CnH2n + H2O

VD: C3H7OH+,-.!

C3H6 + H2O

3).PH9N .NG TRNG H#P

nCH2 =CH2 ( CH2 CH2 )n poli etilen hay P.E

nCH2=CH-CH3 ( CH2 CH )n poli propilen hay P.P

CH3

4).Ch : pc bit to etilen oxit

CH2= CH2 + 0 2 O21!

CH2 CH2 (ETILEN OXIT)

O

CH2=CH-CH3+ O21!

CH2 CH - CH3 propilen oxit

O

Cch lm :thng khi anken tham gia pcng HX ta sthu c 2sp (chnh v

ph") nhng nu 2 sp ny trung nhua th chcn 1 bi ny ta i xt tng p n

lm

Xt p n A: CH3-CH2-CH CH2

CH3-CH2-CH= CH2 + HCl H Cl

CH3-CH2-CH CH2

Cl H

T/ng tp n B: CH3-CH2-CH=CH-CH3cho 2 sp

p n C: CH3-CH CH=CH2c,ng cho 2 sp

CH3

Ch&c p n D cho 1 sp v: CH3-CH-CH-CH3

CH3 CH = CH CH3 + HCl : H Cl

CH3-CH-CH-CH3

Cl H

Trng h(p ny ch&cho 1 sp v 2 CTCT trng nhau quay ng(c li l th t*

cng thc ny n sra cng thc kia

Cu 3-B-2012:Hirat ha 2-metylbut-2-en (iu kin nhit , xc tc thch hp)

thu c sn ph#m chnh lA. 2-metylbutan-2-ol. B. 3-metylbutan-2-ol.

C. 3-metylbutan-1-ol. D. 2-metylbutan-3-ol.

Cch lm :Hirat ha l pcng H2O vo 2 metylbut 2-en

OH

CH3 C = CH CH3 + H2O CH3 C CH2 CH3

CH3 CH3

Sp chnh OH nh C bc cao

Cu 4-A-2010:Anken X hp nc to thnh 3-etyl pentan-3-ol. Tn ca X l

A. 3-etylpent-1-en. B. 2-etylpent-2-en.C. 3-etylpent-3-en. D. 3-etylpent-2-en.

Suy lun : OH

Anken + H2O CH3 CH2 C CH2 CH3

C2H5

Suy lun theo cu trc mch cacbon nhcu 7 ta sc anken cn xc &nh l

CH3 CH2 C = CH CH3

C2H5

C gng !fn

2 C34 5 K36 32C H3U C4 TH H3U C4

t phn ng h/n v c/C I1U khi bn hc l thuyt h5u c/th c1 pbn phi n'm (c 3 dng vit pt ca n.

Mt l CCH VIT 0dng t6ng qut )lm bi tp lin quan n xc nh CTPT, hai l 0dng phn t)lmbi tp lin quan n tnh ton khi l(ng smol, th) tch., ba l CCH VIT 0dng CTCT )lm bi tp lin

quan n l thuyt p


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LM\L=S]M:CnH2n-2 ( n73 ; c 28t/ng ng vi 2 ni = )1) PH9N.NG OXI HA

1.1) Hon ton CnH2n-2+ O2CO2+ H2O .

1.2) Ko ho tan: lm mt mu dung d&ch thuc tm KMnO4

2) PH9N .NG C!NG

Do ankadien c 2 lk ,nn n c thcng theo tl(1:1) ph vi mt lk , ho-c

(1:2) ph v2lk ,

a ) cng H2: CnH2n-2+ H2 CnH2n (xc tc : Ni, Pt, Pd )

CnH2n-2+ 2H2 CnH2n+2

VD 1 : C4H6+ H2 C4H8

CH2=CH-CH=CH2+ H2 CH2= CH CH CH2 (hayCH2=CH CH2-CH3)

H H

Cng (1,2) xy ra nhit thp

CH2=CH-CH=CH2+ H2 CH2 CH = CH CH2(hay CH3CH=CHCH3)

H H

Cng (1,4) nhit caoVD 2:

C4H6+ 2H2 C4H10

CH2=CH-CH=CH2+ 2H2 CH2CH-CH-CH2 (hay CH3-CH2-CH2-CH3)

H H H H

b ) cng X2: CnH2n-2+ X2 CnH2n-2X2 ( dn xut ihalozen )

CnH2n-2+ 2X2 CnH2n-2X4 ( dn xut tetra halozen )

VD1: C4H6 + Br2(:) C4H6Br2

Thhin dng cu to

CH2=CH-CH=CH2+ Br2(34!") CH2= CH CH CH2

(Buta-1,3-ien) Br Br

CH2=CH-CH=CH2+ Br2(34!6) CH2 CH = CH CH2

Br Br

CH2= C = CH CH3 + Br2 CH2 C = CH CH3

(buta-1,2-ien) Br Br

CH2= C = CH CH3 + Br2 CH2= C CH CH3

Br Br

VD2 : C4H6+ 2Br2 C4H6Br4

CH2=CH-CH=CH2+ 2Br2 CH2 CH CH CH2

Br Br Br Br

Ch : pti to li ankaien tdn xut i halozen

CnH2nX2 + 2KOH78 CnH2n-2 + 2KX + H2O

VD : C3H6Cl2+ 2KOH78 C3H4 + 2KCl + H2O

c ) cng HX----theo quy tc maccopnhicop

CnH2n-2+ HX CnH2n-1X

CnH2n-2+ 2HX CnH2nX2

VD: C4H6+ HCl9 C4H7Cl

Cu 1 :Hiro ho ankaien X thu c 2,3 i metyl butan . X l :

A.2,3 imetylbuta-1,3-ien B.2,3 imetyl penta 1,3 ien

C.2,3 imetyl buta 1,2 ien D.isopren

Suy lun :

Ankaien X + H2 CH3 CH CH CH3

CH3 CH3

Pcng khng lm thay i cu trc mch cacbon nn cu trc mch ca

ankaien sl C C C C

CH3 CH3

l ankaien l phi c 2 ni i in 2 ni i vo cu trc mch trn ta s

c ankaien cn tm : CH2= C - C = CH2

CH3CH3

2,3 imetyl buta-1,3-ien

Cu 2-A-2012: Hiro ha hon ton hirocacbon mch hX thu c

isopentan. Scng thc cu to c thc ca X l

A. 6. B. 7. C. 4. D. 5.

Suy lun : X + H2 iso pentan ( CH3 CH CH2 CH3)

CH3Chng t*X phi l hirocabon khng no v cu trc ca X phi l

C C C C

CH3

TH1 : X c thl anken

CH2= C CH2 CH3 CH3- C = CH CH3 CH3- CH CH = CH2

CH3 CH3 CH3

TH2: X c thl ankaien

CH2= C CH = CH2 CH3- C = C = CH2

CH3 CH3

TH3 : X c thl ankinCH3 - C C .CH

CH3

TH4: X l mt hiocacbon khng no bt k th*a mn cng thc ha tr&

CH = C C .CH

CH3

Cu 3-A-2012:hirocacbon no sau y khi pvi dung d&ch brom thu c

1,2- ibrombutan ?

A. But-1-en B.BUtan C.Buta-1,3-ien D.But 1 in.

Suy lun :bi ny chng ta i khai trin tn gi ra CTCT ri cho pvi dung

d&ch Br2CTCT no khi pvi Br2m to ra 1,2- ibrm butan :

CH3 CH2 CH CH2

Br Br

p n ng A v CH3 CH2 CH = CH2 + Br2CH3 CH2 CH CH2

Br Br

Cu 4-A-2011:Cho buta-1,3-ien phn ng cng vi Br2theo tlmol 1:1.

Sdn xut ibrom (ng phn cu to v ng phn hnh hc) thu c l

A. 2. B. 4. C. 1. D. 3.CH2=CH-CH=CH2+ Br2

(34!") CH2= CH CH CH2

(Buta-1,3-ien) Br Br

CH2=CH-CH=CH2+ Br2(34!6) CH2 CH = CH CH2 ( CT ny c p

Br Br cis v trans)


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CH2= CH CH CH2

H Cl

CH2=CH-CH=CH2 + HCl9 CH2= CH CH CH2

Cl H

CH2 CH = CH CH2

H Cl

d ) cng H2O (x/t axit H+)theo quy tc maccopnhicop

CnH

2n-2+ H

2O C

nH

2n-1OH

CnH2n-2+ 2H2O CnH2n (OH)2

3.PH9N .NG TRNG H#P

nCH2=CH-CH=CH28 ;< = ( CH2-CH=CH-CH2 )n

(buta-1,3-ien) cao su buna

nCH2=CH-CH=CH2>? @;< A ( CH2 - CH )

CH=CH2

nCH2=C CH=CH

2 ( CH

2-C = CH-CH

2) n

CH3 CH3

(iso pren) cao su thin nhin ( poli isopren.)

Cu 5 :Isopren pcng vi Br2th thu c bao nhiu sn ph#m hu c?

A.8 B.7 C.6 D.5

Cch lm:tnh ctl1:1 v1:2 p n ng l D

CH2= C CH(Br) - CH2(Br)

CH3

CH2= C CH = CH2 + Br2 CH2(Br) C(Br) CH = CH2

CH3 CH3

(Br)CH2- C = CH - CH2(Br) ( cis v trans)

CH3

CH2= C CH = CH2 + Br2(:") CH2(Br) C(Br) CH(Br) CH2(Br)

CH3 CH3

LM\SM:CnH2n-2 ( n72 ; c 28t/ng ng vi 1 ni


10/172

Ch : pti to li ankin tdn xut i halozen

CnH2nX2 + 2KOH78 CnH2n-2 + 2KX + H2O

VD : C3H6Cl2+ 2KOH78 C3H4 + 2KCl + H2O

C2H4Br2 + KOH78 C2H2 + 2KCl + H2O

c ) cng HX (tun theo quy tc m/ccopnhicop)

CnH2n-2+ HX CnH2n-1X

CnH2n-2+ 2HX CnH2nX2

VD1 : C3H4 + HCl(:) C3H5Cl

CH3- C = CH2 ( sp chnh )

CH3-C.CH + HCl(:) Cl

CH3- CH = CH (sp ph")

Cl

VD 2:C3H4 + 2HCl(:") C3H6Cl2

d ) cng H2O :(tun theo quy tc m/ccpnhicp)

VD 1 : CH.CH + H2O +,K

/+,-. CH2= CHL@M4? @

CH3CHO

OH

( ko bn)

VD2:

CH3 C .CH + H2O+,K/+,-. CH3 C = CH2

L@M4? @ CH3CO-CH3

OH

( ko bn)

CH3 C .CH + H2O+,K/+,-. CH3-CH=CH

L@M4? @ CH3 CH2CHO

OH(ko bn)

3).PH9N .NG TRNG H#P

2CH.CH CH2= CH C .CH

( vinyl axetilen)

3CH.CH C6H6 ( benzen )

4) phn ng thion kim loi ( pvi AgNO3/ NH3hoc CuCl/ NH3)

Ch : chc nhng ankin c ni 3 u mnh mi tham gia phn ng ny

CH.CH+ 2AgNO3+ 2NH3 CAg.CAg++ NH4NO3(vng )

( C2H2 + 2AgNO3+ 2NH3 C2Ag2++ NH4NO3

CH3 C .CH + AgNO3+ NH3CH3 C .CAg + + NH4NO3

( vng )

(C3H4 + + AgNO3+ NH3C3H3Ag + + NH4NO3 )

CH.CH + 2CuCl + 2NH3 CCu .CCu ++ NH4Cl

(*gch )

Ch :phn ng ti to li ankin tsn ph#m ca phn ng thion kim loi

ha tr&1

AgC


11/172

NgNgNgNgy thy thy thy th3 :3 :3 :3 : CCCCdNH BNH BNH BNH BV !!!!OOOO

TH>C TRNG KH?C PH@C

DY NG NG BENZEN HP CHT THMA.Khi nim v cch c tn :

1.Khi nim:

-Dy ng 0ng ca benzen cn gi l azen l hp cht thm khng no

mch vng cha nhn benzen L

C CT tng qut : CnH2n-6( n 16)

2.Tn gi :

C6H6 : : benzene

C7H8: CH3 : metyl benzene ( toluen)

C8H10: C2H5 CH3: etyl benzen CH3

*1,2 i metyl benzene

* 0 i metyl benzene

* 0 xilen

Cu 1-B-2014:Cho cc cht sau: etilen, axetilen, phenol (C6H5OH), buta-1,3-

ien, toluen, anilin. Scht lm mt mu nc brom iu kin thng l

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

Tli :nhng cht lm mt mu dung d&ch nc Br2gm C2H4etilen; C2H2axetilen;

Phenol ; CH2=CH-CH=CH2 buta-1,3-ien ; anilin C6H5NH2 ( xem pphn amin )

Cu 2-A-2012:Cho dy cc cht: cumen, stiren, isopren, xiclohexan, axetilen,

benzen. Scht trong dy lm mt mu dung d&ch brom l

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

Trli :C6H5CH=CH2(stiren) ; CH2= C(CH3) CH =CH2(isopren) ;

C2H2(axetilen)

Cu 3-A-2011: Cho schuyn ha sau

Benzen + C2H4/ xt,t0 X + Br2/ as.Tl1:1 Y + KOH/C2H5OH Z X,Y,Z

l sn ph#m chnh )

Tn gi ca Y, Z ln lt lA. 1-brom-1-phenyletan v stiren. B. 1-brom-2-phenyletan v stiren

C. 2-brom-1-phenylbenzen v stiren D. benzylbromua v toluen


12/172

CH3 * 1,3 i metyl-benzen

*. m-imetyl benzen

CH3 *.m-xilen

CH3 * 1,4 i metyl-benzen

*. p-imetyl benzen

*.p-xilen

CH3

Ch : C8H8: CH=CH2 : vinyl benzen

(stizen)

B. TNH CHAT HA H2C

1.Poxi ha Hon ton ( t chy )

Azen + O2CO2+ H2O

Khng hon ton( lm mt mu KMnO4) (tr*benzene)

R COOK COOH

+KMnO4 + H3O+(axit)

VD: CH3 COOH

CH3 + KMnO4 + H2SO4 COOH + K2SO4 + MnSO4+ H2O

2.P"thnhn bezen

Nguyn t'c th:

*) Nu trn nhn benzen nh cc gc #y e nh:Gc R no; -OH ; -NH2;

-OCH3; -X khi t/g p/thn u tin thvo v&tr O ho-c P

*) Nu trn nhn benzen c nh cc nhm ht e nhgc R ko no : -NO2;

CHO ; COOH ; -COOkhi t/g p/thn u tin vo v&tr m

a) Pthhalogen ( nguyn cht ) c Fe xc tc mi xy ra

Cl

+ Cl2Q (:) + HCl

CH3

CH3 Cl

+ Cl2Q (:) + HCl

CH3

Cl

Cch lm: CH2CH3

C6H6 + C2H4R! J C6H5C2H5 ( hay )

CH2-CH3 CH(Br)CH3

+ Br2!S TU : + HBr

CH(Br)CH3 CH=CH2

+ KOH ,+E+ + KBr + H2O

Cu 4-B-2011:Cho cc pht biu sau:

(a) Khi t chy hon ton mt hirocacbon X bt k, nu thu c smol

CO2 b$ng smol H2O th X l anken.

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

(c) Lin kt ho hc chyu trong hp cht hu cl lin kt cng ho tr&.

(d) Nhng hp cht hu ckhc nhau c cng phn t!khi l ng phn ca nhau.

(e) Phn ng hu cthng xy ra nhanh v khng theo mt hng nht &nh.

(g) Hp cht C9H14BrCl c vng benzen trong phn t!.Spht biu ng l

A.2 B5 C.4 D.3Suy lun :

a) Pht biu ny cha chnh xc v c thl xiclo ankanb) ngc) ngd) Cha chnh xc v d"nhC2H5OH v HCOOH u b$ng 46e) Sai : v n xy ra chm v theo cchxc &nhf) Sai v k=2 ; nu cha nhn benzene k =4 ( K l slk ,ho-c vng )

Cu 5)BenzenB+FG(:) A1

BC>,/ Q (:) A2

A2l :

A.1-nitro-3-brom benzen B.1-brom-4-nitro benzen

C.m-brom nitro benzen D.p-brom nitro benzen

cch lm: NO2

+ HNO3+,-. + H2O

NO2 NO2

+ Br2QP(:) Br + HBr

Cu 6:CH3COONa metan axetilen benzen tolulen axit

benzoic natri benzoat benzen

Cch lm :

CH3COONa + NaOH CH4 + Na2CO3

2CH4POOF C2H2 + 3H2 (C2H2hay CH.CH)

3C2H2>? @7A C6H6

C6H6+ CH3Cl1TTG C6H5CH3 + HCl ( xem phn iu ch benzene )

CH3

hay ( + CH3Cl1TTG + HCl ))

CH3 COOH

+ KMnO4 + H2SO4 + K2SO4+ MnSO4+ H2O


13/172

CH = CH2 CH = CH2

+ Cl2Q (:) Cl + HCl

a) Thnitro (-NO2)

NO2

+ HNO3+,-.VL (:) + H2O

CH3

CH3 NO2

+ HNO3+,-.VL (:) + H2O

CH3

NO2

CH = CH2 CH = CH2

+ HNO3+,-.VL (:) NO2 + H2O

3.P"thhalogen vo nhnh (nh sng lm xc tc )CH2 CH3 CH2 CH2(Cl)

+ Cl2?@ ?

+ HCl

CH(Cl) CH3

4. P.C!NG PH VBLIN KT 8Benzen v dy ng 0ng benzen rt kh tham gia phn ng cng chtham gia

vo phn ng cng vi H2v Cl2

C6H6 + 3H2F3 C6H12 (xiclohexan)

+ 3H2

C6H6 + 3Cl2 C6H6Cl6 ( thuc trsu 6.6.6)

III). I1U CH1.iu chbenzene

3C2H2>? @7A C6H6 ( ) )

COOH COONa

+ Na + H22

COONa

+ NaOH + Na2CO3

Cu 7:C2H5COONa C2H6C2H2C6H6toluenTNT

Cch lm :

C2H5COONa + NaOH C2H6 + Na2CO3

C2H6J!R C2H2 + 2H2 ( xem ptch loi H2ca ankan )

3C2H2>? @7A C6H6


CH3

hay ( + CH3Cl 1TTG + HCl ))

CH3CH3

+ 3HNO3+,-.VL (:W) NO2 NO2 + 3H2O

NO2 (T.N.T 2,4,6 tri nitro toluene)

Cu 8 : vi vi sng t en axetilen benzentoluen

C6H5CH2Br2C6H5CH2OH

Cch lm :

CaCO3J CaO + H2O

CaO + CJ CaC2 + CO

CaC2 + H2O Ca(OH)2 + C2H22

3C2H2>? @7A C6H6


CH3

hay ( + CH3Cl1TTG + HCl ))

C6H5CH3 + Br2 C6H5CH2Br + HBr

(xem phn thhalozen vo nhnh )

C6H5CH2Br + NaOHJ C6H5CH2OH + NaBr

Cu 9) C2H2C2Ag2C2H2CH3CHO

C2H4Br2C2H2C6H6C6H6Cl6

Fc ti%p n(


14/172

2.iu ch$ng Cng bezen

C2H5

+ C2H5Cl1TTG + HCl

CH3

+ CH3Cl1TTG + HCl

NgNgNgNgy thy thy thy th4444 :::: hhhh+c khuyac khuyac khuyac khuyaLp 12 hc chnh ri hc thm ngn ht thi

gian ca bn . Ch c b!"i m c#c bn mi c nhi$!

thi gian % &n t'p c &n t'p mi ()ng *in th+c

,-c.

n /0ng ca % th+c tnh gi3c ng 4 *ch thch

n5 ph#t 6inh /5pamin7 (8 g9!tamat7 . : gi;p bn

tnh t#5 minh mh /'? @/0 ng t nh,ng 6#ng /'? r3t *hA7 . B3n $ 98 chD

*hi ng bn ng mEt nhEm nh,ng n5 thF (


15/172

C3H8pJ!R C3H6+ H2

x x x

C3H8 pJ!R CH4+ C2H4

y y y .

C3H8(cn d) C3H8(cn d)0,02 0,02

H= 90% nZW[\(phn ng) = x + y = 0,18(mol)

nZW[\(d)= 0,2 0,18 = 0,02(mol)

Ta c: msau = mtrc= 8,8 (g)

nsau= x + x + y + y + 0,02= 2 . 0,18 + 0,02 = 0,38=> Msau =

\!\!W\

Bi 2:Nhit phn 8,8(g) propan thu c h)n hp kh X gm CH4, C2H4, C3H6, H2, C3H8. Tnh khi lng oxi cn dng t chy va ht

X, bit c 30% lng propan b&nhit phn. A)16(g) B)32 (g) C)8 (g) D) 4 (g)

Cch lm : Thay v i t chy h)n hp sau phn ng th ta i t C3H8ban u kt qutnh ko thay i v snguyn t!C , H trc vsau pko thay i C3H8 + 5O2 3CO2 + H2O

0,2 1 mol

m]"= 1 . 32 = 32 (g)

Bi 3-B-2011:Cho butan qua xc tc (nhit cao) thu c h)n hp X gm C4H10, C4H8,C4H6v H2. Tkhi ca X so vi butan l

0,4. Nu cho 0,6 mol X vo dung d&ch brom (d) th smol brom ti a phn ng l

A. 0,24 mol. B. 0,36 mol. C. 0,60 mol. D. 0,48 mol

C4H10 C4H8+ H2x x x

C4H10 C4H6+ 2H2y y y

C4H10(cn d) C4H10(cn d)z z

C4H10 h2X C4H8 + Br2 C4H8Br2

x xMX= 23,2 C4H6 + 2Br2 C4H6Br2nX = 0,6 y 2y

C4H10(cn d)z

Ta c : Mtrc = nsau

Msau ntrc

=>\

"W!" =

!^

?_`abc => ntrc = 0,24

=> nd6e0f(b)= x + y + z = 0,24 (1)

ng"

X = x + x + y + 2y + z = 0,6

(2)

y l bi ton th#n thiu phng trnh nn ta mnh dn lp tip

mt pt theo smol Br2cn tnh: nhi"= x + 2y = ??? ri x!l pt (1)v (2) sc pt cn tnhLy (2) tr(1) x + 2y = 0,36

Vy nhi"= x + 2y = 0,36

CDSWX DYOMZ 9:

FS Hj. kW .e =H CDVbCxHy + O2 : CO2 + H2O

LBT: nO (j") = nO (Zj") + nO (["O)nC (Zk [l) = nC (Zj")nH (Zk [l) = nH (["O)

C =mZj2

?Zn [o ; H =

"m[2j

?Zn [o

1) KHDI L"#NG BNH TENG

Ankan : CnH2n+2 + O2 nCO2 + (n+1) H2OAnken : CnH2n + O2 nCO2 + n H2OAnkin / Ankaien : CnH2n-2 + O2 nCO2 + (n-1) H2O

1. - Nu t chy 1 hirocacbon m thu (c np$O > nqr$ n l ankan. Ta c :

nankan = n["O - nZj" C =mZj2

?stust

- Nu t chy 1 hirocacbon mch h=m thu (c np$O =

nqr$ n c 1 8 n l anken. Ta c

C =mZj2

mstuvt

- Nu t chy 1 hirocacbon m thu (c nqr$> np$O n c 2 8 tr0ln . Mc nh 28:n l ankin/ ankaien. Ta c

nankin = nZj"- n["O wx =m,

?yz{|z


16/172

2) S>TENG GI9M KHDI L"#NG DUNG DFCH

2. t h%n h(p g$m Ankan v Anken : th sthu c

n["O > nZj"n["O - nZj" = nankan

3. t h%n h(p g$m Ankin v Anken : th sthu c

nZj"> n["OnZj"- n["O = nankin

4. t h%n h(p g$m Ankin v Ankan : th sthu c :n["O > nZj" nu nankan> nankinn["O = nZj" nu nankan= nankinnZj"> n["O nu nankin > nankan

Bi 1 )t chy hon ton mt h)n hp gm 2 hidrocacbon ktip nhau trong dy ng 0ng dn sn ph#m chy qua H2SO4-c v KOH

dthy khi lng bnh t/ng ln lt l 5,4 gam v 7,92 gam. Xc &nh 2 hidrocacbon v khi lng ca chng

Cch lm :

m["j = 5,4 (g) 0,3 (mol) nhn thy n["j > nZj"=> n l 2 ankanmZj"= 7,92 (g) 0,18 (mol) CnH2n+2=0,3 0,18 = 0,12

C =n = !\!"

= 1,5. V 2 ankan ktip nhau 2 ankan lCH4: x mol

C2H6: y mol

Mun xc &nh khi lng ca 2 ankan th phi tm x v yCch tm x v y nhsau:

* Cch1:

CH4 + O2 CO2 + H2Ox xC2H6 + O2 2CO2 + H2Oy 2y

Ta c: x + y = 0,12 x = 0,06x + 2y = 0,18 y = 0,06

* Cch 2:

CH4: x nh)n hp= x + y = 0,12C2H6 : y C =

}RB"}M

RBM= 1,5

x = 0,06y = 0,06

Bi 2).t chy 0,05 mol hidrocacbon X sau hp th"ton bsn ph#m vo dung d&ch nc vi trong dthy xut hin 10 gam kt ta

trng v thy khi lng dung d&ch gim 3,8 gam so vi khi lng dung d&ch trc phn ng.

A.C2H6 B.C2H4 C.C2H2 D.C4H10

H2O

CO2 + Ca(OH)2d CaCO3~ + H2O0,1 30,1

hp th" tch ra

mdung d&ch gim = mZZjW - m(Zj" ["j)3,8 = 0,1. 100 ( 0,1.44 + mH2O)

Suy ra mH2O= 1,8 gam nH2O = 0,1 mol

Nhn thy nZj" = n["j= 0,1 n l anken : CnH2nsnguyn t!trong anken l C =

!

! 2 C2H4

Bi 3): .t chy hon ton 2 hdrocacbon A, B ktip nhau thu c sn ph#m cho i qua dung d&ch Ca(OH)2dsau phn ng thu c 80

gam kt ta .Xc &nh CTPT ca A,B v tnh % khi lng tng hidrocacbon A,B. Bit khi lng CO2v H2O hp th"vo bnh l 44,2

gam.

CO2 + Ca(OH)2d CaCO3~ + H2OnZZjW = 0,8 (mol) :nZj" = 0,8 (mol)

M : m(Zj" ["j) = 44,2 (g)mH2O = 44,2 0,8.44= 9 gam n["j = 0,5 (mol)

Nhn thy nZj" > n["j : n l ankin/ ankaienCnH2n-2 = 0,3 (mol)

C2H2: x mol

=> C =n = !\!W

= 2,67C3H4 : y mol

Ta c h:nh)n hp= x + y = 0,3 x = 0,1C =

"}RBW}M

RBM=

!\

!W y = 0,2


17/172

Bi 4) :.t chy 11,2 lt h)n hp kh gm 2 hidrocacbon A,B thuc cng dy ng 0ng cn 40,32 lt O2bit pto ra 26,88 lt CO2 ktc.

Xc &nh cng thc ca A,B. V tm khi lng tng cht

CxHy + O2 CO2 + H2O1,8mol 1,2mol

p d"ng LBTNT (O) : 1,8 . 2 = 1,2 . 2 + nO (["j) nO (["j) = 1,2 n["j = 1,2 mol

Nhn thy nCO2=nH2O c1 , C2H4

CnH2n C =n =!"

!= 2,67

V A, B ko cho ktip nhaunn B c thl C3H6ho-c C4H8( B ko thln hn v A, B thkh C1C4)TH1: C2H4v C3H6TH2: C2H4v C4H8

m)i TH cc bn d-t #n v tm smol ca tgf cht ging nhbi 1 ri tnh kl

Bi 5).t chy ht mt hidrocacbon mch hht 8,96 lt O2thu c 6,72 lt CO2ktc. Xc &nh CTPT v CTCT ca hidrocacbon .

CxHy + O2 CO2 + H2O0,4mol 0,3 mol

p d"ng LBTNT (O) : nO (["O) = 0,2 n["O = 0,2 mol

Nhn thy nw]" = 0,3 > ne"] 0,2hidrocacbon ko no c slin kt ,12.

M-c &nh 2,CnH2n-2= 0,3-0,2= 0,1 (mol) C =!W

!C3H4

Bi 6).Mt h)n hp X gm 2 hidrocacbon mch h'thuc cng dy ng 0ng thkh ktc cn 20,16 lt O2.t chy ht h)n hp X

sau phn ng thu c 7,2 gam H2O .Xc &nh CTCT ca A,B v % khi lng tng cht.

X + O2 CO2 + H2O0,9 mol 0,4 mol

p d"ng LBTNT (O) : nO (Zj") = 1,4 nZj" = 0,7

Nhn thy nCO2 =0,7 > nH20 = 0,4 12,. C2H2

M-c &nh 2,CT: CnH2n-2 =0,3 mol C = 0,7/0,3 = 2,333C3H4ho-c C4H6

Bi 7) .t chy ht h)n hp kh X gm 2 hidrocacbon thuc cng dy ng 0ng thu c 19,712 lt CO2ktc v 10,08 gam H2O .Xc &nh

CTPT ca A,B v smol tng cht.

Nhn thy nZj" = 0,88 > n["O = 0,56 CnH

2n-2 = 0,1 (mol) C = n = 2,75 .Suy ra A l: C2H2cn B l C3H4ho-c C4H6

Bi 8):.H)n hp X gm ankan A v anken B thkh. t chy hon ton 6,72 lt h)n hp X ktc thu c 15,68 lt CO2ktc v 14,4 gam

H2O. XCTPT ca A,B v tnh % khi lng ca chng.

* Cch 1:

CnH2n+2 + O2 nCO2+ (n+1)H2Ox nx (n+1)x

CmH2m + O2 mCO2 + mH2Oy my my

nh)n hp= x + y = 0,3 x = 0,1Ta c : nZj" = nx + my = 0,7 y = 0,2

n["O = (n+1)x + my = 0,8 0,1n + 0,2m = 0,7

y l bi ton tha #n thiu pt lin quan n vic xc &nh cng thcphn t!nn ta bin lun b$ng cch bin i nhtrn ri a vpt cmi lin hgia n v m l 0,1n + 0,2m = 0,7 ( rt gn i ta c n +2m = 7 Sau bin lunM 2 3 4

N 3 1 m

- TH1: Anken : C2H40,2 (mol) Ankan : C3H80,1 (mol) . TH2: Anken : C3H60,2 (mol)

Ankan : CH40,1 (mol)

Cch 2:

nankan = n["O - nZj" = 0,8 0,7 = 0,1 (mol) nanken = 0,3 0,1 = 0,2 (mol)

CnH2n+2 nCO20,1 0,1n

CmH2m nCO20,2 0,2m

=> nw]" = 0,1n + 0,2m = 0,7Sau lm tng tcch 1 tm n v m


18/172

Bi 9).t chy hon ton 0,1 mol h)n hp X gm propan v anken A sau cho ton bsp chy vo dung d&ch Ca(OH)2dthy c 26 gam

kt ta . Xc &nh cng thc phn t!ca anken.

CO2 + Ca(OH)2d CaCO3~ + H2O0,26 0,26

* Cch 1: C3H8 3CO2x 3x

CnH2n nCO2y ny

nh)n hp X= x + y = 0,1 y =!6?W

nCO2 = 3x + ny = 0,26

nhbi ton ny ta c thbin lun theo gii hn v y > o nn n-3 Vi n=2,5 m = 3 C3H8

Bi 12).H)n hp X gm 2 anken A1 , A2 ( MA2 = MA1) v ankadien B . Hidro ha hon ton h)n hp X thu c h)n hp Y gm 2 ankan E1,

E2. em t chy 0,1 mol h)n hp X thu c 6,272 lt kh CO2v 4,68 gam H2O . Xc &n cng thc ca ankadien

Ta c : n =?,

?|`JcycJz=

!"\

!= 2,8

Phi c 1 anken c C < 2,8 n l C2H4

V M"= 2 M anken cn li l C4H8

Cho cng H2vo anken v ankadienchthu c 2 ankan

Ankadien l C4H6


19/172

BN THNL nhng a hay chi by cng nhauL nhng a a cho vui n khi ta tc x khiL nhng a d ta c chi mng nhng .vn chi!"i m kh#ng h$ th%&ng tic

L c'i b(n m g)* nhi$u th d+ gh,t nhng vng th!"i bu-n"n th/n ! nhng a hi0u ta nh1t ..

P/s: l nhng a bit qu nhiu b mt ca bnn ni nu khng git n! th" bn #h$i th%n &'i n!n su(t )i*

Hi%n !fn th?i

.NG D@NG GI9I 1:Cu 1-A-2012:t chy hon ton 4,64 gam mt hirocacbon X (cht kh iu kin thng) ri em ton bsn ph#m chy hp th"ht

vo bnh ng dung d&ch Ba(OH)2. Sau cc phn ng thu c 39,4 gam kt ta v khi lng phn dung d&ch gim bt 19,912 gam. Cng

thc phn t!ca X l

A. CH4. B. C3H4. C. C4H10. D. C2H4

Xl sliu :

t X sthu c CO2v H2O hp th"ht vo bnh ng dung d&ch

hp th"

H2O tch ra 39,4(g)

CO2 + Ba(OH)2 BaCO3+ + H2OCO2 + Ba(OH)2 Ba(HCO3)2 + H2O

mdung d&ch gim = mCaCO3 - mCO2 + H2O

mCO2+H2O = 39,4 19,912 = 19,488 (gam)

Ti sao khi CO2pvi Ca(OH)2li xy ra 2 p?

V bi khng cho ta bit Ca(OH)2 dhay khng nn ta phi

lng trng h(p n to ra c mui axit do phi vit c 2

phn ng Cn nu cho bazo dth ch&cn vit 1 p to mui

trung ha thi.

nh hng cch lm :

i vi nhng bi ton bit khi lng ca hp cht hu cm bi

cho lin quan n phn ng t chy th khi lm ta nn p d"ng

BTNT lm sgip ta tnh rt nhanh ra smol CO2v H2O . Qua

ta srt ra c kt lun vhidrocacbon X

Cch lm:

6!^6

+ O2 w]"R T

+ e"]M T

19,488 (gam)

p d"ng BTNT : nC(X)= nC(CO2)= x mol

nH(X)= nH(H2O)= 2y mol

mX= 12x + 2y = 4,64 x= 0,348

mCO2+ H2O= 44x + 18y = 19,912 y = 0,232

nhn thy nCO2= 0,348 > nH2O = 0,232 X c 2,CT X: CnH2n-2

ta c Snguyn t!C ca X =?,

? !W6\

!W6\!"W" X l C3H4


20/172

Cu 2-B-2012:t chy hon ton h)n hp X gm hai hirocacbon (tlsmol 1 : 1) c cng thc n gin nht khc nhau, thu c 2,2

gam CO2v 0,9 gam H2O. Cc cht trong X l

A. mt ankan v mt ankin. B. hai ankaien. C. hai anken. D. mt anken v mt ankin

Cch lm : xem li mc s5 phn cng thc ton t chy

V nCO2= nH2O = 0,5 mol m hn hp X em t gm 2 hirocacbon c cng thc n gin khc nhau nn n khng thl 2 anken c chc thl 1 ankan v 1

ankin vi smol 2 cht b$ng nhau. p n ng A.

Cu 3-A-2010:t chy hon ton mt lng hirocacbon X. Hp th" ton bsn ph#m chy vo dung d&ch Ba(OH)2(d) to ra

29,55 gam kt ta, dung d&ch sau phn ng c khi lng gim 19,35 gam so vi dung d&ch Ba(OH)2ban u. Cng thc phn t!ca X l

A. C2H6. B. C3H6. C. C3H8. D. C3H4

t X sthu c CO2v H2O hp th"ht vo bnh ng dung d&ch

hp th"

H2O tch ra 39,4(g)

CO2 + Ba(OH)2d BaCO3+ + H2O

0,15 3 0,15 molmdung d&ch gim = mCaCO3 - m (CO2 + H2O

mCO2+H2O = 29,55 19,35 = 10,2 (gam)

Vi mCO2= 0,15.44 = 6,6 (gam) mH2O = 10,2 6,6 = 3,6 (gam)0,2 mol H2O

Nhn thy smol H2O = 0,2 > smol CO2= 0,15 X l hirocacbon no hay

chnh l ankan : CnH2n+2

Snguyn t!C = ?,?yz{yz

!!"!

C3H8

Cu 4-B-2010:H)n hp kh X gm mt ankan v mt anken. Tkhi ca X so vi H2b$ng 11,25. t chy hon ton 4,48 lt X, thu

c 6,72 lt CO2(cc thtch kh o ktc). Cng thc ca ankan v anken ln lt l

A. CH4v C4H8. B. C2H6v C2H4. C. CH4v C2H4. D. CH4v C3H6

CnH2n+2 + O2 nCO2 + (n+1)H2O

x nx

CmH2m + O2 mCO2 + mH2O

y my

h)n hp= 22,5 0,3 mol

nh)n hp= 0,2 mol

Cch 1: lm ging cch 1 ca bi 8 (cc bn nn xem li )

h)n hp =R}(6?B")B M}6

RBM 22!5 x= 0,15

nh)n hp = x + y = 0,2 mol y = 0,05

nCO2 = nx + my = 0,3 mol 0,15n + 0,05m = 0,3

rt gon pt s3 ta c : 3n + m = 6 . Bin lun ta c n= 1, m = 3

suy ra an kan l CH4 v an ken l C3H6

Cch 2:t gg : !

z 7 ""! } !" 6! ()+ O2 w]"

!W T

+ e"]R T

Bo ton ny ta bit khi lng ca 1 nhm cht hu cnn ta c thp d"ng

BTNT lm cho n nhanh:

nC(hhX)= nC(CO2) = 0,3 mol

nH(hhX)= nC(H2O)= 2x mol

y l bi ton t chy h)n hp ankan v anken thu c CO2= 0,3 mol v

H2O= 0,45 mol . nankan = 0,45-0,3 = 0,15 mol ; nanken = 0,2 0,15 = 0,05 mol

Thit lp ra smol CO2ta c 0,15n + 0,05m = 0,3 3n + m = 6

Bin lun ta c n= 1, m = 3Suy ra an kan l CH4 v an ken l C3H6

Cu 4-B-2014:t chy hon ton 0,2 mol h)n hp X gm mt ankan v mt anken, thu c 0,35 mol CO2 v 0,4 mol H2O. Phn tr/m s

mol ca anken trong X l

A.50%. B.40%. C.25%. D.75%.

Ta c :p d"ng cng thc vt chy h)n hp ankan v an ken nankan = 0,4 0,35 = 0,05 ; nanken = 0,2 0,05 = 0,15 %nanken=!!"

} 0ff 5

mh)n hp X = 0,3.12 + 2x = 4,5x= 0,45


21/172

NgNgNgNgy thy thy thy thnnnnommmm7 8Chc con ngi no ) tng tri *ua sgi vca a ngc th mi c sc mnh x9 :ng c thi+n ng.

CDSWX DYOMZ : FS Hj. kW .e CMZ

A.Anken + H2/X2/HX/H2O th xy ra phn ng cng theo tl(1:1)

B.Ankin/ankadien + H2/X2/HX/H2O th xy ra phn ng cng theo tl(1:1) ho-c (1:2)

Trong qu trnh lm bi tp cn phi bit n xy ra phn ng no.

* Cch xc nh:

ntc nhn

- Lp tl: T =

nankin/ankadien

- Sau i chiu T vi cc mc

1 2

(1:1) (1:2)

+ Nu T < 1 chxy ra phn ng theo tl(1:1).


22/172

+ Nu 1 < T < 2 xy ra phn ng theo tl(1:1) v (1:2).

+ Nu T > 1 chxy ra phn ng theo tl(1:2).

- Ho-c :

+ Nu cho tc nhn d chxy ra phn ng theo tl(1:2).

+ Nu cho ankin + H2vi xc tc l Pd th chxy ra phn ng theo tl(1:1).

C. Cng thc gii nhanh cn nhi vi phn ng cng H2

* Gi A l h)n hp trc phn ng

B l h)n hp sau phn ng

Th ta lun c : mtrc= msau

Mtrc nsau

Msau ntrc

ntrc- nsau= ne"(phn ng)

* Nu bt t h)n hp sau phn ng th ta i t h)n hp trc phn ng, kt quth ra O2; CO2; H2O khng thay i.

D.CH : TGLH8Pcng lm ph v'lin kt , chnh v vy slk ,c trong h)n hp ban u sb$ng smol lk ,b&ph v'hay chnh l b$ng s mol tcnhn cng (H2, Br2) tham gia phn ng ph v',

VD: gis!cht X tham gia p cng theo tl( 1:1 ) nX : nH2 = 1: 1 ph v'1,( nhvy smol , b&ph v' sb$ng smol H2 p)gis!cht X tham gia p cng theo tl( 1:2 ) nX : nH2 = 1: 2ph v'2,( nhvy smol , b&ph v' sb$ng smol H2 p

n8ban u = nph v= + n8d ( ch n8ph v== ntc nhn cng ) nu ko c dth n8ban u = ntc nhn cng

Cu 1-A-2014:H)n hp kh X gm 0,1 mol C2H2; 0,2 mol C2H4v 0,3 mol H2. un nng X vi xc tc Ni, sau mt thi gian thu c h)n

hp kh Y c tkhi so vi H2b$ng 11. H)n hp Y phn ng ti a vi a mol Br2trong dung d&ch. Gi tr&ca a l

A.0,1. B.0,2. C.0,4. D.0,3.

Cch 1:

C2H2 + H2F3 C2H4

P: x x x

C2H2 + 2 H2F3 C2H6

P: y 2y y

C2H4 + H2F3 C2H6

P: z z z

TM T4T :

C2H2: 0,1 mol

C2H4:0,2 mol Ni,to

H2: 0,3 mol

H)n hp X trc p

H)n hp Y sau pc 22

BTKL : mY(sau)= mX(trc)= 0,1.26 + 0,2.28 + 0,3.2 = 8,8 gam

nY = 8,8/22 = 0,4 mol

Ta c : nH2 p= ntrc - nsau x + 2y + z = (0,1+0,2+0,3) - 0,4 = 0,2 (1)

nBr2 p = 0,2 2(x+y) + x + (0,2 z ) = 0,4 (x + 2y + z )

Nhn vo pt (1) ta ssuy ra c nBr2 p = 0,4 0,2 = 0,2 mol

Cch 2:

BTKL : mY(sau)= mX(trc)= 0,1.26 + 0,2.28 + 0,3.2 = 8,8 gam

nY = 8,8/22 = 0,4 mol

nH2 p= ntrc - nsau = (0,1+0,2+0,3) - 0,4 = 0,2 (1)

p dng t&l8:

Cho C2H2=0,1 mol c 2,, C2H4= 0,2 mol c 1 , lng ,b&ph v'hon ton qua

hai ln tham gia pcng l khi tham gia pcng H2v sau l Br2.

Ta c n,ban u = nH2 p+ nBr2 p

0,1.2 + 0,2.1 = 0,2 + xx = 0,2 mol . Vy smol Br2pl 0,2 mol

.hi ti%n !fn nqa ch

C2H2cn d + 2Br2 C2H2Br4

0,1-(x+y) 0,2 - 2(x+y)

C2H4 cn d + Br2 C2H4Br2

x+(0,2-z) x + (0,2-z)

C2H6

(y+z)

H2cn d


23/172

Cu 2-B-2014:Mt bnh kn chcha cc cht sau: axetilen (0,5 mol), vinylaxetilen (0,4 mol), hiro (0,65 mol) v mt t bt niken. Nung nng

bnh mt thi gian, thu c h)n hp kh X c tkhi so vi H2b$ng 19,5. Kh X phn ng va vi 0,7 mol AgNO3trong dung d&ch NH3,

thu c m gam kt ta v 10,08 lt h)n hp kh Y (ktc). Kh Y phn ng ti a vi 0,55 mol Br2trong dung d&ch. Gi tr&ca m l

A.76,1. B.75,9. C.91,8. D.92,0.

Tm t't :

C2H2: 0,5 molC4H4: 0,4 mol +Ni,t

oh2X

H2: 0,65 mol

Ta c ntrc p= 0,5 + 0,4 + 0,65 = 1,55 molmtrc p= 0,5.26 + 0,4.52 + 0,65.2 = 35,1 (gam)

Cch lm :

p d"ng BTKL : mX(sau) = mtrc = 35,1 gam

nH2p = ntrc - nsau = 1,55 0,9 = 0,65 molkt lun : H2pht

Ta c

nAgNO3= 2x + y + z = 0,7 mol

nX- nY = x + y + z = 0,9 0,45

n,ban u = nH2(p) + n,d + nBr2p

hay 0,5.2 + 0,4.3 = 0,65 + (2x+3y+2z) + 0,55

gi htm c

x= 0,25; y =z = 0,1 mol

Vy khi lng += 0,25.C2Ag2 + 0,1. C3H3Ag +0,1.C4H5Ag = 92 gam

Cu 3-A-2012H)n hp X gm H2, C2H4 c tkhi so vi H2l 7,5. Dn X i qua Ni nung nng thu c h)n hp X c tkhi so vi H2l

12,5. Tnh hiu sut hidro ha. A)70% B)80% C)60% D)50%

Cu 4. H)n hp kh X gm H2v C2H4c tkhi so vi He l 3,75. Dn X qua Ni nung nng , thu c h)n hp kh Y c tkhi so vi He

l 5 . Hiu sut ca phn ng hidro ha l :

A. 20% B. 25% C. 50% D. 40%

Cch lm :Cch lm tng tch cho He =4

Cu 5-A-2013 :H)n hp X gm H2, C2H4v C3H6c tkhi so vi H2l 9,25. Cho 22,4 lt X (ktc) vo bnh kn c s5n mt t bt Ni .unnng bnh mt thi gian , thu c h)n hp kh Y c t6khi so vi H2b$ng 10.Tng smol H2 phn ng l

A.0,07 B.0,05 C.0,015 D..0,075

Cch lm :

i vi nhng bi tom ko cho sliu c"thnhkhi lng, smol, thtchth ta quy h)n hp ban u v1 mol i vi cht kh lm , 100g i vi cht

rn lm .Nguyn tc l ngay sau khi quy phi tim c smol tng chttronng h)n hp ban u .

C2H4 : x nh)n hp = x + y = 1 x = 0,5Quy h)n hp v1 mol: => =>

H2 : y M h)n hp="\kB"l

kBl= 15 y = 0,5

Cch 2Mtrc nsau

Msau ntrc

=>

"=

ms => nsau = 0,6 (mol) => np.= 1- 0,6 = 0,4 (mol)=> n = 80%

* Cch 1C2H4 + H2 C2H6

B: 0,5 0,5

P.: x x x

d: 0,5 x 0,5 x

C2H4 d: 0,5 x (mol)

H)n hp sau phn ng : H2 d: 0,5 x (mol)

C2H6 : x (mol)

M =

(!k )}"\ B (!k)}" B W}k

! k B !k B k = 25 x = 0,4

Hp.=!6 }

!= 80%

C2H2 cn d hay CH.CH : x mol

C4H4 cn dhay CH2=CH-C.CH :y mol + AgNO3(=0,7 mol)

C4H6 hay CH3-CH2-C.CH :z mol

C4H8

C4H6 (loi c hai ni i)

C2H4 + Br2(=0,55mol)

C2H6

C4H10


24/172

C2H4 Ta c : Mtrc nsau

Msau ntrcH)n hp X trc phn ng H2Ni,to H)n hp Y sau phn ng c

MX= 18,5 , nX = 1 (mol) C3H6 MY= 20 => nsau = 0,925 => np. = 1 0,925 = 0,075

(mol)

Cu 6Trn 1 mol anken X vi 1,6mol H2dn h)n hp qua Ni nung nng thu c h)n hp Y. Sau pdn h)n hp Y vo dung d&ch Br2d

thy c 0,2mol Br2 p. Cho bit hiu sut pHidro ha .

CnH2n + H2 CnH2n+2B: 1 mol 1,6 molP.: x x xd: (1 x) (1,6 x)

CnH2n (d) + Br2 CnH2nBr2H)n hp sau phn ng: ( 1 x ) ( 1 x )

H2 (d) ; CnH2n +2

Ta c: nBr2 = (1 x ) = 0,2 => x = 0,8 => np.=

!\}

= 80%

Cu 7.H)n hp X gm 0,15mol C2H4 ; 0,1mol propen v 0,3mol H2. Cho h)n hp X i qua Ni nung nng thu c h)n hp Y c thtch

l 8,4 lt. Cho h)n hp Y c thtch l 8,4lit . Cho h)n hp Y qua dung d&ch Br2dthy khi lng bnh ng dung d&ch Br2t/ng 2,45(g) .

Hy la chon hiu sut phidro ha ca C2H4(H1) v ca C3H6(H2) ln lt l

A)60% v 75% B) 66,67% v 60% C) 75% v 66,67% D) 66,67% v 75%

C2H4 + H2 C2H6P: x x x

C3H6 + H2 C3H8P: y y y

C2H4 (d) + Br2 C2H4Br2( 0,15 x )

H)n hp sau phn ng: C3H6(d) + Br2 C3H6Br2Cho tc d"ng vi d2Br2 (0,1 y )Th: H2 d: 0,3 (x + y );

C2H6: x (mol);C3H8: y (mol)

Ta c : nY= (0,15 x ) + (0,1 xy) + 0,3 (x + y) + x + y = 0,375

mbnh Br2 t/ng = mC2H4 d+ C3H6 d= 28 . (0,15 x ) + 42 . (0,1 y ) = 2,45

x = 0,1 n C2H4 p. =!}

!W= 66,67%

y = 0,225 n C3H6 p. =!""}

!W= 75%

Cu 8:Cho 3,12g ankin X phn ng vi 0,1mol H2( xc tc Pd/PdCO3; t0) thu c h)n hp Y chc 2 hidrocacbon . Cng thc phn t!

ca X l :

A.C2H2 B. C4H6 C. C5H8 D. C3H4

CnH2n - 2 + H2 CnH2nP: 0,1 0,1

V sau p.thu c h)n hp 2 hirocacbon => Ankin phi d

mp. < mb hay 0,1.(14n 2 ) < 3,12 => n < 2 Ankin l C2H2

Cu 9Cho 2,24 lt C2H2hp th"ht vo 100(g) dung d&ch Br224% thy c kh thot ra. Tnh khi lng tng sn ph#m

T =m,

m-,,=

!

!= 1,5 => 1 < T < 2 => xy ra 2 phn ng

C2H2 + Br2 C2H2Br2x x

C2H2 + 2 Br2 C2H2Br4y 2y

nZ"["= x + y = 0,1 x = 0,05 nZ"["hi"= 0,05=> =>

nhi" = x + 2y = 0,15 y = 0,05 nZ"["hi6 = 0,05

Tsmol stm ra c khi lng.

Cu 10- A-2010un nng h)n hp kh X gm 0,02 mol C2H2v 0,03 mol H2trong mt bnh kn (xc tc Ni), thu c h)n hp kh

Y. Cho Y li ttvo bnh nc brom (d), sau khi kt thc cc phn ng, khi lng bnh t/ng m gam v c 280 ml h)n hp kh Z

(ktc) thot ra. Tkhi ca Z so vi H2 l 10,08. Gi tr&ca m l

A. 0,328. B. 0,620. C.0,585. D. 0,205


25/172

T =m,

m,,=

!W!"

= 1,5

C2H2 + H2 C2H4

C2H2 + 2 H2 C2H6

C2H2: 0,02 (mol) C2H2(d) v C2H4 pvi Br2

Tm t't: h2ban u => h2 sau p. C2H6 ko p l Z thot ra MhhZ= 20,16

H2 : 0,03 (mol) H2cn d nhhZ= 0,0125

p d"ng LBTKL: mban u= msau => 0,02.26 + 0,03.2 = mC2H4 + C2H2 cn d+ 20,16. 0,0125

Suy ra mC2H4 + C2H2 cn d = 0,328 (g) mbnh brom t/ng= 0,328 gam

Cu 11-A-2011:H)n hp X gm C2H2v H2c cng smol. Ly mt lng h)n hp X cho qua cht xc tc nung nng, thu c h)n

hp Y gm C2H4, C2H6, C2H2v H2. S"c Y vo dung d&ch brom (d) th khi lng bnh brom t/ng 10,8 gam v thot ra 4,48 lt h)n

hp kh (ktc) c tkhi so vi H2l 8. Thtch O2(ktc) cn t chy hon ton h)n hp Y l

A. 33,6 lt. B. 22,4 lt. C.26,88 lt. D. 44,8 lt.

C2H2 + H2 C2H4C2H2 + 2 H2 C2H6

C2H2: x (mol) C2H2(d) v C2H4 pvi Br2 m(C2H4+ C2H2(d)) = mbnh brom t/ng = 10,8 (g)h2ban u => h2 sau p. C2H6 ko p l Z thot ra Mhh = 16

H2 : x (mol) H2cn d nhh = 0,2 (mol)

p d"ng LBTKL : mban u= msau 26x + 2x = 10,8 + 0,2.16 => x = 0,5Thay v t hh sau phn ng ta i t hh ban u ( kt qukhng thay i)

C2H2 +

" O2 2 CO2 + H2O

0,5 1,25

H2 +

" O2 H2O

0,5 0,25

Vj"= ( 0,25 + 1,25 ).22,4 = 33,6 (l)

Cu 12)un nng 7,6g h)n hp X gm C2H2; C2H4; H2trong bnh kn vi xc tc Ni thu c h)n hp Y. t chy hon ton h)n hp Y dn sn

ph#m chy thu c ln lt qua bnh 1 ng H2SO4-c , bnh 2 ng Ca(OH)2dthy khi lng bnh 1 t/ng 14,4g , khi lng bnh 2 t/ng ln l :

A.35,2g B.22g C.24,93g D.17,6g

C2H2 + O2 CO2 + H2O

C2H4 + O2 CO2 + H2O

H2 + O2 H2O7,6gam ? 0,8 mol

Khi lng bnh 1 t/ng chnh l khi lng ca H2O mH2O= 14,4 (gam) 0,8 mol

-t nCO2= x mol . P d"ng bo ton nguyn tnC(hhX) = nC(CO2) = x mol

nH(hh X) = nH(H20) = 0,8.2 = 1,6 molTa c mhhX = x.12 + 1,6.1 = 7,6 x= 0,5 mol mbnh 2 t/ng = mCO2= 0,5.44 = 22 gam

Cu 13-A-2013:Trong mt bnh kn cha 0,35mol C2H2; 0,65mol H2v mt t bt Ni. Nung nng bnh mt thi gian, thu c h)n hp kh X c t

khi i vi H2b$ng 8. S"c X vo lng ddung d&ch AgNO3trong NH3n phon ton thu c h)n hp kh X v 24 gam kt ta . H)n hp Y phn

ng va vi bao nhiu mol Br2trong dung d&ch

A.0,2mol B.0,1mol C.0,25mol D.0,15mol

Gi :

C2H2 + H2 C2H4

P: x x x

C2H2 + 2 H2 C2H6

P: y 2y y


26/172

C2H2: 0,35 (mol) C2H2dAgNO

3/ NH

3 C2Ag2~

h2ban u => h2 sau p. 0,35 (x+y) 0,35 (x+y)

H2 : 0,65 (mol) Msau= 16 C2H4 + Br2 C2H4Br2

x

C2H6: y (mol);

H2 d: 0,65 (x + y) (mol)

n~= 0,35 (x + y) ="6

"6

Ta c

Mhh sau p = y

? y=

!W}"^B!^}"

!W(kBl)B kBlB!^(kBl)= 16 (ch : BTKL cho ta msau = mtrc )

Cu 14: Cho 0,5 mol H2v 0,15 mol vinyl axetilen vo bnh kn c m-t xc tc Ni ri nung nng . Sau phn ng thu c h)n hp kh X c

tkhi so vi CO2b$ng 0,5 . Cho h2X tc d"ng vi dung d&ch Br2dthy c m(g) Br2 tham gia phn ng . Ga tr&ca m l :

A.40g B. 24g C. 16g D. 32g

T =m,

m-..=

!

!= 3,33

* Cch 1 :C4H4 + 3 H2 C4H10

B: 0,15 0,5P: x 3x xd: (0,15 x) (0,5 3x)

C4H4 + 3 Br2 C4H4Br6H2sau phn ng X: (0,15 x ) ?

H2 d: (0,5 3x)C4H10 : x

MX =(!k)}"B"}(!Wk)B \k

! k B } Wk B k = 22 => x =

"

Br2 = (0,15 x ) . 3 = (0,15 -

") . 3

Cch 2: C4H4 + 3 H2 C4H10B: 0,15 0,5P: x 3x xD (0,15 x) (0,5 3x)

Ta cmsau = mtrc= mC4H4+ mH2 = 0,15.52 + 0,5.2 = 8,8 gamnsau = 8,8/0,5.44 = 0,4 mol

Suy ra nH2 p = ntrc - nsau 3x = (0,5 + 0,15) 0,4 x = 1/12

Vy smol Br2= (0,15 -

") . 3

Cu 15-B-2012: H)n hp X gm 0,15 mol vinylaxetilen v 0,6 mol H2. Nung nng h)n hp X (xc tc Ni) mt thi gian, thu c h)n hp Y c tkhi so

vi H2b$ng 10. Dn h)n hp Y qua dung d&ch brom d, sau khi phn ng xy ra hon ton, khi lng brom tham gia phn ng l

A. 7,0 gam. B. 24 gam. C. 8 gam. D. 16 gam

Cch lm : Lm tng tbi 13

rt tt , phi nh #y ch .


27/172

NgNgNgNgY ThY ThY ThY The5555 :::: HHHH_Y BIY BIY BIY BIT QUT QUT QUT QUTRTRTRTRNG THNG THNG THNG TH !!!!I GIANI GIANI GIANI GIAN TRONGTRONGTRONGTRONG

TTTTNH YNH YNH YNH Yv ;;;;UUUU CUCUCUCUC SC SC SC S ;;;;NGNGNGNG CCCCwNG VING VING VING VIlCCCC vvvvHHHHC TC TC TC Tj ....PPPP =SSSSNGNGNGNG

!H!H!H!HwNG HNG HNG HNG HI TII TII TII TICCCC

Hy sng nhhm nay l ngy cui cng !

u ci cht - l ch n ca tt cmi ngi, l quy lut ca to ha th cg phi s, sao kh"ng sng nh

nay l ngy cui cng ca bn c phi cuc sng sc nhiu / ngh0a v hu ch hn kh"ng1 gh0 nhthbn s

cm thy nh7nhng, bi bn - trtr"i ri, bn ch#t b*c tt cp lc, gnh n-ng trc mang th%o.

2n ch0ng cn g phi $u hc. 2n hon ton thot ra ngoi khu"n khb h7p hng ngy lm nhng vic

m bn cho l quan trng nht i mnh, cho l v0 i. 2n cn phi lm n ngay trnh lan man vi nhng g v"

bkh"ng cn thit.

u ngh0 nhth, bn slm c nhiu vic c / ngh0a hn, dnh cho nhng ngi thn ca bn nhiu tnh

y!u thng hn, quan tm ch3m sc hnhiu hn v sng3n li c nhng li ni, nhng hnh ng lm tn

th

ng n h

.

u ngh0 nh

th

, b

n trn tr

ng h

n nh

ng g m cu

c s

ng - ban t

-ng cho b

n v b

n bi

t

cn phi lm g t-ng li cho cuc sng. u ngh0 nhth, bn ss8chia nhiu hn, bi nhngi ta ni nim

vui sc nhn "i v n)i bun svi i mt n!a khi ta s8chia n vi mt ngi. 2n cng sbit tha th

cho nhng l)i lm ca ngi khc v bn chcn mt ngy, thi gian u dnh cho hn th, gin d)i na.

4hi ch#ng ta sinh ra, ch#ng ta l ci mi nhng ri ch#ng ta gi i v cht nhng ci mi khc c sinh

ra. 5 l stht ng tin nht, l quy lut ca cuc sng ny. & vy ch#ng ta chc mt qu9thi gian nht

&nh hon thnh tt ccc m"c ti!u ca i mnh n!n ng ph phm b$ng cch sng cuc i ca ngi

khc, suy ngh0 v quan im sng ca ngi khc. 6ng ng ph phm thi gian qu/ bu y lm nhng vic

v" ngh0a, hay sng tm bkh"ng m"c ti!u, kh"ng l/ tng72n l mt c nhn c lp, c c m, c l/ tngv kht vng ca ri!ng bn m chc bn mi bit. 2n c quyn ly ai lm m"c ti!u hng ti v vt

qua chkh"ng phi bn phn u trthnh h. 5 l mt ssao ch8p ng $u hd bn c t c iu

i ch3ng na. 9-y hnh ng th%o nhng g tri tim v trc gic bn mch bo chkh"ng phi nhng rung

ng tttng b!n ngoi, bi ai cng chsng c mt ln. 4hi - cht ri d bn c mun cng kh"ng th


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lm nhng g m khi sng tri tim bn mch bo, du r$ng ngi ta vn ni c mt thgii no b!n kia

nhng c ai t trvni vi ch#ng ta r$ng ra sao u1

Li cui, chc cnh tht vui v hy sng nhnay l ngy cui cng!

CDSWX DYOMZ ;: FS Hj. kW .e HD SUM \S NUzS D{L HT| / E

LgMU9QMD9Bi tp vphn ng thion kim loi ha tr& I -AgNO3/ NH3

Chc nhng hirocacbon c ni 3 u mch mi tham gia phn ng ny

Vd1 : CH CH + AgNO3 + NH3 CAg CAg~+ NH4NO3 (C2H2 C2Ag2+)

CH3 C CH + AgNO3 + NH3 CH3 C CAg~+ NH4NO3 ( C3H4 C3H3Ag+)

Vd2 : CH2 = CH CCH + AgNO3 + NH3 CH2 = CH C CAg~+ NH4NO3 (C4H4 C4H3Ag +)

CH C C CH + AgNO3 + NH3 Z Z Z CAg~+ NH4NO3 (C4H2 C4Ag2+)

Bi 1 )H)n hp X gm etylen v axetilen tc d"ng va ht vi 200ml dung d&ch AgNO31M trong NH3. M-t khc c(ng lng h)n hp X

trn lm mt mu va ht 0,3mol dung d&ch Br2. Phn tr/m thtch ca axetilen trong h)n hp A l :

A.50% B. 20% C. 25% D. 33,33%

TN1: C2H4

X

C2H2 + 2AgNO3 + NH3 C2Ag2+ + NH4NO3

y 2y

TN2: C2H4 + Br2 C2H4Br2

x x

C2H2 + 2Br2 C2H2Br4

y 2y

Ta c

nAgNO3= 2y = 0,2 y = 0,1

nBr2 = x + 2y = 03 x= 0,1

%VC2H2 = 0,1/(0,1 + 0,1) = 50%

Bi 2-A-2014:H)n hp kh X gm etilen v propin. Cho a mol X tc d"ng vi lng ddung d&ch AgNO3 trong NH3, thu c 17,64 gam

kt ta. M-t khc a mol X phn ng ti a vi 0,34 mol H2. Gi tr&ca a l

A.0,46. B.0,22. C.0,34. D.0,32.

Cch lm : t/ng tbi 1

Bi 3,H)n hp A gm CH4; C2H4; C3H4. Nu cho 13,4g h2X tc d"ng vi dung d&ch AgNO3/NH3dth thu c 14,7g kt ta . Nu cho

16,8 lit h)n hp X ( ktc) tc d"ng vi dung d&ch Br2th thy c 108g brom phn ng . % thtch CH4trong h2X l :

A.30% B. 25% C. 35% D. 40%

Cch lm: bi ton y thc hin 2 TN vi 2 lng khc nhau ca cng h)n hp nn ta -t TN2 b$ng n ln TN1 lm

TN1: CH4 : x mol

C2H4 : y mol

C3H4 + AgNO3 + NH3 C3H3Ag + + NH4NO3

z mol z mol13,4 gam 14,7 gam

mhh X= 16x + 28y + 40z = 13,4 (1)

m+= (36+3+108).z = 14,7 (2)

TN 2: CH4: nx mol

C2H4 + Br2 C2H4Br2

ny ny

C3H4 + 2Br2 C3H4Br4nz 2nz

16,8/22,4 mol 108 gam

nhh X= nx + ny + nz = 16,8/22,4 (3)

mBr2= (ny + 2nz).160 = 108 (4)


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)gii h4 In ny cc bn thng ly pt (3) chia cho (4) (c mt pt mi sau ghp pt mi ny vi pt 1,2 (c h3 pt 3 In x, y, z gii ta s(c

nghim

Bi 4) Cho h)n hp X gm CH4, C2H4v C2H2. Ly 8,6gam X tc d"ng ht vi dung d&ch brom dth khi lng Brom phn ng l 48gam .

M-t khc nu cho 13,44 lt ktc h)n hp kh X tc d"ng vi lng dAgNO3trong NH3, thu c 36 gam kt ta. Phn tr/m thtch ca

CH4c trong X l

A.20% B.50% C.25% D.40%

Cch lm : t/ng tbi 2

Bi 5).t chy hon ton h)n hp X gm C2H2, C3H4v C4H4( c smol m)i cht b$ng nhau ) thu c 0,09 mol CO2. Nu ly cng mt

lng h)n hp X nhtrn tc d"ng vi mt lng ddung d&ch AgNO3trong NH3th khi lng kt ta thu c ln hn 4 gam. Xc &nh

cng thc cu to ca C3H4v C4H4

Suy lun: C3H4c 2 CTCT : CH2= C = CH2 v CH3 C .CH ; C4H4c 2 CTCT : CH2=C=C=CH2v CH2 = CH - C.CH . bi ny l l bt ta i tm CTCT c"thca C3H4v C4H4trong bi ny sl g ?

TN1: C2H2 + O2 2 CO2 + H2O

x 2x

C3H4 + O2 3CO2 + H2O

x 3x

C4H4 + O2 4CO2 + H2O

x 4x

nCO2= 2x + 3x + 4x = 0,09 x = 0,01 mol

TN2: C2H2 C2Ag2+ mC2Ag2+= 2,4 gam

0,01 0,01

C3H4 C3H3Ag+ mC3H3Ag+= 1,47 gam

0,01 0,01

C4H4 C4H3Ag+ mC4H3Ag+ = 1,59 gam

0,01 0,01

Nu chc mnh C2H2pth kt ta thu c chln hn 4 chng t*

C3H4phi p( vy CTCT ca n l CH3 C .CH ) nhng tng kt

ta do C2H2v C3H4to ra vn cha nn C4H4phi pnn CTCT

ca C4H4l CH2 = CH - C.CH

Bi 6)Cho 13,8 gam cht hu cX c cng thc phn t!l C7H8tc d"ng vi mt lng ddung d&ch AgNO3/NH3thu c 45,9 gam kt

ta . X c bao nhiu ng phn cu to th*a mn tnh cht trn.

A.5 B.6 C.4 D.2

Cch gi :C7H8 n = 7 TH1 : c 4 ni i = ; = ; = ; =

k= 4 TH2 : c 2 ni i , 1 ni 3 = ; = ; .

CnH2n +2 2k 2n + 2 -2k = 8 TH3 : c 2 ni ba .; .

Ta c C7H8(:R)1FG /F+G C7H8-xAgx+

0,15 0,15 mol

m+= 0,15.(12.7 + 8-x + 108.x ) = 45,9 x = 2 . Vy C7H8thAg+

trong AgNO3theo tl1:2 C7H8c 2 ni 3 u mch

Vy cc CTCT c thca C7H8lHC .C CH2 CH2CH2 C .CH

HC .C CH CH2 C .CHCH3

CH3HC .C CH C .CH HC.C C C .CH

C2H

5CH

3

Bi 7).t chy h)n hp X gm 2 hidrocacbon thuc cng dy ng 0ng thu c 19,712 lt kh CO2ktc v 10,08 gam H2O. Xc &nh

cng thc cu to ng ca A,B bit khi cho 1 lng h)n hp X tc d"ng vi AgNO3/NH3dth thu c 48 gam kt ta . Bit A, B u l

thkh.

C gng !fn


30/172

nCO2= 0,88 mol > nH2O= 0,56 mol .Suy ra X l CnH2n-2 = 0,88 -0,56 = 0,32 mol

c C = n =?,

?=

!\\!W"

= 2,75 V A,B thkh khng ktip nhau nn A l C2H2cn

B c thl C3H4ho-c C4H6.

bi ny ngi ta bt phi xc &nh chnh xc CTCT ng ca A v B

TH1 : A l C2H2 (CH.CH) : x mol v B l C3H4 (CH3-C.CH) : y mol

Ta c nhh X= x + y = 0,32 mol x= 0,08 mol

C ="}RBW}M

RBM 2,75 y= 0,24 mol

Cho h)n hp X tc d"ng vi AgNO3/NH3th c2 cht u phn ng

C2H2 C2Ag2+0,08 0,08

C3H4 C3H3Ag+

0,24 0,24

Khi lng kt ta thu c l : m+= 0,08. C2Ag2+ 0,24. C3H3Ag = 54,48 gam > 48

gam bi cho nn loi

TH2 :A l C2H2( CH.CH) cn B l C4H6( CH3-CH2-C.CH ho-c

CH3-C.C-CH3)


C ="}RB6}M

RBM 2,75 y= 0,12 mol

Cho h)n hp X tc d"ng vi AgNO3/NH3th

C2H2 C2Ag2+ mC2Ag2+= 0,2.240 = 48 (g)

0,2 0,2 mol

Vy chng t*C4H6phi khng phn ng nn CTCT ca C4H6phi l

CH3-C.C-CH3.

Cu 8:H)n hp kh X gm 0,3mol H2v 0,1mol vinyl axetilen . Nung X mt thi gian vi xc tc Ni thu c h)n hp kh Y c tkhi so

vi khng kh l 1 . Nu cho ton bY s"c ttvo dung d&ch AgNO3/NH3(d) th thu c m(g) kt ta . Ga tr&ca m l bao nhiu

A. 16 B. 5,3 C. 8 D. 32

Cch lm: Lm tng tbi 13 thuc chiu hng 2 ( ch C4H4 : CH2=CH-C.CH + AgNO3+ NH3CH2=CH-C.CAg + + NH4NO3 )

Cu 9 B- 2013: Cho 3,36 lt kh hirocacbon X (ktc) phon ton vi lng ddung d&ch AgNO3trong NH3, thu c 36 gam kt ta .

Cng thc phn t!ca X l


Cch lm: thay th0ng p n vo lm


31/172

PHN 2 : DN XUT HALOZEN - RU

HP CHT PHENOL

B MB MB MB MT CCT CC AAAA THITHITHITHI I HI HI HI HCCCC

B D/E D/.D FGHC IE2


32/172


33/172

nCO2= 0,88 mol > nH2O= 0,56 mol .Suy ra X l CnH2n-2 = 0,88 -0,56 = 0,32 mol

c C = n =

=

,

,= 2,75 V A,B thkh khng ktip nhau nn A l C2H2cn

B c thl C3H4hoc C4H6.

bi ny ngi ta bt phi xc nh chnh xc CTCT ng ca A v B

TH1 : A l C2H2 (CHCH) : x mol v B l C3H4 (CH3-CCH) : y mol


C =..

= 2,75 y= 0,24 mol

Cho hn hp X tc dng vi AgNO3/NH3th c2 cht u phn ng

C2H2 C2Ag20,08 0,08

C3H4 C3H3Ag

0,24 0,24

Khi lng kt ta thu c l : m= 0,08. C2Ag2+ 0,24. C3H3Ag = 54,48 gam > 48

gam bi cho nn loi

TH2 :A l C2H2( CHCH) cn B l C4H6( CH3-CH2-CCH hoc

CH3-CC-CH3)


C =..

= 2,75 y= 0,12 mol

Cho hn hp X tc dng vi AgNO3/NH3th

C2H2 C2Ag2 mC2Ag2= 0,2.240 = 48 (g)

0,2 0,2 mol

Vy chng tC4H6phi khng phn ng nn CTCT ca C4H6phi l

CH3-CC-CH3.

Cu 8:Hn hp kh X gm 0,3mol H2v 0,1mol vinyl axetilen . Nung X mt thi gian vi xc tc Ni thu c hn hp kh Y c tkhi so

vi khng kh l 1 . Nu cho ton bY sc ttvo dung dch AgNO3/NH3(d) th thu c m(g) kt ta . Ga trca m l bao nhiu

A. 16 B. 5,3 C. 8 D. 32

Cch lm: Lm tng tbi 13 thuc chiu hng 2 ( ch C4H4 : CH2=CH-CCH + AgNO3+ NH3CH2=CH-CCAg + NH4NO3 )

Cu 9 B- 2013: Cho 3,36 lt kh hirocacbon X (ktc) phon ton vi lng ddung dch AgNO3trong NH3, thu c 36 gam kt ta .

Cng thc phn tca X l


Cch lm: thay th!ng p n vo lm

NiNiNiNim tinm tinm tinm tin

Zgy th A: [lng 'c !n 'n 3Khi bn cm thy kh hiu hy g"i cho mnh

012 555 08 999 . mnh s#gii thch .


34/172

PHN 2 : DN XUT HALOZEN - RU HP

CHT PHENOL

CC C/ETG /oZI f2 ST D/E ./qZ foGChiu hng 4: !" thuy%t phn ng 5 )


35/172

3.pvi Mghp cht ckim

VD : CH3-CH2 Br + Mg CH3 CH2 Mg Br

( etyl magie brombua)

CH3- CH2 CH2 Br + Mg CH3 CH2 CH2 Mg Br

( propyl magie brombua )

Ch : hp cht cMg tan dtrong este pnhanh vi cc hp cht c hir

linh ng H2O , ancol v tc dng c vi CO2

VD: C2H5MgBr + CO2 C2H5COOMgBr

C2H5COOMgBr + HBr C2H5COOH + MgBr2

II) ng dng

D'n xut halozen c hot tnh sinh h"c rt a dng

- &c dng lm cht gy m trong ph(u thut. V d CHCl3,

(Cl)(Br)CH-Cl

- Lm ha cht di%t su b" C6H6Cl6

- V rt nhiu cht khc c cha halozen c dng phng trdch hi,

di%t c, kch thch sinh trng thc vt

Ch : CFCl3 v CF2Cl2trc y c dng ph*bin trong cc my lnh , hp

xt ngy nay ang bcm sdng , do chng gy hi cho t+ng ozon

Cu 57-a-2013:Trng hp sau y khng xy ra phn ng ?

(a). CH2=CH-CH2-Cl + H2O

(b). CH3-CH2-CH2-Cl + H2O

(C). C6H5-Cl + NaOH(c),

( vi C6H5- l gc phenyl )

(d). C2H5-Cl + NaOH

A.(b) B.(a) C.(d) D.(c)

Cu 41-B-2010:Pht biu no sau y ng?

A. Khi un C2H5Br vi dung dch KOH chthu c etilen.

B. &un ancol etylic 140oC (xc tc H2SO4c) thu c imetyl ete.

C. Dung dch phenol lm phenolphtalein khng mu chuyn thnh mu hng.

D. Dy cc cht: C2H5Cl, C2H5Br, C2H5I c nhi%t si t,ng d+n ttri sang phi

BI TP:

1).&un nng 13,875 gam mt ankyl clorua Y vi dung dch NaOH d, axit ha

dung dch thu c b)ng dung dch HNO3, nhtip vo dung dch AgNO3thy

to thnh 21,525 gam kt ta .Cng thc phn tca Y l

A.C2H5Cl B.C3H7Cl C.C4H9Cl D.C5H11Cl2) &un nng 1,91 gam hn hp X gm C3H7Cl v C6H5Cl vi dung dch

NaOH long va , sau thm tip dung dch AgNO3n d vo hn hp

sau pthu c 1,435 gam kt ta . Tnh khi lng mi cht trong hn hp ban

+u.

3) &un nng 24,7 gam CH3CH(Br)CH2CH3vi KOH dtrong C2H5OH , sau

khi phn ng xy ra hon ton thu c hn hp kh X gm 2 anken trong sn

ph(m chnh chim 80% , snph(m ph chim 20%. &t chy hon ton X thu

c bao nhiu lt CO2(ktc) Bit cc pxy ra vi hi%u sut l 100% .

4) &un si 15,7 gam C3H7Cl vi hn hp KOH/C2H5OH d, sau pd'n kh

sinh ra qua dung dch Br2d thy c X gam Br2tham gia p.Tm x, bit hi%usut pban +u l 80%.

- forGCC LO!I RU KHNG B"N

1 ) R CH = CH!"# $% &' $( !)* !+ R CH2 CHO

OH2 ) R C = CH2

!"# $% &' $( !)* !+ R CO CH3

OH

3 ) R CH OH!"# $% &' $( !)* !+ R CHO + H2O

OH

OH

1) R1 C R2!"# $% &' $( !)* !+ R1 CO R2+ H2O

OH

OH

5 ) R1 C OH

!"# $% &' $( !)* !+

RCOOH + H2OOH

A).pxi ho1.1)hon ton (t chy)

Cu 1-B-2014:Ancol no sau y c snguyn tcacbon b)ng snhm -OH?

A.Propan-1,2-iol. B.Glixerol. C.Ancol etylic. D.Ancol benzylic

Trl#i : p n ng B

A.CH3-CH(OH)-CH2(OH) B.CH2(OH)-CH(OH)-CH2(OH)

C.CH3-CH2-OH D.C6H5CH2OH

Cu 2-A-2014:Ancol X no, mch h, c khng qu 3 nguyn tcacbon trong phn

t. Bit X khng tc dng vi Cu(OH)2iu ki%n thng. Scng thc cu to

bn ph hp vi X l

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

.

Trl#i : ru X c sng tC-3 v ko tc dng vi Cu(OH)2tc l nu n l ru a

chc th n phi cha cc nhm OH khng lin k. Vy n c thl

1) CH3OH 2) CH3CH2OH 3) CH3CH2CH2OH

4).CH2(OH) CH2 CH2(OH) 5) CH3CH(OH)CH3

Cu 3-B-2013:Tn g"i ca anken (sn ph(m chnh) thu c khi un nng

ancol c cng thc (CH3)2CHCH(OH)CH3vi dung dch H2SO4c l

A.3-metylbut-2-en. B.2-metylbut-1-en.

C.2-metylbut-2-en. D.3-metylbut-1-en.


36/172

Ru + O2 t0

CO2+ H2O

1.2)ko hon ton ( CuO)

Ru bc 1 + CuO t0

Anhit + Cu + H2O

VD: CH3-CH2-OH + CuO t0

CH3-CHO + Cu + H2O

Ru bc 2 + CuO t0

Xeton + Cu + H2O

VD: CH3-CH-CH3 + CuO t0

CH3 CO CH3 + Cu + H2O

OH

Ru bc 3 ko b$xi ho

B)p%nhm chc (OH)2.1) Ru + ( Na,K) Mui ( Na,K) + H2&

VD : C2H5OH + Na C2H5ONa + H2.

C2H4(OH)2+ Na C2H4(ONa)2 + H2

Ch : ti to li ru tmui ( Na, K) .Cho tc dng vi axit v c

VD : C2H5ONa + HCl C2H5OH + NaCl

2.2.) Ru + Axit v ceste v c+ H2O

VD: C2H5OH + HCl C2H5Cl + H2O

+) Ru + Axit hu ceste hu c+ H2OVD: C2H5OH + CH3COOH CH3COOC2H5 + H2O

C2H4(OH)2 + CH3COOH C2H4(OOCCH3)2 + H2O

CH2(COOH)2 + CH3OH CH2(COOCH3)2 + H2O

2.3). Ru loi H2O %1700, H2SO4'c cho hnh thnh ni i ho'c lk (

Nguyn tc tch nc 170oC l OH s#tch cng H ca nguyn tC kcnh

nu tch cng H ca nguyn tC bc cao s#cho ra sp chnh cn bc thp ra sp

ph; tch ti vtr no th hnh thnh ni i ti vtr - xem vd2

VD1 : C2H5OH-, /0123 C2H4 + H2O

CH3 CH2 OH-, /0123 CH2= CH2+ H2O

VD2 : C4H9OH-, /0123 C4H8 + H2O

CH3 CH=CH CH3 (sp chnh)

CH3 CH2 CH - CH3-, /0123 + H2O

OH CH3 CH2 CH=CH2 ( sp ph)

Ru loi H2O %1400; H2SO4'c cho hnh thnh ete

VD : 2CH3OH -

, /0123 CH3OCH3+ H2O

CH3O + C2H5OH-, /0123 CH3OC2H5+ H2O

2.4).Ru cha pci Cu(OH)2fc )ng mu xanh

Ch : chc ru c cc nhm (OH) lin kmi tham gia pny

VD: CH2 CH2 CH2 + Cu(OH)2ko p

C3H8O2 OH OH

CH3 CH CH2 + Cu(OH)2p to fc ng mu xanh.

OH OH

C) phn ng %gc R* Nu gc R ko no th c thm pcng ( H2;X2lm mt mu dung d$ch

Br2; KMnO4; trng hp )

VD : CH2= CH CH2OH + H2 Ni

CH3 CH CH2OH

Suy lun :ptch nc thu c anken da trn nguyn tc OH tch cng H ca

nguyn tcc bon kcnh , nu tch cng nguyn tH ca ccbon bc cao th cho ra

sn ph(m chnh , bc thp cho ra sp ph( xem ph+n l thuyt)

CH3 CH CH - CH3-, /012 CH3 C = CH - CH2 + H2O

CH3 OH CH3

Vy tn ca sp chnh l : 2 metyl but 2- en

Cu 4-A-2012:Trong ancol X, oxi chim 26,667% vkhi lng. &un nng X

vi H2SO4c thu c anken Y. Phn tkhi ca Y l

A.42. B. 70. C. 28. D. 56

Cch lm :ru loi nc thu c anken chng t l ru no n chc

&t cng thc ca n l CnH2n+1OH

suy ra % O =-4

-- .566= 26,667 n= 3 anken Y l C3H6 = 42

Cu 5-A-2012:&t chy hon ton mt lng ancol X to ra 0,4 mol CO2v 0,5

mol H2O. X tc dng vi Cu(OH)2to dung dch mu xanh lam. Oxi ha X b)ng

CuO to hp cht h/u ca chc Y. Nhn xt no sau y ng vi X?

A. Trong X c 3 nhm -CH3.B.Hirat ha but-2-en thu c X.

C. Trong X c 2 nhm -OH lin kt vi hai nguyn tcacbon bc hai.

D. X lm mt mu nc brom.

Cch lm:

Nhn thy nH2O > nCO2 ancol X no : CnH2n+2Oxhoc CnH2n+2-x(OH)x

Ta c nru no = 0,5-0,4 = 0,1 ( xem ph+n cng thc t chy )

X c Snguyn tC =,,-= 7

X tc dng c vi Cu(OH)2X l ru a chc

Khi pvi CuO cho hp cht h/u ca chc

Vy X l CH3 CH CH CH3OH OH

&p n ng C.

Cu 6-B-2010: C bao nhiu cht h/u c mch h dng iu ch 4-

metylpentan-2-ol chb)ng phn ng cng H2(xc tc Ni, to)?

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

Suy lun : X + H2 to ra ru no n chc bc 2

4 metyl pentan 2 ol : CH3 CH CH2- CH CH3

CH3 OH

Vy X c thl1) Ru ko no n chc bc 2: CH2 = C CH2- CH CH3

CH3 OH

CH3 C = CH - CH CH3

CH3 OH

2) Xeton no n chc : CH3 CH CH2- CO CH3

CH3

3) Xeton ko no n chc c mt ni i gc R

CH2 = C CH2 CO CH3

CH3

CH3 C = CH - CO CH

3

CH3

Cu 7-B- 2010:Pht biu no sau y ng?

A. Khi un C2H5Br vi dung dch KOH chthu c etilen.

B. &un ancol etylic 140oC (xc tc H2SO4c) thu c imetyl ete.


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CH2= CH CH2OH + Br2 CH2-CH-CH2OH

Br Br

Nu gc R thm th c thm pthvo nhn bengen

CH2OH CH2OH

+ Cl289 Cl + HCl

III. *iu ch1).*iu chC2H5OH trong cng nghi+p

Hidrat ha etilen xuc tc axit: C2H4 + H2O/012 C2H5OH

Ln men tinh bt : (C6H10O5)n + n H2O9:;< nC6H12O6

Tinh bt glucozo

C6H12O69:;< 2C2H5OH + 2CO2

2).*iu chmetanol trong cng nghi+p

Ch : ru metanol l cht rt c , chc+n mt lng nhvo cthc0ng c

thgy m la , lng ln hn c thgy tvong

Cch 1: CH4 + H2O

, CO + H2

CO + 2H2, , 2CH3OH

Cch 2: 2CH4 + O2, , 2CH3OH

C. Dung dch phenol lm phenolphtalein khng mu chuyn thnh mu hng.

D. Dy cc cht: C2H5Cl, C2H5Br, C2H5I c nhi%t si t,ng d+n ttri sang phi

Trl#i :

A.

Sai v : C2H5Br + KOHC2H5OH + KBr

B. Sai v n phi thu c ietyl ete : 2C2H5OH C2H5OC2H5+ H2O

C.

Sai v phenol c tnh axit yu v ko lm *i my qu1tm c0ng nh

phenolphthalein

D. &ng v cng loi hp cht h/u cth khi lng ln nhi%t si cao

Cu 8-A-2010:Cho schuyn ho

C3H6 dung dch Br2 X NaOH Y CuO,t0 Z O2/xt T

CH3OH/xt E

Bit E l este a chc. Tn g"i ca Y l

A. glixerol. B. propan-2-ol.

C. propan-1,2-iol. D. propan-1,3-iol.

Suy lun :

C3H6c thl anken hoc xiclo ankan tha mn iu ki%n bi tp ny l to ra

este a chc E th C3H6phi l xicol ankan

+dung d$ch Br2 CH2-CH2-CH2 + NaOH CH2-CH2-CH2

Br Br OH OH+ CuO HOC CH2 CHO + O2/xt Mn

2+ HOOC CH2 COOH

+ CH3OH CH3OOC-CH2-COOCH3

Cn nu l anken th n sko to ra c este a chc m l hp cht hu c

tp chc

CH3 CH = CH2CH3CH = CH2 CH3 CH = CH2

Br Br OH OH

CH3 CO CHOCH3 CO COOHCH3 CO COOCH3

Zgy th 7 : h+c !m g z t{i h+c # s| t} m} # nhlng iu mnh

cha 'i%t , t{i h+c t| mnh c th i 6hm ph =c nhlng n~i mnh

cha t chn ti m 6h{ng cn %n phin )ch #in, t{i h+c c th

hiu =c nhlng c(n ngi t{i gp h+ sng # !m #ic th% n(, t{i

h+c c th hiu =c nhlng iu phi tri `ng sai, t{i h+c c

th t| ra =c $uy%t nh ch( s phn ca mnh Cui cng t{i h+c

c th hiu =c ch"nh 'n thn mnh , h+c tha mn sng- #

6hi ni nhlng !i ny ra th` th|c #i cc 'n hi cp > nhiu 6hi t{i


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h+c cng ( th 'i%t =c ! !m g z nhng n%u 6( h+c th mnh

!m g z mnh !m =c g z mi ! #n !n - .h`c (p.a 3

Chiu hng ; : ND phn ng t chy*$nh dng cng thc Cch 't cng thc Gii thch

1).Nu t chy 1 ru m thu c

nH2O > nCO2 l ru no

nru no= nH2O-nCO2

>

?@ABC CnH2n+2 - x(OH)x (1)CnH2n + 2 Ox (2)

Mt nh dng th s# c 2 cch

t cng thc dng chi tit .

Cng thc (1) dng cho cc bi

ton lin quan n p xy ra

nhm chc. Cng thc (2) thng

dng cho cc bi ton t chy.

C=FGH

; H=IFGJH

2).Nu t chy mt ru m thu c

nH2O = nCO2 ru c 12

>-K

?@ABC CnH2n-x(OH)x (1)CnH2nOx (2)

3).Nu t chy mt ru m thu c

nCO2> nH2Oru c slk 322.

Nhng khi lm th mc nh l 22lm

nru 22= nCO2 nH2O

>K

?@ABC CnH2n-2-x(OH)x (1)CnH2n 2Ox (2)

*,LM *C BI TP RU HAY BT K LO!I BI TP V"HP CHT H-U C.C CH/A NHM CH/C TH VI0C

QUAN TR1NG NHT L PH2I BI3T CCH *4T *C CNG TH/C SAO CHO *NG V PH HP V5I T6NG LO!IPTP/. KHI * M5I VI3T V CN B7NG *C PTP/. BI TON S8TR9NN *.N GI2N *I.

VY *4T CNG TH/C LM SAO CHO *NG NGUYN T:C L CN PH2I *;NH D!NG *C N R


39/172

Bi 2,&t chy hon ton 10,6g h22 ru n chc l ng !ng ktip thu c 11,2 lit CO2v 12,6g H2O. X&CTPT ca 2 ru v % s

mol tng ru

Nhn thy nH2O = 0,7 mol > nCO2= 0,5 mol. & l 2 ru no n chc

thuc cng dy ng !ng: CnH2n+2Oxn 2ru no= 0,7 0,5 = 0,2 mol

C= n =,M,

= 2,5 hn hp 2 ru l C2H5OH x mol v C3H7OH y mol

Ta c : nhh 2 ru = x + y = 0,2

nhh 2 ru=..

= 2,5

x=0,1 mol ; y = 0,1 mol

Bi 3,&t chy hon ton 1,52g mt ru X thu c 1,344 lt CO2ktc , 1,44g H2O . X&CTCT ca X bit X c khn,ng ha tan Cu(OH)2

k thng. Cch lm ging bi 1

nH2O= 0,08 mol > nCO2= 0,06 mol X l ru no CnH2n+2Ox

nru X= 0,02 mol C= 0,06/0,02 = 3 . X l C3H8Ox

mru= (44 + 16x ).0,02 = 1,52 x = 2 vy n l C3H8O2hay chnh l C3H6(OH)2Nu n ko cho khi lng th ta phi suy lun nhvy

V X c khn,ng ha tan Cu(OH)2nn X phi l ru achc X phi c snhm OH 32, ktip nhau v koc vt qu 3 (V quy nh ca ru l snhm chcphi nhhn snguyn tC)X l CH3-CH-CH2 hoc CH2 CH CH2

OH OH OH OH OH

Bi 4,&t chy hon ton ru X thu c CO2 v H2O c tl%smol 3:4 . Hi c bao nhiu CTCT c thc ca X :

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

Ly lun smol CO2= 3 mol ; H2O b)ng 4 mol lm. Khi thng qua

smol ca CO2v H2O ta kt lun y l ru no : CnH2n+2Ox c smol =

4-3 = 1 mol snguyn tC= 3/1 = 3 C3H8Ox. V snhm chc ca

ru khng bao givt qua snguyn ttcc bon nn ta c cc cng

thc ca ru c thtn ti vi 3 nguyn tcc bon l

1)ru n chc: CH3-CH2-CH2(OH) ; CH3-CH(OH)-CH3

2)ru 2 chc: CH3- CH(OH)-CH2(OH) ; CH2(OH)-CH2-CH2(OH)

3)ru 3 chc : CH2(OH)-CH(OH)-CH2(OH)

&p n A

Cu 5-A-2010:&t chy hon ton m gam hn hp 3 ancol n chc, thuc cng dy ng !ng, thu c 3,808 lt kh CO2(ktc) v 5,4 gam

H2O. Gi trca m l A. 5,42. B. 4,72. C. 7,42. D. 5,72

&t 3 ancol n chc thuc cng dy ng !ng thu c H2O = 0,3 mol > CO2= 0,17 mol

l 3 ancol no n chc : CNOH2NO+2O c smol = 0,3 0,17 = 0,13 mol

snguyn tPQ = NO= ,-,-

Vy khi lng ca 3 ancol = 0,13.(14.,-,-+ 18 ) = 4,72 (g)

Cu 6-B-2010:&t chy hon ton mt lng hn hp X gm 2 ancol (u no, a chc, mch h, c cng snhm -OH) c+n va V lt

kh O2, thu c 11,2 lt kh CO2v 12,6 gam H2O (cc thtch kh o ktc). Gi trca V l

A. 11,20. B. 14,56. C. 4,48. D. 15,68.

Cch l :&t 2 ancol (u no, a chc, mch h, c cng snhm -OH) thu c

H2O = 0,7 mol v CO2= 0,5 mol .

V l 2 ru no nn smol 2 ru no = 0,7 0,5 = 0,2 mol

snguyn tPQ = ,M,

= R,S nhvy phi c 1 ru c snguyn tC < 2,5 .

&y l ru a chc nn ru b hn 2,5 chc thl CH2 CH2

OH OH

B hn 2,5 chc th= 2. M l ru a chc th khng thc snhm OH = 1

nn c phi b)ng 2

V 2 ru c cng snhm OH nn t cng thc chung 2 ru lCNOH2NO+2O2hay CNOH2NO(OH)2

TUVWUVXC ,


40/172

NgNgNgNgy thy thy thy th8 :8 :8 :8 : KHKHKHKHNGNGNGNG

KHUKHUKHUKHUAAAAT PHT PHT PHT PH=CCCC

C l# tng trong i chng ta thy cuc sng c qu

nhiu kh khn, qu nhiu chng gai th thch v i

khi tht bt cng b)ng, ngi th c qu nhiu, ngi

th khng c gCng c nhiu c m c vit ra,c ni !n v c"ng ti 4# v $ ngh%a nhng ri n

v'n ch gic m thi thong b!n ch&n tr ngi ta

ni vi nhau' 5c m, kht v"ng suy cho c"ng (t nhiu

ai ch!ng c nhng khng c nhiu ngi hi%n thc c

c m, c kht v"ng ca mnh bi nh/ng kh khn, nh/ng ththch m ngi ta g# #hi tr!n ng i n

n) C nh/ng kh khn, c nh/ng ththch chng ta vt qua, nhng c nh/ng kh khn chng ta tng chng

nhkhng thvt qua v chng ta tb

Cuc sng vn khng chc mu hng, kh khn un un mt #h+n tt y!u ca cuc sng, kh khn ai cng g# #hi ch*u #hi +nh ri!ng cho ta' hng c hng ngn hng tri%u ngi tr!n thgii ny v'n thnh

cng v h" un un nc, h"khng bao gi+u hng) h/ng ngi thnh cng y c thnh #h+n tngi

bnh thng ti nh/ng ngi g# khuyt tt nng nnhick -ujicic.ick mt ngi c bi%t, v anh khng c

ch*n, khng c tay khi sinh ra v thm anh c bi%t +anh /s"+a0 hng vt !n tt c, anh sng mt cuc

sng tuy%t vi1'

ng ngi than vn vi nh/ng kh khn m mnh ang gp phi, bi bn c than c6

no kh khn cng khng c gii quyt v kht v"ng v'n chl kht v"ng. y ng l!n

hnh ng v ng bao gi+u hng"

Chiu hng >: NME DQ. [T . [E BEP L1JE BETP1) R(OH)x+ xNa R(ONa)x+ x/2 H2&

Bi tp cho phn ng xy ra %nhm chc nh>m 2 mc ch chnh

sau : - xc $nh snhm chc

- xc $nh smol c?a hp cht hu c

B>ng cch so snh t@l+vsmol

Nu bi cho bit smol c?a ru v smol c?a Na kh ta lp

nru : nNa= 1:x ru ny sc x nhm chc

Ch :Rt t trng hp cho l xc nh cng thc nht l i vi bitp cho trong thi

VD1: ROH + Na ROH + H2

nRu: nNa= 1:1 1 nhm chc

nRu= 2^_` 1 nhm chc

VD2: R(OH)2+ 2Na R(ONa)2+ H2

nRu: nNa= 1:2 2 nhm chc

nRu:^_`

= 1:1 2 nhm chc

Ch :

- Nu nhn thy (1:2) < n2 ru : nNa< (1:1)

hay 0,5 < n2 ru : nNa < 1 .Th suy ra trong 2 ru phi c 1

ru n chc, 1 ru 2 chc

VD : nru:nNa= 0,2 : 0,3 = 0,6666667

- Tng tAi vi t@l+vhiro.

2) *i vi nhng bi ton khi ng#i ta cho ru xo ho'c dung d$ch

ru th ta hiu trong g)m 2 thnh phn l ru v H2O

Chnh v vy khi cho kim loi kim (Na, K, Li) pvi dung d$ch ru

ho'c ru xo th n xy ra 2 p

VD : Cho Na vo dung d$ch ru C2H5OH ho'c ru C2H5OH 90o th

sxy ra 2 psau

Na + C2H5OH C2H5ONa + H2&

Na + H2O NaOH + H2&

*ru =aG ?bcB

add G ?bcB. 100

Ch :m= D.V Trong m : khi lng (g).

D : khi ln ring g/ml .

V l thtch (ml)

Ch :D H2O= 1 g/ml


41/172

Cu 1-B-2013:Hn hp X gm ancol metylic, etylen glicol. Cho m gam X phn ng hon ton vi Na d, thu c 2,24 lt kh H2(ktc).

&t chy hon ton m gam X, thu c a gam CO2. Gi trca a l

A.2,2. B.4,4. C.8,8. D.6,6.

Cch lm :

TN1:

CH3OH + Na CH3ONa + H2.

x x/2

C2H4(OH)2 + 2 Na C2H4(ONa)2 + H2.

y y

0,1 mol

Ta c nH2 = x/2 + y = 0,1 x + 2y = 0,2 (1)

TN2:

CH3OH + O2 CO2 + H2O

x x

C2H4(OH)2 + O2 2CO2 + H2O

y 2y

nCO2= x + 2y = ?

Nhn vo (1) ta suy ra c nCO2= 0,2 mol mCO2= 0,2.44 = 8,8 gam

Cu 2-B-2012:Cho hn hp X gm ancol metylic, etylen glicol v glixerol. &t chy hon ton m gam X thu c 6,72 lt kh CO2

(ktc). C0ng m gam X trn cho tc dng vi Na dthu c ti a V lt kh H2 (ktc). Gi trca V l

A. 3,36. B. 11,20. C. 5,60. D. 6,72

Cch lm :TN1 :

CH3OH + O2 CO2 + H2O

x x

C2H4(OH)2 + O2 2CO2 + H2O

y 2y

C3H5(OH)3 + O2 3CO2 + H2O

z 3z

0,3 mol

ta c : nCO2 = x + 2y + 3z = 0,3 mol (1)

TN2:CH3OH + Na CH3ONa + H2&

x x/2

C2H4(OH)2 + Na C2H4(ONa)2 + H2&

y y

C3H5(OH)3 + Na C3H5(ONa)3 + 3/2 H2&

z:

nH2=e f e :

= ? hay nH2 =

:

= ?

Nhn vo (1) ta suy ra nH2=,

= 6,5Smol

Cu 3-B-2012:&t chy hon ton m gam hn hp X gm hai ancol, thu c 13,44 lt kh CO2(ktc) v 15,3 gam H2O. Mt khc, cho m gam X

tc dng vi Na (d), thu c 4,48 lt kh H2(ktc). Gi trca m l

A. 12,9. B. 15,3. C. 12,3. D. 16,9.

TN2 :

gV(OH)hQ + Na gV(ONa)hQ + Q H2.

,Q

70,2 mol

n(O) ru= 0,4 mol

TN1:

&t: ( hai ru ) + @[


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Cu 5,Ly 1 lng Na dtc dng vi 18,7g h2X gm 3 ru n chc th thu c 29,7g sn ph(m rn . Tm CTPT ca ru c phn ng khi nh

nht A.C2H5OH B. CH3OH C. C3H7OH D. C3H5OH

Lm tng tbi1

Cu 6,Cho 15,6g h22 ru no n chc ktip nhau trong dy ng !ng tc dng ht 9,2g Na thu c 24,5g cht rn khi c cn . Hai ru l :

A.C2H5OH v C3H7OH B. CH3OH v C2H5OH C. C4H9OH v C3H7OH D. Kt qukhc

CnH2n + 1OH + Na CnH2n+1ONa + H2

x x/2Na d Nad

15,6 (g) 9,2(g) 24,5(g)

BTKL mH2= 15,6 + 9,2 24,5 = 0,3 ( g )2. x/2 = 0,3 x = 0,3

mru = x. (14n + 18 ) = 15,6 n = 2,4

V 2 ru k stip nhau nn 2 ru l C2H5OH v C3H7OH

Ch thch: (1)vnguyn tc i xc nh cng thc phn ttheo ptpth

phi t km theo smol nn ta t l x(2)V ru pht vi Na nhng Na c ht hay ko th mnh cha

bit c v vy c+n phi phng trng hp Na cn d. Nn bi ny cch

thit lp pt ko cn ging bi 1 n/a

Cu 7,Hn hp X gm 2 ru no n chc A , B c tl%smol 1:4 . Cho 9,4g h2X vo bnh ng Na dthy KL bnh tng 9,15g . X&CTCT v g"i

tn A , B bit chng cng bc cabon :

A.metylic v etylic B. metylic v propylic C. metylic v anlylic D. etylic v propylic

CnH2n + 1OH + Na CnH2n + 1ONa + H2.

x x/2

CmH2m + 1OH + Na CmH2m + 1ONa + H2.

4x 4x/2

mru= x.(14n + 18) + 4x(14m + 18) = 9,4 x = 0,1 mol

mH2 = (x/2 + 4x/2). = 9,4 9,15 n + 4m = 7

Xt :

m 1 2

n 3 -1

Vy 2 ru c+n xc nh l C3H7OH v CH3OH . &p n ng B

Ch thch :

(1)Vnguyn tc i xc nh cng thc phn tphi t km theo s

mol

(2)Sau khi t nhn thy bi ton c 3 (n. nhng c 2 sli%u Tha (n

thiu pt i xc nh cng thc ta bi%n lun tm n v m

(3)Cho 9,4 gam ru vo bnh ng Na dl8ra khi lng bnh phi

t,ng ln 9,4 gam nhng n cht,ng ln 9,15 gam chng tn bmt

i mt ph+n do H2thot ra mH2= 9,4- 9,15 = 0,25 (g)

Cu 8,H2X gm 2 ru n chc A. B hn km nhau 2 nguyn tC trong phn tt chy hon ton 12,2g h2X thu c 22g CO2v 12,6g H2O .

Mt khc nu cho 12,2g h2X trn vo bnh ng Na dthy KL bnh t,ng 11,9g. CT ca 1 trong 2 ru l :

A.CH3(CH2)2OH B. CH3(CH2)3OH C.CH2=CH-CH2-OH D. CH2=CH-CH2-CH2-OH

TN2 : mH2 = 12,2 11,9 = 0,3 (g) 0,15 mol

ROH + Na RONa + H2

0,3 mol 70,15 mol

TN 1: ROH + O2 CO2 + H2O0,3mol 0,5mol 0,7mol

12,2 (gam)

Snguyn tC = 0,5/0,3 = 1,6

Nhvy phi c 1 ru c snguyn tC nhhn 1,6 n phi b)ng 1 CH3OH

Cu 9,X l ru no n chc bc 1 , Y l ru no a chc cho h2A gm 0,1mol ru v 9,2g ru Y vo bnh ng Na dthu c 4,48 lt H2ktc .

Mt khc t chy hon ton h2A thu c 26,4g CO2v 14,4g H2O . X&tn g"i X , Y

A.X l metylic , Y l etylen glicol B. X l etylic , Y l etylen glicol C.X l propanol , Y l glixerol D. X l etylic , Y l glixerol

TN2: CnH2n+1OH + O2 nCO2 + H2O

CmH2m+2-x(OH)x+ O2 mCO2 + H2O

0,6mol 0,8 mol

n2ru no= 0,8 - 0,6 = 0,2 mol nru Y = 0,2 nru X= 0,1 mol

TN1:CnH2n+1OH + Na CnH2n+1ONa + H2.

0,1 mol 0,05 mol

CmH2m+2-x(OH)x + Na CmH2m+2-x(ONa)x + x/2 H2.

0,1 mol 0,05x mol


43/172

Ta c nCO2= 0,1n + 0,1m = 0,6 (*) Ta c nH2= 0,05 + 0,05x = 0,2 x= 3

mru Y= (14m + 2 +16x).0,1 = 9,2 (g). m=3 Y l C3H5(OH)3

Thay vo (*) suy ra n = 3 X l C3H7OH . Vy p n ng C

Cu 10,TN1: Trn 0,015mol ru no X vi 0.02mol ru no Y ri cho h2tc dng ht vi Na c 1,008 lt H2(kct)

TN2: Trn 0,02mol ru no X vi 0,015mol ru no Y ri cho h2tc dng ht vi Na thu c 0,952 lt H2

TN3: &t chy hon ton 1 lng ru nhtrong TN1 ri cho tt csn ph(m chy i qua bnh ng CaO mi nung dthy KL bnh t,ng 6,21g .

Bit V o ktc . X&CTCT 2 ru

TN1:CnH2n + 2 x(OH)x + Na CnH2n+2-x(ONa)x + x/2 H2.

0,015 0,0075x

CmH2m + 2 y(OH)y + Na CmH2m+2-y(ONa)y + y/2 H2.

0,02 0,01y

Ta c nH2= 0,0075x + 0,01y = 0,045 mol (1)

TN2: CnH2n + 2 x(OH)x+ Na CnH2n+2-x(ONa)x + x/2 H2.

0,02 0,01x

CmH2m + 2 y(OH)y + Na CmH2m+2-y(ONa)y + y/2 H2.

0,015 0,0075y

Ta c nH2= 0,01x + 0,0075y = 0,0425 mol (2)

Gii (1) v (2) ta c x= 2 v y = 3

TN 3: &t chy 2 ru TN1 vi x=2 ; y=3 ta c CT mi ca 2 ru l

CnH2n(OH)2 + O2 nCO2 + (n+1)H2O

0,015 0,015n 0,015(n+1)

CmH2m-1(OH)3 + O2 mCO2 + (m+1)H2O

0,02 0,02m 0,02(m+1)

m bnh ng CaO t,ng chnh l t*ng khi lng ca CO2 v H20 == (0,015n +0,02m).44 +

[0,015(n+1) + 0,02(m+1)].18 = 6,21

Lp bng bi%n lun n v m ta xc nh c n = 2 v m = 3

Vy cng thc 2 ru l C2H4(OH)2 v C3H5(OH)3

Cu 11 ,Cho 1 lt cn etylic 920tc dng vi Na dbit KL ring ca ru etylic D = 0,8(g/ml) . Tnh i/thot ra

A.22,4 lit B. 228,96 lt C. 289,8 lt D. 822,9 lt

Vrou C2H5OH = 920ml mC2H5OH =D.V = 736(g)nC2H5OH =16

1000ml cn C2H5OH 92o

*ru =aG ?bcB

add G ?bcB. 100 VH2O= 1000-920=80ml m H2O= D.V = 80 (g) nH2O= 40/9 mol

Ch : Dca H2Occ bn phi ng+m hiu = 1 g/ml . Mc d bi ko cho cc bn phi ta vo lm.

Ta c p

C2H5OH + Na C2H5ONa + H2.

16mol 8 mol

H2O + Na NaOH + H2.

40/9 40/18 mol

VH2=(8+ 40/18).22,4

Cu 12,Cho 10,1g dung dch ru etylic tc dng vi Na dthu c 2,8 lt kh ktc . X&ru . Bit D = 0,8g/ml

A.92,70 B. 79,20 C. 86,90 D. 90,20

C2H5OH + Na C2H5ONa + H2.

x x/2

H2O + Na NaOH + H2.

y y/210,1(g) 0,125mol

Ta c : 46x + 18y = 10,1 v x/2 + y/2 = 0,125x= 0,2 ; y = 0,05mru= 0,2.46=9,2(g); mH2O= 0,9(g

mC2H5OH =D.V Vrou C2H5OH = 9,2/0,8 = 11,5 ml

m H2O= D.V VH2O= 0,9/1 = 0,9 ml

*ru =aG ?bcB

add G ?bcB. 100 =jj,k

jj,kl,m . jll =92,7o

Cu 13,Ngi ta iu chetylen b)ng cch un nng ru C2H5OH 920vi H2SO4c , tnh in/o1/92

0c+n a vo phn ng thu c 2,24 lt etylen

ktc ., bit Hp/= 62,5% , D = 0,8g/ml

A.10ml B. 15ml C. 20ml D. kt qukhc

C2H5OH/012p3,- C2H4 + H2O

B: ?

Hp=62,5% P: 0,1 mol 70,1

Vi Hp=qG$p

.566 nb= 0,1.100/62,5 = 0,16 mol

mC2H5OH = 0,16.46 = 7,36 (g)

mC2H5OH =D.V Vrou C2H5OH = 7,36 /0,8 = 9,2 (g)

*ru = aG ?bcBadd G ?bcB

. 100 Vru 92o = 9,2.100/92 = 10 ml


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Cu 14,Ngi ta tin hnh iu chru etylic ttinh bt . Hy cho bit KL tinh bt c+n ly iu ch100 lt cn etylic 920. Bit H ca ton bqu

trnh l 80% . Bit rn/o1/= 0,8g/ml

A.129,6kg B. 162kg C. 202,5kg D. kt qukhc

100 lt cn etylic 92o . p dng cng thc tnh ru suy ra

Vrou C2H5OH = 92 lt (=92 000 ml) mC2H5OH =D.V = 92000.0,8= 73600 (g)

nC2H5OH =1600 mol

Cch 1: (C6H10O5)n C6H12O6 2 C2H5OH

B: ?H= 80% P: 800mol 71600 mol

Hqu trnh= 80% n (C6H10O5)n b= 800.100/80 = 1000 mol

mtinh bt = 1000.162 =162000 (g) = 162 kg

Cch 2: (C6H10O5)n + nH2O nC6H12O6

P: 800 7800 mol

C6H12O6 2 C2H5OH + 2 CO2

P: 800 71600 mol

Hqu trnh= 80% n (C6H10O5)n b= 800.100/80 = 1000 molmtinh bt = 1000.162 =162000 (g) = 162 kg

Cu 15-A-2011:Ancol etylic c iu chttinh bt b)ng phng php ln men vi hi%u sut ton bqu trnh l 90%. Hp thton blng

CO2sinh ra khi ln men m gam tinh bt vo nc vi trong, thu c 330 gam kt ta v dung dch X. Bit khi lng X gim i so vi khi lng

nc vi trong ban +u l 132 gam. Gi trca m l

A. 324. B. 405. C. 297. D. 486

mdung dch gim= mtch ra - mhp thmdung dch gim = mCaCO3 mCO2

hay 132 = 330 mCO2 mCO2 = 198 gam 4,5 mol CO2

(C6H10O5)n C6H12O6 2 C2H5OH + 2CO2B: ?

H= 90% P: 4,5/2 74,5 mol

Hqu trnh= 90% n (C6H10O5)n b=,M.

. -

s= 2,5 mol

mtinh bt = 2,5 .162 = 405 (gam)

Cu 16-A-2013:Ln men m gam glucozo to thnh ancol etylic (hi%u sut phn ng b)ng 90%). Hp thhon ton lng kh CO2sinh ra

vo dung dch Ca(OH)2dthu c 15(gam) kt ta. Gi trm l

A.7,5 B.15 C. 18,5 D.45

Cch lm : lm tng t- t pt ny ( CO2 + Ca(OH)2d CaCO3 + H2O) ta s#tnh c CO2; cn pt ln men ny ( C6H12O6Z9


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C/iu hng ?: NME DQ. [T . DC/L1JE ZopC1) Ru tch loi nc %170oc H2SO4'c xc tc

Nu 1 ru no tch nc thu c anken chng

bi mat de thi quoc gia tap 2 huu co

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