Bi Mat de Thi Quoc Gia Tap 2 Huu Co
-
Upload
nganhanguyen -
Category
Documents
-
view
215 -
download
0
Transcript of Bi Mat de Thi Quoc Gia Tap 2 Huu Co
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
1/172
Cu n 2)
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
2/172
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
3/172
.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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
4/172
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
5/172
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
6/172
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
7/172
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
8/172
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)
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
9/172
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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$ng 46e) Sai : v n xy ra chm v theo cchxc &nhf) Sai v k=2 ; nu cha nhn benzene k =4 ( K l slk ,ho-c vng )
Cu 5)BenzenB+FG(:) A1
BC>,/ Q (:) A2
A2l :
A.1-nitro-3-brom benzen B.1-brom-4-nitro benzen
C.m-brom nitro benzen D.p-brom nitro benzen
cch lm: NO2
+ HNO3+,-. + H2O
NO2 NO2
+ Br2QP(:) Br + HBr
Cu 6:CH3COONa metan axetilen benzen tolulen axit
benzoic natri benzoat benzen
Cch lm :
CH3COONa + NaOH CH4 + Na2CO3
2CH4POOF C2H2 + 3H2 (C2H2hay CH.CH)
3C2H2>? @7A C6H6
C6H6+ CH3Cl1TTG C6H5CH3 + HCl ( xem phn iu ch benzene )
CH3
hay ( + CH3Cl1TTG + HCl ))
CH3 COOH
+ KMnO4 + H2SO4 + K2SO4+ MnSO4+ H2O
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
13/172
CH = CH2 CH = CH2
+ Cl2Q (:) Cl + HCl
a) Thnitro (-NO2)
NO2
+ HNO3+,-.VL (:) + H2O
CH3
CH3 NO2
+ HNO3+,-.VL (:) + H2O
CH3
NO2
CH = CH2 CH = CH2
+ HNO3+,-.VL (:) NO2 + H2O
3.P"thhalogen vo nhnh (nh sng lm xc tc )CH2 CH3 CH2 CH2(Cl)
+ Cl2?@ ?
+ HCl
CH(Cl) CH3
4. P.C!NG PH VBLIN KT 8Benzen v dy ng 0ng benzen rt kh tham gia phn ng cng chtham gia
vo phn ng cng vi H2v Cl2
C6H6 + 3H2F3 C6H12 (xiclohexan)
+ 3H2
C6H6 + 3Cl2 C6H6Cl6 ( thuc trsu 6.6.6)
III). I1U CH1.iu chbenzene
3C2H2>? @7A C6H6 ( ) )
COOH COONa
+ Na + H22
COONa
+ NaOH + Na2CO3
Cu 7:C2H5COONa C2H6C2H2C6H6toluenTNT
Cch lm :
C2H5COONa + NaOH C2H6 + Na2CO3
C2H6J!R C2H2 + 2H2 ( xem ptch loi H2ca ankan )
3C2H2>? @7A C6H6
C6H6+ CH3Cl1TTG C6H5CH3 + HCl ( xem phn iu ch benzene )
CH3
hay ( + CH3Cl 1TTG + HCl ))
CH3CH3
+ 3HNO3+,-.VL (:W) NO2 NO2 + 3H2O
NO2 (T.N.T 2,4,6 tri nitro toluene)
Cu 8 : vi vi sng t en axetilen benzentoluen
C6H5CH2Br2C6H5CH2OH
Cch lm :
CaCO3J CaO + H2O
CaO + CJ CaC2 + CO
CaC2 + H2O Ca(OH)2 + C2H22
3C2H2>? @7A C6H6
C6H6+ CH3Cl1TTG C6H5CH3 + HCl ( xem phn iu ch benzene )
CH3
hay ( + CH3Cl1TTG + HCl ))
C6H5CH3 + Br2 C6H5CH2Br + HBr
(xem phn thhalozen vo nhnh )
C6H5CH2Br + NaOHJ C6H5CH2OH + NaBr
Cu 9) C2H2C2Ag2C2H2CH3CHO
C2H4Br2C2H2C6H6C6H6Cl6
Fc ti%p n(
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
14/172
2.iu ch$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 (
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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).
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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$ng 11. H)n hp Y phn ng ti a vi a mol Br2trong dung d&ch. Gi tr&ca a l
A.0,1. B.0,2. C.0,4. D.0,3.
Cch 1:
C2H2 + H2F3 C2H4
P: x x x
C2H2 + 2 H2F3 C2H6
P: y 2y y
C2H4 + H2F3 C2H6
P: z z z
TM T4T :
C2H2: 0,1 mol
C2H4:0,2 mol Ni,to
H2: 0,3 mol
H)n hp X trc p
H)n hp Y sau pc 22
BTKL : mY(sau)= mX(trc)= 0,1.26 + 0,2.28 + 0,3.2 = 8,8 gam
nY = 8,8/22 = 0,4 mol
Ta c : nH2 p= ntrc - nsau x + 2y + z = (0,1+0,2+0,3) - 0,4 = 0,2 (1)
nBr2 p = 0,2 2(x+y) + x + (0,2 z ) = 0,4 (x + 2y + z )
Nhn vo pt (1) ta ssuy ra c nBr2 p = 0,4 0,2 = 0,2 mol
Cch 2:
BTKL : mY(sau)= mX(trc)= 0,1.26 + 0,2.28 + 0,3.2 = 8,8 gam
nY = 8,8/22 = 0,4 mol
nH2 p= ntrc - nsau = (0,1+0,2+0,3) - 0,4 = 0,2 (1)
p dng t&l8:
Cho C2H2=0,1 mol c 2,, C2H4= 0,2 mol c 1 , lng ,b&ph v'hon ton qua
hai ln tham gia pcng l khi tham gia pcng H2v sau l Br2.
Ta c n,ban u = nH2 p+ nBr2 p
0,1.2 + 0,2.1 = 0,2 + xx = 0,2 mol . Vy smol Br2pl 0,2 mol
.hi ti%n !fn nqa ch
C2H2cn d + 2Br2 C2H2Br4
0,1-(x+y) 0,2 - 2(x+y)
C2H4 cn d + Br2 C2H4Br2
x+(0,2-z) x + (0,2-z)
C2H6
(y+z)
H2cn d
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
23/172
Cu 2-B-2014:Mt bnh kn chcha cc cht sau: axetilen (0,5 mol), vinylaxetilen (0,4 mol), hiro (0,65 mol) v mt t bt niken. Nung nng
bnh mt thi gian, thu c h)n hp kh X c tkhi so vi H2b$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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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 .
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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 $u hd bn c t c iu
i ch3ng na. 9-y hnh ng th%o nhng g tri tim v trc gic bn mch bo chkh"ng phi nhng rung
ng tttng b!n ngoi, bi ai cng chsng c mt ln. 4hi - cht ri d bn c mun cng kh"ng th
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
28/172
lm nhng g m khi sng tri tim bn mch bo, du r$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)
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
29/172
)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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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)
Ta c nhh X= x + y = 0,32 mol x= 0,2 mol
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
A.C4H6 B.C2H2 C.C4H4 D.C3H4
Cch lm: thay th0ng p n vo lm
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
32/172
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
Ta c nhh X= x + y = 0,32 mol x= 0,08 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)
Ta c nhh X= x + y = 0,32 mol x= 0,2 mol
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
A.C4H6 B.C2H2 C.C4H4 D.C3H4
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 .
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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 )
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
!"# $% &' $( !)* !+
RCOOH + H2OOH
A).pxi ho1.1)hon ton (t chy)
Cu 1-B-2014:Ancol no sau y c snguyn tcacbon b)ng snhm -OH?
A.Propan-1,2-iol. B.Glixerol. C.Ancol etylic. D.Ancol benzylic
Trl#i : p n ng B
A.CH3-CH(OH)-CH2(OH) B.CH2(OH)-CH(OH)-CH2(OH)
C.CH3-CH2-OH D.C6H5CH2OH
Cu 2-A-2014:Ancol X no, mch h, c khng qu 3 nguyn tcacbon trong phn
t. Bit X khng tc dng vi Cu(OH)2iu ki%n thng. Scng thc cu to
bn ph hp vi X l
A.4. B.2. C.5. D.3
.
Trl#i : ru X c sng tC-3 v ko tc dng vi Cu(OH)2tc l nu n l ru a
chc th n phi cha cc nhm OH khng lin k. Vy n c thl
1) CH3OH 2) CH3CH2OH 3) CH3CH2CH2OH
4).CH2(OH) CH2 CH2(OH) 5) CH3CH(OH)CH3
Cu 3-B-2013:Tn g"i ca anken (sn ph(m chnh) thu c khi un nng
ancol c cng thc (CH3)2CHCH(OH)CH3vi dung dch H2SO4c l
A.3-metylbut-2-en. B.2-metylbut-1-en.
C.2-metylbut-2-en. D.3-metylbut-1-en.
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
36/172
Ru + O2 t0
CO2+ H2O
1.2)ko hon ton ( CuO)
Ru bc 1 + CuO t0
Anhit + Cu + H2O
VD: CH3-CH2-OH + CuO t0
CH3-CHO + Cu + H2O
Ru bc 2 + CuO t0
Xeton + Cu + H2O
VD: CH3-CH-CH3 + CuO t0
CH3 CO CH3 + Cu + H2O
OH
Ru bc 3 ko b$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.
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
37/172
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 $uy%t nh ch( s phn ca mnh Cui cng t{i h+c
c th hiu =c ch"nh 'n thn mnh , h+c tha mn sng- #
6hi ni nhlng !i ny ra th` th|c #i cc 'n hi cp > nhiu 6hi t{i
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
38/172
h+c cng ( th 'i%t =c ! !m g z nhng n%u 6( h+c th mnh
!m g z mnh !m =c g z mi ! #n !n - .h`c (p.a 3
Chiu hng ; : ND phn ng t chy*$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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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 ,
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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 ) + @[
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
42/172
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
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
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$p
.566 nb= 0,1.100/62,5 = 0,16 mol
mC2H5OH = 0,16.46 = 7,36 (g)
mC2H5OH =D.V Vrou C2H5OH = 7,36 /0,8 = 9,2 (g)
*ru = aG ?bcBadd G ?bcB
. 100 Vru 92o = 9,2.100/92 = 10 ml
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
44/172
Cu 14,Ngi ta tin hnh iu chru etylic ttinh bt . Hy cho bit KL tinh bt c+n ly iu ch100 lt cn etylic 920. Bit H ca ton bqu
trnh l 80% . Bit rn/o1/= 0,8g/ml
A.129,6kg B. 162kg C. 202,5kg D. kt qukhc
100 lt cn etylic 92o . p dng cng thc tnh ru suy ra
Vrou C2H5OH = 92 lt (=92 000 ml) mC2H5OH =D.V = 92000.0,8= 73600 (g)
nC2H5OH =1600 mol
Cch 1: (C6H10O5)n C6H12O6 2 C2H5OH
B: ?H= 80% P: 800mol 71600 mol
Hqu trnh= 80% n (C6H10O5)n b= 800.100/80 = 1000 mol
mtinh bt = 1000.162 =162000 (g) = 162 kg
Cch 2: (C6H10O5)n + nH2O nC6H12O6
P: 800 7800 mol
C6H12O6 2 C2H5OH + 2 CO2
P: 800 71600 mol
Hqu trnh= 80% n (C6H10O5)n b= 800.100/80 = 1000 molmtinh bt = 1000.162 =162000 (g) = 162 kg
Cu 15-A-2011:Ancol etylic c iu chttinh bt b)ng phng php ln men vi hi%u sut ton bqu trnh l 90%. Hp thton blng
CO2sinh ra khi ln men m gam tinh bt vo nc vi trong, thu c 330 gam kt ta v dung dch X. Bit khi lng X gim i so vi khi lng
nc vi trong ban +u l 132 gam. Gi trca m l
A. 324. B. 405. C. 297. D. 486
mdung dch gim= mtch ra - mhp thmdung dch gim = mCaCO3 mCO2
hay 132 = 330 mCO2 mCO2 = 198 gam 4,5 mol CO2
(C6H10O5)n C6H12O6 2 C2H5OH + 2CO2B: ?
H= 90% P: 4,5/2 74,5 mol
Hqu trnh= 90% n (C6H10O5)n b=,M.
. -
s= 2,5 mol
mtinh bt = 2,5 .162 = 405 (gam)
Cu 16-A-2013:Ln men m gam glucozo to thnh ancol etylic (hi%u sut phn ng b)ng 90%). Hp thhon ton lng kh CO2sinh ra
vo dung dch Ca(OH)2dthu c 15(gam) kt ta. Gi trm l
A.7,5 B.15 C. 18,5 D.45
Cch lm : lm tng t- t pt ny ( CO2 + Ca(OH)2d CaCO3 + H2O) ta s#tnh c CO2; cn pt ln men ny ( C6H12O6Z9
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
45/172
NgNgNgNgy thy thy thy th9999 :::: HHHHy giy giy giy gilllllyyyy c mc mc mc m~Ti c mt ngi bn tn Monty Robert, hin l ch nhn mttri nui ng ! "n #$i%ro& 'nh () cho *h+* ti %ng nh cnh (- t. ch/c nh0ng bu.i gy 1u2 nh3m ti tr4 cho c5c % 5n(6u t c t7nh ri ro co %o thnh nin thc hin&
Mt hm, nh (8n ng9i cnh ti : ni;
< Ti mu=n >- cho bn bi8t ti $o ti (- bn $? %@ng nh cti (- lm nAi t. ch/c gy 1u2&
Bhuyn CDy r c5ch (y nhiEu nFm& B mt cGu b+ $=ng cng:Hi ch c mInh, mt ngi lm nghE huJn luyn ng& Ko
cng :ic, ngi ch *hDi $=ng nh mt >L %u m@c& ng (i tN trng tri ny (8n trng tri >h5c (- huJn luync5c chO ng ch (4c thu6n ho5& P8t 1uD l :ic hQc hnh c cGu b+ >hng (4c .n (nh lSm& Mt hm,th6y gi5o bDo cGu b+ :E :i8t mt bi luGn :Fn :Hi (E ti UHn ln em mu=n lm nghE gIV&
Wm (, cGu b+ () :i8t bXy trng giJy m tD >h5t :Qng ngy no ( $Y lm ch mt trng tri nui ng& Zm
%i[n (t Hc mA c mInh thGt chi ti8t& ThGm ch7 em c\n :Y cD $A (9 tri nui ng tAng li :Hi %in t7ch>hoDng ]^^ m_u, trong ( em ch` ra ch no Cy nh, ch no (t lm (ng chy cho ng&
di8t Cong, cGu b+ (em bi n* th6y gi5o& di ngy $u, cGu b+ nhGn li bi lm c mInh :Hi mt (i-m totHng : mt %\ng bOt *h (f chi c th6y W8n g* ti $u gi hQc&
Th8 l cu=i gi cGu b+ (8n g* th6y : hfi;
< Th th6y, ti $o em li b (i-m V
< Zm () hoch (nh mt :ic m em >hng th- lm (4c& Hc mA c em >hng c cA $! thc t8& Zm >hng c
tiEn, li CuJt thn tN mt gi (Inh >hng c ch ! .n (nh& i chung, em >hng c (4c mt ngu9n lc >hD %no (- thc hin nh0ng % t7nh c mInh& Zm c bi8t (- lm ch mt tri nui ng thI c6n *hDi c rJt nhiEutiEn >hngV jy gi ti cho em :E lm li bi :Fn& 8u em $? ch0 cho n thc t8 hAn thI ti $Y c/u C+t (8n(i-m $= c em& Ra chV
km (, cGu b+ :E nh : ngh ng4i m)i& Bu=i cng cGu g* ch (- hfi >i8n&
hngV B5ch (y hi nFm, : th6y gi5o ( () tInh c %_n ^ hQc tr\ c mInh (8n (y (- cSm tri& Th8 l th6ytr\ ti nhGn r nhu& B6m ty ti, th6y ni;
Monty ny, >hi nh c\n hQc :Hi ti, ti () (5nh cS* Hc mA c nh, : $u=t bo nhiu nFm 1u ti cng ()m th8 :Hi bo (/ trL >h5c, ti rJt n hGn :E (iEu (&
ghe th6y ni th8, ti :i (5*;
Phng, th th6y, th6y >hng c li gI cD, chng 1u th6y ch` mu=n nh0ng gI t=t (p* $Y (8n :Hi hQc tr\ cmInh m thi& B\n em ch` mu=n theo (u.i tHi cng nh0ng >h5t :Qng c (i mInh&
-
7/24/2019 Bi Mat de Thi Quoc Gia Tap 2 Huu Co
46/172
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