Chuyen de 5 Dan Xuat Halogen Cua Hidrocacbon

7/26/2019 Chuyen de 5 Dan Xuat Halogen Cua Hidrocacbon

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http://bloghoahoc.comChuyn trang thi th ha Trang 1

CHUYN 5: DN XUT HALOGEN CA HIROCACBON

Cu 1 :S ng phn ca C4H9Br lA.4. B. 2. C. 3. D. 5.

Xem li chuyn 1 + k thut xc nh ng phn c bi C4H9Cl => tng t C4H9Br

mch i xngCCCC ; CCC Mi tn l v tr Br => 4 ng phn =>A

C

Cu 2:S ng phn dn xut halogen bc I c CTPT C4H9Cl lA.2. B.3. C.4. D.5.

S ng phn bc I => Cl gn vi C bc I ; mch i xng

III ICCCC ; CCC Mi tn l v tr Cl => 2 ng phn =>A

Bc I II II I IC

Cu 3:S ng phn mch h (k c ng phn hnh hc) ca cht c CTPT l C3H5Br lA. 2. B. 3. C. 4. D.5.

C3H5Br c k = (3.2 5 +1)/2 = 1 => c 1 lin kt i hidrocacbon v mch h khng phi vng nh xicloCH(Br) = CHCH3 c p hnh hc xem li K p hnh hc chuyn 1 => 2CH2=CH-CH2Br ; CH2=C(Br)-CH3 khng c p hnh hc => 2=> Tng c 4 p C3H5Br mch h. => CCu 4:Mt hp cht hu c Z c % khi lng ca C, H, Cl ln lt l: 14,28% ; 1,19% ; 84,53%. CTPT ca Z l

A. CHCl2. B. C2H2Cl4. C. C2H4Cl2. D.mt kt qu khc.Nhn thy %C + %H + %CL = 100% => Khng cn cht no naGi CTG

: (CxHyClz)n => x : y : z = %C/12 : %H/1 : %Cl/35,5 = 1,02 : 1,19 : 2,38 = 1 : 1 : 2=> (CHCl2)n Hay CnHnCl2nCch 1:lin kt pi + vng = (2 -1 +2-2).n / 2 = n/2 v s lin k pi + vng lun nguyn

n = 2 => C2H2Cl4 => B Cch 2 xem iu kin ca H - n gn ging cch 1Cu 5:Dn xut halogen khng c ng phn cis-trans lA.CHCl=CHCl. B.CH2=CH-CH2F. C.CH3CH=CBrCH3.D.CH3CH2CH=CHCHClCH3.Xem K ng phn nh ; R1#R2 v R3#R4 => B sai v R1 ging R2 HCu 6:Danh php IUPAC ca dn xut halogen c cng thc cu to : ClCH2CH(CH3)CHClCH3 l

A.1,3-iclo-2-metylbutan. B.2,4-iclo-3-metylbutan.C.1,3-iclopentan. D.2,4-iclo-2-metylbutan.Xem quy tc gi tn v cch lm chuyn 1 => nh s th t gn halogen nht

1 2 3 4

ClCH2CH(CH3)CHClCH3 Hay ClCH2CHCHCH3

CH3 Cl=> 1,3iclo 2metylbutan v tr 1 ,3 cha clo ; i v 2clo ; meyl v tr th 2; butan v mch chnh c 4CCu 7:Cho cc cht sau:C6H5CH2Cl ; CH3CHClCH3 ; Br2CHCH3 ; CH2=CHCH2Cl. Tn gi ca cc cht trnln lt l

A.benzyl clorua ; isopropyl clorua ; 1,1-ibrometan; anlyl clorua.B.benzyl clorua ; 2-clopropan ; 1,2-ibrometan;1-cloprop-2-en.C.phenyl clorua ; isopropylclorua ; 1,1-ibrometan; 1-cloprop-2-en.D.benzyl clorua ; n-propyl clorua ; 1,1-ibrometan; 1-cloprop-2-en.

Xem chuyn 4 => C6H5CH2 Bezyl => Loi C ;xem chuyn 1 => dng CH3 CH(CH3)l iso=> Loi B v D =>A Dng ny loi p n l chnh

Khng th bao qut nu mnh khng vng l thuyt n - l mch thng nhCH2=CHCH2l anlyl nh ; CH2=CH l vinyl ; SGK 11NC 211


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Cu 8:Cho cc dn xut halogen sau : C2H5F (1) ; C2H5Br (2) ; C2H5I (3) ; C2H5Cl (4) th t gim dn nhit si l

A.(3)>(2)>(4)>(1). B.(1)>(4)>(2)>(3). C.(1)>(2)>(3)>(4). D.(3)>(2)>(1)>(4).

Xem chuyn nhit si => i vi dn xut R-X , nu khng c lin kt hidro , nhit si s cng cao khi Xht e cng mnh.

Ta c Tnh phi kim t FBr>Cl>F => (3)>(2)>(4)>(1) =>A

Cu 9:Nh dung dch AgNO3 vo ng nghim cha mt t dn xut halogen CH2=CHCH2Cl, lc nh. Hin tngxy ra lA.Thot ra kh mu vng lc. B.xut hin kt ta trng.C.khng c hin tng. D.xut hin kt ta vng.

SGK 11 NC212 ci bng => Phi c un nng => Sinh ra kt ta i vi CH2=CHCH2CL Anlylclorua=> C khng c hin tng gCu 10: a. Sn phm chnh ca phn ng tch HBr ca CH3CH(CH3)CHBrCH3 l

A. 2-metylbut-2-en. B.3-metylbut-2-en. C.3-metyl-but-1-en. D.2-metylbut-1-en.Xem c ch p SGK 11NC 214 => Halogen tchcng H C bn cnh bc cao

Bc III I 1 2 3 4CH3CH(CH3)CHBrCH3 hay CH3CHCHCH3 => CH3C=CHCH3

CH3 Br CH3nh s th t gn lin kt i + gn mch nhnh nht=> 2metybut2en (CH3 v ni i v tr 3 v mchchnh c 4C)

b. Sn phm chnh to thnh khi cho 2-brombutan tc dng vi dung dch KOH/ancol, un nngA. metylxiclopropan. B.but-2-ol. C.but-1-en. D.but-2-en.

Bc I II 1 2 3 42-brombutan ; CH3CHCH2CH3 => CH3CH = CHCH3=> but2en

BrCu 11:un nng 13,875 gam mt ankyl clorua Y vi dung dch NaOH, tch b lp hu c, axit ha phn cn li

bng dung dch HNO3, nh tip vo dd AgNO3 thy to thnh 21,525 gam kt ta. CTPT ca Y lA.C2H5Cl. B.C3H7Cl. C.C4H9Cl. D.C5H11Cl.

Ankyl :CnH2n+1 => ankylclorua : CnH2n+1ClMo chng cn bit p g => on kt ta l AgCl v ly 21,525 chia cho 143,5MAgCl = 0,15 mol pBTNT Cl => nCnH2n+1Cl = nAgCl = 0,15 mol MY = 14n + 36,5 = 13,875/0,15n = 4 => C4H9Cl

Hoc p ban utora HCl => HCl + AgNO3 => AgCl kt ta trng+ HNO3Cu 12:S tch hiro halogenua ca dn xut halogen X c CTPT C4H9Cl cho 3 olefin ng phn, X l cht notrong nhng cht sau y ?

A.n- butyl clorua. B. sec-butyl clorua. C.iso-butyl clorua. D. tert-butyl clorua.Mo Nu khng bit cht l cht g copy paste vo google phn hnh nh => S hin ra cht Cch hcn- l mch thng tut A .CH3CH2CH2CH2 - ClSec c dng CH3 CH(X)-CH3 - => B . CH3CH(Cl)CH2CH3

Iso c dng CH3 CH(CH3) => C. CH3CH(CH3)CH2 - ClTert c dng CH3 (CH3)C(CH3) - D. CH3(CH3)C(CH3)-Cl A , C , D c Cl u mch => Ch tch cng vi duy nht C bn canh Nu C c H => Ch to ra ti ra 2 ng phn k c hnh hc nu c - i thi hy tim ra s khc bit gia cc p n cth loi v chn p n. => B ng V cho 3 ng phn khi thch HCl

I IICH3CHCH2CH3 => 2 sn phm : CH2=CH-CH2-CH3 ko c p hnh hc

Cl CH3-CH=CH-CH3 => c p hnh hc => 3pkhi tc HClCu 13: Cho hp cht thm : ClC6H4CH2Cl + dung dch KOH (long,d, to) ta thu c cht no ?

A. HOC6H4CH2OH. B. ClC6H4CH2OH. C. HOC6H4CH2Cl. D. KOC6H4CH2OH.SGK 11NC212 => Halogen gn vi vng benzen khng phn ng vi kim un nng nhit cao v p sut

cao mi pDn xut ankyl halogen Hay Halogen gnvi hidrocacbonno s p vi kim nhit un nngPT: ClC6H4CH2Cl + KOH => ClC6H4CH2OH + KCl => B


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Cu 14: Cho hp cht thm : ClC6H4CH2Cl + dung dch KOH (c, d, to, p) ta thu c cht no?A. KOC6H4CH2OK. B. HOC6H4CH2OH. C. ClC6H4CH2OH. D. KOC6H4CH2OH.

Cu 13 => Phn ng nhit v p sut PT : ClC6H4CH2Cl + 2KOH => OHC6H4CH2OH + 2KCl

Xem phn phn biu gia anol thm v phenol l thuyt chuyn 5 Ancol thm OH gn vi C ngoi vng benzen khng phn ng vi kim Phenol OH gn viC vng benzen c p vi kim

Do KOH d => PT : OHC6H4CH2OH + KOH => KOC6H5CH2OH+ H2O => sn phm p n DL thuyt n gin SGK nhng khng hc s rt kh lmCu 15: Thy phn dn xut halogen no sau y s thu c ancol ?

(1) CH3CH2Cl. (2)CH3CH=CHCl. (3) C6H5CH2Cl. (4) C6H5Cl.A.(1), (3). B.(1), (2),(3). C.(1), (2), (4). D.(1), (2), (3), (4).

SGK 11 NC 212 Halogen gn vi hidoracbon no => B thy phn to ra ancolPTTQ : RCH2Cl R c th l ng ng ankyl hoc ng ng ankanRCH2Cl + OH- => RCH2OH + Cl- P u tin SGK => (1) v (3) tha mnXt (2) . CH3 CH = CHCl + OH- => CH3CH = CHOH + HClxem li iu kin h bin ca ru trong phn l thuyt Dng R CH=CHOH tch H2O => andehit : RCH2CHO => (2) to ra andehit Xem bi ca

Nguyn Tn Trung trn htv4.vn hoc thaytro.vn

(4) SGK c p => C6H5OK khng phi ru C kim d

Hoc C6H5OH kim p va - y l phenolCu 16: a.un si dn xut halogen X vi nc mt thigian, sau thm dung dch AgNO3 vo thy xut hinkt ta. X l

A. CH2=CHCH2Cl. B. CH3CH2CH2Cl. C. C6H5CH2Br. D. A hoc C.SGK 11 NC212 => Dun si cht no => to ra HX => S c phn ng to kt ta khi cho AgNO3 X l Cl,Br,ITa c A v C c to ra khi un si => A , C to ra HX => c p kt taA ng ; C ng v Br gn vi C ngoi vng benzen => km bnB khng c p un nng Tng qut ankyl halogen khng p vi H2O iu kin thng hoc khi un nng=> D

b.un si dn xuthalogen X vi dung dch NaOH long mt thi gian, sau thm dung dch AgNO3 vo thyxut hin kt ta. X khng th l

A. CH2=CHCH2Cl. B. CH3CH2CH2Cl. C. C6H5CH2Cl. D. C6H5Cl.SGK => D chp vi kim nhit cao v p sut caoto ra NaCL hay c p vi AgNO3 sinh ra kt ta.chng khng phn ngvi NaOH nhit thng v khi un si

=> D . A,B,C nh Av un vi dung dch NaOHCu 17: Khi un nng dn xut halogen X vi dung dch NaOH to thnh hp cht anehit axetic. Tn ca hp chtX l

A.1,2- ibrometan. B.1,1- ibrometan. C.etyl clorua. D. A v B ng.=> to ra andehit axetic SGK 11 NC -239 CH3-CHOXem iu kin h bin ca ruthy Nguyn Tn Trung C trng hp 2 nhm OH gn vo cng 1 C TQ : RCH(OH)-OH => RCHO andehit + H2O R l hidrocacbon : H,CnH2n+1

B tha mn : CH2 CH(Br)-Br +NaOH => CH2-CH(OH)-OH b h bin to ra andehit nh trnA ch to ra OH-CH2-CH2-OH etilen glicol nu bo to thnhC ch to ra CH3-CH2-OHNh trnghp OH gn vi C khng no 1 ni i cng b h bin to ra andehitRCH=CH2-OH => RCH2-CHOCu 18:Hp cht X c cha vng benzen v c CTPT l C7H6Cl2. Thy phn X trong NaOH c(to cao, p cao) thu c cht Y c CTPT l C7H7O2Na. Hy cho bit X c bao nhiu CTCT ?

A.3. B.5. C.4. D.2.C7H6Cl2 + 2NaOH => C7H7O2Na + NaCl Xem li bi 14 p n D C7H7O2K=> X l ClC6H4CH2Cl Ly clo l v tr th 1 trong vng benzen => CH2Cl chy v tr o,m,p => 3 CTCT thamn Xem li bi benzen thy r hn v tr o,m,pv mt s bi xc nh ng phn benzenCu 19:Cho s phn ng sau:

CH3

XBr2/as Y

Br2/Fe, to

Zdd NaOH

TNaOH n/c, to, p


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X, Y, Z, T c cng thc ln lt lA.p-CH3C6H4Br, p-CH2BrC6H4Br, p-HOCH2C6H4Br, p-HOCH2C6H4OH.B.CH2BrC6H5, p-CH2Br-C6H4Br, p-HOCH2C6H4Br, p-HOCH2C6H4OH.C.CH2Br-C6H5, p-CH2Br-C6H4Br, p-CH3C6H4OH, p-CH2OHC6H4OH.D.p-CH3C6H4Br, p-CH2BrC6H4Br, p-CH2BrC6H4OH, p-CH2OHC6H4OH.

Kt hp vi bi benzen v chng ny P Br2/as => Br th vo H C ngoi vng benzen => BrCH2C6H5X=> loi A v D

Phn ng Br2/Fe,to => Br th vo vng benzen c th o hocp sn phm chnh p=> Y : pCH2Br-C6H4Br phn ng NaOH => Ch th Halogen gn vi C ngoi vng benzenXem bi 13 => p OHCH2C6H4Br => B V C p n khc

Cu 20:Cho s phn ng sau : CH4 X Y Z T C6H5OH. (X, Y, Z l cc cht hu c khc nhau).Z l

A. C6H5Cl. B.C6H5NH2. C.C6H5NO2. D.C6H5ONa.SGK 11 NC 231 => T l C6H5ONa Da vo T => C6H5OH => phn ng iu ch phenol

Xem SGK 11NC218 => Z lC6H5Cl => A Thm mnh ch on X l C2H2 ; Y l C6H6 ; Z l C6H6Cl ; T l C6H5ONa

Cu 21:X l dn xut clo ca etan. un nng X trong NaOH d thu c cht hu c Y va tc dng vi Na vatc dng vi Cu(OH)2 nhit thng. Vy X l

A.1,1,2,2-tetracloetan. B.1,2-icloetan. C.1,1-icloetan. D.1,1,1-tricloetan.

Va tc dng vi Na va tc dng vi Cu(OH)2 nhit thng => C nhiu gc OH lin k nhau (Hay gn viC k nhau VD: C2H4(OH)2 ; OHCH2-CH2-OH ; C3H5(OH)3 ; OHCH2CH(OH)CH2OHP vi Na => c H linh ng (VD; R - OH ;R - COOH ; C6H5OH)P vi Cu(OH)2 iu kin thng => R OH ; R - COOHXem s h bin ca ru na nh A . Cl-CH(Cl)- CH(Cl)-Cl + NaOH => CHOCHO (v to ra OH CH(OH)CH(OH)-OH ) + NaCl B . ClCH2CH2Cl + NaOH => OHCH2CH2OH + NaCL (THa mn 2OH lin k) =>B

C . CH3CH(Cl)Cl +NaOH => CH3CHO + NaL (V to ra CH3 CH(OH)OH h bin) D . CH3(Cl)C(Cl)Cl + NaOH => CH3COOH + NaCl (Dang R(OH)C(OH)-OH h bin to ra

Axit )

Nhng do NaOH d => CH3COOH + NaOH => CH3COONa + H2O => khng p vi Na v Cu(OH)2=> B

Cu 22:Cho 5 cht: CH3CH2CH2Cl (1); CH2=CHCH2Cl (2); C6H5Cl (3); CH2=CHCl (4);C6H5CH2Cl (5). un tng cht vi dung dch NaOH long, d, sau gn ly lp nc v axit ho bng dung dchHNO3, sau nh vo dung dch AgNO3 th cc cht c xut hin kt ta trng l

A. (1), (3), (5). B. (2), (3), (5). C. (1), (2), (3), (5). D. (1), (2), (5).Cht to kt ta trng l cht sinh ra NaCliu kin p l un vi dd NaOH d=> (1) , (2) , (5) c to ra NaCl (Bi 16 b)(3) ch p vi nhit cao , p sut caoHalogen gn vi C ni i khng phn ng vi NaOH cng nh H2O nhit thng v khi ung nng(Mnhon)Cu ny ng cc p n khng cha (4)

Cu 23:Cho s chuyn ho : Benzen ABCA axit picric. B lA.phenylclorua. B. oCrezol. C.Natri phenolat. D. Phenol.

Axit picric . Seach google nu khng bit cc chthttp://vi.wikipedia.org/wiki/Ax%C3%ADt_picricAxit picric tng hp t benzen hoc phenolC6H5OH +3HNO3 => C6H2(OH)(NO2)3 (c c xt H2SO4 )=> B Natri phenolat : C6H5ONa iu chphenol C

Cu 24:Cho s phn ng : NaOHCCl YX0

2 500, ancol anlylic. X l cht no sau y ?A.Propan. B.Xiclopropan. C.Propen. D.Propin.

Anlyl => CH2=CH-CH2- => ancol anlylic : CH2=CHCH2OH Y l CH2 = CH CH2Cl => X l CH2 = CH CH3
http://vi.wikipedia.org/wiki/Ax%C3%ADt_picrichttp://vi.wikipedia.org/wiki/Ax%C3%ADt_picric


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V Ch mt s Anken u mch mi c kh nng tham gia phn ng th 500-600 C

CH2=CH-CH3 +CL2 -->CH2=CH-CH2CL => C propen

Cu 25:Cho s sau : C2H5Br ete,Mg

A 2CO

B HCl

C. C c cng thc lA. CH3COOH. B. CH3CH2COOH. C. CH3CH2OH. D. CH3CH2CH2COOH.

SGK 11 NC214 => C2H5Br + Mg => C2H5-MgBr AC2H5MgBr + CO2 => C2H5COOMgBrC6H5CO2MgBr + HCl C6H5CO2H + MgBrCl

Xem p y :http://en.wikipedia.org/wiki/Benzoic_acid Tm seach googleCu 26: Cho bt Mg vo ietyl ete khan, khuy mnh, khng thy hin tng g. Nh t t vo etyl bromua,khuy u th Mg tan dn thu c dung dch ng nht. Cc hin tng trn c gii thchnh sau:

A.Mg khng tan trong ietyl ete m tan trong etyl bromua.B.Mg khng tan trong ietyl ete, Mg phn ng vi etyl bromua thnh etyl magiebromua tan trong ete.C.Mg khng tan trong ietyl ete nhng tan trong hn hp ietyl ete v etyl bromua.D. Mg khng tan trong ietyl ete, Mg phn ng vi etyl bromua thnh C2H5Mg tan trong ete.

Phn 3 SGK 11 NC 214 => BCu 27:Cho s : C6H6 X Y Z m-HOC6H4NH2. X, Y, Z tng ng l

A. C6H5NO2, m-ClC6H4NO2, m-HOC6H4NO2. B. C6H5NO2, C6H5NH2, m-HOC6H4NO2.

C.

C6H5Cl, m-ClC6H4NO2, m-HOC6H4NO2.D.

C6H5Cl, C6H5OH, m-HOC6H4NO2.

Cu 28:Cng thc dy ng ng ca ancol etylic lA. CnH2n + 2O. B. ROH. C. CnH2n + 1OH. D. Tt c u ng.SGK 11 NC220 => C ancol etylic C2H5OH l ancol non chc

Cu 29:Cng thc no di y l cng thc ca ancol no, mch h chnh xc nht ?A. R(OH)n. B. CnH2n + 2O. C. CnH2n + 2Ox. D. CnH2n + 2x

(OH)x.Xem chuyn 1 . Cch tm CT tng qut cch 1Ancol no => k = 0 => CnH2n+22.0x(OH)x hay CnH2n+2x (OH)x => D

Cu 30:un nng mt ancol X vi H2SO4 c nhit thch hp thu c mt olefin duy nht. Cng thc tngqut ca X l (vi n > 0, n nguyn)

A. CnH2n + 1OH. B. ROH. C. CnH2n + 2O. D. CnH2n +1CH2OH.

Tch H2O => OH vi Cbc I bn cnh => Do duy nht => Dhp l nhtA sai v khng th bit OH gn vi C bc I , II ,IIIB khng bit l ru no hay khng noC khng bit c phi l ru khng. C th andehit , xeton ,t

Cu 31:Tn quc t ca hp cht c cng thc CH3CH(C2H5)CH(OH)CH3 lA. 4-etyl pentan-2-ol. B. 2-etyl butan-3-ol. C. 3-etyl hexan-5-ol. D. 3-metyl pentan-2-ol.

5 4 3 2 1CH3CH(C2H5)CH(OH)CH3 hay CH3CH2 CH CHCH3

CH3 OH

=> 3metyl pentan2olCu 32:Mt ancol no c cng thc thc nghim l (C2H5O)n. CTPT ca ancol c th l

A. C2H5O. B. C4H10O2. C. C4H10O. D. C6H15O3.(C2H5O)n => C2nH5nOn ; ancol no => Tn pi + vng = 0 = (2.2n 5n + 2)/2n = 2 => C4H10O2Cu 33:Ancol no, n chc c 10 nguyn t H trong phn t c s ng phn l

A. 5. B. 3. C. 4. D. 2.Ancol no n chc => CT : CnH2n+1OH hay CnH2n+2O => S H = 2n + 2 = 10 => n =4=> C4H10O xem CT tnh ancol no trong chuyn ng phn 2n-2 =22 = 4 =>CCu 34:Mt ancol no n chc c %H = 13,04% v khi lng. CTPT ca ancol l

A. C6H5CH2OH.B. CH3OH. C. C2H5OH. D. CH2=CHCH2OH.Ancol no n chc => CnH2n+1OH hay CnH2n+2O=> %H = (2n+2).100%/(14n + 18) = 13,04 => n = 2 => C2H5OH => C

Cu 35:Mt ancol no n chc c %O = 50% v khi lng. CTPT ca ancol lA. C3H7OH. B. CH3OH. C. C6H5CH2OH. D.CH2=CHCH2OH.

Ancol no n chc => CnH2n+1OH => %O = 16.100% / (14n + 18) = 50% => n = 1 => CH3OH => B
http://en.wikipedia.org/wiki/Benzoic_acidhttp://en.wikipedia.org/wiki/Benzoic_acidhttp://en.wikipedia.org/wiki/Benzoic_acid


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Cu 36: C bao nhiu ru (ancol) bc 2, no, n chc, mch hl ng phn cu to ca nhau m phn t cachng c phn trm khi lng cacbon bng 68,18% ?

A. 2. B. 3. C. 4. D. 5.Ru no n chc => CnH2n+1OH hay CnH2n+2O

%C = 12n.100% / (14n+18) = 68,18% => n = 5 => C5H11OH Ru bc 2 => OH gn vi C bc2 - (C)C(OH) -CCCCC ; CCCCC ; CCC(C)C => 3 => B

OH OH OHCu 37: C bao nhiu ng phn c cng thc phn t l C4H10O ?A. 6. B. 7. C. 4. D. 5.

Dng CnH2n+2O k=0 Xem phn cch vit ng phn =>Cc trng hp : Ancol no n chc ; ete -O-Ancol no n chc CT: 2n-2 => n = 4 => 4 pEte : COCCC ; CCOCC ; COC(C)C => 3p=> Tng = 7 p => BCu 38:C bao nhiu ancol bc III, c cng thc phn t C6H14O ?

A.1. B.2. C.3. D.4.Ancol bc 3 => OH gn vi C bc 3 : - C - C(C)C -

OHC - C(C)CCC ; CC(C)C(C)C ; CCC(C)CC => 3p => C

OH OH OHCu 39:C bao nhiu ancol thm, cng thc C8H10O ?

A.5. B.6. C.7. D.8.C8H10O c k = 4tha mn 1 vng benzen c 1 vng + 3 piAncol thm => OH gn vi C ngoi vng benzenC6H5C - OHC6H5CHCH3 ; C6H5CH2CH2OH ; =>2

OHCH3C6H4CH3OH CH3 v tr 1 , CH3-OH v tr o,m,p => 3Tng => c 5 p => ACu 40:C bao nhiu ancol thm, cng thc C8H10O khi tc dng vi CuO un nng cho ra anehit?

A.2. B. 3. C.4. D.5.Ancol thm => OH gn vi C ngoi , Tc dng CuO => andehit => OH gn vi C bc 1

Bi 39 => Loi i trng hp bc II ; C6H5 CHCH3 => 4p => COH

Cu 41:C bao nhiu ancol C5H12O khi tch nc ch to mt anken duy nht?A. 1. B.2. C.3. D.4.

Tch nc thu c anken duy nht => OH gn viC bc I R CH2-OHv Ancol i xngOH gn vi C bc I ; C CCCCOH ; CC(C)CCOH ; C(C)C(C)COHAncol i xng : C CCCC => Tng = 4 =>D

OHCu 42:S ng phn ancol ng vi CTPT C5H12O l

A. 8. B. 7. C. 5. D. 6.p Ancol => 2n-2 = 8 v n = 5=> ACu 43:S ng phn ancol ti a ng vi CTPT C3H8Ox l

A. 4. B. 5. C. 6. D. khng xc nh c.Xt x = 1 => C3H8O ; p ancol => gc OH gn vi C CCCOH ; CC(OH)CXt x = 2 =>C3H8O2; p OH- CC(OH)C ; OHCCCOHXt x = 3 => C3H8O2 OHCC(OH)COHTng = 5 =>BCu 44:X l ancol mch h c cha 1 lin kt i trong phn t. khi lng phn t ca X nh hn 60. CTPT caX l

A. C3H6O. B. C2H4O. C. C2H4(OH)2. D. C3H6(OH)2.Ancol mch h cha 1 lin kt i => k = 1 => CnH2n+2 2.1 Oz hay CnH2nOz Xem chuyn 1 cch tm CTcch 2=> Loi C v D v C, D u no c dng CnH2n+2Oz

B loi v OH gn vi C bc 2 h bin => andehitCu 45:A, B, D l 3 ng phn c cng cng thc phn t C3H8O. Bit A tc dng vi CuO un nng cho raandehit, cn B cho ra xeton. Vy D l

A.Ancol bc III. B. Cht c nhit si cao nht.


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C.Cht c nhit si thp nht. D.Cht c kh nng tch nc to anken duy nht.C3H8O => dng CnH2n+2O => ng phn ancol v ete => c 2 p ancol v 1 p eteA to ra andehit =>A l ancol OH gn vi C bc 1B to ra xeton =>B l ancol bc OH gn vi C bc 2=> D l ete ; C OCC => Loi A v D v ete khng tch H2O ch c ruXem phn nhit si => Cht c H linh ng => Nhit si cao => Ancol nhit si > Ete V c H linh ngtrong OH

=> Loi B => C nhit si thp nhtCu 46:X, Y, Z l 3 ancol lin tip trong dy ng ng, trong MZ = 1,875MX. X c c im lA.Tch nc to 1 anken duy nht. B.Ha tan c Cu(OH)2.C.Cha 1 lin kt trong phn t. D.Khng c ng phn cng chc hoc khcchc.

Cng ng ng => Hn km nhau 14.k vc k l s C ca cht ln s C ca cht nhVD C2H5OH c M = 46 => C4H9OH c M = 46 + 14.2 = 74

MZ = MX + 28 ; => MZ = 1,875MX MZ = 60 ; MX = 32 => X : CH3OH Duy nht=> D

A sai v phi c 2C tr ln ; B sai v phi c t nht 2 nhm OH lin k C sai v CH3OH c k = 0 ;

Cu 47:Ancol X n chc, no, mch h c t khi hi so vi hiro bng 37. Cho X tc dng vi H2SO4 c unnng n 180oC thy to thnh mt anken c nhnh duy nht. X l

A.propan-2-ol. B.butan-2-ol. C.butan-1-ol. D. 2-metylpropan-2-ol.Cch 1: => to ra anken c nhnh duy nht => Ancol phi c nhnh => Dduy nht 2 metylCch 2:Ancol n chc , no , mch h => CnH2n+1OH c M = 14n + 18 = 37.2 => n = 4 ; C4H9OHun nng 1800 => ancol c nhnh duy nht => OH gn vi C bc I hoc i xng + Mch chnh c nhnh

3 2 1 1 2 3=> CH3CH(CH3)CH2OH hoc CH3 C(CH3)CH3 => p n c trng hp D ng

OHCu 48:Mt ancol n chc X mch h tc dngvi HBr c dn xut Y cha 58,4% brom v khi lng. un Xvi H2SO4 c 170oC c 3 anken. Tn X l

A.pentan-2-ol. B.butan-1-ol. C.butan-2-ol. D. 2-metylpropan-2-ol.P vi HBr SGK 11NC 226p n => Ancolno n chc V c ui an ankan ; v ol

CnH2n+1OH + HBr => CnH2n+1Br Y + H2O %Br = 80.100%/(14n + 81) = 58,4 => n = 4 => C4H9OH

Loi A v c 5C ; B v ch to ra 1 anken OH gn vi C bc I - 1ol Xt C v D ; D loi v ch to ra 1 anken mch i xng C C(C)C

OHC : CC(OH)CC => anken C = CCC ko p hnh hc ; C C = CC p hnh hcCu 49:Mt cht X c CTPT l C4H8O. X lm mt mu nc brom, tc dng vi Na. Sn phm oxi ha X biCuO khng phi l anehit. Vy X l

A.but-3-en-1-ol. B.but-3-en-2-ol. C. 2-metylpropenol. D. tt c u sai.C4H8O => k = 1 => c 1 ni i ; P vi Na => c H linh ng Ancol , phenol, axit hay c H linh ng c nhmOH , COOH=> Ch c th l Ancol v k=1 ; Nu l axit phi c 2 Oxi ; pheno l c k 4Oxi ha X bi CuO khng phi l andehit => c nhm OHT 3 d kin trn => Ancol n chcc 1 lin kt ic nhm OH khng gn vi C bc I=> B But3en2ol ; CC(OH)C = C Cch c tn lun => nh s gn OH hay gn H linh ng nhtCu 50:Bc ca ancol l

A.bc cacbon ln nht trong phn t. B.bc ca cacbon lin kt vi nhm -OH.C. s nhm chc c trong phn t. D. s cacbon c trong phn t ancol.

SGK 11 NC220 => B phn di cng *Cu 51:Bc ancol ca 2-metylbutan-2-ol l

A.bc 4. B.bc 1. C.bc 2. D.bc 3.1 2 3 4

2metyl butan2ol CC(C)CC => Bc 3OH

Cu 52:Cc ancol c phn loi trn c sA. s lng nhm OH. B. c im cu to ca gc hirocacbon.

C.bc ca ancol. D. Tt c cc c s trn.SGK 11 NC220 => S lng OH ; B c im gc no , khng no, vng, bezen bc ancol I, II, III => D


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Cu 53:Cc ancol (CH3)2CHOH ; CH3CH2OH ; (CH3)3COH c bc ancol ln lt lA. 1, 2, 3. B. 1, 3, 2. C. 2, 1, 3. D. 2, 3, 1.

(CH3)2CHOH hay CH3CHCH3 => bc 2OH

CH3CH2OH hay CH3CH2OH => bc 1 ; (CH3)3COH hay CH3C(CH3)CH3 => bc 3OH

=> CCu 54:

Cu no sau y l ng ?A. Hp cht CH3CH2OH l ancol etylic. B. Ancol l hp cht hu c trong phn t nhm -OH.C. Hp cht C6H5CH2OH l phenol. D. Tt c u ng.

CH3CH2OH => ancol etylic => AB sai v tp chc hocCacbonhidrat C sai v l ancol thm OH gn vi C ngoi vng benzenCu 55:Ancol etylic tan tt trong nc v c nhit si cao hn hn so vi ankan v cc dnxut halogen c khilng phn t xp x vi n v

A. Trong cc hp cht trn ch c ancol etylic tc dng vi Na.B. Trong cc hp cht trn ch c ancol etylic c lin kt hiro vi nc.C. Trong cc hp cht trn ch c ancol etylic c lin kt hiro lin phn t.D. B v C u ng.

SGK 11 NC222 => D c k phn A l do lin kt hidro lin phn tCu 56: A, B, C l 3 cht hu c c cng cng thc CxHyO. Bit % O (theo khi lng) trong A l 26,66%. Chtc nhit si thp nht trong s A, B, C l

A. propan-2-ol. B.propan-1-ol. C.etylmetyl ete. D.propanal.Chnh l bi 45Xem lion t p n => C3H8Ov A, B , C c cng CTvy D lC3H6O andehit c CT : CnH2nOTm ra CT xem chuyn 1 Phn tm CT xem trn thaytro.comGi CT: CxHyO => %O = 16.100%/(12x+y+16) = 26,66%12x + y = 44 vi A,B,C c CT l CnH2n+2O=> 12n + (2n+2) = 44n = 3 => C3H8O 3 p bi 45 + da vo nhit nh nht => CCu 57:Ancol etylic c ln mt t nc, c th dng cht no sau y lm khan ancol ?

A. CaO. B. CuSO4 khan. C. P2O5. D. tt c u c.

Xem li chuyn 1 bi hp th H2O vo bnh 1 => DCaO + H2O => Ca(OH)2 ;

P2O5 + H2O => H3PO4 ; CuSO4 khan + H2O => CuSO4 dung dchCu 58:Phng php iu ch ancol etylic t cht no sau y l phng php sinh ha ?

A. Anehit axetic. B. Etylclorua. C. Tinh bt. D. Etilen.Phng php sinh ha => C . Tinh bt (C6H10O5)n Ln men => SGKCu 59:Anken thch hp iu ch 3-etylpentan-3-olbng phn ng hirat ha l

A. 3,3-imetyl pent-2-en. B. 3-etyl pent-2-en. C. 3-etyl pent-1-en. D. 3-etyl pent-3-en.V Khi phn ng hidrat ha Phn ng + H2O bi anken OH v H cng vo ni i=> mch chnh khng thay i => vn l 3 etylpent , v tr ol d nguyn hoc thay i

OH 1 2 3 4 5CCCCC Mch i xngqua C trung tm=> anken duy nht CC = CCC

C CC C

Cu60:Hirat ha 2-metyl but-2-en thu c sn phm chnh lA. 2-metyl butan-2-ol. B. 3-metyl butan-1-ol. C. 3-metyl butan-2-ol. D. 2-metyl butan-1-ol.

Nh bi trn => mch chnh khng thay i => 2 metylbut => loi B v CThu c sn phm chnh . => ni i gn vi C v tr th 2 OH gn vi C bc 2 khng th l bc I V sn phm chnh => OH gn vi C bc cao

I 2 II OH2metyl but2en : CC = CC => CC - CC => A

C CCu 61:Hirat ha propen v mt olefin A thu c 3 ancol c s C trong phn t khng qu 4. Tn ca A l

A.

etilen.B.

but-2-en.C.

isobutilen.D.

A, B u ng.Propen : C = CC => To ra 2 ancol ; OHCCC hoc C C(OH)CC Ancol cn li to ra 1 ancol => Mch i xng A v B u tha mn : C = C ; C C = CC => D


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D : isobutilen : C = C(C)C => OHC - C(C)C hoc C (C)C(OH)C Loi

Cu 62:X l hn hp gm hai anken ( th kh trong k thng). Hirat ha X c hn hp Y gm 4 ancol(khng c ancol bc III). X gm

A.propen v but-1-en. B.etilen v propen.C.propen v but-2-en. D.propen v 2-metylpropen.

Bi 61 => Propen to ra 2 sn phm ; etilen , but21 sn phm => Loi B v C2metylpropen l isobutilen => to ra ancol bc III C (C)C(OH)C => Loi V bi

=> A hoc nhn thy anken c ni i v tr 1 lun to ra 2 ancol But1en : C = CCCCu 63:Hirat ha 2 anken c hn hp Z gm 2 ancol lin tip trong dy ng ng. t chy hon ton 0,53gam Z ri hp th ton b sn phm chy vo 2 lt dung dch NaOH 0,05M c dung dch T trong nng ca

NaOH l 0,025M (Gi s th tch dung dch thay i khng ng k). Cng thc cu to ca2 anken lA.CH2=CH2 v CH2=CHCH3. B.CH2=CHCH3 v CH2=CHCH2CH3.C.CH2=CHCH3 v CH3CH=CHCH3. D.CH2=CHCH3 v CH2=C(CH3)2.

CM cn li = 0,025M => CM p = 0,05 0,025 = 0,025 v NaOH d => Xy ra p trung haCO2 + 2NaOH => Na2CO3 + H2O => nCO2 = nNaOH p / 2 = 2.0,025/2 = 0,025 molAncol l ancol no n chc da vo Anken c 1 ni iCu 64:Mt chai ng ancol etylic c nhn ghi 25oc ngha l

A. c 100 ml nc th c 25 ml ancol nguyn cht.B. c 100 gam dung dch th c 25 ml ancolnguyn cht.

C. c 100 gam dung dch th c 25 gam ancol nguyn cht.D. c 75 ml nc th c 25 ml ancol nguyn cht.

ru = ( V ru x100 ) / V dd ru; V dung dch ru = V ru + V H2O=> DCu 65:Pha a gam ancol etylic (d = 0,8 g/ml) vo nc c 80 ml ancol 25o. Gi tr a l

A.16. B.25,6. C.32. D. 40.Bi 64: ru = ( V ru x100 ) / V dd ru 25 = (Vruou.100)/80V ru = 20 => mRuou = V.d =20.0,8 = 16 g => ACu 66: Dy gm cc cht u tc dng vi ancol etylicl

A. HBr (to), Na, CuO (to), CH3COOH (xc tc). B. Ca, CuO (to), C6H5OH (phenol), HOCH2CH2OH.C.NaOH, K, MgO, HCOOH (xc tc). D.Na2CO3, CuO (to), CH3COOH (xc tc), (CHCO)2O.

SGK 11 NC 225 Tnh cht ha hc=> A CH3COOH l p iu ch este SGK 11 254Nu khng lm c thng dng cch loi p n

Loi B v thy ngay Ca khng p Ch p vi Na,KLoi C v c NaOH ru ch p vi NaLoi D v c Na2CO3 ru khng p vi muiCu 67: Cho cc hp cht sau :

(a) HOCH2CH2OH. (b) HOCH2CH2CH2OH. (c) HOCH2CH(OH)CH2OH.(d) CH3CH(OH)CH2OH. (e) CH3CH2OH. (f) CH3OCH2CH3.Cc cht u tc dng c vi Na, Cu(OH)2 lA. (a), (b), (c). B. (c), (d), (f). C. (a), (c), (d). D. (c), (d), (e).

Tc dngNa, Cu(OH)2 => ru c nhmOH lin k : dng C(OH)C(OH) C th l axit nh:(a) , (c), (d) u c dng trn => CB loi v OH cch nhau 1 CH2 ; e sai v ch c 1 nhm OH , f sai v khng phi ru

Cu 68: a.Cho s chuyn ha sau (mi mi tn l mt phng trnh phn ng) :Tinh bt X Y Z metyl axetat. Cc cht Y, Z trong s trn ln lt l A. CH3COOH, CH3OH. B. C2H4, CH3COOH.C. C2H5OH, CH3COOH. D. CH3COOH, C2H5OH.

SGK 11 NC227 => X l glucozo , Y l C2H5OH => CZ l CH3COOH v C2H5OH + O2 => CH3COOH + H2O ; Metylaxetat este CH3COOCH3CH3COOH + CH3OH => CH3COOCH3 + H2O phn ng iu ch este

b.Cho s chuyn ho : Glucoz X Y CH3COOH. Hai cht X, Y ln lt lA. CH3CH2OH v CH=CH. B. CH3CH2OH v CH3CHO.C. CH3CHO v CH3CH2OH. D. CH3CH(OH)COOH v CH3CHO.

X l CH3CH2OH=> loi C v D => Xt A, B ; vi CH=CH khng c p iu ch CH3COOHVi CH3CHO + O2 => CH3COOH => BCu 69:

Cho Na tc dng va vi 1,24 gam hn hp 3 ancol n chc X, Y, Z thy thot ra 0,336 lt kh H2(kc). Khi lng mui natri ancolat thu c lA. 2,4 gam. B. 1,9 gam. C. 2,85 gam. D. 3,8 gam.

Xem phng php tng gim khi lng . Ancol ROH => Mui RONa => M tng = 22.x g Vi x l nROH


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Ta lun c x.nRuou = 2nH2Vi x l s nhm OH , tng qut hn l s H linh nghay x.nRuou = mH2nRuou = 2nH2 v ru n chc => 1 nhm OH= 0,03 mol

mROH = mROH + 22x = 1,24 + 22.0,03 = 1,9 gCch khc: BT e => nNa = 2nH2nNa = 0,03 molBT KL => m Ru + mNa = m mui+ mH21,24 + 0,03.23 = m mui + 0,015.2m mui = 1,9 gCu 70:Cho 7,8 gam hn hp 2 ancol n chc k tip nhau trong dy ng ng tc dng ht vi 4,6 gam Na

c 12,25 gam cht rn. l 2 ancolA.CH3OH v C2H5OH. B.C2H5OH v C3H7OH.C.C3H5OH v C4H7OH. D.C3H7OH v C4H9OH.

p n => Ru no n chc V c dng CnH2n+1OH

Gi CT : CnH2n+1OH ; BT Khi lng => m hn hp ancol + mNa = m Rn mui + mH2

mH2 = 0,15 g => n H2 = 0,075 mol => nRuou = 2nH2 = 0,15 mol AD 69

M ru = 14 n + 18 = 7,8/0,15 n = 2,43 => BCu 71: 13,8 gam ancol A tc dng vi Na d gii phng 5,04 lt H2 ktc, bit MA < 100. Vy A c cng thccu to thu gn l

A.CH3OH. B. C2H5OH. C.C3H6(OH)2. D.C3H5(OH)3.Cch 1 : Ly 13,8 chia cho cc p n => B v D tha mn => Xt B hoc DCch 2:MA < Cc dng bi thng ly 2 s gn nht hay 2 p n gn nht

=> C, D m loi C v tnh l => DCch 3:p n ru no => CnH2n+2m(OH)m Hay CnH2n+2OmTm CT theo cch 2 chuyn 1 m. n Ru = 2nH2n Ru = 2nH2/m = 0,45/m => M ru = 14n + 2 + 16m = 13,8/(0,45/m)14n + 2 = 44m/3

Ci s / di chnh l s OH , ng vi trng hp /2 v / 344m/3 => /3 => c 3 nhm OH => DCch 4: 14n + 2 = 44m/3 V 14n + 2 nguyn => 44m/3 nguyn => m = 3 tng qut hn s chia ht cho 3=> 3 nhm OH => D Hoc th m =3 => n = 3 => D Cu 72: C hai th nghim sau :TN 1: Cho 6 gam ancol no, mch h, n chc A tc dng vi m gam Na, thu c 0,075 gam H2.

TN 2: Cho 6 gam ancol no, mch h, n chc A tc dng vi 2m gam Na, thu c khng ti 0,1 gam H2. A ccng thc lA.CH3OH. B.C2H5OH. C.C3H7OH. D.C4H7OH.

Ch bi ny khng dng cch chia c v khng bit n p ht hay khng htDa vo phn 1 l m g Na ;Phn 2 l 2m g Na Ci ny ch l kinh nghim, nhiu bi tp n thng cho vy, 20102009 u c dng nm gia.V phn 2 thng l p htTH1 => Ancol d , Na htTH2 => ancol ht , Na d V s mol H2 phn H2 khng gp i phn 1 nu Na ht=> n Ru TH2 < mH2 CT => M ru > 6/0,1 = 60 => DCu 73: Cho 12,8 gam dung dch ancol A (trong nc) c nng 71,875% tc dng vi lng Na d thu c 5,6lt kh (ktc). Cng thc ca ancol A l

A.CH3OH. B.C2H4 (OH)2. C.C3H5(OH)3. D.C4H7OH.

Cu ny hay thi th 2010. Mnh mong n c trong 2011 Xem phn li gii cc nm thi H khi A , B Cch 1: m ru = 12,8.71,875/100% = 9,2 g . Mo ly 9,2 chia cc p n => C pCch 2: => m H2O = mddm ru = 12,8 9,2 = 3,6 gV p vi Na d => H2O c p to ra H2 v ru cng p to ra H2PT : 2Na + 2H2O => 2NaOH + H2 => nH2 to thnh do H2O= nH2O / 2 = 0,1 mol nH2 to ra do ancol = 0,25 0,1 = 0,15 mol

x . n Ru = 2nH2n Ru = 0,3 / x => M ru = 9,2/(0,3/x) = 92x / 3 => x = 3 V M nguyn

M = 92 => CCu 74: Ancol A tc dng vi Na d cho s mol H2 bng s mol A dng. t chy hon ton A c mCO2 =1,833mH2O. A c cu to thu gn l

A.C2H4(OH)2. B.C3H6(OH)2. C. C3H5(OH)3. D.C4H8(OH)2.n Ru = nH2 => Ru c 2 nhm OH => Loi C v CT : CnH2n(OH)2hay CnH2n+2O2

mCO2 = 1,833mH2O => chn mH2O = 18g => mCO2 = 33 g V ancol no , p n=> n = nCO2 / (nH2OnCO2) = 0,75 / (10,75) = 3 => C3H6(OH)2 => B Xem CT tnh nhanh trong chuyn gii nhanh bng ha hu c


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Cu 75: un 12 gam axit axetic vi 13,8 gam etanol (c H2SO4c lm xc tc) n khi phn ng t titrng thicn bng thu c 11 gam este. Hiusut ca phn ng este ho l

A. 55%. B. 50%. C. 62,5%. D. 75%.PT : CH3COOH (Axit axetic) + C2H5OH (etanol) => CH3COOC2H5 + H2O

Ban u 0,2 mol 0,3molP 0,2 mol => 0,2 mol (V C2H5OH d) mPT CH3COOC2H5 (este) = 0,2.88=17,6 g

H% sn phm = m thc t . 100% / mPT = 11.100% / 1 7,6 = 62,5 % => CCu 76: Khi thc hin phn ng este ho 1 mol CH3COOH v 1 mol C2H5OH, lng este ln nht thu c l 2/3mol. t hiu sut cc i l 90% (tnh theo axit) khi tin hnh este ho 1 mol CH3COOH cn s mol C2H5OHl (bit cc phn ng este ho thc hin cng nhit )

A. 0,342. B. 2,925. C. 2,412. D. 0,456. i hc phn nng caoXt ban u

PT : CH3COOH + C2H5OH => CH3COOC2H5 + H2OBan u 1 mol 1molP 2/3 mol 2/3 molSau p 1/3 mol 1/3 mol 2/3 mol 2/3 mol=> Kcb K cn bng Xem SGK bi cui cng = [CH3COOC2H5].[H2O] / [CH3COOH].[C2H5OH] K hi [X]

l nng M ca cht X = ((2/3).(2/3))/((1/3).(1/3)) = 4Xt khi H% cc i = 90%PT : CH3COOH + C2H5OH => CH3COOC2H5 + H2O

Ban u 1mol x molP 0,9 mol V H%=90 0,9 mol => 0,9mol 0,9 molSau p 0,1 x0,9 0,9 0,9=> Kcb = 0,9.0,9 / 0,1.(x-0,9) = 4 V Kcb ca p khng thay i => gii ra x = 2,925 => BCu 77:Khi un nngbutan-2-ol vi H2SO4 c 170oC th nhn c sn phm chnh l

A.but-2-en. B. ibutyl ete. C. ietyl ete. D.but-1-en. nhit 170 => To ra anken => Loi B v DSn phm chnh => OH tch cng vi H C bccao I IIButan2ol : CC(OH)CC => CC = CC => A

Cu 78:Khi un nng 2 trong s 3 ancol CH4O, C2H6O, C3H8O vi xc tc, nhit thch hp ch thu c 1olefin duy nht th 2 ancol l

A. CH4O v C2H6O. B. CH4O v C3H8O. C. A, B ng. D. C3H8O v C2H6O.CH4O khng to ra olefin anken v C2H6O to ra 1 anken duy nht CCOH => C = CC3H8O hay OHCCC => Ch to ra 1 anken C = C - CHoc C C(OH)-C => C = CC => ging nhau => ch to ra 1 anken=> A v B tha mn to ra 1 anken => CCu 79: Khi tch nc ca ancol C4H10O c hn hp 3 anken ng phn ca nhau (tnh c ng phn hnh hc).Cng thc cu to thu gn ca ancol l

A.CH3CHOHCH2CH3. B.(CH3)2CHCH2OH.C.(CH3)3COH. D.CH3CH2CH2CH2OH.

3anken ng phn => c p hnh hc=> OH gn vi C bc IIhoc IIIv OH gn vi C bc I => ch to ra 1 anken

Loi B v DXt A hoc C ; CH3 CHCH2CH3 => CH2=CH-CH2-CH3 v CH3 CH=CH-CH3 p hnh

hc OH => A , C ch to ra 1 ankenCu 80:Hp cht hu c X c cng thc phn t l C5H12O, khi tch nc to hn hp 3 anken ng phn (k cng phn hnh hc). X c cu to thu gn l

A.CH3CH2CHOHCH2CH3. B.(CH3)3CCH2OH.C. (CH3)2CHCH2CH2OH. D.CH3CH2CH2CHOHCH3.

Bi 79 => OH gn vi C bc II hoc bc III=> Loi B v CXt A. CH3 CH2CH(OH)CH2CH3 => CH3CH2 =CHCH2CH3 p hnh hc => 2 anken=>Loi=> D OH gn vi C trung tm ca mch hay i xng qua nCu 81:Khi un nng hn hp ancol etylic v ancol isopropylic vi H2SO4 c 140oC c th thu c s ete ti

a l A. 2. B. 4. C. 5. D. 3.CT tnh ete : = n(n+1)/2 Vi n l s ru tham gia p => n = 2 => S ete = 3=> DCu 82:Khi un nng hn hp gm C2H5OH v C3H7OH vi H2SO4 c 140oC c th thu c s ete ti a l


12/26


A. 6. B. 4. C. 5. D. 3.Bi 81 => DCu 83:un nng hn hp gm 3 ancol l AOH, BOH v ROH vi H2SO4 c 140oC th thu c ti a baonhiu ete ?

A.3. B.4. C.5. D.6.Bi 81 => DCu 84:un nng hn hp n ancol n chc khc nhau vi H2SO4 c 140oC th s ete thu c ti a l

A.2

1)n(n . B.2

1)2n(n . C.2

2n . D.n!

Bi 81 => A

Cu 85:Cho s chuyn ha : But-1-en HCl A NaOH B C170,SOHo

c42 E

Tn ca E lA.propen. B. ibutyl ete. C.but-2-en. D. isobutilen.

Cc phn ng trung gian => khng lm thay i s C v v tr mch chnh V ch l p cng v th => Loi A,D nhit 170 => Loi B v to ra anken=> CP : But 1en : C = CCC + HCl => CC(Cl)CC +NaOH => CC(OH)CC => CC = CC=> C p u l cng , p 2 lphn ng th bi dn xut, p 3 l tch H2O => to ra sn phm chnh l CCu 86: un nng hn hp gm hai ru (ancol) n chc, mch h, k tip nhau trong dy ngng vi H2SO4

c 140oC. Sau khi cc phn ng kt thc, thu c 6 gam hn hp gm ba ete v1,8 gam nc. Cng thc phnt ca hai ru trn l

A. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.C. C3H5OH v C4H7OH. D. C3H7OH v C4H9OH.

A, B , D l ancol no n chc => Xt Ancol no n chc : CnH2n+1OH hn hpBT khi lng => m ancol p = m ete + mH2O v p ancol => ete + H2O= 7,8 gn hh ancol p = 2nH2O V lun c 2 ancol p => 1 H2O VD; CH3OH + CH3OH => CH3 OCH3 + H2O

=> n ancol p = 0,2 mol => M ancol = 14 n + 18 = 7,8/0,2 = 1,5 => ACu 87:un nng t t hn hp etanol vpropan-2-ol vi H2SO4 c c th thu c ti a s sn phm hu c l

A. 3. B. 2. C. 5. D. 4.2 anken v 3 ete => 5 cht => C etanol C2H5OH => to ra 1 anken ; Propan - 2ol CC(OH)C => to ra 1

anken ; s ete = cng thc = 3Cu 88:C bao nhiu ng phn ng vi cng thc phn t C8H10O, u l dn xut ca benzen, khi tch nccho sn phm c th trng hp to polime ?

A.1. B.2. C.3. D.4.Sn phm c th trng hp to polime => C ni i C ngoi vng benzenp : C6H5 CH2CH2OH => C6H5CH=CH2 StirenC6H5CH(OH)CH2 => C6H5CH =CH2 Stiren => c 2 p=> BCu 89:A l ancol n chc c % O (theo khi lng) l 18,18%. A cho phn ng tch nc to 3 anken. A c tnl

A.Pentan-1-ol. B.2-metylbutan-2-ol. C.pentan-2-ol. D.2,2-imetyl propan-1-ol.p n => Ancol no n chc v u l an hay ankan , v ol hay 1 ol => CnH2n+1OH %O = 16.100% / (14n + 18) = 18,18%n = 5 => C5H11OH

To ra 3 anken => OH gn vi C bc II hoc III => Loi A, D Xt B : 2 metylbutan2ol ; I II

CC(C)CC => CC(C) = CC v C = C(C) CCC 2 cht u khng c p hnh hc=> loi B => C OHCu 90: hiratha 14,8 gam ancol thu c 11,2 gam anken. CTPT ca ancol l

A. C2H5OH. B. C3H7OH. C. C4H9OH. D. CnH2n + 1OH.Cch 1 mo => Ly 14,8 chia cc p n => C hidrat => Tch H2O => mancol = manken + mH2OmH2O = 3,6 g => nH2O = nRuou = 0,2 mol=> M ruou = 14n + 18 = 14,8 / 0,2n = 4 => CCu 91:un nng hn hp X gm 2 ancol n chc lin tip trong dy ng ng vi H2SO4 c 140oC. Sau

phn ng c hn hp Y gm 5,4 gam nc v 19,4 gam 3 ete. Hai ancol ban u lA.

CH3OH v C2H5OH.B.

C2H5OH v C3H7OH.C.C3H5OH v C4H7OH. D.C3H7OH v C4H9OH.

Ging bi 86 => m Ancol = m H2O + m hh ete ; n Ancol = 2nH2O => n = 1,67 => A


13/26


Cu 92:un nng hn hp X gm 0,1 mol CH3OH v 0,2 mol C2H5OH vi H2SO4 c 140oC, khi lng etethu c l

A.12,4 gam. B.7 gam. C.9,7 gam. D.15,1 gam.Ta lun c n hh Ancol = 2nH2O => nH2O = (0,1 + 0,2)/2 = 0,15 mol=> m hn hp ete = m hn ru m H2O = 0,1.32 + 0,2 .460,15.18 = 9,7 g => CCu 93:un nng ancol n chc X vi H2SO4 c 140oC thu c Y. T khi hi ca Y i vi X l 1,4375. Xl

A.

CH3OH.B.

C2H5OH.C.

C3H7OH.D.

C4H9OH.Dng bi ny c mo.p n => Ancol c dng : CnH2n+1OH140oC => To ra ete => 2X => Y + H2OTa c 2MX = MY + 18 Bo ton Nguyn t khi PT => MY = 1,4375MX gii h => MX = 32 => CH3OH

Rt ra mo . Dng bi 140oC => MY = k.MX k>1 do 2 MX = MY + 18 v MX 32 vi Ancol thp nht CH3OH s cho => Th vo 2MX = MY + 18 => MX (2k) = 18 => MX = 18/(2-k)

Nh bi trn => k = 1,4375 => M = 32Cu 94: un nng hn hp hai ancol n chc, mch h vi H2SO4 c, thu c hn hp gm cc ete. Ly 7,2gam mt trong cc ete em t chy hon ton, thu c 8,96 lt kh CO2 (ktc) v 7,2 gam H2O. Hai ancol l

A.C2H5OHv CH2=CHCH2OH. B. C2H5OH v CH3OH.C. CH3OH v C3H7OH. D. CH3OH v CH2=CHCH2OH.

H: nCO2 = nH2O => ete c cha 1 lin kt i => 1 trong 2 ancol c 1 lin kt i=> Loi B v CEte c 1 lin kt i => c dng CnH2nO=> nCnH2nO = nCO2 / n = 0,4 / n

M CnH2nO = 14n + 16 = 7,2 / (0,4/n) 4 = Tng s C ca 2 ancol => D tha mn A loi v = 5

Ch khi p to ra ete nu cng tt c C , H ca cc ete => s ra sn phm l p gia 2 ancol Cu 95:Khi un nng mt ancol n chc no A vi H2SO4 c iu kin nhit thch hp thu c sn phmB c t khi hi so vi A l 0,7. Vy cng thc ca A l

A. C4H7OH. B. C3H7OH. C. C3H5OH. D. C2H5OH.MB = 0,7 MA => sn phm to ra Anken > 1 l to ra ete V p tch nc t ruA => B + H2O => MA = MB + 18 ; MB = 0,7.MA => th vo => MA = 60 => B

T => CT : MA = 18 / (1-k)Cu 96: un nng mt ru (ancol) n chc X vi dung dch HSO4c trong iu kin nhit thch hp sinh racht hu c Y, t khi hi ca X so vi Y l 1,6428. Cng thc phn t ca X l

A. C3H8O. B. C2H6O. C. CH4O. D. C4H8O.

MX = 1,6428MY => MY = 0,6087MX => k < 1 => AD 95 => MX = 46 => BCu 97:Ch ra dy cc cht khi tch nc to 1 anken duy nht ?

A. Metanol ; etanol ; butan -1-ol.B. Etanol; butan -1,2-iol ; 2-metylpropan-1-ol.C. Propanol-1; 2-metylpropan-1-ol; 2,2 imetylpropan-1-ol.D. Propan-2-ol ; butan -1-ol ; pentan -2-ol.

Ancol tch H2O => anken duy nht => OH gn vi v tr C bc I => C v OH gn vi C v tr 1A loi v metanol CH3OH khng to ra anken, B sai v butan 1,2iol c OH gn vi v tr C th 2 hay bc IID loi v pentan 2olCu 98:Ancol X n chc tch nc ch to mt anken duy nht. t chy mt lng X c 11 gam CO2 v 5,4gam H2O. X c th c bao nhiu cng thc cu to ph hp ?

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

nCO2 = 0,25 mol ; nH2O = 0,3 mol => Ru no v nH2O > nCO2 . xem phn pp gii nhanh ha hu cn = nCO2 / (nH2OnCO2) = 5 => C5H11OH => Xem bi 41 => C : 4 CT cu to tha mn.Cu 99:un nng hn hp X gm 2 ancol n chc no (c H2SO4 c lm xc tc) 140oC. Sau khi phn ngc hn hp Y gm 21,6 gam nc v 72 gam ba ete c s mol bng nhau. Cng thc 2 ancol ni trn l

A. CH3OH v C2H5OH. B.C2H5OH v C3H7OH. C.C2H5OH v C3H7OH. D.C3H7OH v C4H9OH.

Xem bi 86 => n = 1,5 => ACu 100:un nng V (ml) ancol etylic 95o vi H2SO4 c 170oC c 3,36 lt kh etilen (ktc). Bit hiu sut

phn ng l 60% v ancol etylic nguyn cht c d = 0,8 g/ml. Gi tr ca V (ml) lA. 8,19. B. 10,18. C. 12. D. 15,13.

Ancol etylic C2H5OH ; nC2H5OH = nAnken = 0,15 mol Phn ng t l 1: 1


14/26


H% p = nPT . 100% / nBn Ru ban u = 0,15.100% / 60 = 0,25 mol=> mC2H5OH = 11,5 g V C2H5OH = 11,5/0,8 = 14,375 ml => ru = V C2H5OH . 100 / V

V = 14,375.100 / 95 = 15,13 ml => DCu 101:Ancol no b oxi ha to xeton ?

A.propan-2-ol. B.butan-1-ol. C. 2-metyl propan-1-ol. D.propan-1-ol.Ancol b oxi ha to ra xeton => Ancol c OH gn vi C bc II hay ancol bc II=> A v 2ol v khng cnhnh

B , C , D u c OH gn vi C bc I 1ol Cu 102:Ancol no n chc tc dng c vi CuO to anehit lA.ancol bc 2. B. ancol bc 3. C. ancol bc 1. D. ancol bc 1 v ancol bc 2.

Ancol n chc phn ng vi CuO to ra andehit => Ancol c OH gn vi C bc I hay ancol bc I =>CCu 103:Oxi ha 6 gam ancol no Xthu c 5,8 gam anehit. CTPT ca ancol l

A. CH3CH2OH. B. CH3CH(OH)CH3. C. CH3CH2CH2OH. D. Kt qu khc.Ly 6 chia cho p n => CCch khc . Tng gim khi lng : T Ru CnH2n+1OH => CnH2nOandehit=> Gim2 g mt i 2 hidroPT : CnH2n+1OH + CuO => CnH2nO + Cu + H2O=> n Ru= (m ru m andehit) / 2 =0,1 mol => MX = 6/0,1 = 60 => CCu 104:Cho m gam ancol n chc, no, mch h qua bnh ng CuO (d) nung nng. Sau khi phn ng xy rahon ton, khi lng cht rn trong bnh gim 0,32 gam. Hn hp thu c c t khi hi i vi H2 l 19. Gi tr

m lA.1,48 gam. B.1,2 gam. C.0,92 gam. D.0,64 gam.

Cch 1: H. T CuO => Cu => M gim 16g Mt i 1 Oxi nCuO = 0,32 / 16 = 0,02 mol

PT : CnH2n+1OH +CuO => CnH2nO + Cu + H2O Ch H2O l hikhng phi lng Hn hp thu c l CnH2nO v H2O

Ta c nCuO = nCnH2nO = nH2O = 0,02 mol=> m hn hp = mCnH2nO + mH2O= 0,04 molBT KL : m Ru + mCuO = mCnH2nO + mCu + mH2Om Ru = (mCnH2nO + mH2O) - (mCuOmCu)

= m hn hp - m gim= 38.0,040,32 = 1,2 gCch 2:Ta lun c m Ru = mAndehit + 2.n Ru V gim 2 H khi oxi ha: t CnH2n+1OH => CnH2nO

M hn hp = mAndehit + mH2O / (nAndehit + nH2O)38 = (m andehit + 0,02.18)/0,04=> m andehit = 1,16 g => m ru = 1,16 + 0,02.2 = 1,2 gCch 3: Tm n ta c M andehit = 14n + 16 = 1,16/0,02 n = 3 => Ru C3H7OH=> m = 0,02.60 = 1,2 gCu 105*:Oxi ha 4 gam ancol n chc A bng oxi khng kh (c xc tc v un nng) thu c 5,6 gam hn hpanehit, ancol d v nc. A c cng thc l

A.CH3OH. B.C2H5OH. C.C3H5OH. D.C3H7OH.Cch 1 : Ly 4 g chia cho p n => A p nht.P : CnH2n+1OH + O => CnH2nO + H2OBT khi lng : m ru A + mO = m hn hp sau pmO = 1,6 g => nO = n Ru = n Andehit = nH2O = 0,1molTa c ru d=> m Ru p < 4 => M ru < 4/0,1 = 40 => A tha mnCu 106:Oxi ha 6 gam ancol n chc A bng oxi khng kh (c xc tc v un nng) thu c 8,4 gam hn hpanehit, ancol d v nc. Phn trm A b oxi ha l

A.60%. B.75%. C.80%. D.53,33%.Tng t bi 105 => n Ru p = nOxi = 0,15 mol => M ru < 6/0,15 = 40 => Ch c CH3OH=> m CH3OH p = 32.0,15 = 4,8g => %A b oxi ha = 4,8.100%/6 = 80%Cu 107:Dn m gam hi ancol n chc A qua ng ng CuO (d) nung nng. Sau khi phn ng hon ton thykhi lng cht rn trong ng gim 0,5m gam. Ancol A c tn l

A.metanol. B.etanol. C.propan-1-ol. D.propan-2-ol.T CuO => CuO => M gim = 16 g Mt 1 oxi=> n Ru = nCuO = 0,5m/16 => M ru = m / (0,5m/16) = 32 => CH3OH hay metanol => ACu 108: Dn hi C2H5OH qua ng ng CuO nung nng c hn hp X gm anehit, ancol d v nc. Cho Xtc dng vi Na d c 4,48 lt H2 ktc. Khi lng hn hp X l (bit ch c 80% ancol b oxi ha)

A.

13,8 gamB.

27,6 gam.C.

18,4 gam.D.

23,52 gam.X gm C2H5OH d , andehit , H2O => Ta lun c nC2H5OH d +nH2O = 2nH2OMt khc ta c nH2O = nC2H5OH p => nC2H5OH ban u = 2nH2O = 0,4 molnC2H5OH p = 0,8.0,4 = 0,32 mol v ch p 80%= nCuO = nCu


15/26


m Ru ban u + mCuO = m hn hp X + mCu 0,4 . 46 + 0,32.80 = m hn hp X + 0,32.64m hn hp X = 23,52

C th dng tng 16 g i vi CuO ,CuCu 109:Dn hi C2H5OH qua ng ng CuO nung nng c 11,76 gam hn hp X gm anehit, ancol d vnc. Cho X tc dng vi Na d c 2,24 lt H2 ( ktc). % ancol b oxi ho l

A.80%. B.75%. C.60%. D. 50%.Ngc li bi 108 => nC2H5OH ban u = 2nH2 = 0,2mol

m Ru ban u + mCuO = m hn hp X + mCu0,2.46 + x.80 = 11,76 + x.64 vi x l nC2H5OH p => x = 0,16 mol=> %C2H5OH b oxi ha = x.100% / 0,2 = 80%Cu 110:t chy mt ancol X c

22 COOHnn . Kt lun no sau y l ng nht?

A. X l ancol no, mch h. B. X l ankaniol.C. X l ankanol n chc. D. X l ancol n chc mch h.

nH2O > nCO2 => ancol no , mch h=> A Ging ca ankanCu 111:Khi t chy ng ng ca ancol n chc thy t l s mol OHCO 22 n:n tngdn. Ancol trn thuc dy

ng ng caA. ancol khng no. B. ancol no. C. ancol thm. D. khng xc nh c.

Xem chuyn hidrocacbon no=> vi ancol no=> nCO2 / nH2O = n / (n+1) = 11/(n+1) bit tng mo xt n

= 1 v n = 2 so sch => Lun tng khi n tng n l s C hay CnH2n+1OHCu 112:t chy hon ton m gam ancol n chc A c 6,6 gam CO2 v 3,6 gam H2O. Gi tr m lA.10,2 gam. B.2 gam. C.2,8 gam. D.3 gam.

Cch 1: nCO2 = 0,15 mol ; nH2O = 0,2 mol => n = nCO2 / (nH2OnCO2) = 3 => C3H7OH v nH2O > nCO2=> c dng CnH2n+2Oz

nC3H7OH = nCO2 / 3 = 0,05 mol => m = 3 gCch 2: nO2 = (2nCO2 + nH2O)/2 = 0,25 mol ; BTKL => m ru + mO2 = mCO2 + mH2O => m ru = 3 gCch 3: Xt x : y = nCO2 / 2nH2O = 3/8 => C3H8O v n chcnn c 1 Oxi => cch 1Cu 113: t chy mt ancol n chc, mch h X thu c CO2 v hi nc theo t l th tch

5:4V:V OHCO 22 . CTPT ca X l

A. C4H10O. B. C3H6O. C. C5H12O. D. C2H6O.Cch 1 :x : y = VCO2 / 2VH2O = 4 / 10 => Ancol C4H10O v n chc => c 1 OxiCch 2 : VCO2 : VH2O = 4 : 5 t l th tch = t l s mol => chn nCO2 = 4 mol ; nH2O = 5 mol n = nCO2 / (nH2OnCO2) = 4 => C4H10O v nH2O > nCO2 => c dng CnH2n+2OzCu 114:t chy mt ancol a chc thuc H2O v CO2 c t l mol 2:3n:n

22 COOH . Vy ancol l

A. C3H8O2. B. C2H6O2. C. C4H10O2. D. tt c u sai.nH2O : nCO2 = 3 : 2 => chn nH2O = 3 ; nCO2 = 2 => n = nCO2 / (nH2O nCO2) = 2 v nH2O > nCO2 => cdng CnH2n+2Oz=> C2H6Oz v a chc + nhn p n => C2H6O2 => BCu 115:Khi t chy mt ancol a chc thu c nc v kh CO2 theo t l khi lng

44:27m:m22 COOH

. CTPT ca ancol l

A. C5H10O2. B. C2H6O2. C. C3H8O2. D. C4H8O2.

mH2O : mCO2 = 27 : 44 => chn mH2O = 27 g => mCO2 = 44 g ; nH2O = 1,5 mol ; nCO2 = 1 mol=> n = nCO2 / (nH2OnCO2) = 2 => B v ch ch 2 CCu 116:t chy hon ton 5,8 gam ancol n chc X thu c 13,2 gam CO2 v 5,4 gam H2O. Xc nh X

A. C4H7OH. B. C2H5OH. C. C3H5OH. D. tt c u sai.Cch 1: ly 5,8 chia p n => CCch 2: nCO2 = nH2O =0,3 mol => Ancol CnH2nO => nAncol = nCO2 / n = 0,3 / n=> M ancol = 14n + 16 = 5,8 / (0,3/n)n = 3 => C3H6O hay C3H5OH => CCu 117:Ba ancol X, Y, Z u bn v c khi lng phn t khc nhau. t chy mi cht u sinh ra CO2 vH2O theo t l mol OHCO 22 n:n = 3 : 4. Vy CTPT ba ancol l

A. C2H6O ; C3H8O ; C4H10O. B. C3H8O ; C3H8O2 ; C3H8O3.C. C3H8O ; C4H10O ; C5H10O. D. C3H6O ; C3H6O2 ; C3H6O3.

V t chy mi cht => n = nCO2 / (nH2OnCO2) = 3 => B v l Ancol no do nH2O > nCO2cch chn nhbi trn nCO2 = 3 ; nH2O = 4Cu 118:t chy ru A bng O2 va nhn thy: nCO2 : nO2 : nH2O = 4 : 5: 6. A c cng thc phn t l

A. C2H6O. B.C2H6O2. C.C3H8O. D.C4H10O.


16/26


nCO2 : nO2 : nH2O = 4 : 5 : 6 => chn nCO2 = 4 mol => nO2 = 5mol ; nH2O = 6 mol n = nCO2 / (nH2OnCO2) = 2 => C2H6Oz

Hoc x : y = nCO2 / 2nH2O = 2 / 6 => C2H6OzCch 1: nC2H6Oz = nCO2 / 2 = 2 mol => AD x + y/4 z/2 = nO2 / nX

2 + 6/4z/2 = 5/2z = 2 => C2H6O2Cch 2 : nC2H6Oz = nCO2 / 2 = 2 molBT NT Oxi : z . nC2H6Oz + 2nO2 = 2nCO2 + nH2O => z = 2

Cch 3: Nh cng thc nO2 / nAncol = n,5 => bi trn nO2 / nC2H6Oz = 5/2 = 2,5 => s C = s OH => z = 2Cu 119:t chy ancol ch cha mt loi nhm chc A bng O2 va nhn thy :nCO2 : nO2 : nH2O = 6: 7: 8. A c c im l

A.Tc dng vi Na d cho nH2 = 1,5nA. B.Tc dng vi CuO un nng cho ra hp cht a chc.C.Tch nc to thnh mt anken duy nht. D.Khng c kh nng ha tan Cu(OH)2.

nCO2 : nO2 : nH2O = 6 : 7 : 8 => chn nCO2 = 6 mol => nO2 = 7 mol ; nH2O = 8 mol n = nCO2 / (nH2OnCO2 ) = 3 => C3H8Oz hoc x : y = nCO2 / 2nH2O = 3 / 8=>C3H8Oz

AD bi 118 => z = 3 => C3H8O3 hay C3H5(OH)3 => 3nC3H5(OH)3 = 2nH2 CT => ANh nO2 / nX = n,5 , n l s C v s OHCu 120:Ancol n chc A chy cho mCO2 : mH2O = 11: 9. t chy hon ton 1 mol A ri hp th ton b sn

phm chy vo 600 ml dung dch Ba(OH)2 1M th lng kt ta lA.11,48 gam. B.59,1gam. C.39,4gam. D.19,7gam.

mCO2 : mH2O = 11 : 9 => chn mCO2 = 11 g ; mH2O = 9 g => nCO2 , nH2O

n = nCO2 / (nCO2nH2O) = 1 => CH3OH Ancol A t 1 mol CH3OH => 1 mol CO2 ADCT : nCO3(2-) hay BaCO3 kt ta = nOH- - nCO2 = 2nBa(OH)2nCO2 = 1,21 = 0,2 mol

m BaCO3 = 39,4 g => CCu 121:X l mt ancol no, mch h. t chy 0,05 mol X cn 4 gam oxi. X c cng thc l

A. C3H5(OH)3. B.C3H6(OH)2. C.C2H4(OH)2. D.C4H8(OH)2.nO2 / nX = 2,5 => C2H4(OH)2 Dng bi s C = s nhm OHchuyn 1 hay bi ging nguyn tn trungCu 122:t chy hon ton ancol X c CO2 v H2O c t l mol tng ng l 3: 4, th tch oxi cn dng tchy X bng 1,5 ln th tch CO2 thu c (o cng k). X l

A.C3H8O. B.C3H8O2. C.C3H8O3. D.C3H4O.x : y = nCO2 / 2nH2O = 3/8 => C3H8Oz

Cch 1:VO2 = 1,5 VCO2 => chn nCO2 = 3 mol => nO2 = 4,5 mol ; => nC3H8Oz = nCO2 / 3 = 1 mol=> (3 + 8/4z/2) = nO2 / nC3H8Oz = 4,5 => z = 1 => C3H8OCch 2: Vit PT C3H8Oz + (3 + 8/4 z/2)O2 = 3CO2 + 4H2O

1,5 mol 1 mol Chn nCO2 = 1mol => nO2 = 1,5 mols=> 3 + 8/4 - z/2 = 1,5.3 => z = 1Cu 123: X l mt ancol (ru) no, mch h. t chy hon ton 0,05 mol X cn 5,6 gam oxi, thu c hi nc v6,6 gam CO2. Cng thc ca X l

A. C3H5(OH)3. B. C3H6(OH)2. C. C2H4(OH)2. D. C3H7OH.

Cch 1 :nO2 / nX = 3,5 => C3H5(OH)3Cch 2: Ru no : CnH2n+2Oz => n = nCO2 / nX = 3 => C3H8Oz(3+8/4z/2) = nO2 / nX = 3,5 => z = 3 => C3H8O3 hay C3H5(OH)3Cu 124*:X l hn hp 2 ancol n chc, cng dy ng ng, c t l khi lng 1:1. t chy ht X c 21,45gam CO2 v 13,95 gam H2O. Vy X gm 2 ancol l

A.CH3OH v C2H5OH. B.CH3OH v C4H9OH.C.CH3OH v C3H7OH. D.C2H5OH v C3H7OH.

Cc p n u no n chc => n hn hp ru = nH2O nCO2 = 0,2875 molBTNT Oxi : n hn hp ru + 2n O2 = 2nCO2 + nH2OnO2 = 0,73125

m hn hp ru = mCO2 + mH2O mO2 = 12 g => m ru 1 = m ru 2 =6 g V t l khi lng 1 : 1Cch 1: M trung bnh hn hp ru = 12 / 0,2875 = 41,74 = 14n + 18 CnH2n+1OHn = 1,69 n trung bnh

nh => Loi D V m ru 1 hoc 2 = 6 g => Chc chn c C3H7OH v chia p => C

Cch 2: th p n tnh ra s mol + th d kin BTNT C => n . nA + m.nB = nCO2 n , m l s C ca 2 ru

Cch 3: Th p n : Nu dng my tnh Fx 570 th nhanh hn 500 bm vo h 2 PTBTNT C => n.x + m.y = nCO2BTNT H => (2n+2)x + (2m + 2)y = 2nH2O Vi ru c CT l : CnH2n+2 ; CmH2m+2


17/26


Th tng p n => C p nht vi n = 1 v m = 3 v tha mn iu kin m 1 = m 2 = 6 g Cch ny khng tm m= 6 g cng cCch 4: p n => Ancol no n chc.n hn hp ancol = nH2O nCO2 = 0,2875 mol=> n tb = nCO2 / (nH2OnCO2) = 1,7 => C CH3OHm hn hp ancol=mH + mC + O = 2.nH2O + 12.nCO2 + (nH2OnCO2).16 = 12 gt l khi lng l 1:1=> mCH3OH=6g, m ancol cn li=6g

mCH3OH=6g => nCH3OH=0.1875mol, n ancol cn li= 0.2875-0.1875=0.1 mol=> M ancol cn li = 60 => C3H7OH.

Cu 125:t chy hon ton a gam ancol X ri hp th ton b sn phm chy vo bnh nc vi trong d thy

khi lng bnh tng b gam v c c gam kt ta. Bit b = 0,71c v c =1,02

ba . X c cu to thu gn l

A.C2H5OH. B.C2H4(OH)2. C.C3H5(OH)3. D.C3H6(OH)2.b = 0,71c ; c = (a + b)/1,02kt ta CaCO3 => Chn c = 100 gMo chn MCaCO3=> b = 71 g th vo => a = 31 gnCaCO3 = nCO2 = 1 mol , m bnh tng = mCO2 + mH2O = 7144 + mH2O = 71mH2O = 27 g=> nH2O = 1,5 mol => n Ancol = nH2OnCO2 = 0,5 mol => M X = 31 / 0,5 = 62 => BCu 126:t chy hon ton a gam hn hp gm metanol v butan -2-ol c 30,8 gam CO2 v 18 gam H2O. Gitr a l

A.30,4 gam. B. 16 gam. C.15,2 gam. D.7,6 gam.

Cch 1: n = nCO2 / (nH2OnCO2) = 7/3 => 7 7 7 202. +2

3 3 3 3

C H O hay C H O v 2 ancol cho u no

n

7 20

3 3

C H O = nCO2 / (7/3) = 0,3 mol => m = 0,3 . 152/3 = 15,2 g => C

Cch 2 : n hn hp ru = nH2O - nCO2 = 10.7 = 0,3 molBTNT Oxi : n hn hp Ru + 2nO2 = 2nCO2 + nH2OnO2 = 1,05 mol v Ru u no => c 1 oxiBTKL => m + mO2 = mCO2 + mH2Om = 15,2 gCu 127: t chy hon ton 0,4 mol hn hp X gm ancol metylic, ancol etylic v ancol isopropylic ri hp thton b sn phm chy vo nc vi trong d c 80 gam kt ta. Th tch oxi (ktc) ti thiu cn dng l

A.26,88 lt. B.23,52 lt. C.21,28 lt. D.16,8 lt.nCO2 = nCaCO3 = 0,8 mol => nH2O = nRuou + nCO2 v ru no = 1,2 molBTNT Oxi => nhn hp ru + 2nO2 = 2nCO2 + nH2O0,4 + 2nO2 = 2.0,8 + 1,2 nO2 = 1,2 mol => V =26,88 lt => ACu 128:t chy hn hp X gm 2 ancol c s mol bng nhau thu c hn hp CO2 v H2O theo l mol tngng 2 : 3. X gm

A. CH3OH v C2H5OH. C.C2H5OH v C2H4(OH)2.B.C3H7OH v C3H6(OH)2. D.C2H5OH v C3H7OH.

Tng x / Tng y = nCO2 / 2nH2O = 1/3 => C tha mnv s C = 4 ; s H = 12 v do n 2 ancol bng nhau.Cu 129:t chy hon ton a mol ancol A c b mol CO2 v c mol H2O. Bit a = c -b. Kt lun no sau yng ?

A.A l ancol no, mch vng. B.A l ancol no, mch h.C.A la 2ancol cha no. C.A l ancol thm.

a = nAncol = nH2OnCO2 => ancol no , mch hCu 130:t chy mt lng ancol A cn va 26,88 lt O2 ktc, thu c 39,6 gam CO2 v 21,6 gam H2O. Ac cng thc phn t l

A.C2H6O. B.C3H8O. C.C3H8O2. D.C4H10O.

n = nCO2 / (nH2OnCO2) = 3 => C3H8Oz ; nC3H8Oz = nCO2 / 3 = 0,33 + 8/4z/2 = nO2 / nC3H8Oz = 4 => z = 2 => C3H8O2Cu 131: Cho hn hp X gm hai ancol a chc, mch h, thuc cng dy ng ng. t chy hon ton hn hpX, thu c CO2 v H2O c t l mol tng ng l 3 : 4. Hai ancol l

A. C3H5(OH)3 v C4H7(OH)3. B. C2H5OH v C4H9OH.C. C2H4(OH)2 v C4H8(OH)2. D. C2H4(OH)2 v C3H6(OH)2.

n = nCO2 / (nH2OnCO2) = 3 => C v chn nCO2 = 3 mol ; nH2O =4 molv do a chc=> c my nhm OHCu 132: Khi t chy hon ton m gam hn hp hai ancol no, n chc, mch h thu c V lt kh CO2( ktc)v a gam H2O. Biu thc lin h gia m, a v V l

A. m = 2a - V/22,4. B. m = 2a - V/11,2. C. m = a + V/5,6. D. m = a - V/5,6.


18/26


Cch 1:Chn 32 g CH3OH => to ra 1 mol CO2 22,4 lt + 2mol H2O 36 g thay s vo tng p n => D ng : 32 = 36 22,4/5,6 Ly

Cch 2:Ancol no , n chc => CnH2n+1OHnAncol = nH2OnCO2 ; BTNT Oxi : nancol + 2nO2 = 2nCO2 + nH2OnH2OnCO2 + 2nO2 = 2nCO2 + nH2OnO2 = 3nCO2 / 2BTKL => m + mO2 = mCO2 + mH2O m + (3V.32)/(22,4.2) = VCO2 . 44/22,4 + a m = aV/5,6

Cch 3:m ancol = mC + mH + mO = 12.nCO2 + 2.nH2O + 16(nH2O nCO2) = 18nH2O4nCO2= 18. mH2O /18 - 4.VCO2 / 22,4 = mH2OVCO2 / 5,6 = aV / 5,6Cu 133: t chy hon ton 0,2 mol mt ancol X no, mch h cn va 17,92 lt kh O2 ( ktc). Mt khc, nucho 0,1 mol X tc dng va vi m gam Cu(OH)2 th to thnh dung dch c mu xanh lam. Gi tr ca m v tngi ca X tng ng l

A. 9,8 v propan-1,2-iol. B. 4,9 v propan-1,2-iol.C. 4,9 v propan-1,3-iol. D. 4,9 v glixerol.

X tc dng vi Cu(OH)2 => dung dch mu xanh lam => X phi c t nht 2 nhm OH lin k ; - C(OH)C(OH)=> Loi CnO2 / nX = 4 # 3,5 => khng th l glixerol C3H5(OH)3 nO2 / nX = n,5=> Loi D => Thy A, B u c Propan1,2iol C3H6(OH)2 => xt khi lng Cu(OH)2

Ta lun c 2Ancol + Cu(OH)2 => nCu(OH)2 = nAncol / 2 = 0,1/2 0,05 mol => mCu(OH)2 = 4,9 g => B

Cu 134: a. Kh CO2 sinh ra khi ln men ru mt lng glucoz c dn vo dung dch Ca(OH)2 d to c40g kt ta. Khi lng ancol etylic thu c lA. 18,4 gam. B. 16,8 gam. C. 16,4 gam. D. 17,4 gam.

P : C6H12O6 => 2C2H5OH + 2CO2 => nC2H5OH = nCO2 = nCaCO3 k ta = 0,4 mol=> mC2H5OH = 18,4 g => A

b.Nu hiu sut phn ng ln men l 80% th khi lng glucoz dng l bao nhiu gam ?A. 45 gam. B. 90 gam. C. 36 gam. D. 40 gam.

nC6H12O6 = nCO2 / 2 = 0,2 mol => mC6H12O6 theo PT = 36 g => H% p = mPT .100% / mBm ban u = 36.100% / 80% = 45 g => ACu 135:Cho m gam tinh bt ln men thnh C2H5OH vi hiu sut 81%, hp th ht lng CO2 sinh ra vo dungdch Ca(OH)2 c 55 gam kt ta v dung dch X. un nng dung dch X li c 10 gam kt ta na. Gi tr m l

A.75 gam. B.125 gam. C.150 gam. D.225 gam

Xem li chuyn 1phn CT : nCO2 = n Kt ta 1 + 2. n Kt ta 2 nung

nCO2 p = 0,55 + 2.0,1 = 0,75 mol=> nC6H10O5 = nCO2 / 2 = 0,325 mol V C6H10O5 => C6H12O6 => 2CO2 + 2C2H5OH => mC6H10O5 theo PT = 0,375.162 = 60,75g

H% p = mPT . 100% / mBmB= 60,75.100% / 81 = 75g => ACu 136:Th tch ancol etylic 92o cn dng l bao nhiu iu ch c 2,24 lt C2H4 (ktc). Cho bit hiu sut

phn ng t 62,5% v d = 0,8 g/ml.A. 8 ml. B.10 ml. C.12,5ml. D.3,9 ml.

nC2H5OH = nC2H4 = 0,1 mol => mC2H5OH theo PT = 4,6 g ; H% p = mPT.100% / mBmB = 4,6.100% / 62,5% = 7,36 g => V C2H5OH = 7,36/0,8 = 9,2 g mPT , mB ca ru=> ru = VC2H5OH .100 / V RuV ru = 9,2.100% / 92 = 10 ml => BCu 137:i t 150 gam tinh bt s iu ch c bao nhiu ml ancol etylic 46o bng phng php ln men ancol?Cho bit hiu sut phn ng t 81% v d = 0,8 g/ml.

A.46,875 ml. B.93,75 ml. C.21,5625 ml. D.187,5 ml.H% p = m PT . 100% / mB81% = mPT . 100% / 150 = mPT Tinh bt = 121,5 g => nTinh bt = 0,75 mol nC2H5OH = 2nTinh bt = 1,5 mol Bi 135=> mC2H5OH = 69 g

V C2H5OH = 69/0,8 = 86,25 g => ru = VC2H5OH . 100 / V ru V ru = 86,25.100 / 46 = 187,5 ml => D

Cu 138: Khi lng ca tinh bt cn dng trong qu trnh ln men to thnh 5 lt ru (ancol) etylic 46 l (bithiu sut ca c qu trnh l 72% v khi lng ring ca ru etylic nguyn cht l 0,8 g/ml)

A. 5,4 kg. B. 5,0 kg. C. 6,0 kg. D. 4,5 kg.Tm c nC2H5OH da vo V ru , ru , d => m Tinh bt theo PT da vo H%=> mTinh bt ban du = 4,5 kg=> DCu 139: Ln men hon ton m gam glucoz thnh ancol etylic. Ton b kh CO2 sinh ra trong qu trnh ny c

hp th ht vo dung dch Ca(OH)2d to ra 40 gam kt ta. Nu hiu sut ca qu trnh ln men l 75% th gi trca m lA. 60. B. 58. C. 30. D. 48.

nCO2 = nCaCO3 = 0,4 mol => nC6H12O6 = nCO2 / 2 = 0,2 mol C6H12O6>2C2H5OH +2CO2


19/26


=>m C6H12O6 theo PT = 36 g ; H% p = mPT . 100% / mBmB = 36.100% / 75% = 48 => DCu 140: Ln men m gam glucoz vi hiu sut 90%, lng kh CO2 sinh ra hp th ht vo dung dch nc vitrong, thu c 10 gam kt ta. Khi lng dung dch sau phn ng gim 3,4 gam so vi khi lng dung dch ncvi trong ban u. Gi tr ca m l

A. 20,0. B. 30,0. C. 13,5. D. 15,0Xem chuyn 1 => m dd gim = m kt ta mCO2 cho vo H3,4 = 10nCO2mCO2 = 6,6 g => nCO2 = 0,15 mol => nC6H12O6 = nCO2 / 2 = 0,075 mol

H% p = mPT .100% / mB

mB = 0,075.180.100% / 90% = 15g => DCu 141:X l hn hp gm phenol v ancol n chc A. Cho 25,4 gam X tc dng vi Na (d) c 6,72 lt H2 (ktc). A l

A. CH3OH. B. C2H5OH. C.C3H5OH. D. C4H9OH.Ta c nhn hp = 2nH2 = 0,6 mol V ta lun c x . nA = 2nH2 vi x l s H linh ng , A l cht cha H linh ng, phenol v ancol nchc u c 1 nhm OH => x = 1

=>25,4

M = 42,330,6

=> M ru < 42,33 V M phenol = 94 >42,33 => Ch c CH3OH M = 32=> A

Cu 142: C bao nhiu hp cht hu c C7H8O va tc dng vi Na, va tc dng vi NaOH ?A.1. B.2. C.3. D. 4.Va tc dng vi Na ,va tc dng vi NaOH=>hp cht cha gc phenol

CH3C6H4OH CH3 gn v tr 1 , gc OH gn v tr m , p , o => 3 chtCu 143:A l hp cht c cng thc phn t C7H8O2. A tc dng vi Na d cho s mol H2 bay ra bng s mol

NaOH cn dng trung ha cng lng A trn. Ch ra cng thc cu to thu gn ca A.A.C6H7COOH. B. HOC6H4CH2OH. C.CH3OC6H4OH. D.CH3C6H3(OH)2.

nA = nH2 => A cha 2 H linh ng hay 2 nhm OH => Loi A v CnA = nNaOH => A cha 1 gc OH gn vi vng benzen => BLoi D v 2 OH gn vi vng benzenCu 144:Khi t chy 0,05 mol X (dn xut benzen) thu c di 15,4gam CO2. Bit 1 mol X phn ng va vi 1 mol NaOH hoc vi 2 mol Na. X c cng thc cu to thu gn l

A.CH3C6H4OH.B.CH3OC6H4OH. C.HOC6H4CH2OH. D.C6H4(OH)2.Tc dng 2 mol Na v 1 mol NaOH =>dn xut ca phenol c 2 nhm OH , 1 nhm gn vi vng benzen , 1 nhmgn vi C ngoi vng benzen nh bi 143 => C

V nX = nNaOH => c 1OH gn vi vng benzen ; nX = 2nNa => c 1 OH gn vi C ngoi vng benzenCu 145:Ha cht no di y dng phn bit 2 l mt nhn cha dung dch phenol v benzen.1. Na. 2. dd NaOH. 3. nc brom.A. 1 v 2. B.1 v 3. C.2 v 3. D.1, 2 v 3.

Na => phenol to ra kh , benzen khng pNaOH phenol c p nhng khng bit hin tngdd Br2 => phenol to ra kt ta , benzen khng p => 1 v 3 => B Mnh khng bit NaOH cngng ko? Cho mnh kin ca bnCu 146:A l hp cht hu c cng thc phn t l C7H8O2. A tc dng vi NaOH theo t l 1 : 2. Vy A thucloi hp cht no di y ?

A. i phenol. B.Axit cacboxylic C.Este ca phenol. D.Va ancol, va phenol.nNaOH = 2nA => c 2 H linh ng

A tha mn v 2OH gn vi vng benzenB sai v ch l n chc do 2OxiC sai v khng c H linh ngD sai v ancol khng p vi NaOHCu 147:C bao nhiu ng phn (cha vng bezen), cng thc phn t C8H10O, khng tc dng vi Na?

A.2. B.3. C.4. D.5.Khng phn ng vi Na => Khng l ancol , khng l phenolK = (8.210 + 2)/2 = 4 m vng benzen c 4 pi=> ete O v ch c ete v ru l c 1 oxi + noC6H5CH2 - O - CH3; C6H5OCH2CH3CH3C6H4OCH3 CH3 v tr 1 , - OCH3 v tr o ,p , m => 3=> Tng = 5 p => DCu 148:A l cht hu c c cng thc phn t CxHyO. t chy hon ton 0,1 mol A ri hp th ton b sn

phm chy vo nc vi trong thy c 30 gam kt ta. Lc b kt ta em un nng phn nc lc thy c 20 gamkt ta na. Bit A va tc dng Na, va tc dng NaOH. Ch ra cng thc phn t ca A.

A.C6H6O. B.C7H8O. C.C7H8O2. D.C8H10O.


20/26


nCO2 = n Kt ta 1 + 2.n kt ta 2 = 0,3 + 2.0,2 = 0,7 mol=> x = nCO2 / nA = 7=> C7H8O Nhn p n v l CxHyOv x = 7Cu 149:Ch ra th t tng dn mc linh ca nguyn t H trong nhm -OH ca cc hp cht sau: phenol,etanol, nc.

A.Etanol < nc < phenol. C.Nc < phenol < etanol.B.Etanol < phenol < nc. D.Phenol < nc < etanol.

Phenol p vi NaOH , nNa => Phenol ln nht

H2O p mnh lit vi Na => H2O th nhEtanol p vi Na bnh thng => Ru cui cng => ACu 150:T 400 gam bezen c th iu ch c ti a bao nhiu gam phenol. Cho bit hiu sut ton b qu trnht 78%.

A.376 gam. B.312 gam. C.618 gam. D.320 gam.C6H6 => C6H6O => nC6H6 = nC6H6O ; H% sp = mTT.100% / mPTmTT = mPT . H% / 100% = nC6H6.94.78%/100% = 376 g => ACu 151:Ha cht no di y c th dng phn bit cc l mt nhn cha cc dung dch : C6H5ONa, NaCl,BaCl2, Na2S, Na2CO3 l

A.dd NaOH. B.dd HCl. C.Na. D dd KCl.KCl trung tnh s khng p vi cc cht v to t KL mnh v gc axit mnh

NaOH khng p vi cc cht v cc cht u cha Na v BaCl2 ko p vi NaOH

Na cng khng p v Na + H2O => NaOH ri ko p vi cc cht

B HCl Phn tchHCl + Na2S => H2S + NaCL => to ra kh H2S mi trng thi => Nhn bit Na2SHCl + Na2CO3 => NaCl + CO2 + H2O => to ra kh => To ra kh nhn bit Na2CO3HCl + C6H5ONa => C6H5OH + NaCl => c p => C6H5ONa

Nhn bit c Na2CO3 cho vo 2 bnh con li khng p => BaCL2 to ra kt ta trng BaCO3Cu 152:So vi etanol, nguyn t H trong nhm -OH ca phenol linh ng hn v :

A.Mt electron vng benzen tng ln, nht l cc v tr o v p. B.Lin kt C-O ca phenol bn vng.C. Trong phenol, cp electron cha tham gia lin kt ca nguyn t oxi tham gia lin hp vo vng

benzen lm lin kt -OH phn cc hn.D.Phenol tc dng d dng vi nc brom to kt ta trng 2, 4, 6-tri brom phenol.

SGK 11 NC232 => CCu 153:C bao nhiu phn ng xy ra khi cho cc cht C6H5OH ; NaHCO3 ; NaOH ; HCl tc dng vi nhau tngi mt ?

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

NaHCO3 + NaOH => Na2Co3 + H2ONaHCO3 + HCl => NaCl + CO2 + H2ONaOH + HCl => NaCl + H2O

C6H5OH + NaOH => C6H5ONa + H2O => 4 => BCu 154: Dy gm cc cht u phn ng vi phenol l

A. dung dch NaCl, dung dch NaOH, kim loi Na. B. nc brom, axit axetic, dung dch NaOH.C. nc brom, anhirit axetic, dung dch NaOH. D. nc brom, anehit axetic, dung dch NaOH.

SGK 11 NC - 231 => C

Cu 155: Hin tng ln lt xy ra khi nh vi git dung dch HCl c vo ng nghim cha mt t dung dchHCOONa v mt t dung dch C6H5ONa ri lc mnh l

A.C s phn lp ; dung dch trong sut ha c.B. Dung dch trong sut ha c.C. C phn lp ; dung dch trong sut. D. Xut hin s phn lp c 2 ng nghim.

Cu 156: nh hng ca nhm -OH n gc C6H5- trong phn t phenol th hin qua phn ng gia phenol viA. dung dch NaOH. B.Na kim loi. C. nc Br2. D. H2(Ni, nung nng).

SGK 11NC231 => Phn ng th Br2 vo phenol d hn so vi benzen => do nh hng canhm OH => CCu 157:Cht c cng thc phn t no di y c th tc dng c c Na, c NaOH ?

A.C5H8O. B.C6H8O. C.C7H10O. D.C9H12O.Tc dng c Na v NaOH => phenol => c sTng vng + pi 4

=> ch c D c tng pi + vng = 4 => D tha mnCu 158:Ba hp cht thm X, Y, Z u c cng thc phn t C7H8O. X tc dng vi Na v NaOH ; Y tc dngvi Na, khng tc dng NaOH ; Z khng tc dng vi Na v NaOH Cng thc cu to ca X, Y, Z ln lt l

A. C6H4(CH3)OH ; C6H5OCH3 ; C6H5CH2OH. B. C6H5OCH3 ; C6H5CH2OH ; C6H4(CH3)OH.


21/26


C. C6H5CH2OH ; C6H5OCH3 ; C6H4(CH3)OH. D. C6H4(CH3)OH ; C6H5CH2OH ; C6H5OCH3.X tc dng Na v NaOH => X l phenol , OH gn vi C trn vng benzen=> Loi B v CY tc dng vi Na khng tc dng vi NaOH => Y l ancol thmOH gn vi C ngoi vng benzen => DZ l eteCu 159:Cho ln lt cc cht C2H5Cl, C2H5OH, C6H5OH, C6H5Cl vo dung dch NaOH long un nng. Himy cht c phn ng ?

A. C bn cht. B. Mt cht. C. Hai cht. D. Ba cht.

C2H5Cl v C6H5OH v p vi NaOH long un nng => dn xut halogen no c p ; v phenol cng c p => CCu 160:a. S ng phn ca C3H5Cl3 lA.5. B.6. C.3. D.4.

K = (3.25 + 23)/2 = 0 => p cu toC(Cl3)CC ; C(Cl)C(Cl2)C ;C(Cl)C(Cl)C(Cl) ; C(Cl2)C(Cl)C ; C(Cl2)CC(Cl)=> Tng = 5 => A

b. Trong s cc ng phn ca C3H5Cl3 c th c bao nhiu ng phn khi thu phn trongmi trng kim chosn phm phn ng c c vi Na v dung dch AgNO3/NH3 to ra Ag ?

A.1. B.4. C.3. D.2.Thy phn mi trng kim => phn ng th Cl thnh OHSn phm p c Na v AgNO3/NH3 to ra Ag => Phi cha gc OH ( p Na) v 1 gc CHO p vi

AgNO3/NH3Nh iu kin h bin ca ru => phi c 2 Cl gn vo 1 C (khi 2 OH gn cng 1 C => h bin to ra andehit)=> ClCCC(Cl2) + NaOH => OHCCCHO + NaCl + H2OCC(Cl)C(Cl2) + NaOH => CC(OH)CHO + NaCl + H2O=> 2 p => DCu 161:Hp cht X c cha vng benzen v c CTPT l C7H6Cl2. Thy phn X trong NaOHdc, to cao, pcao thu c cht Y c CTPT l C7H6O. Hy cho bit X c bao nhiu CTCT?

A.3. B.1. C.4. D.2.V Y l C7H6O hay C6H5 CH2OH v nu l phenol th s p vi NaOH d=> CH c trng hp duy nht C6H5 CH(Cl2) +NaOH => C6H5-CHO + H2O + NaCL iu kin h bin khi 2OH gn cng 1 CCu 162:Cho cc hp cht sau : (I) CH3CH2OH. (II) C6H5OH. (III) NO2C6H4OH.

Chn pht biu saiA. C 3 cht u c nguyn t H linh ng.B. C 3 u phn ng c vi dung dch baz iu kin thng.C. Cht (III) c nguyn t H linh ng nht.D. Th t linh ng ca nguyn t H c sp xp theo chiu nh sau : III > II > I.

CH3CH2OH l ru khng p vi bazo => B sai => BCu 163:Cho cc cht sau A : CH4O ; B: C2H6O2 ; C: C3H8O3.iu no sau y lun ng ?

A. A, B, C l cc ancol no, mch h. B. A, B, C u lm mt mu dd thuc tm.C. A, B, C l cc hp cht hu c no. D. A, B, C u l este no, n chc.

A,B,C u c dng CnH2n+2Oz => u l hp cht no , mch h => ACu 164: Cho 2 phn ng :(1) 2CH3COOH + Na2CO3 2CH3COONa + H2O + CO2

(2) C6H5ONa + CO2 + H2O C6H5OH + NaHCO3Hai phn ng trn chng t lc axit theo th t CH3COOH, H2CO3, C6H5OH, HCO3- l

A.Tng dn. B.Gim dn. C.Khng thay i. D.Va tng va gim.P (1) to ra H2CO3 (CO2+H2O) => Tnh axit ca CH3COOH > H2CO3P (2) C6H5ONa to ra NaHCO3 hay HCO3- => Tnh axit caC6H5ONa > HCO3-

Ngoi ra CH3COOH c H linh ng mnh => Ln nht => Lc axit gim dn => BCu 165: Cho dy cc cht : phenol, anilin, phenylamoni clorua, natri phenolat, etanol. S cht trong dy phn ngc vi NaOH (trong dung dch) l

A. 4. B. 3. C. 1. D. 2.Cu 166:X l hn hp gm phenol v metanol. t chy hon ton X c nCO2 = nH2O. Vy % khi lngmetanol trong X l

A.25%.

B.59,5%.

C.50,5%.

D.20%.Phenol C6H5OH ; metanol CH3OH

nCO2 = nH2O => chn nCO2 = 1 mol => nH2O = 1molBTNTC => 6nC6H5OH + nCH3OH = nCO2 ; 6nC6H5OH + 4nCH3OH = 2nH2O


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Gii h => nC6H5OH = 1/9 ; nCH3OH = 1/3=> %m CH3OH = (32/3).100%/(94/9 + 32/3) = 50,5 % => CCu 167: Mt hp cht X cha ba nguyn t C, H, O c t l khi lng mC : mH : mO = 21 : 2 : 4. Hp cht X ccng thc n gin nht trng vi cng thc phn t. S ng phn cu to thuc loi hp cht thm ng vi cngthc phn t ca X l

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

mC : mH : mO = 21 : 2 : 4 => chn mC = 21g => mH = 2g ; mO = 4 g

x : y : z = 21/12 : 2/1 : 4/16 = 1,75 : 2 : 0,25 = 7 : 8 : 1 => C7H8O v cng thc n gin nht trng CTPT

p thm => ch cn cha vng benzen C6H5CH2OH ; C6H5OCH3 ; CH3C6H4OH CH3 gn v tr 1 , OH gn v tr o,m,p

Tng = 5 => DCu 168: Cho X l hp cht thm ; a mol X phn ng va ht vi a lt dung dch NaOH 1M. Mt khc, nu cho amol X phn ng vi Na (d) th sau phn ng thu c 22,4a lt kh H2 ( ktc). Cng thc cu to thu gn ca X l

A. HOC6H4COOCH3. B. CH3C6H3(OH)2. C. HOC6H4COOH. D. HOCH2C6H4OH.Ta c nX = nNaOH => X c 1 gc OH gn vi vng benzennX = nH2=> X c 2 H linh ng m X c 1 gc OH gn vng benzen=> X c thm 1 gc OH gn vi C ngoi vng benzen => D tha mn

Cu 169: Hp cht hu c X (phn t c vng benzen) c cng thc phn t l C7H8O2, tc dng c vi Na v vi

NaOH. Bit rng khi cho X tc dng vi Na d, s mol H2 thu c bng s molX tham gia phn ng v X tc dngc vi NaOH theo t l s mol 1 : 2. Cng thc cu to thu gn ca X lA. C6H5CH(OH)2. B. CH3C6H3(OH)2. C. CH3OC6H4OH. D. C. HOCH2C6H4OH.

X tc dng vi Na => nX = nH2 => x cha 2 H linh ng hay 2 gc OHX tc dng vi NaOH => 2nX = nNaOH => X cha 2 gc OH gn vi vng benzen => BCu 170:Cho hn hp hai anken ng ng k tip nhau tc dng vi nc (c H2SO4lm xc tc) thu c hnhp Z gm hai ru (ancol) X v Y. t chy hon ton 1,06 gam hn hp Z sau hp th ton b sn phm chyvo 2 lt dung dch NaOH 0,1M thu c dung dch T trong nng ca NaOH bng 0,05M. Cng thc cu tothu gn ca X v Y l (Th tch dung dch thay i khng ng k)

A. C4H9OH v C5H11OH. B. C3H7OH v C4H9OH.C. C2H5OH v C3H7OH. D. C2H5OH v C4H9OH.

NaOH dv CM NaOH cn li = 0,05 M => CM NaOH p = 0,1 0,05 = 0,05 M

nNaOH p = 0,1 mol => nCO2 p = nNaOH / 2 = 0,05 mol V NaOH d => xy ra p trung ha CO2 + 2NaOH => Na2CO3 + H2O

p n => ru no n chc : C n H2 n +1OH => n Hn hp ru = nCO2 / n = 0,05 / n

M ru = 14 n + 18 = 1,06 / (0,05/ n )n = 2,5 => C

Cu 171: t chy hon ton hn hp M gmhai ru (ancol) X v Y l ng ng k tip ca nhau, thu c 0,3mol CO2v 0,425 mol H2O. Mt khc, cho 0,25 mol hn hp M tc dng vi Na (d), thu c cha n 0,15 molH2. Cng thc phn t ca X, Y l

A. C3H6O, C4H8O. B. C2H6O, C3H8O. C. C2H6O2, C3H8O2. D. C2H6O, CH4O.n = nCO2 / (nH2OnCO2) = 2,4Ta c 0,25 mol M => nH2 < 0,15 mol=> Tha mn nM = 2nH2 => Ru n chc=> BCu 172:Oxi ho m gam etanol thu c hn hp X gm axetanehit, axit axetic, nc v etanol d. Cho ton b X

tc dng vi dung dch NaHCO3 (d), thu c 0,56 lt kh CO2( ktc). Khi lng etanol b oxi ho to ra axitl

A. 1,15 gam. B. 4,60 gam. C. 2,30 gam. D. 5,75 gam.nAxit = nCO2 = 0,025 mol P CH3COOH + NaHCO3 => CH3COONa + CO2 + H2OnAxit = n Ru= 0,025 mol P : C2H5OH + O2 => CH3COOH=> m ru b oxi ha to ra axit = 0,025.46 = 1,15 gCu 173: Khi phn tch thnh phn mt ru (ancol) n chc X th thu c kt qu : Tng khi lng ca cacbonv hiro gp 3,625 ln khi lng oxi. S ng phn ru (ancol) ng vi cng thc phn t ca X l

A. 3. B. 4. C. 2. D. 1.Ancol n chc => CxHyOchn 1 mol CxHyO : mC + mH = 3,625mO12x + y = 3,625.16 = 58 y = 5812x ; ta c 0< y 2x + 20 < 5812x 2x + 2 4 x < 4,833 => x = 4 => y = 10 => C4H10O ; ru no n chc

=>CT tnh s p : 2n-2=> C4H10O c 4 p ru=> BCu 174: Oxi ho ancol n chc X bng CuO (un nng), sinh ra mt sn phm hu c duy nht l xeton Y (tkhi hi ca Y so vi kh hiro bng 29). Cng thc cu to ca X l


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A. CH3CHOHCH3. B. CH3COCH3. C. CH3CH2CH2OH. D. CH3CH2CHOHCH3.

Xeton :CnH2nO => M = 58 = 14n + 16 => n = 3 hay C3H7OH => Loi B v DV to ra xeton => Ru bc 2 hay OH gn vi C bc 2 => ACu 175:Mt hp cht hu c A gm C, H, O c 50% oxi v khi lng. Ngi ta cho A qua ng ng 10,4 gamCuO nung nng thu c 2 cht hu c v 8,48 gam cht rn. Mt khc cho hn hp 2 cht hu c trn tc dng vidung dch AgNO3(d) trong NH3 to ra hn hp 2 mui v 38,88 gam Ag. Khi lng ca A cn dng l

A. 1,28 gam. B. 4,8 gam. C. 2,56 gam. D. 3,2 gam.

Chuyn 1 cch tm C,H,O khi bit % ca thnh phnGi CxHyOz => %O = 16z .100% (12x + y + 16z) = 50%12x + y = 16z => z = 1 , x = 1 , y = 4=> CH4O HayCH3OHCu 176:un nng ancol A vi hn hp NaBr v H2SO4 c thu c cht hu c B, 12,3 gam hi cht B chimmt th tch bng th tch ca 2,8 gam N2 cng nhit 560oC ; p sut 1 atm. Oxi ho A bng CuO nung nngthu c hp cht hu c c kh nng lm mt mu dung dch nc brom. CTCT ca A l

A. CH3OH. B. C2H5OH. C. CH3CHOHCH3. D. CH3CH2CH2OH. => A l ancol no n chc: CnH2n+1OHCh nh B l CnH2n+1Brta c nB = nN2 v th tch bng nhaumB / MB = 2,8 / 28 => MB = 123 = 14n + 81 n = 3 =>A l C3H7OH

Ta c oxi ha to ra sn phm lm mt mu dung dch Br2 => Ch c to ra andehit => OH gn vi C bc I=> D

Cu 177:un mt ancol A vi dung dch hn hp gm KBr v H2SO4 c th trong hn hp sn phm thu c ccht hu c B. Hi ca 12,5 gam cht B ni trn chim 1 th tch ca 2,80 gam nit trong cng iu kin. Cng thccu to ca A l

A.C2H5OH. B.CH3CH2CH2OH. C.CH3OH. D.HOCH2CH2OH.A , B , C l ancol no , n chc => Xt CnH2n+1OHTng t 176 => B l CnH2n+1Br c M = 14n + 81 = 125 v nB= nN2 = 0,1mol=> khng c n nguyn D ng v A , B , C sai

C th khi cho vo hn hp KBr v H2SO4th s c mt OH b thay th bi Br

CnH2n+2O2 => CnH2n+1O Br => c M = 14n + 17 + 80 = 125 => n = 2 => DCu 178:Anken X c cng thc phn t l C5H10. X khng c ng phn hnh hc. Khi cho X tc dng viKMnO4 nhit thp thu c cht hu c Y c cng thc phn t l C5H12O2. Oxi ha nh Y bng CuO d thuc cht hu c Z. Z khng c phn ng trng gng. Vy X l

A.2-metyl buten-2. B.But-1-en. C.2-metyl but-1-en. D.But-2-en.

Cu 179:t chy hon ton 1 th tch hi ancol no n chc A thu c CO2 v H2O c tng th tch gp 5 lnth tch hi ancol A dng ( cng iu kin). Vy A l

A. C2H5OH. B. C4H9OH. C. CH3OH. D. C3H7OH.T l th tch = t l s mol => Chn 1 mol Ancol => 5 mol hn hp CO2 v H2O nCO2 + nH2O = 5 mol ; nH2OnCO2 = n Ru = 1 => nH2O = 3 ; nCO2 = 2

n = nCO2 / (nH2OnCO2) = 2 => C2H5OH => ACu 180: Cho 30,4 gam hn hp gm glixerolC3H5(OH)3v mt ru n chc, no A phn ng vi Na th thuc 8,96 lt kh (ktc). Nu cho hn hp trn tc dng vi Cu(OH)2 th ho tan c 9,8 gam Cu(OH)2. Cng thcca A l

A.C2H5OH. B.C3H7OH. C.CH3OH. D. C4H9OH.

Ch c glixerol mi p vi Cu(OH)2 => nGlixerol = 2nCu(OH)2 = 0,2 mol V lun c 2 ancol + Cu(OH)2 m ru cn li = 30,4 m glixerol = 30,40,2.92 = 12 g

Ta c 3nC3H5(OH)3 + nRuou = 2nH2 CT : x . nX = 2nH2 vi x l s H linh ng. ..3.0,2 + n Ru = 2.0,4n Ru = 0,2=> M ru = 12/0,2 = 60 => B Hoc = 14n + 18 =>n = 3

Cu 181: Hn hp X gm ancol metylic v mt ancol no, n chc A, mch h. Cho 2,76 gam X tc dng vi Nad thu c 0,672 lt H2 (ktc), mt khc oxi ha hon ton 2,76 gam X bng CuO (to) thu c hn hp anehit.Cho ton b lng anehit ny tc dng vi dung dch AgNO3/NH3 d thu c 19,44 gam cht kt ta. Cng thccu to ca A l

A.C2H5OH. B.CH3CH2CH2OH.C.CH3CH(CH3)OH. D.CH3CH2CH2CH2OH.

Cch 1 th p n vo gii h m hn hp ru = 2,76 g v n hn hp = 2nH2=> vi C3H7OH pDa vo p to andehit => OH gn vi C bc I =>BCch 2: ta c gi x , y ln lt l s mol ca tng ru x + y = 2nH2 = 0,06mol


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Ta c CT : 2x.nAndehit = nAg vi x l s nhm CHO - ch HCHO c 2 gc CHO do CH3OH oxi ha to ra 4x + 2y = nAg = 0,18 mol V nAndehit = nancol

Gii h => x = y= 0,03 => mAncol = 2,76 mCH3OH = 1,8 g => M = 1,8/0,03 = 60 => C3H7OHDa vo p to ra andehit => B

Cu 182:Ancol no mch h A cha n nguyn t C v m nhm OH trong cu to phn t. Cho 7,6 gam A tc dnght vi Na cho 2,24 lt H2 (ktc). Mi quan h gia n v m l

A. 2m = 2n + 1. B. m = 2n + 2. C. 11m = 7n + 1. D. 7n = 14m + 2.

Ancol no c n nguyn t C v m nhm OH => c CT : CnH2n+2OmTa c m . nAncol = 2nH2 CT = 0,2 molnAncol = 0,2 / m=> M ancol = 7,6/(0,2/m) = 38m = 14n + 2 + 16m11m = 7n + 1 => CCu 183:Cht hu c X mch h c to ra t axit no A v etylen glicol. Bit rng a gam X th hichim thtch bng th tch ca 6,4 gam oxi cng iu kin nhit v p sut; a gam X phn ng ht vi xt to ra 32,8gam mui. Nu cho 200 gam A phn ng vi 50 gam etilenglicol ta thu c 87,6 gam este. Tn ca X v hiu sut

phn ng to X lA. Etylen glicol iaxetat ; 74,4%. B. Etylen glicol ifomat ; 74,4%.C. Etylen glicol iaxetat ; 36,3%. D. Etylen glicol ifomat ; 36,6%.

Cu 184:Oxi ho ancol etylic bng xc tc men gim, sau phn ng thu c hn hp X (gi s khng to raanehit). Chia hn hp X thnh 2 phn bng nhau. Phn 1 cho tc dng vi Na d, thu c 6,272 lt H2 (ktc).

Trung ho phn 2 bng dung dch NaOH 2M thy ht 120 ml. Hiu sut phn ng oxi ho ancol etylic l:A. 42,86%. B. 66,7%. C. 85,7%. D. 75%.

X gm ancol d + Axit to thnh+ H2O PT : C2H5OH + O2 => CH3COOH + H2OPhn 1 => nC2H5OH d + nAxit + nH2O = 2nH2 = 0,56 molPhn 2 => nAxit = nNaOH = 0,24 mol = nC2H5OH p v nCH3COOH = nC2H5OH pth vo phn 1 v 2 phn bng nhau

nC2H5OH d+ nH2O = 0,560,24 = 0,32 mol = nC2H5OH ban u v nH2O = nC2H5OH p H% oxi ha = nC2H5OH p .100% / nC2H5OH ban u = 0,24/0,32 = 75%=> D

Cu 185:t chy hon ton mt lng hn hp 2 ancol no n chc X, Y l ng ng lin tip thu c 11,2 ltCO2 cng vi lng hn hp trn cho phn ng vi Na d th thu c 2,24 lt H2 ( ktc). Cng thc phn t ca2 ancol trn l

A. C2H5OH; C3H7OH. B. CH3OH; C3H7OH.

C. C4H9OH; C3H7OH. D. C2H5OH ; CH3OH.P vi Na => 0,1 mol H2 => n hn hp ru = 2nH2 = 0,2mol

p n => Ancol no n chc : C n H2 n +1OH

=> n = nCO2 / n hn hp = 0,5 / 0,2 = 2,5 => ACu 186*: Oxi ho 9,2 gam ancol etylic bng CuO un nng thu c 13,2 gam hn hp gm anehit, axit, ancold v nc. Hn hp ny tc dng vi Na sinh ra 3,36 lt H2 ( ktc). Phn trm ancol b oxi ho l

A.25%. B.50%. C.75%. D. 90%.Phn ng vi Na => nAxit CH3COOH + n ru d C2H5OH + nH2O = 2nH2= 0,3 molTa c n Ru p = nH2O (1)nCH3COOH + nC2H5OH ban u= 0,3nCH3COOH = 0,1 mol

V p t l 1 : 1 gia C2H5OH vi H2O C2H5OH + CuO => CH3CHO + Cu + H2O (I)

C2H5OH + O2 => CH3COOH + H2O (II)

nC2H5OH PT II = nCH3COOH = nH2O PT II = 0,1 mol Gi x l s mol C2H5OH PT I => x = nCH3CHO = nH2O Pti => hn hp = 13,2 g = mCH3CHO + mH2O (PT I) + mCH3COOH + mH2O (PT 2)+ C2H5OH d 13,2 = 44x + 18x + 6 + 1,8 + (9,20,1.46x.46)x = 0,05 mol

nC2H5OH p = x + 0,1 = 0,15 mol => H% = 0,15.100% / 0,2 = 75%=>CCu 187:Thc hin cc th nghim sau:TN 1 : Trn 0,015 mol ru no X vi 0,02 mol ru no Y ri cho tc dng ht vi Na th thu c 1,008 lt H2.TN 2 : Trn 0,02 mol ru X vi 0,015 mol ru Y ri cho hp tc dng ht vi Na th thu c 0,952 lt H2.Th nghim 3 : t chy hon ton mt lng hn hp ru nh trong th nghim 1 ri cho tt c sn phm chy iqua bnh ng CaO mi nung, d thy khi lng bnh tng thm 6,21 gam. Bit th tch cc khi o ktc. Cngthc 2 ru l

A. C2H4(OH)2 v C3H6(OH)2. B. C2H4(OH)2 v C3H5(OH)3.C. CH3OH v C2H5OH. D. Khng xc nh c.

Gi x , y l s nhm OH ca X v Y


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TH1 => x .nX + y.nY = 2nH20,015x + 0,02y = 0,09

TH2 => 0,02x + 0,015y = 0,085 Gii h => x = 2 ; y = 3=> B i thi ko c p nko xc nh X l CnH2n(OH)2; Y l CmH2m-1(OH)3 Xem chuyn 1 cch xc nh cht

CnH2n+22am(OH)m a l tng b + vng , m l s chcTa c c 2 ru u no => nH2O nCO2 = n hn hp ru = 0,015 + 0,02 = 0,035 molm bnh tng = mH2O + mCO2 = 18.nH2O + 44.nCO2 = 6,21

Gii h =>nH2O = 0,125 ; nCO2 = 0,09 => n = nCO2 / (nH2OnCO2) = 2,57 => B


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CHUYN 5 : DN XUT HALOGEN-PHENOL-ANCOL

1A 2A 3C 4B 5B 6A 7A 8A 9C 10AD

11C 12B 13B 14D 15A 16D 17B 18A 19B 20A

21B 22D 23C 24C 25B 26B 27A 28C 29D 30D

31D 32B 33C 34C 35B 36B 37B 38C 39A 40C

41D 42A 43B 44A 45C 46D 47D 48C 49B 50B

51D 52D 53C 54A 55D 56C 57D 58C 59B 60A61D 62A 63A 64D 65A 66A 67C 68CB 69B 70B

71D 72D 73C 74B 75C 76B 77A 78C 79A 80D

81D 82A 83D 84A 85C 86A 87C 88B 89C 90C

91A 92C 93A 94D 95B 96B 97C 98C 99A 100D

101A 102C 103C 104B 105A 106C 107A 108D 109A 110A

111B 112D 113A 114B 115B 116C 117B 118B 119A 120C

121C 122A 123A 124C 125B 126C 127A 128C 129B 130C

131C 132D 133B 134AA 135A 136B 137D 138D 139D 140D

141A 142C 143B 144C 145D 146A 147D 148B 149A 150A

151B 152C 153B 154C 155B 156C 157D 158D 159C 160AD

161B 162B 163A 164B 165D 166C 167D 168D 169B 170C

171B 172A 173B 174A 175D 176D 177D 178A 179A 180B

181B 182C 183A 184D 185A 186A 187B

p n khng phi ng 100% u nh c th 1 s p n saiBn c cho kin v cu . Mnh v mt s ngi s xem li.

Cm n bn gip .Chc bn thnh cng.

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