chuyÊn ĐỀ 5 bÀi tẬp dẪn xuẤt halogen - phenol - ancol

HOCHOAHOC.COM Chuyn trang hc ha hc

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CHUYN 5 : DN XUT HALOGEN - PHENOL - ANCOL

Cu 1 : S ng phn ca C4H9Br l A. 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 xng C C C C ; C C C Mi tn l v tr Br => 4 ng phn =>A C

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

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

C C C C ; C C C Mi tn l v tr Cl => 2 ng phn =>A Bc I II II I I C

Cu 3: S ng phn mch h (k c ng phn hnh hc) ca cht c CTPT l C3H5Br l A. 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 xiclo CH(Br) = CH CH3 c p hnh hc xem li K p hnh hc chuyn 1 => 2 CH2=CH-CH2Br ; CH2=C(Br)-CH3 khng c p hnh hc => 2 => Tng c 4 p C3H5Br mch h. => C Cu 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 na Gi 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 CnHnCl2n

Cch 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 1

Cu 5: Dn xut halogen khng c ng phn cis-trans l A. 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 H Cu 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 Cl CH2 CH CH CH3 CH3 Cl

=> 1,3 iclo 2 metylbutan v tr 1 ,3 cha clo ; i v 2clo ; meyl v tr th 2; butan v mch chnh c 4C Cu 7: Cho cc cht sau: C6H5CH2Cl ; CH3CHClCH3 ; Br2CHCH3 ; CH2=CHCH2Cl. Tn gi ca cc cht trn ln 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 nh CH2=CHCH2 l anlyl nh ; CH2=CH l vinyl ; SGK 11NC 211 Cu 8: Cho cc dn xut halogen sau : C2H5F (1) ; C2H5Br (2) ; C2H5I (3) ; C2H5Cl (4) th t gim dn nhit si l


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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 X ht 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 tng xy ra l

A. 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 NC 212 ci bng => Phi c un nng => Sinh ra kt ta i vi CH2=CHCH2CL Anlylclorua => C khng c hin tng g Cu 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 tch cng H C bn cnh bc cao Bc III I 1 2 3 4 CH3CH(CH3)CHBrCH3 hay CH3 CH CH CH3 => CH3 C=CH CH3 CH3 Br CH3

nh s th t gn lin kt i + gn mch nhnh nht => 2 metybut 2 en (CH3 v ni i v tr 3 v mch chnh c 4C) b. Sn phm chnh to thnh khi cho 2-brombutan tc dng vi dung dch KOH/ancol, un nng

A. metylxiclopropan. B. but-2-ol. C. but-1-en. D. but-2-en.

Bc I II 1 2 3 4 2-brombutan ; CH3 CH CH2 CH3 => CH3 CH = CH CH3=> but 2 en Br

Cu 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 l

A. C2H5Cl. B. C3H7Cl. C. C4H9Cl. D. C5H11Cl.

Ankyl :CnH2n+1 => ankylclorua : CnH2n+1Cl

Mo chng cn bit p g => on kt ta l AgCl v ly 21,525 chia cho 143,5MAgCl = 0,15 mol p BTNT Cl => nCnH2n+1Cl = nAgCl = 0,15 mol

MY = 14n + 36,5 = 13,875/0,15 n = 4 => C4H9Cl Hoc p ban u to ra HCl => HCl + AgNO3 => AgCl kt ta trng + HNO3 Cu 12: S tch hiro halogenua ca dn xut halogen X c CTPT C4H9Cl cho 3 olefin ng phn, X l cht no trong 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 hc n- l mch thng tut A .CH3 CH2 CH2 CH2 - Cl Sec c dng CH3 CH(X)-CH3 - => B . CH3 CH(Cl) CH2 CH3 Iso c dng CH3 CH(CH3) => C. CH3 CH(CH3) CH2 - Cl Tert 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 c th loi v chn p n. => B ng V cho 3 ng phn khi thch HCl I II

CH3 CH CH2 CH3 => 2 sn phm : CH2=CH-CH2-CH3 ko c p hnh hc Cl CH3-CH=CH-CH3 => c p hnh hc => 3p khi tc HCl Cu 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 11NC 212 => Halogen gn vi vng benzen khng phn ng vi kim un nng nhit cao v p sut cao mi p Dn xut ankyl halogen Hay Halogen gn vi hidrocacbon no s p vi kim nhit un nng PT: ClC6H4CH2Cl + KOH => ClC6H4CH2OH + KCl => B

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


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Ancol thm OH gn vi C ngoi vng benzen khng phn ng vi kim Phenol OH gn vi C vng benzen c p vi kim Do KOH d => PT : OHC6H4CH2OH + KOH => KOC6H5CH2OH + H2O => sn phm p n D

L thuyt n gin SGK nhng khng hc s rt kh lm Cu 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 ancol PTTQ : R CH2 Cl R c th l ng ng ankyl hoc ng ng ankan R CH2 Cl + OH- => R CH2 OH + Cl- P u tin SGK => (1) v (3) tha mn Xt (2) . CH3 CH = CH Cl + OH- => CH3 CH = CH OH + HCl xem li iu kin h bin ca ru trong phn l thuyt Dng R CH=CH OH tch H2O => andehit : R CH2 CHO => (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 phenol

Cu 16: a. un si dn xut halogen X vi nc mt thi gian, sau thm dung dch AgNO3 vo thy xut hin kt ta. X l

A. CH2=CHCH2Cl. B. CH3CH2CH2Cl. C. C6H5CH2Br. D. A hoc C. SGK 11 NC 212 => Dun si cht no => to ra HX => S c phn ng to kt ta khi cho AgNO3 X l Cl,Br,I Ta c A v C c to ra khi un si => A , C to ra HX => c p kt ta A ng ; C ng v Br gn vi C ngoi vng benzen => km bn B 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 xut halogen X vi dung dch NaOH long mt thi gian, sau thm dung dch AgNO3 vo thy xut hin kt ta. X khng th l

A. CH2=CHCH2Cl. B. CH3CH2CH2Cl. C. C6H5CH2Cl. D. C6H5Cl. SGK => D ch p vi kim nhit cao v p sut cao to ra NaCL hay c p vi AgNO3 sinh ra kt ta. chng khng phn ng vi NaOH nhit thng v khi un si => D . A,B,C nh A v un vi dung dch NaOH Cu 17: Khi un nng dn xut halogen X vi dung dch NaOH to thnh hp cht anehit axetic. Tn ca hp cht X 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-CHO Xem iu kin h bin ca ru thy Nguyn Tn Trung C trng hp 2 nhm OH gn vo cng 1 C TQ : R CH(OH)-OH => R CHO 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 trn

A ch to ra OH-CH2-CH2-OH etilen glicol nu bo to thnh C ch to ra CH3-CH2-OH Nh trng hp OH gn vi C khng no 1 ni i cng b h bin to ra andehit R CH=CH2-OH => R CH2-CHO Cu 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 tha mn Xem li bi benzen thy r hn v tr o,m,p v mt s bi xc nh ng phn benzen Cu 19: Cho s phn ng sau:

CH3

XBr2/as Y

Br2/Fe, to

Zdd NaOH

TNaOH n/c, to, p

X, Y, Z, T c cng thc ln lt l A. 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.


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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 => BrCH2C6H5 X => loi A v D Phn ng Br2/Fe,to => Br th vo vng benzen c th o hoc p sn phm chnh p => Y : p CH2Br-C6H4Br phn ng NaOH => Ch th Halogen gn vi C ngoi vng benzen Xem 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 11NC 218 => Z l C6H5Cl => 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 va tc 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 vi C k nhau VD : C2H4(OH)2 ; OH CH2-CH2-OH ; C3H5(OH)3 ; OH CH2 CH(OH) CH2 OH P vi Na => c H linh ng (VD ; R - OH ;R - COOH ; C6H5OH) P vi Cu(OH)2 iu kin thng => R OH ; R - COOH Xem s h bin ca ru na nh A . Cl-CH(Cl)- CH(Cl)-Cl + NaOH => CHO CHO (v to ra OH CH(OH) CH(OH)-OH ) + NaCl B . Cl CH2 CH2 Cl + NaOH => OH CH2 CH2 OH + NaCL (THa mn 2OH lin k ) =>B C . CH3 CH(Cl) Cl +NaOH => CH3 CHO + 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 dch HNO3, 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 NaCl iu 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 cao Halogen gn vi C ni i khng phn ng vi NaOH cng nh H2O nhit thng v khi ung nng(Mnh on) Cu ny ng cc p n khng cha (4)

Cu 23: Cho s chuyn ho : Benzen A B C A axit picric. B l A. phenylclorua. B. o Crezol. C. Natri phenolat. D. Phenol.

Axit picric . Seach google nu khng bit cc cht http://vi.wikipedia.org/wiki/Ax%C3%ADt_picric Axit picric tng hp t benzen hoc phenol

C6H5OH +3HNO3 => C6H2(OH)(NO2)3 (c c xt H2SO4 ) => B Natri phenolat : C6H5ONa iu ch phenol C

Cu 24: Cho s phn ng : NaOHCCl YX

02 500, ancol anlylic. X l cht no sau y ?

A. Propan. B. Xiclopropan. C. Propen. D. Propin.

Anlyl => CH2=CH-CH2- => ancol anlylic : CH2=CH CH2 OH Y l CH2 = CH CH2 Cl => X l CH2 = CH CH3 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 l

A. CH3COOH. B. CH3CH2COOH. C. CH3CH2OH. D. CH3CH2CH2COOH.

SGK 11 NC 214 => C2H5Br + Mg => C2H5-Mg Br A

C2H5 Mg Br + CO2 => C2H5COO Mg Br


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C6H5CO2MgBr + HCl C6H5CO2H + MgBrCl Xem p y : http://en.wikipedia.org/wiki/Benzoic_acid Tm seach google Cu 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 thch nh 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 => B Cu 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 l A. CnH2n + 2O. B. ROH. C. CnH2n + 1OH. D. Tt c u ng. SGK 11 NC 220 => C ancol etylic C2H5OH l ancol no n 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 + 2 x

(OH)x.

Xem chuyn 1 . Cch tm CT tng qut cch 1 Ancol no => k = 0 => CnH2n+2 2.0 x(OH)x hay CnH2n+2 x (OH)x => D

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

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

Tch H2O => OH vi C bc I bn cnh => Do duy nht => D hp l nht A sai v khng th bit OH gn vi C bc I , II ,III B khng bit l ru no hay khng no C 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 l A. 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 1

CH3CH(C2H5)CH(OH)CH3 hay CH3 CH2 CH CH CH3 CH3 OH => 3 metyl pentan 2 ol

Cu 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)/2 n = 2 => C4H10O2 Cu 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 =>C Cu 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 l A. C3H7OH. B. CH3OH. C. C6H5CH2OH. D. CH2=CHCH2OH.

Ancol no n chc => CnH2n+1OH => %O = 16.100% / (14n + 18) = 50% => n = 1 => CH3OH => B Cu 36: C bao nhiu ru (ancol) bc 2, no, n chc, mch h l ng phn cu to ca nhau m phn t ca chng 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 bc 2 - (C)C(OH) - C C C C C ; C C C C C ; C C C(C) C => 3 => B OH OH OH


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Cu 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 p Ete : C O C C C ; C C O C C ; C O C(C) C => 3p => Tng = 7 p => B Cu 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 - OH

C - C(C) C C C ; C C(C) C(C) C ; C C C(C) C C => 3p => C OH OH OH

Cu 39: C bao nhiu ancol thm, cng thc C8H10O ? A. 5. B. 6. C. 7. D. 8.

C8H10O c k = 4 tha mn 1 vng benzen c 1 vng + 3 pi Ancol thm => OH gn vi C ngoi vng benzen C6H5 C - OH C6H5 CH CH3 ; C6H5 CH2 CH2 OH ; =>2 OH

CH3 C6H4 CH3 OH CH3 v tr 1 , CH3-OH v tr o,m,p => 3 Tng => c 5 p => A Cu 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 CH CH3 => 4p => C OH

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 vi C bc I R CH2-OH v Ancol i xng OH gn vi C bc I ; C C C C C OH ; C C(C) C C OH ; C (C)C(C) C OH Ancol i xng : C C C C C => Tng = 4 =>D OH Cu 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 => A Cu 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 C C C OH ; C C(OH) C Xt x = 2 => C3H8O2; p OH- C C(OH) C ; OH C C C OH Xt x = 3 => C3H8O2 OH C C(OH) C OH Tng = 5 =>B Cu 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 ca X 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 CT cch 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 => andehit Cu 45: A, B, D l 3 ng phn c cng cng thc phn t C3H8O. Bit A tc dng vi CuO un nng cho ra andehit, cn B cho ra xeton. Vy D l

A. Ancol bc III. B. Cht c nhit si cao nht. 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 ete A to ra andehit => A l ancol OH gn vi C bc 1 B to ra xeton =>B l ancol bc OH gn vi C bc 2 => D l ete ; C O C C => Loi A v D v ete khng tch H2O ch c ru Xem phn nhit si => Cht c H linh ng => Nhit si cao => Ancol nhit si > Ete V c H linh ng trong OH


- 7 -

=> Loi B => C nhit si thp nht Cu 46: X, Y, Z l 3 ancol lin tip trong dy ng ng, trong MZ = 1,875MX. X c c im l

A. 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 khc chc.

Cng ng ng => Hn km nhau 14.k vc k l s C ca cht ln s C ca cht nh VD 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 un nng 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 => D duy nht 2 metyl Cch 2:Ancol n chc , no , mch h => CnH2n+1OH c M = 14n + 18 = 37.2 => n = 4 ; C4H9OH un 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

=> CH3 CH(CH3) CH2 OH hoc CH3 C(CH3) CH3 => p n c trng hp D ng OH Cu 48: Mt ancol n chc X mch h tc dng vi HBr c dn xut Y cha 58,4% brom v khi lng. un X vi 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 226 p 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 - 1 ol Xt C v D ; D loi v ch to ra 1 anken mch i xng C C(C) C

OH C : C C(OH) C C => anken C = C C C ko p hnh hc ; C C = C C p hnh hc Cu 49: Mt cht X c CTPT l C4H8O. X lm mt mu nc brom, tc dng vi Na. Sn phm oxi ha X bi CuO 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 nhm OH , COOH => Ch c th l Ancol v k=1 ; Nu l axit phi c 2 Oxi ; phenol c k 4 Oxi ha X bi CuO khng phi l andehit => c nhm OH T 3 d kin trn => Ancol n chc c 1 lin kt i c nhm OH khng gn vi C bc I => B But 3 en 2 ol ; C C(OH) C = C Cch c tn lun => nh s gn OH hay gn H linh ng nht Cu 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 NC 220 => 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

2 metyl butan 2 ol C C(C) C C => Bc 3 OH

Cu 52: Cc ancol c phn loi trn c s A. 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 NC 220 => S lng OH ; B c im gc no , khng no, vng, bezen bc ancol I, II, III => D Cu 53: Cc ancol (CH3)2CHOH ; CH3CH2OH ; (CH3)3COH c bc ancol ln lt l

A. 1, 2, 3. B. 1, 3, 2. C. 2, 1, 3. D. 2, 3, 1. (CH3)2CHOH hay CH3 CH CH3 => bc 2 OH

CH3CH2OH hay CH3 CH2 OH => bc 1 ; (CH3)3COH hay CH3 C(CH3) CH3 => bc 3 OH


- 8 -

=> C

Cu 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 => A

B sai v tp chc hoc Cacbonhidrat C sai v l ancol thm OH gn vi C ngoi vng benzen Cu 55: Ancol etylic tan tt trong nc v c nhit si cao hn hn so vi ankan v cc dn xut halogen c khi lng 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 NC 222 => D c k phn A l do lin kt hidro lin phn t Cu 56: A, B, C l 3 cht hu c c cng cng thc CxHyO. Bit % O (theo khi lng) trong A l 26,66%. Cht c 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 45 Xem li on t p n => C3H8O v A, B , C c cng CT vy D l C3H6O andehit c CT : CnH2nO Tm ra CT xem chuyn 1 Phn tm CT xem trn thaytro.com Gi 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) = 44 n = 3 => C3H8O 3 p bi 45 + da vo nhit nh nht => C Cu 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 => D CaO + H2O => Ca(OH)2 ; P2O5 + H2O => H3PO4 ; CuSO4 khan + H2O => CuSO4 dung dch Cu 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 => SGK Cu 59: Anken thch hp iu ch 3-etylpentan-3-ol bng 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 5

C C C C C Mch i xng qua C trung tm => anken duy nht C C = C C C C C

C C

Cu 60: Hirat ha 2-metyl but-2-en thu c sn phm chnh l A. 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 C Thu 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 OH

2 metyl but 2 en : C C = C C => C C - C C => A C C Cu 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 = C C => To ra 2 ancol ; OH C C C hoc C C(OH) C C Ancol cn li to ra 1 ancol => Mch i xng A v B u tha mn : C = C ; C C = C C => D D : isobutilen : C = C(C) C => OH C - 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.


- 9 -

Bi 61 => Propen to ra 2 sn phm ; etilen , but 2 1 sn phm => Loi B v C 2 metylpropen 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 But 1 en : C = C C C Cu 63: Hirat ha 2 anken c hn hp Z gm 2 ancol lin tip trong dy ng ng. t chy hon ton 0,53 gam 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 l

A. 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 ha CO2 + 2NaOH => Na2CO3 + H2O => nCO2 = nNaOH p / 2 = 2.0,025/2 = 0,025 mol Ancol l ancol no n chc da vo Anken c 1 ni i Cu 64: Mt chai ng ancol etylic c nhn ghi 25o c ngha l

A. c 100 ml nc th c 25 ml ancol nguyn cht. B. c 100 gam dung dch th c 25 ml ancol nguyn 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 => D Cu 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)/80 V ru = 20 => mRuou = V.d = 20.0,8 = 16 g => A Cu 66: Dy gm cc cht u tc dng vi ancol etylic l

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 11 254 Nu khng lm c thng dng cch loi p n Loi B v thy ngay Ca khng p Ch p vi Na,K Loi C v c NaOH ru ch p vi Na Loi D v c Na2CO3 ru khng p vi mui Cu 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 l A. (a), (b), (c). B. (c), (d), (f). C. (a), (c), (d). D. (c), (d), (e).

Tc dng Na, Cu(OH)2 => ru c nhm OH lin k : dng C(OH) C(OH) C th l axit nh: (a) , (c), (d) u c dng trn => C B 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 NC 227 => X l glucozo , Y l C2H5OH => C Z l CH3COOH v C2H5OH + O2 => CH3COOH + H2O ; Metylaxetat este CH3COOCH3 CH3COOH + CH3OH => CH3COOCH3 + H2O phn ng iu ch este b. Cho s chuyn ho : Glucoz X Y CH3COOH. Hai cht X, Y ln lt l

A. 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 CH3COOH Vi CH3CHO + O2 => CH3COOH => B Cu 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 l

A. 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 Ta lun c x.nRuou = 2nH2 Vi x l s nhm OH , tng qut hn l s H linh ng hay x.nRuou = mH2 nRuou = 2nH2 v ru n chc => 1 nhm OH = 0,03 mol mROH = mROH + 22x = 1,24 + 22.0,03 = 1,9 g

Cch khc: BT e => nNa = 2nH2 nNa = 0,03 mol


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BT KL => m Ru + mNa = m mui + mH2 1,24 + 0,03.23 = m mui + 0,015.2 m mui = 1,9 g Cu 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 ancol

A. 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 => B Cu 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 thc cu 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 D Cch 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 => D

Cch 3:

p n ru no => CnH2n+2 m(OH)m Hay CnH2n+2OmTm CT theo cch 2 chuyn 1 m. n Ru = 2nH2 n 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 / 3 44m/3 => /3 => c 3 nhm OH => D

Cch 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 c cng thc l

A. CH3OH. B. C2H5OH. C. C3H7OH. D. C4H7OH. Ch bi ny khng dng cch chia c v khng bit n p ht hay khng ht Da 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, 2010 2009 u c dng nm gia. V phn 2 thng l p ht TH1 => Ancol d , Na ht TH2 => 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 => D Cu 73: Cho 12,8 gam dung dch ancol A (trong nc) c nng 71,875% tc dng vi lng Na d thu c 5,6 lt 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 p Cch 2: => m H2O = mdd m ru = 12,8 9,2 = 3,6 g V p vi Na d => H2O c p to ra H2 v ru cng p to ra H2 PT : 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 = 2nH2 n Ru = 0,3 / x => M ru = 9,2/(0,3/x) = 92x / 3 => x = 3 V M nguyn M = 92 => C

Cu 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)2 hay CnH2n+2O2 mCO2 = 1,833mH2O => chn mH2O = 18g => mCO2 = 33 g V ancol no , p n => n = nCO2 / (nH2O nCO2) = 0,75 / (1 0,75) = 3 => C3H6(OH)2 => B Xem CT tnh nhanh trong chuyn gii nhanh bng ha hu c Cu 75: un 12 gam axit axetic vi 13,8 gam etanol (c H2SO4 c lm xc tc) n khi phn ng t ti trng thi cn bng thu c 11 gam este. Hiu sut ca phn ng este ho l

A. 55%. B. 50%. C. 62,5%. D. 75%.

PT : CH3COOH (Axit axetic) + C2H5OH (etanol) => CH3COOC2H5 + H2O


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Ban u 0,2 mol 0,3mol P 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 % => C

Cu 76: Khi thc hin phn ng este ho 1 mol CH3COOH v 1 mol C2H5OH, lng este ln nht thu c l 2/3 mol. t hiu sut cc i l 90% (tnh theo axit) khi tin hnh este ho 1 mol CH3COOH cn s mol C2H5OH l (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 cao Xt ban u PT : CH3COOH + C2H5OH => CH3COOC2H5 + H2O

Ban u 1 mol 1mol P 2/3 mol 2/3 mol Sau 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)) = 4 Xt khi H% cc i = 90% PT : CH3COOH + C2H5OH => CH3COOC2H5 + H2O

Ban u 1mol x mol P 0,9 mol V H%=90 0,9 mol => 0,9mol 0,9 mol Sau p 0,1 x 0,9 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 => B Cu 77: Khi un nng butan-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 D Sn phm chnh => OH tch cng vi H C bc cao I II Butan 2 ol : C C(OH) C C => C C = C C => A Cu 78: Khi un nng 2 trong s 3 ancol CH4O, C2H6O, C3H8O vi xc tc, nhit thch hp ch thu c 1 olefin 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 C C OH => C = C C3H8O hay OH C C C => Ch to ra 1 anken C = C - C Hoc C C(OH)-C => C = C C => ging nhau => ch to ra 1 anken => A v B tha mn to ra 1 anken => C Cu 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 II hoc III v OH gn vi C bc I => ch to ra 1 anken Loi B v D Xt A hoc C ; CH3 CH CH2 CH3 => CH2=CH-CH2-CH3 v CH3 CH=CH-CH3 p hnh

hc OH => A , C ch to ra 1 anken Cu 80: Hp cht hu c X c cng thc phn t l C5H12O, khi tch nc to hn hp 3 anken ng phn (k c ng 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 C Xt A. CH3 CH2 CH(OH) CH2 CH3 => CH3 CH2 =CH CH2 CH3 p hnh hc => 2 anken =>Loi => D OH gn vi C trung tm ca mch hay i xng qua n Cu 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 => D Cu 82: Khi un nng hn hp gm C2H5OH v C3H7OH vi H2SO4 c 140oC c th thu c s ete ti a l

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

Bi 81 => D


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Cu 83: un nng hn hp gm 3 ancol l AOH, BOH v ROH vi H2SO4 c 140oC th thu c ti a bao nhiu ete ?

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

Bi 81 => D Cu 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 , SOH o

c 42 E

Tn ca E l A. 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 => C P : But 1 en : C = C C C + HCl => C C(Cl) C C +NaOH => C C(OH) C C => C C = C C => C p u l cng , p 2 l phn ng th bi dn xut, p 3 l tch H2O => to ra sn phm chnh l C

Cu 86: un nng hn hp gm hai ru (ancol) n chc, mch h, k tip nhau trong dy ng ng vi H2SO4 c 140oC. Sau khi cc phn ng kt thc, thu c 6 gam hn hp gm ba ete v 1,8 gam nc. Cng thc phn t 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 hp BT khi lng => m ancol p = m ete + mH2O v p ancol => ete + H2O = 7,8 g n hh ancol p = 2nH2O V lun c 2 ancol p => 1 H2O VD; CH3OH + CH3OH => CH3 O CH3 + H2O

=> n ancol p = 0,2 mol => M ancol = 14 n + 18 = 7,8/0,2 = 1,5 => A Cu 87: un nng t t hn hp etanol v propan-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 - 2 ol C C(OH) C => to ra 1 anken ; s ete = cng thc = 3 Cu 88: C bao nhiu ng phn ng vi cng thc phn t C8H10O, u l dn xut ca benzen, khi tch nc cho 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 benzen p : C6H5 CH2 CH2 OH => C6H5 CH=CH2 Stiren C6H5 CH(OH) CH2 => C6H5 CH =CH2 Stiren => c 2 p => B Cu 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 tn l

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 metylbutan 2 ol ; I II

C C(C) C C => C C(C) = C C v C = C(C) C C C 2 cht u khng c p hnh hc => loi B => C OH Cu 90: hirat ha 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 + mH2O mH2O = 3,6 g => nH2O = nRuou = 0,2 mol => M ruou = 14n + 18 = 14,8 / 0,2 n = 4 => C Cu 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 l


Ging bi 86 => m Ancol = m H2O + m hh ete ; n Ancol = 2nH2O => n = 1,67 => A Cu 92: un nng hn hp X gm 0,1 mol CH3OH v 0,2 mol C2H5OH vi H2SO4 c 140oC, khi lng ete thu c l

A. 12,4 gam. B. 7 gam. C. 9,7 gam. D. 15,1 gam.


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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 .46 0,15.18 = 9,7 g => C Cu 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. X l

A. CH3OH. B. C2H5OH. C. C3H7OH. D. C4H9OH. Dng bi ny c mo. p n => Ancol c dng : CnH2n+1OH 140oC => To ra ete => 2X => Y + H2O Ta 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 (2 k) = 18 => MX = 18/(2-k) Nh bi trn => k = 1,4375 => M = 32 Cu 94: un nng hn hp hai ancol n chc, mch h vi H2SO4 c, thu c hn hp gm cc ete. Ly 7,2 gam 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. C2H5OH v 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 C Ete 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 phm B 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 ru A => 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 HSO4 c trong iu kin nhit thch hp sinh ra cht 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 => B Cu 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 1 A loi v metanol CH3OH khng to ra anken , B sai v butan 1,2 iol c OH gn vi v tr C th 2 hay bc II D loi v pentan 2 ol Cu 98: Ancol X n chc tch nc ch to mt anken duy nht. t chy mt lng X c 11 gam CO2 v 5,4 gam 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 c n = nCO2 / (nH2O nCO2) = 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 ng c 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


Xem bi 86 => n = 1,5 => A Cu 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) l

A. 8,19. B. 10,18. C. 12. D. 15,13.

Ancol etylic C2H5OH ; nC2H5OH = nAnken = 0,15 mol Phn ng t l 1 : 1 H% p = nPT . 100% / nB n 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 => D

Cu 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.


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Ancol b oxi ha to ra xeton => Ancol c OH gn vi C bc II hay ancol bc II => A v 2 ol v khng c nhnh B , C , D u c OH gn vi C bc I 1 ol Cu 102: Ancol no n chc tc dng c vi CuO to anehit l

A. 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 =>C Cu 103: Oxi ha 6 gam ancol no X thu 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 => C Cch khc . Tng gim khi lng : T Ru CnH2n+1OH => CnH2nO andehit => Gim 2 g mt i 2 hidro PT : CnH2n+1OH + CuO => CnH2nO + Cu + H2O => n Ru = (m ru m andehit) / 2 =0,1 mol => MX = 6/0,1 = 60 => C Cu 104: Cho m gam ancol n chc, no, mch h qua bnh ng CuO (d) nung nng. Sau khi phn ng xy ra hon 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 l

A. 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 hi khng 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 mol BT KL : m Ru + mCuO = mCnH2nO + mCu + mH2O m Ru = (mCnH2nO + mH2O) - (mCuO mCu) = m hn hp - m gim = 38.0,04 0,32 = 1,2 g Cch 2:

Ta lun c m Ru = m Andehit + 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 g Cch 3: Tm n ta c M andehit = 14n + 16 = 1,16/0,02 n = 3 => Ru C3H7OH => m = 0,02.60 = 1,2 g Cu 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 hp anehit, 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 + H2O BT khi lng : m ru A + mO = m hn hp sau p mO = 1,6 g => nO = n Ru = n Andehit = nH2O = 0,1 mol Ta c ru d => m Ru p < 4 => M ru < 4/0,1 = 40 => A tha mn Cu 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 hp anehit, 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 thy khi 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 => A Cu 108: Dn hi C2H5OH qua ng ng CuO nung nng c hn hp X gm anehit, ancol d v nc. Cho X tc 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 gam B. 27,6 gam. C. 18,4 gam. D. 23,52 gam. X gm C2H5OH d , andehit , H2O => Ta lun c nC2H5OH d +nH2O = 2nH2O Mt khc ta c nH2O = nC2H5OH p => nC2H5OH ban u = 2nH2O = 0,4 mol nC2H5OH p = 0,8.0,4 = 0,32 mol v ch p 80% = nCuO = nCu m Ru ban u + mCuO = m hn hp X + mCu 0,4 . 46 + 0,32.80 = m hn hp X + 0,32.64 m hn hp X = 23,52

C th dng tng 16 g i vi CuO ,Cu Cu 109: Dn hi C2H5OH qua ng ng CuO nung nng c 11,76 gam hn hp X gm anehit, ancol d v nc. 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 = 0,2 mol


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m Ru ban u + mCuO = m hn hp X + mCu 0,2.46 + 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 COOH

n n . 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 ankan

Cu 111: Khi t chy ng ng ca ancol n chc thy t l s mol OHCO 22 n : n tng dn. Ancol trn thuc dy

ng ng ca A. 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) = 1 1/(n+1) bit tng mo xt n = 1 v n = 2 so sch => Lun tng khi n tng n l s C hay CnH2n+1OH Cu 112: t chy hon ton m gam ancol n chc A c 6,6 gam CO2 v 3,6 gam H2O. Gi tr m l

A. 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 / (nH2O nCO2) = 3 => C3H7OH v nH2O > nCO2 => c dng CnH2n+2Oz nC3H7OH = nCO2 / 3 = 0,05 mol => m = 3 g

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

5 : 4 V : 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 Oxi Cch 2 : VCO2 : VH2O = 4 : 5 t l th tch = t l s mol => chn nCO2 = 4 mol ; nH2O = 5 mol n = nCO2 / (nH2O nCO2) = 4 => C4H10O v nH2O > nCO2 => c dng CnH2n+2Oz

Cu 114: t chy mt ancol a chc thu c H2O v CO2 c t l mol 2 : 3 n : n22 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 => c dng CnH2n+2Oz => C2H6Oz v a chc + nhn p n => C2H6O2 => B Cu 115: Khi t chy mt ancol a chc thu c nc v kh CO2 theo t l khi lng

44 : 27 m : 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 / (nH2O nCO2) = 2 => B v ch ch 2 C Cu 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 => C Cch 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 => C Cu 117: Ba ancol X, Y, Z u bn v c khi lng phn t khc nhau. t chy mi cht u sinh ra CO2 v H2O

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 / (nH2O nCO2) = 3 => B v l Ancol no do nH2O > nCO2 cch chn nh bi trn nCO2 = 3 ; nH2O = 4 Cu 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.

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

Hoc x : y = nCO2 / 2nH2O = 2 / 6 => C2H6Oz Cch 1: nC2H6Oz = nCO2 / 2 = 2 mol => AD x + y/4 z/2 = nO2 / nX 2 + 6/4 z/2 = 5/2 z = 2 => C2H6O2

Cch 2 : nC2H6Oz = nCO2 / 2 = 2 mol

BT NT Oxi : z . nC2H6Oz + 2nO2 = 2nCO2 + nH2O => z = 2


- 16 -

Cch 3: Nh cng thc nO2 / nAncol = n,5 => bi trn nO2 / nC2H6Oz = 5/2 = 2,5 => s C = s OH => z = 2 Cu 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 / (nH2O nCO2 ) = 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 => A Nh nO2 / nX = n,5 , n l s C v s OH Cu 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 l

A. 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 / (nCO2 nH2O) = 1 => CH3OH Ancol A t 1 mol CH3OH => 1 mol CO2 ADCT : nCO3(2-) hay BaCO3 kt ta = nOH- - nCO2 = 2nBa(OH)2 nCO2 = 1,2 1 = 0,2 mol m BaCO3 = 39,4 g => C

Cu 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 OH chuyn 1 hay bi ging nguyn tn trung Cu 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 t chy 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 + 8/4 z/2) = nO2 / nC3H8Oz = 4,5 => z = 1 => C3H8O Cch 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 = 1

Cu 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 v 6,6 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)3

Cch 2: Ru no : CnH2n+2Oz => n = nCO2 / nX = 3 => C3H8Oz (3+8/4 z/2) = nO2 / nX = 3,5 => z = 3 => C3H8O3 hay C3H5(OH)3 Cu 124*: X l hn hp 2 ancol n chc, cng dy ng ng, c ty l khi lng 1:1. t chy ht X c 21,45 gam 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 mol BTNT Oxi : n hn hp ru + 2n O2 = 2nCO2 + nH2O nO2 = 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 : 1

Cch 1:

M trung bnh hn hp ru = 12 / 0,2875 = 41,74 = 14n + 18 CnH2n+1OH n = 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 PT BTNT C => n.x + m.y = nCO2

BTNT H => (2n+2)x + (2m + 2)y = 2nH2O Vi ru c CT l : CnH2n+2 ; CmH2m+2 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 c

Cch 4: p n => Ancol no n chc. n hn hp ancol = nH2O nCO2 = 0,2875 mol => n tb = nCO2 / (nH2O nCO2) = 1,7 => C CH3OH m hn hp ancol= mH + mC + O = 2.nH2O + 12.nCO2 + (nH2O nCO2).16 = 12 g t l khi lng l 1:1=> mCH3OH=6g, m ancol cn li=6g


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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)/1,02

kt ta CaCO3 => Chn c = 100 g Mo chn MCaCO3 => b = 71 g th vo => a = 31 g nCaCO3 = nCO2 = 1 mol , m bnh tng = mCO2 + mH2O = 71 44 + mH2O = 71 mH2O = 27 g => nH2O = 1,5 mol => n Ancol = nH2O nCO2 = 0,5 mol => M X = 31 / 0,5 = 62 => B Cu 126: t chy hon ton a gam hn hp gm metanol v butan-2-ol c 30,8 gam CO2 v 18 gam H2O. Gi tr a l

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

Cch 1: n = nCO2 / (nH2O nCO2) = 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 7 203 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 = 1 0.7 = 0,3 mol BTNT Oxi : n hn hp Ru + 2nO2 = 2nCO2 + nH2O nO2 = 1,05 mol v Ru u no => c 1 oxi BTKL => m + mO2 = mCO2 + mH2O m = 15,2 g Cu 127: t chy hon ton 0,4 mol hn hp X gm ancol metylic, ancol etylic v ancol isopropylic ri hp th ton 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 mol BTNT Oxi => nhn hp ru + 2nO2 = 2nCO2 + nH2O 0,4 + 2nO2 = 2.0,8 + 1,2 nO2 = 1,2 mol => V = 26,88 lt => A

Cu 128: t chy hn hp X gm 2 ancol c s mol bng nhau thu c hn hp CO2 v H2O theo l mol tng ng 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 mn v 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 y ng ?

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 = nH2O nCO2 => ancol no , mch h Cu 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. A c cng thc phn t l

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

n = nCO2 / (nH2O nCO2) = 3 => C3H8Oz ; nC3H8Oz = nCO2 / 3 = 0,3 3 + 8/4 z/2 = nO2 / nC3H8Oz = 4 => z = 2 => C3H8O2 Cu 131: Cho hn hp X gm hai ancol a chc, mch h, thuc cng dy ng ng. t chy hon ton hn hp X, 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 / (nH2O nCO2) = 3 => C v chn nCO2 = 3 mol ; nH2O =4 mol v do a chc => c my nhm OH Cu 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.

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+1OH nAncol = nH2O nCO2 ; BTNT Oxi : nancol + 2nO2 = 2nCO2 + nH2O nH2O nCO2 + 2nO2 = 2nCO2 + nH2O nO2 = 3nCO2 / 2 BTKL => m + mO2 = mCO2 + mH2O


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m + (3V.32)/(22,4.2) = VCO2 . 44/22,4 + a m = a V/5,6 Cch 3: m ancol = mC + mH + mO = 12.nCO2 + 2.nH2O + 16(nH2O nCO2) = 18nH2O 4nCO2 = 18. mH2O /18 - 4.VCO2 / 22,4 = mH2O VCO2 / 5,6 = a V / 5,6 Cu 133: t chy hon ton 0,2 mol mt ancol X no, mch h cn va 17,92 lt kh O2 ( ktc). Mt khc, nu cho 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 tn gi 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 C nO2 / nX = 4 # 3,5 => khng th l glixerol C3H5(OH)3 nO2 / nX = n,5 => Loi D => Thy A, B u c Propan 1,2 iol 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 c 40g kt ta. Khi lng ancol etylic thu c l

A. 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% / mB m ban u = 36.100% / 80% = 45 g => A Cu 135: Cho m gam tinh bt ln men thnh C2H5OH vi hiu sut 81%, hp th ht lng CO2 sinh ra vo dung dch 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 1 phn 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% / mB mB = 60,75.100% / 81 = 75g => A

Cu 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% / mB mB = 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 Ru V ru = 9,2.100% / 92 = 10 ml => B Cu 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% / mB 81% = 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 (bit hiu 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=> D Cu 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)2 d to ra 40 gam kt ta. Nu hiu sut ca qu trnh ln men l 75% th gi tr

ca m l A. 60. B. 58. C. 30. D. 48.

nCO2 = nCaCO3 = 0,4 mol => nC6H12O6 = nCO2 / 2 = 0,2 mol C6H12O6>2C2H5OH +2CO2 =>m C6H12O6 theo PT = 36 g ; H% p = mPT . 100% / mB mB = 36.100% / 75% = 48 => D Cu 140: Ln men m gam glucoz vi hiu sut 90%, lng kh CO2 sinh ra hp th ht vo dung dch nc vi trong, thu c 10 gam kt ta. Khi lng dung dch sau phn ng gim 3,4 gam so vi khi lng dung dch nc vi trong ban u. Gi tr ca m l

A. 20,0. B. 30,0. C. 13,5. D. 15,0


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Xem chuyn 1 => m dd gim = m kt ta mCO2 cho vo H 3,4 = 10 nCO2 mCO2 = 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 => D Cu 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 n chc 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 CH3 C6H4 OH CH3 gn v tr 1 , gc OH gn v tr m , p , o => 3 cht Cu 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 C nA = nNaOH => A cha 1 gc OH gn vi vng benzen => B Loi D v 2 OH gn vi vng benzen Cu 144: Khi t chy 0,05 mol X (dn xut benzen) thu c di 15,4 gam 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 nhm gn 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 benzen Cu 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 p NaOH phenol c p nhng khng bit hin tng dd Br2 => phenol to ra kt ta , benzen khng p => 1 v 3 => B Mnh khng bit NaOH cng ng ko? Cho mnh kin ca bn Cu 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 thuc loi 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 benzen B sai v ch l n chc do 2Oxi C sai v khng c H linh ng D sai v ancol khng p vi NaOH Cu 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 phenol K = (8.2 10 + 2)/2 = 4 m vng benzen c 4 pi => ete O v ch c ete v ru l c 1 oxi + no C6H5 CH2 - O - CH3; C6H5 O CH2 CH3 CH3 C6H4 O CH3 CH3 v tr 1 , - O CH3 v tr o ,p , m => 3 => Tng = 5 p => D Cu 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 gam kt 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.

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 CxHyO v x = 7 Cu 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.


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B. Etanol < phenol < nc. D. Phenol < nc < etanol. Phenol p vi NaOH , nNa => Phenol ln nht H2O p mnh lit vi Na => H2O th nh Etanol p vi Na bnh thng => Ru cui cng => A Cu 150: T 400 gam bezen c th iu ch c ti a bao nhiu gam phenol. Cho bit hiu sut ton b qu trnh t 78%.

A. 376 gam. B. 312 gam. C. 618 gam. D. 320 gam. C6H6 => C6H6O => nC6H6 = nC6H6O ; H% sp = mTT.100% / mPT

mTT = mPT . H% / 100% = nC6H6.94.78%/100% = 376 g => A Cu 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 tch

HCl + Na2S => H2S + NaCL => to ra kh H2S mi trng thi => Nhn bit Na2S HCl + Na2CO3 => NaCl + CO2 + H2O => to ra kh => To ra kh nhn bit Na2CO3 HCl + C6H5ONa => C6H5OH + NaCl => c p => C6H5ONa Nhn bit c Na2CO3 cho vo 2 bnh con li khng p => BaCL2 to ra kt ta trng BaCO3 Cu 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 NC 232 => C

Cu 153: C bao nhiu phn ng xy ra khi cho cc cht C6H5OH ; NaHCO3 ; NaOH ; HCl tc dng vi nhau tng i mt ?

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

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

NaOH + HCl => NaCl + H2O

C6H5OH + NaOH => C6H5ONa + H2O => 4 => B Cu 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 dch HCOONa 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 vi A. dung dch NaOH. B. Na kim loi. C. nc Br2. D. H2 (Ni, nung nng).

SGK 11 NC 231 => Phn ng th Br2 vo phenol d hn so vi benzen => do nh hng ca nhm OH => C Cu 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 s Tng vng + pi 4 => ch c D c tng pi + vng = 4 => D tha mn Cu 158: Ba hp cht thm X, Y, Z u c cng thc phn t C7H8O. X tc dng vi Na v NaOH ; Y tc dng vi 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.

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 C Y tc dng vi Na khng tc dng vi NaOH => Y l ancol thm OH gn vi C ngoi vng benzen => D Z l ete


- 21 -

Cu 159: Cho ln lt cc cht C2H5Cl, C2H5OH, C6H5OH, C6H5Cl vo dung dch NaOH long un nng. Hi my 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 => C Cu 160: a. S ng phn ca C3H5Cl3 l

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

K = (3.2 5 + 2 3)/2 = 0 => p cu to C(Cl3) C C ; C(Cl) C(Cl2) C ; C(Cl) C(Cl) C(Cl) ; C(Cl2) C(Cl) C ; C(Cl2) C C(Cl) => Tng = 5 => A b. Trong s cc ng phn ca C3H5Cl3 c th c bao nhiu ng phn khi thuy phn trong mi trng kim cho sn 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 OH Sn phm p c Na v AgNO3/NH3 to ra Ag => Phi cha gc OH ( p Na) v 1 gc CHO p vi AgNO3/NH3

Nh 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) => Cl C C C(Cl2) + NaOH => OH C C CHO + NaCl + H2O C C(Cl) C(Cl2) + NaOH => C C(OH) CHO + NaCl + H2O => 2 p => D Cu 161: Hp cht X c cha vng benzen v c CTPT l C7H6Cl2. Thy phn X trong NaOH d c, to cao, p cao 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 CH2 OH 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 2 OH gn cng 1 C Cu 162: Cho cc hp cht sau : (I) CH3CH2OH. (II) C6H5OH. (III) NO2C6H4OH. Chn pht biu sai

A. 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 => B Cu 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 => A

Cu 164: Cho 2 phn ng :(1) 2CH3COOH + Na2CO3 2CH3COONa + H2O + CO2

(2) C6H5ONa + CO2 + H2O C6H5OH + NaHCO3 Hai 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 > H2CO3 P (2) C6H5ONa to ra NaHCO3 hay HCO3- => Tnh axit ca C6H5ONa > HCO3- Ngoi ra CH3COOH c H linh ng mnh => Ln nht => Lc axit gim dn => B Cu 165: Cho dy cc cht : phenol, anilin, phenylamoni clorua, natri phenolat, etanol. S cht trong dy phn ng c 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 lng metanol trong X l

A. 25%. B. 59,5%. C. 50,5%. D. 20%.

Phenol C6H5OH ; metanol CH3OH

nCO2 = nH2O => chn nCO2 = 1 mol => nH2O = 1mol BTNTC => 6nC6H5OH + nCH3OH = nCO2 ; 6nC6H5OH + 4nCH3OH = 2nH2O Gii h => nC6H5OH = 1/9 ; nCH3OH = 1/3 => %m CH3OH = (32/3).100%/(94/9 + 32/3) = 50,5 % => C


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Cu 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 c cng thc n gin nht trng vi cng thc phn t. S ng phn cu to thuc loi hp cht thm ng vi cng thc 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 C6H5 CH2 OH ; C6H5 O CH3 ; CH3 C6H4 OH CH3 gn v tr 1 , OH gn v tr o,m,p Tng = 5 => D

Cu 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 a mol 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 benzen nX = 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 mol X tham gia phn ng v X tc dng c vi NaOH theo t l s mol 1 : 2. Cng thc cu to thu gn ca X l

A. 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 OH X tc dng vi NaOH => 2nX = nNaOH => X cha 2 gc OH gn vi vng benzen => B Cu 170: Cho hn hp hai anken ng ng k tip nhau tc dng vi nc (c H2SO4 lm xc tc) thu c hn hp 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 chy vo 2 lt dung dch NaOH 0,1M thu c dung dch T trong nng ca NaOH bng 0,05M. Cng thc cu to thu 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 d v 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 gm hai ru (ancol) X v Y l ng ng k tip ca nhau, thu c 0,3 mol CO2 v 0,425 mol H2O. Mt khc, cho 0,25 mol hn hp M tc dng vi Na (d), thu c cha n 0,15 mol H2. Cng thc phn t ca X, Y l

A. C3H6O, C4H8O. B. C2H6O, C3H8O. C. C2H6O2, C3H8O2. D. C2H6O, CH4O.

n = nCO2 / (nH2O nCO2) = 2,4 Ta c 0,25 mol M => nH2 < 0,15 mol => Tha mn nM = 2nH2 => Ru n chc => B Cu 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 axit l


nAxit = nCO2 = 0,025 mol P CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O nAxit = n Ru = 0,025 mol P : C2H5OH + O2 => CH3COOH => m ru b oxi ha to ra axit = 0,025.46 = 1,15 g Cu 173: Khi phn tch thnh phn mt ru (ancol) n chc X th thu c kt qu : Tng khi lng ca cacbon v 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 => CxHyO chn 1 mol CxHyO : mC + mH = 3,625mO 12x + y = 3,625.16 = 58 y = 58 12x ; ta c 0< y 2x + 2 0 < 58 12x 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 => B Cu 174: Oxi ho ancol n chc X bng CuO (un nng), sinh ra mt sn phm hu c duy nht l xeton Y (t khi hi ca Y so vi kh hiro bng 29). Cng thc cu to ca X l

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


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Xeton :CnH2nO => M = 58 = 14n + 16 => n = 3 hay C3H7OH => Loi B v D V to ra xeton => Ru bc 2 hay OH gn vi C bc 2 => A Cu 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 gam CuO 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 vi dung dch AgNO3(d) trong NH3 to ra hn hp 2 mui v 38,88 gam Ag. Khi lng ca A cn dng l


Chuyn 1 cch tm C,H,O khi bit % ca thnh phn Gi CxHyOz => %O = 16z .100% (12x + y + 16z) = 50% 12x + y = 16z => z = 1 , x = 1 , y = 4=> CH4O Hay CH3OH

Cu 176: un nng ancol A vi hn hp NaBr v H2SO4 c thu c cht hu c B, 12,3 gam hi cht B chim mt th tch bng th tch ca 2,8 gam N2 cng nhit 560oC ; p sut 1 atm. Oxi ho A bng CuO nung nng thu 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+1OH Ch nh B l CnH2n+1Br ta c nB = nN2 v th tch bng nhau mB / 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 c cht 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 thc cu to ca A l

A. C2H5OH. B. CH3CH2CH2OH. C. CH3OH. D. HOCH2CH2OH.

A , B , C l ancol no , n chc => Xt CnH2n+1OH Tng 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 H2SO4 th s c mt OH b thay th bi Br CnH2n+2O2 => CnH2n+1O Br => c M = 14n + 17 + 80 = 125 => n = 2 => D

Cu 178: Anken X c cng thc phn t l C5H10. X khng c ng phn hnh hc. Khi cho X tc dng vi KMnO4 nhit thp thu c cht hu c Y c cng thc phn t l C5H12O2. Oxi ha nh Y bng CuO d thu c 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 ln th 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 ; nH2O nCO2 = n Ru = 1 => nH2O = 3 ; nCO2 = 2 n = nCO2 / (nH2O nCO2) = 2 => C2H5OH => A

Cu 180: Cho 30,4 gam hn hp gm glixerolC3H5(OH)3 v mt ru n chc, no A phn ng vi Na th thu c 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 thc ca 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 = 30,4 0,2.92 = 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 = 2.0,4 n 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 Na d 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 thc cu 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 p Da vo p to andehit => OH gn vi C bc I =>B Cch 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 => C3H7OH Da 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 dng ht 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+2Om Ta c m . nAncol = 2nH2 CT = 0,2 mol nAncol = 0,2 / m => M ancol = 7,6/(0,2/m) = 38m = 14n + 2 + 16m 11m = 7n + 1 => C Cu 183: Cht hu c X mch h c to ra t axit no A v etylen glicol. Bit rng a gam X th hi chim th tch 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 32,8 gam 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 l

A. 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 ra anehit). 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 + H2O Phn 1 => nC2H5OH d + nAxit + nH2O = 2nH2 = 0,56 mol Phn 2 => nAxit = nNaOH = 0,24 mol = nC2H5OH p v nCH3COOH = nC2H5OH p th vo phn 1 v 2 phn bng nhau nC2H5OH d + nH2O = 0,56 0,24 = 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 lt CO2 cng vi lng hn hp trn cho phn ng vi Na d th thu c 2,24 lt H2 ( ktc). Cng thc phn t ca 2 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 => A Cu 186*: Oxi ho 9,2 gam ancol etylic bng CuO un nng thu c 13,2 gam hn hp gm anehit, axit, ancol d 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 mol Ta c n Ru p = nH2O (1) nCH3COOH + nC2H5OH ban u = 0,3 nCH3COOH = 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 + (9,2 0,1.46 x.46) x = 0,05 mol nC2H5OH p = x + 0,1 = 0,15 mol => H% = 0,15.100% / 0,2 = 75% =>C

Cu 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 i qua bnh ng CaO mi nung, d thy khi lng bnh tng thm 6,21 gam. Bit th tch cc khi o ktc. Cng thc 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 = 2nH2 0,015x + 0,02y = 0,09 TH2 => 0,02x + 0,015y = 0,085 Gii h => x = 2 ; y = 3 => B i thi ko c p n ko xc nh X l CnH2n(OH)2 ; Y l CmH2m-1(OH)3 Xem chuyn 1 cch xc nh cht CnH2n+2 2a m(OH)m a l tng b + vng , m l s chc

Ta c c 2 ru u no => nH2O nCO2 = n hn hp ru = 0,015 + 0,02 = 0,035 mol m bnh tng = mH2O + mCO2 = 18.nH2O + 44.nCO2 = 6,21

Gii h => nH2O = 0,125 ; nCO2 = 0,09 => n = nCO2 / (nH2O nCO2) = 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 16DD 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 60A

61D 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 sai Bn c cho kin v cu . Mnh v mt s ngi s xem li.

Cm n bn gip . Chc bn thnh cng.

chuyÊn ĐỀ 5 bÀi tẬp dẪn xuẤt halogen - phenol - ancol

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