btHoahochuucobtHoahochuuco Phng php gii bi tp ho hc Hu c Mc lc

I H NG DN CHUNG 1. PHN LOI BI TP HA HCHin nay c nhiu cch phn loi bi tp khc nhau trong cc ti liu gio khoa. V vy, cn c cch nhn tng qut v cc dng bi tp da vo vic nm chc cc c s phn loi. 1. Phn loi da vo ni dung ton hc ca bi tp: Bi tp nh tnh (khng c tnh ton) Bi tp nh lng (c tnh ton) 2. Phn loi da vo hot ng ca hc sinh khi gii bi tp: Bi tp l thuyt (khng c tin hnh th nghim) Bi tp thc nghim (c tin hnh th nghim) 3. Phn loi da vo ni dung ha hc ca bi tp: Bi tp ha i cng - Bi tp v cht kh - Bi tp v dung dch - Bi tp v in phn Bi tp ha v c - Bi tp v cc kim loi - Bi tp v cc phi kim - Bi tp v cc loi hp cht oxit, axit, baz, mui, Bi tp ha hu c - Bi tp v hydrocacbon - Bi tp v ru, phenol, amin - Bi tp v andehyt, axit cacboxylic, este, 4. Da vo nhim v v yu cu ca bi tp: Bi tp cn bng phng trnh phn ng Bi tp vit chui phn ng Bi tp iu ch Bi tp nhn bit Bi tp tch cc cht ra khi hn hp Bi tp xc nh thnh phn hn hp Bi tp lp CTPT. Bi tp tm nguyn t cha bit 5. Da vo khi lng kin thc, mc n gin hay phc tp ca bi tp: Bi tp dng c bn Bi tp tng hp 6. Da vo cch thc tin hnh kim tra: Bi tp trc nghim Bi tp t lun 7. Da vo phng php gii bi tp: Bi tp tnh theo cng thc v phng trnh.

Bi tp bin lun Bi tp dng cc gi tr trung bnh 8. Da vo mc ch s dng: Bi tp dng kim tra u gi Bi tp dng cng c kin thc Bi tp dng n tp, n luyn, tng kt Bi tp dng bi dng hc sinh gii Bi tp dng ph o hc sinh yu, Mi cch phn loi c nhng u v nhc im ring ca n, ty mi trng hp c th m gio vin s dng h thng phn loi ny hay h thng phn loi khc hay kt hp cc cch phn loi nhm pht huy ht u im ca n. Thng gio vin s dng bi tp theo hng phn loi sau: Bi tp gio khoa: Thng di dng cu hi v khng tnh ton nhm lm chnh xc khi nim; cng c, h thng ha kin thc; vn dng kin thc vo thc tin. Cc dng hay gp: vit phng trnh phn ng, hon thnh chui phn ng, nhn bit, iu ch, tch cht, gii thch hin tng, bi tp v tnh cht ha hc cc cht, C th phn thnh 2 loi : + Bi tp l thuyt (cng c l thuyt hc) + Bi tp thc nghim : va cng c l thuyt va rn luyn cc k nng, k xo thc hnh, c ngha ln trong vic gn lin l thuyt vi thc hnh. Bi tp ton: L nhng bi tp gn lin vi tnh ton, thao tc trn cc s liu tm c s liu khc, bao hm 2 tnh cht ton hc v ha hc trong bi. Tnh cht ha hc: dng ngn ng ha hc & kin thc ha hc mi gii c (nh va , hon ton, khan, hidrocacbon no, khng no, ) v cc phng trnh phn ng xy ra. Tnh cht ton hc: dng php tnh i s , qui tc tam sut, gii h phng trnh, Ha hc l mt mn khoa hc t nhin, tt yu khng trnh khi vic lin mi vi ton, l, c im ny cng gp phn pht trin t duy logic cho hc sinh. Hin nay, hu ht cc bi tp ta ha nh nhn vic rn luyn t duy ha hc cho hc sinh, gim dn thut ton.

2.MT S PHNG PHP GII BI TP1Tnh theo cng thc v phng trnh phn ng 2- Phng php bo ton khi lng 3- Phng php tng gim khi lng 4- Phng php bo ton electron 5- Phng php dng cc gi tr trung bnh Khi lng mol trung bnh Ha tr trung bnh S nguyn t C, H, trung bnh S lin kt p trung bnh G hydrocacbon trung bnh S nhm chc trung bnh, 6- Phng php ghp n s 7- Phng php t chn lng cht 8- Phng php bin lun

3.IU KIN HOC SINH GII BI TP C TT:Nm chc l thuyt: cc nh lut, qui tc, cc qu trnh ha hc, tnh cht l ha hc ca cc cht. 2. Nm c cc dng bi tp c bn, nhanh chng xc nh bi tp cn gii thuc dng bi tp no. 3. Nm c mt s phng php gii thch hp vi tng dng bi tp 4. Nm c cc bc gii mt bi ton hn hp ni chung v vi tng dng bi ni ring 5. Bit c mt s th thut v php bin i ton hc, cch gii phng trnh v h phng trnh bc 1,2, 1.

4.CC B C GII BI TP TRN LP:Tm tt u bi mt cch ngn gn trn bng. Bi tp v cc qu trnh ha hc c th dng s . 2. X l cc s liu dng th thnh dng cn bn (c th bc ny trc khi tm tt u bi) 3. Vit cc phng trnh phn ng xy ra (nu c) 4. Gi v hng dn hc sinh suy ngh tm li gii: Phn tch d kin ca bi xem t cho ta bit c nhng g Lin h vi cc dng bi tp c bn gii Suy lun ngc t yu cu ca bi ton 5. Trnh by li gii 6. Tm tt, h thng nhng vn cn thit, quan trng rt ra t bi tp (v kin thc, k nng, phng php) 1.

5.C S THC TIN:

Thc t nhiu trng ph thng, s tit ha trong tun t, phn ln dng vo vic ging bi mi v cng c cc bi tp c bn trong sch gio khoa. Bi tp gio khoa m rng v cc bi tp ton ch c cp mc thp. Khi c bi tp ha nhiu hc sinh b lng tng khng nh hng c cch gii, ngha l cha hiu r bi hay cha xc nh c mi lin h gia gi thit v ci cn tm. Cc nguyn nhn lm hc sinh lng tng v sai lm khi gii bi tp ha hc: Cha hiu mt cch chnh xc cc khi nim, ngn ng ha hc (v d nh : nng mol, dd long, c, va , ) Cha thuc hay hiu c th vit ng cc phng trnh phn ng, cha nm c cc nh lut c bn ca ha hc Cha thnh tho nhng k nng c bn v ha hc, ton hc (cn bng phn ng, i s mol, V, nng , lp t l, ) Khng nhn ra c mi tng quan gia cc gi thit, gi thit vi kt lun c th la chn v s dng phng php thch hp i vi tng bi c th.

II C S L THUYT CN NM VNG 1 THUYT CU TO HA HC :Ni dung : 1. Trong phn t cht hu c, cc nguyn t lin kt vi nhau theo ng ha tr v theo mt th t

nht nh. Th t lin kt gi l cu to ha hc. S thay i th t s to nn cht mi. 2. Trong phn t hp cht hu c, cacbon c ha tr IV. Nhng nguyn t cacbon c th kt hp khng nhng vi nhng nguyn t ca cc nguyn t khc m cn kt hp trc tip vi nhau thnh nhng mch cacbon khc nhau (mch khng nhnh, c nhnh, mch vng). 3. Tnh cht ca cc hp cht ph thuc vo thnh phn phn t (bn cht v s lng cc nguyn t) v cu to ha hc (th t lin kt cc nguyn t) ng ng : - ng ng l hin tng cc cht c cu to v tnh cht tng t nhau, nhng v thnh phn phn t khc nhau mt hay nhiu nhm CH2. Nhng cht c gi l nhng cht ng ng vi nhau, chng hp thnh mt dy ng ng. 2. ng phn : - ng phn l hin tng cc cht c cng CTPT, nhng c cu to khc nhau nn c tnh cht ha hc khc nhau. Cc cht c gi l nhng cht ng phn. 1. Vic nm vng ngha ca mi loi cng thc ha hu c c vai tr rt quan trng. iu ny cho php nhanh chng nh hng phng php gii bi ton lp CTPT, dng ton c bn v ph bin nht ca bi tp hu c. Cc bi ton lp CTPT cht hu c nhn chung ch c 2 dng : - Dng 1 : Lp CTPT ca mt cht - Dng 2 : Lp CTPT ca nhiu cht. Vi kiu 1, c nhiu phng php khc nhau gii nh : tm qua CTG, tm trc tip CTPTKiu 2 ch yu dng phng php tr s trung bnh (xem phn tr s trung bnh). Nhng d dng phng php no chng na th cng vic u tin l t cng thc tng qut ca cht , hoc cng thc tng ng cho hn hp mt cch thch hp nht ,vic t cng thc ng chim 50% yu t thnh cng. 1. Cng thc thc nghim : cho bit thnh phn nh tnh, t l v s lng cc nguyn t trong phn t. V d : (CH2O)n (n 1, nguyn dng nhng cha xc nh ) 2. Cng thc n gin : c ngha nh cng thc thc nghim nhng gi tr n = 1 3. Cng thc phn t : cho bit s lng nguyn t ca mi nguyn t trong phn t, tc l cho bit gi tr n 4. Cng thc cu to : ngoi vic cho bit s lng nguyn t ca mi nguyn t trong phn t cn cho bit trt t lin kt gia cc nguyn t trong phn t. C nhiu loi CTCT khc nhau, chng hn CTCT y , CTCT vn tt, CTCT bn khai trinNguyn tc chung vit CTCT bn khai trin l c th bt cc lin kt n gia cc nguyn t cc nguyn t, cc lin kt bi trong nhm chc (nu thy khng cn thit) nhng nht thit khng c b lin kt bi gia cc C-C. Cc loi cng thc CTTN, CTG, CTPT trng hau khi gi tr n = 1. Cng thc tng qut : cho bit thnh phn nh tnh cht c cu to nn t nhng nguyn t no, i vi CTTQ ca mt dy ng ng c th th cn cho bit thm t l nguyn t ti gin hoc mi lin h gia cc thnh phn cu to . V d : CTTQ ca hydrocacbon l CxHy hoc CnH2n+2-2k nhng vi hydrocacbon c th l ankan th CTTQ l : CnH2n+2, anken l : CnH2n ,

2 NG NG NG PHN :

3 CC LOI CNG THC HA HU C

4 TM TT HA TNH CC HYDROCACBON

ANKAN : - Hydrocacbon no, mch h, trong phn t ch c lin kt n gia C-C v C-H - CTTQ : CnH2n +2 , n1, nguyn a) Tnh cht ho hc : 1. Phn ng oxiha : + Phn ng oxy ha hon ton : CnH2n +2 + (3n +1)/2 O2 n CO2 + (n+1)H2O Nu thiu oxi : CnH2n +2 + (n +1)/2 O2 n C + (n+1)H2O + Phn ng oxy ha khng hon ton : nu c xc tc th ankan s b oxi ha to nhiu sn phm : andehyt, axit CH4 + O2 HCHO + H2O (andehyt fomic) n-C4H10 + 5/2 O2 2CH3COOH + H2O 2. Phn ng phn hy + Bi nhit : CnH2n +2 n C + (n+1)H2 + Bi Clo : CnH2n +2 + (n +1)Cl2 n C + 2(n+1)HCl 3. Phn ng th vi cc halogen : CnH2n +2 + mX2 CnH2n+2-m Xm + mHX 4. Phn ng hidro ha (tch hydro) : to sn phm c th c mt hay nhiu ni i hoc khp vng. CnH2n +2 CnH2n + H2 (n 2) V d : CH3CH3 CH2CH2 + H2 n-hexanxiclohexan + H2 (C6H14) (C6H12) 5. Phn ng cracking (b gy mch cacbon) CnH2n +2 CmH2m + CxH2x+2 iu kin : n 3, m 2, nguyn x1 n=m+x Tng qut : Ankan (3C) Ankan + anken C3H8CH4 + C2H4 XICLOANKAN - L hydrocacbon no, mch vng, trong phn t ch tn ti lin kt n. - CTTQ : CnH2n , n3 nguyn Xicloankan c y tnh cht ca mt hydrocacbon no (vng C5 tr ln ), ngoi ra cn c tnh cht ca vng:cc vng nh c sc cng ln, km bn, d tham gia phn ng cng m vng (vng C3, C4 ) : ANKEN : - L nhng hydrocacbon mch h c mt ni i trong phn t - CTTQ : CnH2n ,n 2, nguyn 1. Phn ng cng CnH2n + H2 CnH2n+2 CnH2n + Br2 CnH2nBr2 CnH2n + HA CnH2n+1A

(vi HA l cc axit nh HCl, HBr, H2SO4) CnH2n + H2O CnH2n +1OH - Phn ng cng ca anken tun th quy tc Maccopnhicop : nguyn t H (hay phn mang in tch dng) cng vo nguyn t Cacbon c nhiu H hn, cn phn m ca tc nhn (nguyn t X)gn vo C ca ni i mang in dng (C t H hn). 2. Phn ng oxiha : + oxiha hon ton : CnH2n + 3n/2 O2 => nCO2 + nH2O + oxi ha khng hon ton bi ddKMnO4 : CnH2n + [O] + H2O CnH2n(OH)2 CH2CH2 + [O] + H2OHOCH2CH2 OH 3. Phn ng trng hp CH2CH2 [ CH2CH2]n (Poly etilen) (nha PE) Tng qut : ANKADIEN : L hydrocacbon khng no, mch h, trong phn t c 2 ni i C=C. CTTQ : CnH2n-2, n 2 nguyn 1. Phn ng cng : 1CH2CH CHCH2 + 2Br2 BrCH2CHBr CHBr CH2Br 2. Phn ng trng hp : nCH2CH CHCH2 [CH2CHCHCH2]n nCH2C CHCH2 [CH2CCHCH2]n CH3 CH3 3. Phn ng oxi ha : + Oxi ha hon ton : CnH2n-2 +(3n-1)/2O2nCO2 + (n-1)H2O + Oxi ha khng hon ton : 3CH2CH CHCH2 + 4KMnO4 + 8H2O CH2OHCHOHCHOHCH2OH + 4MnO2 + 4KOH ANKIN - L nhng hydrocacbon khng no, mch h c mt ni ba trong phn t - CTPTTQ : CnH2n-2, n 2 , nguyn 1. Phn ng cng : CnH2n-2 + H2 CnH2n CnH2n-2 + 2H2CnH2n+2 CnH2n-2 + X2 CnH2nX2 CnH2nX4 Vi X : halogen HCCH + X2XHC=CHXX2HC-CHX2 CnH2n-2 + HACnH2n-1A Vi HA : cc axit nh : HCl, HCN, H2SO4 HCCH + H2O CH3CHO Lu : trong phn ng cng gia ankin bt i v tc nhn bt i, sn phm chnh c xc nh theo quy tc Maccopnhicop. 2. Phn ng oxi ha : CnH2n-2 +(3n-1)/2O2nCO2 + (n-1)H2O 3C2H2 + 8KMnO43K2C2O4 +8MnO2 + 2KOH + 2H2O

C2H2 + 2KMnO4 + 3H2SO4 2CO2 + 2MnSO4 + K2SO4 + 4 H2O 5CH3CCH +8KMnO4 + 12H2SO4 5CH3COOH + 5CO2 + 8MnSO4 + 4K2SO4 + 12H2O (Hin tng mu tm dung dch nht dn hoc mt hn) 3. Phn ng trng hp 2HCCHCH2CH CCH (Trng hp) (Vinylaxetyl hay vinylaxetilen) 3HCCHC6H6 (benzen) (Tam hp) nHCCH(CHCH) (Cupren) 4. Phn ng bi kim loi ca Ankin-1 : HCCH + 2AgNO3 + 2NH3AgCCAg + 2NH4NO3 RCCH + AgNO3 + NH3 RCCAg + NH4NO3 Vit tt : HCCH + Ag2OAgCCAg + H2O 2RCCH + Ag2O2RCCAg + H2O => Trong dy ng ng ankin, ch c axetilen c th th hai ln vi ion kim loi HCCH + 2Na NaCCNa + H2 HYDROCACBON THM : Aren hay hydrocacbon thm l loi hydrocacbon c c trng trong phn t bi s c mt mt hay nhiu vng benzen 1. Phn ng th Vi Halogen : (Brombenzen) Vi axit nitric (xc tc H2SO4, toC) (phn ng nitro ha) (1,3-dinitrobenzen) Vi axit H2SO4, bo ha SO3 (phn ng sunfo ha) ng ng ca benzen cng cho phn ng th C mch nhnh vi Halogen trong iu kin chiu sng : 2. Phn ng cng : 3. Phn ng oxi ha : CnH2n-6 + (3n-3)/2O2 nCO2 + (n-3)H2O C6H5CH3 + 2KMnO4 C6H5COOK + 2MnO2 + KOH +H2O Toluen Kalibenzoat * Benzen bn, khng b oxiha bi ddKMnO4, ch c mch nhnh ca vng benzen mi b oxiha => phn ng dng phn bit benzen v cc ng ng ca n.

1. iu ch ankan : Nguyn liu ly t thin thin nh kh than , kh du m Tng hp t cc dn xut halogen hoc cc mui ca cc axit hu c RX + 2Na + XR RR + 2NaX C2H5Cl + 2Na + ClCH3 C2H5CH3 + 2NaCl R1(COONa)m + mNaOH(r) R1Hm + mNa2CO3 iu ch Metan : C + 2H2 CH4 CO + 3H2 CH4+ H2O CH3COONa + NaOHr CH4 + Na2CO3 Al4C3 + 12 H2O 4Al(OH)3 + 3CH4

5 IU CH CC HYDROCACBON

2. iu ch anken : + Phn ng cracking v phn ng hydro ha : CH3CH2OHCH2CH2 + H2O RCHXCH2R RCHCHR + HX RCHXCHXR + Zn RCHCHR + ZnX2 RCHOHCH2R RCHCHR + H2O RCCR + H2 RCHCHR CnH2n+2 CmH2m + CxH2x+2 CnH2n+2 CmH2m + (n + 1 - m)H2 3. iu ch Ankin, Ankadien : RCHXCHXR RCCR + 2HX RCHX2 CHX2R +2Zn RCCR RCCH + Na RCCNa +1/2H2 RCCNa + X-R RCCR (Phn ng tng mch C) CaC2 + 2H2OCa(OH)2 + C2H2 2CH4 C2H2 + 3H2 2C + H2 C2H2 4. iu ch ankadien 2CH3CH2OH CH2CH CHCH2 + 2H2O 2HCCHCHCCHCH2 CHCCHCH2 + H2 CH2CH CHCH2 CH2CHOHCHOHCH3 CH2CH CHCH2 +2H2O CH2CH CH2CH3 CH2C CHCH2 CH3 CH3 (Isopren) CHCCHCH2 + HCl CH2C CHCH2 Cl (Cloropren) 5. iu ch hydrocacbon thm v cc hydrocacbon khc : 3C2H2C6H6 C6H12(xicloankan) C6H6 + 3H2 C6H14C6H6 + 4H2 C6H5COOH + 2NaOH C6H6 + Na2CO3 + H2O C6H5 X +2Na +X-CH3 C6H5CH3 + 2NaX C6H6 + CH3X C6H5CH3 + HX Nhn xt : Hydrocacbon no (ankan), phn ng c trng l phn ng th, khng c phn ng cng v kh b oxiha bi dd KMnO4 Hydrocacbon khng no (anken, ankadien, ankin) phn ng c trng l phn ng cng. (anken c phn ng th nhit cao, th ) Phn ng cng Hidro : + xt Ni/toC th xicloankan (C3, C4), anken v ankin, ankadien cng H2 c ankan; aren cng H2 c xicloankan + xt Pd/toC th ankin, ankadien cng H2 c anken Phn ng cng HX vo anken, ankadien, ankin phi ch sn phm chnh ph v s lng sn phm. t chy CxHy: t th :

T>1 => CxHy l ankan, CTTQ : CnH2n+2 T = 2 => CxHy l CH4 T=1 => CxHy l anken, xicloankan CTTQ : CnH2n T CxHy l ankadien, ankin, CTTQ : CnH2n-2 hoc l aren, CTTQ : CnH2n-6 T = 0,5 => CxHy l C2H2 hoc C6H6.

III BI TP GIO KHOA 1 BI TP V CNG THC CU TO NG NG NG PHN DANH PHP1.1 Bi tp v ng ng v Phng php : C 2 cch xc nh dy ng ng ca cc hydrocacbon : Da vo nh ngha ng ng Da vo electron ha tr xc nh Lu : C lun c ha tr IV tc l c 4e ha tr nC s c 4ne ha tr H lun c ha tr I tc l c 1e ha tr - Parafin chnh l ankan, dy ng ng parafin chnh l dy ng ng ca CH4. - Olefin chnh l anken, dy ng ng olefin chnh l dy ng ng ca C2H4 - Ankadien cn c gi l ivinyl - Aren : dy ng ng ca benzen. - Hydrocacbon : CxHy : y chn, y < 2x + 2 v Bi tp v d : V d 1: Vit CTPT mt vi ng ng ca CH4. Chng minh cng thc chung ca dy ng ng ca CH4 l CnH2n+2. GII : Da vo nh ngha ng ng, CTPT cc ng ng ca CH4 l C2H6, C3H8, C4H10,, C1+kH4+2k Chng minh CTTQ dy ng ng metan CH4 l CnH2n+2 : Cch 1: Da vo nh ngha ng ng th dy ng ng ca metan phi l: CH4 + kCH2 = C1+kH4+2k Tm mi lin h gia s nguyn t C v s nguyn t H t SnC = 1 + k = n SnH = 4 + 2k = 2(k + 1) + 2 = 2n + 2 Vy dy ng ng farafin l CnH2n+2 (n 1) Cch 2: Da vo s electron ha tr : - S e ha tr ca nC l 4n - S e ha tr ca 1C dng lin kt vi cc C khc l 2 => S e ha tr ca nC dng lin kt vi cc C khc l [2(n-2)+2] = 2n2 (v trong phn t ch tn ti lin kt n) (S d +2 v 1C u mch ch lin kt vi 1C nn dng 1e ha tr, 2C u mch dng 2e ha tr. - S e ha tr dng lin kt vi H: 4n2n-2 = 2n + 2 - V mi nguyn t H ch c 1 e ha tr nn s e ha tr ca (2n +2)nguyn t H trong phn t l 2n + 2. => Cng thc chung ca ankan l CnH2n+2 (n 1) Cch 3:

Metan c CTPT CH4 dng CnH2n+2 => dy ng ng ca ankan l CnH2n+2 V d 2: CT n gin nht ca 1 ankan l (C2H5)n. Hy bin lun tm CTPT ca cht trn. GII : CT n gin ca ankan l (C2H5)n. Bin lun tm CTPT ankan : Cch 1: Nhn xt: CT n gin trn l 1 gc ankan ha tr 1 tc c kh nng kt hp thm vi 1 gc nh vy na => n = 2 => CTPT ankan C4H10 Cch 2: Cch 3: CTPT ca ankan trn : (C2H5)n = CxH2x+2 => 2n = x v 5n = 2x + 2 => 5n = 2.2n + 2 => n = 2. => CTPT ankan : C4H10 Ankan trn phi tha iu kin s H < 2.s C + 2 => 5n < 2.2n + 2 =>n < 2 n =1 th s H l => loi n= 2 => CTPT ankan l C4H10 (nhn) Vy CTPT ankan l C4H10 V d 3 : Phn bit ng phn vi ng ng. Trong s nhng CTCT thu gn di y, nhng cht no l ng ng ca nhau? Nhng cht no l ng phn ca nhau.? GII : Phn bit ng phn vi ng ng : xem I.2.2/12 Nhng cht l ng ng ca nhau l : 1 v 5 hoc 1 v 3(ankan); 6 v 7 hoc 6 v 9 (anken). Nhng cht l ng phn ca nhau : 2 v 4; 3 v 5; 6 v 9 v 8. v Bi tp tng t : 1) Vit CTPT mt vi ng ng ca C2H4. Chng minh CTTQ ca dy ng ng ca etilen l CnH2n , n 2 nguyn 2) Vit CTPT mt vi ng ng ca C2H2 . Chng minh CTTQ ca dy ng ng ca axetilen l CnH2n-2, n 2 nguyn 3) Vit CTPT mt vi ng ng ca C6H6. Chng minh CTTQ ca cc aren l CnH2n-6, n 6 nguyn 1.2 Bi tp v ng phn danh php : v Phng php vit ng phn : Bc 1: - T CTPT suy ra cht thuc loi hydrocacbon hc no. - Vit cc khung cacbon Bc 2 :- ng vi mi khung cacbon, di chuyn v tr lin kt bi (nu c), di chuyn v tr cc nhm th (nu c). - Nu c ni i hoc vng trong CTCT ca cht th xt xem c ng phn hnh hc khng. Bc 3 : - in Hidro. Lu : lm xong phi kim tra li xem cc nguyn t ng ha tr cha. v Bi tp v d : V d 1 : a) Nu iu kin mt phn t c ng phn hnh hc? b) Vit tt c cc CTCT cc ng phn ca C5H10; Trong cc ng phn , ng phn no c

ng phn hnh hc? c tn cc ng phn . GII : a) iu kin mt phn t c ng phn hnh hc (ng phn cis-trans) : Xt ng phn : iu kin : a d v b f - Nu a > d v b>f (v kch thc phn t trong khng gian hoc v phn t lng M)* ta c ng phn cis. - Nu a > d v b to 3 sn phm (loi) (2) c 4 v tr th (C1, C2, C3, C4) => to 4 sn phm (nhn) (3) c 1 v tr th (C1 hoc C3) => to mt sn phm (loi) Vy CTCT ca pentan l (2) : 2-metylbutan (isopentan) Ptp : b) Tng t : CTCT ca pentan l (3): 2,2-dimetylpropan (neopentan), khi cracking ch cho 2 sn phm : V d 3 : Vit CTCT ca cc cht c tn sau v gi tn li cho ng nu cn : a) 3,3-Diclo-2-etyl propan b) 4-Clo-2-isopropyl-4-metylbuten-2 c) 2-isopropyl penten-1 GII : i vi loi bi ny th nguyn tc l t tn gi vit CTCT ca cht . Sau xt xem ngi ta gi tn ng cha bng cch chn mch chnh, nh s ch v tr nhnhnu sai th gi tn li. a) 3,3-Diclo-2-etylpropan

(tn sai do chn mch chnh 3C cha phi l mch di nht) b) 4-Clo-2-isopropyl-4-metylbuten-2 Cch gi tn trn sai v chn mch chnh sai. Tn ng l : 5-Clo-2,3-Dimetylhexen-3 c) 2-isopropylpenten-1 Cch c tn trn l ng. Nu chn mch chnh l 6C (ii) l sai v mch ny khng cha ni i. v Bi tp tng t : 1) Vit CTCT ca cht X c CTPT C5H8. Bit rng khi hydro ha cht X, ta thu c isopren. Mt khc, cht X c kh nng trng hp cho ra cao su tng hp. c tn danh php IUPAC cc ng phn mch h ca X 2) Cho aren c CTPT C8H10. vit CTCT v gi tn cc ng phn ca A. 3) Vit CTCT v gi tn li cho ng nu cn. Xt xem ng phn no c ng phn hnh hc. a) 1,2- Diclo-1-metyl hexan b) 2,3,3-Tri metyl butan c) 1,4-Dimetyl xiclobutan. c) Diallyl d) 3-allyl-3-metylbuten-1 e) 2,2,5,5- tetrametylhexin-3 f) 3-metylpentin-1

2 CHUI PHN NG IU CHv Phng php : 1) Mun lm bi tp chui phn ng cn lu : - Mi mi tn ch vit mt phng trnh phn ng. - Bt u t phn ng trong c CTCT ca mt cht ta bit chnh xc (phn ng khng c sai CTCT ca cht) da vo cc iu kin phn ng suy lun tm ra cc cht cn li. - Xem trong chui c phn ng no ct bt mch hay tng mch cacbon khng. 2) Cc phn ng ct bt mch hoc ct t mch cacbon th dng cc phn ng : - Ct bt mch th dng cch nhit phn mui : R COONa + NaOH(r) RH + Na2CO3 - Cch t th dng phng php cracking 3) Ni di thm (tng mch) cacbon : dng mt trong hai cch n gin ca chng trnh ha hc ph thng : a) Trng hp : 2HCCH CH2=CH-CCH b) to Ni hai gc ankyl : RCl + 2Na + RCl => RR + 2NaCl 3) Bi tp iu ch l mt dng khc ca chui phn ng, y bi ch cho bit nguyn liu ban

u v yu cu iu ch mt cht no . lm c bi ny, hc sinh phi nh v vit cc ptp trung gian c ghi km y iu kin phn ng. C nhiu cch iu ch khc nhau vi cng mt bi iu ch. Lu : nu bi yu cu vit s iu ch (hoc s tng hp) th ta ch cn vit di dng mt chui phn ng t nguyn liu n sn phm, trn cc mi tn c ghi km iu kin phn ng. * Thnh phn ch yu ca : - Kh thin nhin : ch yu l Metan (90%), cn li l etan, propan, butan v mt s ng ng cao hn. - Kh cracking : Hydrocacbon cha no (C2H4, C3H6, C4H8), ankan (CH4, C2H6, C4H10) v H2. - Kh than : ch yu l H2(60%), CH4 (25%) cn li l CO, CO2, N2 - Kh l cao : CO2, CO, O2, N2, v Bi tp v d : V d 1 : Chui phn ng cho bit CTPT cc cht : Hon thnh chui phn ng sau : C2H5COONaC2H6 C2H5ClC4H10CH4CO2 GII : Nhn xt : bi cho bit CTPT cc cht, ta ch cn nh v vit phn ng c y iu kin hon thnh phn ng khng cn suy lun nhiu. Loi bi ny thng c dng tr bi hoc lm bi tp c bn trong tit bi tp. (1) ct bt mch => nhit phn mui. (3) tng mch cacbon => ni hai gc ankyl. Ptp : (1) C2H5COONa + NaOH (r) C2H6 + Na2CO3 (2) C2H6 + Cl2 C2H5Cl + HCl (3) C2H5Cl + 2Na + C2H5Cl C4H10 (4) C4H10 CH4 + C3H6 (5) CH4 + 2O2 CO2 + 2H2O V d 2 : bi khng cho bit CTPT ca cc cht nhng cho bit iu kin phn ng. + Vit phng trnh phn ng, xc nh CTCT cc cht : AlC3 + L => E + X (1) E Y + Z (2) CH3COOH + Y A (3) nA B (4) GII : Phn tch : iu kin phn ng chnh l du hiu suy lun tm CTCT cc cht. - Da vo (2) => E : CH4 - Y hoc l C2H2 hoc H2 (4)A c phn ng trng hp => trong phn t A c C=C; (3) CH3COOH + Y => A => Y l C2H2 v Z : H2. (1) => L : H2O; X : Al(OH)3 (3) CH3COOH + C2H2(Y) => CH3COOCH =CH2 (A) (4) => (B) : Ptp : Al4C3 + 12H2O => 3CH4 + 4Al(OH)3 2CH4 C2H2 + 3H2 CH3COOH + HCCHCH2=CHOCOCH3

V d 3: bi khng cho iu kin phn ng, ch cho bit duy nht CTPT ca mt cht + B tc chui phn ng sau : A => D + F D => F + C F + Br2 => G G + KOH => J + + J => B (tam hp) B + Cl2 => C6H6Cl6 J + C => D 2J => X X + C => E GII : Nhn xt : gia cc phn ng u c mi lin h vi nhau, mi ch ci ng vi mt cht nht nh v cc cht khng trng nhau. bi ny, t phn ng to 666, ta tm c B, da vo cc du hiu khc, suy lun tm ra cc cht cn li Phn tch : B + Cl2=>C6H6Cl6 => B : C6H6 J => B => J : C2H2 F + Br2 => G => G c hai nguyn t Brom trong phn t. M G + KOH => C2H2 (J) => G : C2H4Br2 => F : C2H4 C2H2(J) + C=>D => C : H2 v D : C2H6 (D khng th l C2H4 c v trng F) D=>C2H4(F) + C A => D(C2H6) + F(C2H4) => A : C4H10 Vy A : C4H10; C:H2 ; D:C2H6 ; F : C2H4; G:C2H4Br2 ; J:C2H2 Ptp : C4H10C2H6 + C2H4 C2H6 C2H4 + H2 Ru C2H4 + Br2 => C2H4Br2 C2H4Br2 + 2KOHC2H2 + 2KBr + 2 H2O 3C2H2 => C6H6 C6H6 + 3Cl2 => C6H6Cl6 C2H2 + 2H2 C2H6 V d 4 : Vit s phn ng tng hp PVC t vi v than . GII : S : v Bi tp tng t : Hon thnh cc chui phn ng sau. Ghi y iu kin phn ng : 1) 2) 3) 4) 5) vi=>vi sng=>canxicacbua=>axetilen=>vinyl axetilen=>Divinyl=>caosu Buna 6*)

p n : A: CH4; A1:C2H2 ; A2 :C6H6 ; A3: C6H5CH3; B1:C2H6 ; B2: C2H5Cl; B3: C2H5OH 7*) p n : X:C2H2; X1:C4H4 (vinyl axetilen); X2 : C4H6 (Butadien-1,3) ; X3: C6H5CH=CH2; X4: C2H4; X5: C2H5OH 8*) Bit A v A3 c cng s C. p n : A:C4H10; A1:C2H4; A2: C2H5OH; A3 :C4H6 (Divinyl); A5:C4H8; A6:CH3-CH(OH)CH3 9*) T kh thin nhin vit phng trnh phn ng iu ch caosu Isopren, cao su Cloropren, Caosu Buna N, CCl4. Cho cc cht v c v iu kin th nghim coi nh . 10) Vit phng trnh phn ng tng hp tng hp caosu t cht u l isopentan. Cc iu kin phn ng v cc cht v c coi nh . 11) Vit phng trnh phn ng iu ch C2H5OH t kh cracking.

3 TCH TINH CH3.1 Tch cc hydrocacbon : v Nguyn tc : Tch ri l tch ring tt c nguyn cht ra khi hn hp bng cch tch dn tng cht mt. Th nghim ny kh, i hi phi chn ho cht thch hp tch v hon nguyn li cht . S : v Phng php: * Phng php vt l : Phng php chng ct tch ri cc cht lng ha ln vo nhau, c th dng phng php chng ct ri ngng t thu hi ha cht. Phng php chit (dng phu chit) tch ring nhng cht hu c tan c trong nc vi cc cht hu c khng tan trong nc (do cht lng s phn thnh 2 lp) Phng php lc (dng phu lc) tch cc cht khng tan ra khi dd. * Phng php ha hc : Chn nhng phn ng ha hc thch hp cho tng cht ln lt tch ring cc cht ra khi hn hp, ng thi ch dng nhng phn ng ha hc m sau phn ng d dng ti to li cc cht ban u. Mt s phn ng tch v ti to: Hidrocacbon Phn ng tch Phn ng ti to Phng php thu hi Anken R-CH=CH2 + Br2 => R-CHBr-CH2Br R-CHBr-CH2Br R-CH=CH2 Thu ly kh anken bay ra (hoc chit ly anken lng phn lp) Etilen CH2=CH2 CH2=CH2 + H2SO4 =>CH3CH2OSO3H

CH3CH2OSO3H CH2=CH2+H2SO4 Ankin-1 v axetilen R-CCH 2R-CCH + Ag2O 2R-CCAg + 2H2O RCCAg + HCl => RCCH + AgCl(kt ta) Lc b kt ta thu hi ankin lng hoc thu ly ankin kh. Benzen v cc ng ng ca benzen Khng tan trong nc v trong cc dd khc nn dng phng php chit tch. Nu c anken v ankin th tch ankin trc bng dd AgNO3/NH3 v ankin cng cho phn ng cng vi dd Br2 nh anken. v Bi tp v d : Tch ring tng kh ra khi hn hp kh gm CH4, C2H4, C2H2 v CO2. GII : Nhn xt: CO2 tan trong dd nc vi trong, CH4, C2H4, C2H2 th khng, nn dng cc phn ng bng trn tch: S tch Li gii v phng trnh phn ng: Dn hn hp kh qua dd Ca(OH)2 d, thu c (kt ta) CaCO3 CO2 + Ca(OH)2 => CaCO3(kt ta) + H2O Thot ra ngoi l hn hp kh CH4, C2H4, C2H2 c dn qua dd AgNO3/NH3 th C2H2 b gi li trong (kt ta) C2Ag2, cc kh CH4, C2H4 thot ra C2H2 + 2AgNO3 (dd) + 2NH3 => C2Ag2(kt ta) + 2NH4NO3 Tip tc dn hn hp kh CH4, C2H4 qua dd nc Br th C2H4 b gi li, CH4 thot ra ta thu c CH4. C2H4 + Br2 => C2H4Br2 Ti to CO2 bng cch nhit phn kt ta CaCO3 Ti to C2H2 bng cch cho kt ta C2Ag2 tc dng vi dd HCl C2Ag2 + 2HCl => C2H2 + 2AgCl(kt ta) Ti to C2H4 bng cch cho cht lng C2H4Br2 tc dng vi Zn/ru: v Bi tp tng t : Tch ri cc kh sau ra khi hn hp gm : a) Benzen, styren, phenol b) NH3, butin-1, butadien v butan c) Kh HCl, butin-1 v butan .2 Tinh ch : v Nguyn tc : Tinh ch l lm sch ha cht nguyn cht no bng cch loi b i tp cht ra khi hn hp (nguyn cht v tp cht). v Phng php : Dng ha cht tc dng vi tp cht m khng phn ng vi nguyn cht to ra cht tan hoc to ra cht kt ta lc b i. S tinh ch : Trong X l ha cht ta phi chn tc dng vi B loi B ra khi hn hp. v Bi tp v d : Cc phng trnh phn ng u l nhng phng trnh phn ng quen thuc gp trn. Do phn hng dn gii ch a ra cc s tinh ch.

V d 1 : Tinh ch (lm sch) Propilen c ln propin, propan v kh sunfur GII : Lu : SO2 v C3H6 u lm cho phn ng vi dd Brom nn phi tch SO2 trc ri mi dng dd Brom tch ly C3H6 ra khi hn hp ri tinh ch. S tinh ch V d 2: Tinh ch C6H6 c ln C6H12, C6H5CH3 GII : S : V d 3: Tinh ch Styren c ln benzen, toluen, hexin-1. GII : S : v Bi tp tng t : 1) Tinh ch C3H8 ln NO2 v H2S, hi nc 2) Tinh ch C2H6 ln NO, NH3, CO2 3) Lm sch etan c ln etilen v lm sch etilen c ln etan. 4) Lm sch etan c ln axetilen v ngc li 5) Lm sch etilen c ln axetilen v ngc li.

4 NHN BIT PHN BITv Phng php: Tng qut: - Lm th nghim vi cc mu th + Ch dng nhng phn ng c trng ca hidrocacbon nhn bit + Cc phn ng dng nhn bit phi n gin, d thc hin v du hiu phn ng quan st c (mu sc, (kt ta), si bt kh, ) - Khi c c cht hu c v v c nn phn bit cht v c trc, nu c. Cch nhn bit vi cht kh v c quen thuc: CO2, SO2 : lm c nc vi trong nhng SO2 to kt ta vng khi sc vo dd H2S hoc lm mt mu nu ca dd nc Brom. 2H2S + SO23S(vng) + H2O SO2 + Br2 + H2O2HBr + H2SO4 H2O (hi) : i mu trng ca CuSO4 khan thnh xanh N2, kh tr : khng chy NH3 : lm xanh mu qu tm m hoc to khi trng (NH4Cl) vi kh HCl HCl (kh) : lm qu tm m ha hoc to khi trng vi NH3(kh) HCl (dd) : lm qu tm , si bt CO2 vi CaCO3. NO : chuyn thnh nu khi gp khng kh (NO + O2 => NO2) nu NO2 : kh mu nu H2 : cho qua CuO nung nng, CuO chuyn t mu en sang mu . CuO + H2Cu + H2O (en) () CO : cho li qua dd PdCl2, sn phm kh thu c cho sc vo dd nc vi trong d th nc vi trong b c.

CO + PdCl2 + H2O CO2 + Pd + 2HCl CO2 + Ca(OH)2CaCO3(kt ta) + H2O Th t tng i nhn bit cc hydrocacbon Hidrocacbon Thuc th Du hiu Phng trnh phn ng Ankin u mch dd AgNO3/NH3 (kt ta)vng nht CHCH + 2AgNO3 + 2NH3 => AgCCAg(kt ta) + 2NH4NO3 CxHy cha no (anken, akin, ankadien, ) Dd Br2 mu nu Mu nu ca dd Br2 b nht hay mt mu CnH2n+2-2k + kBr2 =>CnH2n+2Br2k Dd KMnO4l (tm) Mu tm ca dd KMnO4 b nht hay mt mu Benzen & ankan Cl2, askt Ch benzen to m trng Toluen Dd KMnO4l Mt mu tm C6H5CH3 + 3[O] C6H5COOH + H2O v Nhng im cn lu thm khi nhn bit cc hydrocacbon : 1) Phn bit anken vi cc hydrocacbon mch h khc c s lin kt p nhiu hn Bng cch ly cng th tch nh nhau ca cc hydrocacbon ri nh tng lng dd Br2 (cng nng ) vo. Mu no c th tch Br2 b mt mu nhiu hn ng vi hydrocacbon c s lin kt p nhiu hn. 2) Phn bit axetilen vi cc ankin-1 khc - Bng cch cho nhng th tch bng nhau ca cc cht th tc dng vi lng d dd AgNO3 trong NH3 ri nh lng kt ta kt lun. CH CH + 2AgNO3 + 2NH3 AgC CAg + 2NH4NO3 R C CH + AgNO3 + NH3 R C CAg + NH4NO3 3) Phn bit ankin-1 vi cc ankin khc Ankin-1 to kt ta vng nht vi dd AgNO3 trong NH3 4) Phn bit benzen v ng ng khc ca benzen Benzen khng lm mt mu dd thuc tm (KMnO4) trong khi cc ng ng ca benzen lm mt mu hoc nht mu dd thuc tm. * Nu hn hp phc tp nn lp bng nhn bit * Lu : t hin tng suy ra cht Vd: Khi lm c nc vi trong v to (kt ta) vng vi dd H2S l SO2 () Kh SO2 lm c nc vi trong v to (kt ta) vng vi dd H2S l SO2 ( ng v mt khoa hc nhng khi nhn bit nh vy l sai qui tc) v Bi tp v d : Nhn bit cc l kh mt nhn : Bi 1:

a)N2, H2, CH4, C2H4, C2H2 b) C3H8, C2H2, SO2, CO2. GII : a) N2, H2, CH4, C2H4, C2H2 C 3 cch gii : Cch 1 : Nhn xt: - N2 : khng cho phn ng chy - H2 : phn ng chy, sn phm chy khng lm c nc vi trong - CH4 : phn ng chy, sn phm chy lm c nc vi trong - Cc kh cn li dng cc phn ng c trng nhn bit. Tm tt cch gii: - Ly mi kh mt t lm mu th. - Dn ln lt cc kh i qua dd AgNO3/NH3. Kh no to c kt ta vng l C2H2 - Dn cc kh cn li qua dd nc Brm (mu nu ). Kh no lm nht mu nc brom l C2H4 H2C=CH2 + Br2 => BrH2CCH2Br - Ln lt t chy 3 kh cn li. Kh khng chy l N2. Sn phm chy ca hai kh kia c dn qua dd nc vi trong. Sn phm chy no lm c nc vi trong l CH4. Mu cn li l H2. CH4 + 2O2 => CO2 + 2H2O CO2 + Ca(OH)2 => CaCO3(kt ta) + H2O H2 + O2 => H2O Cch 2 : - Dn 5 kh trn ln lt qua dd Brom, c 2 kh lm mt mu dd nc Brom (nhm 1) gm C2H4 v C2H2. 3 kh cn li khng c hin tng g thot ra ngoi (nhm 2) gm CH4 v CO2, H2. - Sau nhn bit cc kh trong mi nhm trn tng t cch 1. Cch 1 ti u hn cch 2. b) C3H8, C2H2, SO2, CO2. Nhn xt: C 3 cch : Cch 1 : - Dn bn kh trn ln lt qua dd nc vi trong d. C 2 kh lm c nc vi trong (nhm 1) v 2 kh kia khng lm c nc vi trong (nhm 2). - Cho 2 kh mi nhm ln lt qua dd nc Brom. Kh nhm 1 lm mt mu nu ca dd Brom l SO2 v kh nhm 2 cng c hin tng nh vy l C2H2. Hai kh cn li l CO2 v C3H8. Cch 2 : - Dng phn ng c trng nhn bit. - Th t nhn bit C2H2, SO2, CO2, C3H8 Cch 3 : - Dn 4 kh trn ln lt vo dd Brom, c 2 kh lm mt mu nu ca dd Brom (nhm 1) v 2 kh kia khng c hin tng g (nhm 2). - Dn ln lt 2 kh nhm 1 qua dd AgNO3/NH3. Kh no to kt ta vng nht l C2H2, kh cn li l SO2. - Dn ln lt 2 kh nhm 2 qua dd nc vi trong. Kh no lm c nc vi trong l CO2, cn li l C3H8. Vy c nhiu cch gii bi ny nhng cch 2 l ti u hn c. v Bi tp tng t :

1) Ch dng 1 thuc th nhn bit 3 cht lng: benzen, toluen, styren 2) Pentan, penten-1, pentin-1, dd AgNO3, nc, dd NH4OH, nc Br, dd HCl, dd HI (ch s dng qu tm) 3) Ch dng 1 ha cht nhn bit : n-butan, buten-2, butadien-1,3 , vinylacetylen. 4) Nhn bit : n-hexan, hexen-2, hexen-1, n-heptan, toluen, styren v benzen 5*) Nhn bit cc l mt nhn sau : a) Kh etan, etylen, acetylen (bng 2 cch) b) Kh metan, etylen, SO2, NO2 v CO2.

5 BI TP VIT PHNG TRNH PHN NG GIA CC CHTNhng ch khi lm loi bi tp ny : - Phi nm vng cc phn ng ha hc ca cc hydrocacbon. - Nh cc im c bit trong cc phn ng, v d : Ankan : - Phn ng th : t C3 tr ln nu th vi Cl2 (askt, 1:1) s thu c hn hp sn phm l ng phn ca nhau. - Phn ng cracking : ch c ankan t C3 tr ln. - Phn ng hidro ha i khi cng c gi l phn ng cracking nhng xc tc l Ni,to - Lu : phn ng cng H2 v H2 u c xc tc l Ni,to . Xicloankan : - Vng C3, C4 ch c phn ng cng m vng khng c phn ng th. Vng C5 tr ln khng c phn ng cng ch c phn ng th. Aken, ankadien, ankin : - Phn ng cng : nu tc nhn bt i cng vi anken bt i th sn phm chnh c xc nh theo quy tc Macopnhicop. Ch n s sn phm. - i vi ankin th cn ch n xc tc bit 1 hoc 2 lin kt s b t. - Phn ng trng hp : cn ch cc phn ng trng hp 1,4 thng to thnh cao su. Aren : - Cn ch n quy tc th vo vng benzen. v Bi tp p dng : Bi 1: a) Vit phng trnh phn ng khi cho propen, propin, divinyl tc dng vi Br2 theo t l mol 1: 1. b) Hi khi cho 3 cht trn tc dng vi HCl (c xt) theo t l 1: 1 th thu c nhng sn phm g? Gi tn chng c) Hy cho bit CTCT v tn gi ca sn phm khi cho isopren v pentadien-1,4 tc dng vi dung dch Br2, HCl theo t l mol 1: 1. Vit CTCT ca polime thu c khi trng hp 2 ankadien cho trn GII : a) Phn ng cng gia hydrocacbon khng no vi tc nhn i xng th tng i n gin. Ty vo t l s mol m 1 hoc 2 kin kt s b t. Ptp : xem phn tm tt ha tnh trn b) Tc dng vi HCl (1:1) p dng quy tc Maccopnhicop * Propen cng HCl cho 2 sn phm * Propin cng HCl to 2 sn phm * Divinyl th c 2 hng cng - Cng 1,2 (hay 3,4) th to 2 sn phm - Cng 1,4 to 1 sn phm duy nht c) CTCT v tn gi ca sn phm khi cho * Isopren tc dng vi HCl (1:1) - Cng 1,2 to 2 sn phm - Cng 3,4 to 2 sn phm - Cng 1,4 to 2 sn phm * Pentadien-1,4 tc dng vi HCl (1:1) - Cng 1,2 hay 3, 4 : tng t nh Divinyl

- Khng c phn ng cng 1,4 do hai lin kt p khng lin hp. - CTCT cc polime thu c khi trng hp 2 ankadien trn : - Pentadien-1,4 khng c sn phm trng hp 1,4 do khng c 2 lin kt lin hp. Bi 2 : a) Pht biu quy tc th vng benzen. b) T benzen vit phng trnh phn ng iu ch ortho-bromnitrobenzen v meta-Bromnitrobenzen (ghi r iu kin phn ng). GII : a) Quy tc th vng benzen : - Khi vng benzen c nhm th y electron(gc ankyl hoc OH, NH2, Cl, Br) phn ng th xy ra d hn so vi benzen v u tin th vo v tr ortho hoc para. - Khi vng benzen c nhm th rt electron (nhm th c lin kt nh NO2, - COOH, -CHO, -SO3H,) phn ng th kh hn (so vi benzen) v u tin th vo v tr meta. b) Cc phng trnh phn ng : * iu ch ortho bromnitrobenzen : * iu ch meta bromnitrobenzen : v Bi tp tng t : 1) Vit phng trnh phn ng ca butin-1, butadien-1,3 vi H2, Br2, HCl, H2O. Gi tn sn phm. 2) Khi trng hp butadien-1,3 vi xc tc Na ta thu c cao su Buna c ln 2 sn phm ph A v B. A l mt cht do khng c tnh n hi, mi mt xch c mt mch nhnh l nhm vinyl. B l hp cht vng c tn l 1-vinyl xiclohexan-3 c phn t bng 108. Vit cc phng trnh phn ng xy ra di dng CTCT. 3) Phn ng cracking l g? Vit cc phng trnh phn ng dng tng qut khi cracking mt ankan. - Khi cracking butan thu c mt hn hp gm 7 cht, trong c H2 v C4H8. Hi CTCT ca butan l n hay iso? Vit cc phng trnh phn ng xy ra? 4) Olefin l g? Vi CTPT CnH2n c th c cc cht thuc dy ng dng no? Nu tnh cht ha hc c bn ca n? Vit phng trnh phn ng khi cho propylen tc dng vi O2; dd Br2; HCl; dd KMnO4; phn ng trng hp. Hp cht C6H12 khi cng hp HBr ch thu c mt sn phm duy nht, nh CTCT c th c ca olefin ny v vit phng trnh phn ng. 5) Vit phng trnh phn ng (nu c) ca cc hp cht sau vi dung dch AgNO3/NH3 a) Axetylen b) Butin-1 c) Butin-2 6) Vit phng trnh phn ng (nu c) gia cc cht sau vi Brom, ghi r iu kin: dd, to, kh (nu c): a) Isopren (1:1) b) Toluen c) Benzen d) Styren 7) Vit phng trnh phn ng (nu c) gia cc cht sau: a) Toluen + dd KMnO4 b) Propylen + AgNO3/NH3 d c) Styren + dd KMnO4 + Ba(OH)2 d) Axetylen + dd KMnO4_+ H2SO4 e) Propin +dd KMnO4_+ H2SO4

6 BI TP SO SNH GII THCH CU TO, TNH CHT HA HC CA CC HYDROCACBONv Nguyn tc : Da vo s so snh v c im cu to cc cht ri suy ra tnh cht ha hc ca cc cht . v Bi tp v d : Bi 1 : So snh v mt CT v ha tnh ca cc hp cht sau, vit phng trnh phn ng minh ha. a) Etan, etylen, axetylen b) hexan, hexen, benzen c) butin-1, butin-2 v butadien-1,3 GII : a) Etan, Etilen, Axetilen : * Ging nhau : - Thnh phn cu to ch gm C v H * Khc nhau : Phn t C2H6 C2H4 C2H2 Cu to Trong phn t ch tn ti cc lin kt n () bn gia C v C, gia C v H Trong phn t c mt lin kt i gm mt kin kt () bn v mt lin kt () linh ng km bn. Trong phn t c mt lin kt ba gm mt lin kt () bn v hai lin kt () linh ng km bn. c im lin kt Lin kt n () rt bn vng rt kh b t khi tham gia phn ng ha hc Lin kt () linh ng km bn rt d b t khi tham gia phn ng ha hc. Tnh cht ha hc Tnh cht ha hc c trng l phn ng th, kh b oxi ha. Ngoi ra cn c phn ng hydro ha nhit thch hp v xc tc thch hp. Tnh cht ha hc c trng l phn ng cng. Ring vi axetilen th khi tham gia phn ng ha hc ty iu kin xc tc m mt hay c hai lin kt () s b t. Ngoi ra cn c phn ng trng hp, oxiha. Phng trnh phn ng Xem I.2.4/14 Xem I.2.4/15 Ring axetilen c hai nguyn t H linh ng nn n cn c kh nng tham gia phn ng th vi ion kim loi. iu ny c gii thch nh sau : do lin kt ba rt ngn nn hai nhn C rt gn nhau, in tch tp trung nhiu v 2 C ny nn cc H gn trc tip vi C ca ni ba tr nn rt linh ng. b) n-hexan, n-hexen, benzen. * Ging nhau :

8) Mun iu ch n-pentan, ta c th hidro ha nhng anken no? Vit CTCT ca chng. 9) Vit phng trnh phn ng iu ch cc hp cht sau y t nhng anken thch hp : a) CH3CHBr CHBrCH3 b) CH3CHBr CBr(CH3)2 c) CH3CHBr CH(CH3)2

- Thnh phn cu to gm C v H * Khc nhau : - n-hexan v n- hexen so snh cu to v tnh cht ha hc tng t cu trn. Ring n-hexan cn c phn ng b gy mch C khi c xc tc nhit cao. Ankan ankan + anken CnH2n+2 CmH2m + 2 + CxH2x m 1, x 2, n = m + x. - Benzen : c im cu to : trong phn t c mt vng kn v 3 lin kt => benzen c phn ng c trng l phn ng cng. nhng 3 lin kt ny li lin hp vi nhau to thnh mt h thm bn vng lm cho kh nng t lin kt tham gia phn ng ha hc b hn ch => benzen kh tham gia phn ng cng, ch c cng vi H2, d tham gia phn ng th v bn vi tc nhn oxiha. Phng trnh phn ng : xem phn ha tnh (I.2.4/14). c) So snh c im cu to ca butin-1, butin-2, divinyl Tng t cu a. Nhng kh nng tham gia phn ng cng ca lin kt i (Divinyl) hi d hn so vi lin kt ba v : lin kt 3 ngn, hai nhn C gn nhau nn lin kt 3 hi bn hn so vi lin kt i. Bi 2 : C7H8 l ng ng ca benzen. Khi cho C6H6 v C7H8 tc dng vi Brom khan (c bt Fe lm xc tc) th phn ng no xy ra d hn? Gii thch (vit phng trnh phn ng theo t l 1:1 v s mol) GII : C7H8 tham gia phn ng th nhn d hn so vi benzen v nhm CH3 y electron v nhn lm nhn giu electron hn. * C6H6 cho 1 sn phm : * C6H5CH3 cho hn hp hai sn phm. Bi 3 : So snh phn ng trng hp v phn ng cng. GII : * Ging nhau : u l phn ng cng hp cc phn t nh thnh mt phn t mi. * Khc nhau : Phn ng cng : Ch n thun cng 2 phn t nh (monome) thnh mt phn t mi cng l monome. Ch cn mt trong hai monome ban u c t nht mt lin kt trong phn t. Phn ng trng hp : khng ch cng hai m cng nhiu phn t ging nhau hoc tng t nhau thnh mt phn t mi c khi lng v kch thc rt ln gi l nhng polime. Cc monome tham gia phn ng trng hp nht thit phi c t nht mt lin kt trong phn t. Bi 4 : So snh di lin kt dC-C trong ankan (C C), anken (C=C), ankin (CC). Gii thch kh nng tham gia phn ng ca ankan km anken nhng ankin km anken.? Mc d phn t ankin c nhiu lin kt hn anken? GII : Thc nghim cho bit dC-C trong etan (C-C) l : 1,54Ao Etilen(C=C) : 1,34Ao Axetilen (CC):1,2 Ao * C th gii thch nh sau : Khi hnh thnh lin kt C-C trong phn t ankan th 2C xy ra s xen ph trc lin nhn lm cho khong cch 2 nhn xa nhau nn dC-C ln.

Khi hnh thnh lin kt C=C trong phn t anken th lin kt c hnh thnh nh cch trn, cn lin kt c hnh thnh do s xen ph bn lm cho khong cch gia 2 nhn C gn nhau hn. Tng t vi ankin c 2 lin kt nn xy ra 2 s xen ph bn lm cho khong cch gia hai nhn cng gn nhau hn. Do dC-C C C > C=C > CC. * Gii thch v kh nng tham gia phn ng : - S xen ph trc xy ra vi mt ln lm cho lin kt bn vng. - S xen ph bn xy ra vi mt nh nn lin kt km bn vng d b t khi c tc nhn tn cng => kh nng tham gia phn ng ca ankan< anken, ankin. - y do lin kt 3 lm cho khong cch 2 nhn C rt gn nhau nn lin kt 3 hi bn hn lin kt i nn kh nng tham gia phn ng ca ankin hi km hn anken. - V cng do khong cch gia hai nhn C b m mt in tch tp trung hu ht nhn nn cc ankin-1 c H linh ng tham gia c phn ng th vi ion kim loi v Bi tp tng t : 1) Gii thch quy tc cng Maccopnhicop? Minh ha bng v d c th. 2) Gii thch ti sao di lin kt n C-C trong butadien-1,3 ch bng 1,46Ao ngn hn lin kt n C-C bnh thng? 3) Ti sao khi nhit phn mui axetat vi xt iu ch ankan tng ng li phi dng xc tc CaO,to? 4) So snh nhit si ca cc hydrocacbon a) Khi khi lng phn t tng dn? b) C cng CTPT nhng khc nhau dng khung Cacbon? 5) Khi thc hin phn ng phn hy ankan bi nhit li c tin hnh nhit trn 1000oC ti sao li nhn mnh trong iu kin khng c khng kh? 6) So snh kh nng tham gia phn ng th ca cc halogen Flo, Clo, Brom, Iod vi cc ankan? 7) Ti sao cao su khi chy li c nhiu khi en? Lm th no khi en t li? 8) Trong phn ng iu ch axetilen t metan c tin hnh nhit 1500oC cn ghi km iu kin lm lnh nhanh? 9) So snh cao su thng v cao su lu ha v thnh phn, bn, ng dng? 10) Gii thch v sao cao su tng hp c tnh n hi km cao su thin nhin? 11) Phn bit cc khi nim: a) CTN, CTG, CTPT v CTCT b) Lin kt , p . Ly propen lm v d c) ng ng, ng phn l g? Nu cc loi ng phn, cho v d? d) C th coi nguyn t Br trong phn t CnH2n+1Br l mt nhm chc c khng? Ti sao?

7 BI TON LP CTPT HYDROCACBON

7.1 CC PHNG PHP LP CNG THC PHN T CA HYDROCACNON 7.1.1 Phng php khi lng hay % khi lng. 1) Phng php gii : Bc 1 : Tm MA : ty theo gi thit bi cho m s dng cc cch tnh sau tm MA Tm MA da trn cc khi nim c bn, cc nh lut c bn. C nhiu cch tm khi lng phn t, ty tng gi thit bi cho m dng cch tnh thch hp. 1. Da vo khi lng ring DA (ktc) =>MA = 22,4 . DA vi DA n v g/l 2. Da vo t khi hi ca cht hu c A MA = MB . dA/B

MA = 29 . dA/KK 3. Da vo khi lng (mA ) ca mt th tch VA kh A ktc MA = (22,4 . mA)/ VA mA: khi lng kh A chim th tch VA ktc 4. Da vo biu thc phng trnh Mendeleep Claperon: Cho mA (g) cht hu c A ha hi chim th tch VA (l) nhit T (oK) v p sut P(atm) PV = nRT => (R = 0,082 atm/ oKmol) 5. Da vo nh lut Avogadro: nh lut: cng iu kin nhit v p sut, mi th tch kh bng nhau u cha cng mt s phn t kh. VA = VB => nA = nB => => MA = mA Bc 2 : t CTPT cht A: CxHy Xc nh thnh phn cc nguyn t trong hydrocacbon. Cch 1 :Dng khi bi -Khng cho khi lng hydrocacbon em t chy -Tnh c mC, mH t mCO2, mH2O * Tnh khi lng cc nguyn t c trong A v mA (g) cht A. - Xc nh C: - Xc nh H - Xc nh mA => mA = mH + mA * Xc nh CTPT cht hu c A: CxHy Da trn CTTQ cht hu c A: CxHy Cch 2 : Khi bi cho bit thnh phn % cc nguyn t trong hn hp * Dng cng thc sau: ; => CTPT A. Cch 3 : * Tm CTG nht => CTN => CTPT A hoc - CTG nht : CaHb => CTTN : (CaHb)n - Xc nh n: bin lun t CTTN suy ra CTPT ng ca A : y < 2x + 2; y chn, nguyn dng ; x 1, nguyn dng. => T xc nh c CTPT ng ca cht hu c A. Lu : Khi bi tan yu cu xc nh CTG nht ca cht hu c A (hay CTN ca A) hoc khi khng cho d kin tm MA th ta nn lm theo cch trn. 2) Cc v d : V d 1 : Mt hydrocacbon A c thnh phn nguyn t: % C = 84,21; %H = 15,79; T khi hi i vi khng kh bng dA/KK = 3,93. Xc nh CTPT ca A GII Bc 1: Tnh MA:

Bit dA/KK => MA = MKK. dA/KK = 29.3,93 = 114 Bc 2 : t A : CxHy Suy ra CTPT A: C8H18 V d 2 : Mt hydrocacbon A th kh c th tch gp 4 ln th tch ca lu hunh ioxit c khi lng tng ng trong cng iu kin. Sn phm chy ca A dn qua bnh ng nc vi trong d th c 1g kt ta ng thi khi lng bnh tng 0,8g. Tm CTPT A. GII * Tm MA : 1VA = 4VSO2( cng iu kin ) =>nA = 4nSO2 => (A v SO2 c khi lng tng ng nhau) => Cch 1 : gii theo phng php khi lng hay % khi lng : t A : CxHy Bnh ng Ca(OH)2 hp th CO2 v H2O m(kt ta) = mCaCO3 = 1g nCO2 = nCaCO3 = 1/100= 0,01mol =>nC = nCO2 = 0,01mol =>mC = 12.0,01=0,12g mCO2 = 0,01.44 = 0,44g rmbnh = mCO2 + mH2O =>mH2O = 0,8-0,44 = 0,36g LBT khi lng (A) :mA = mC + mH = 0,12 +0,04 = 0,16 Ta c Vy CTPT A : CH4 Cch 2 : Bin lun da vo iu kin y < 2x + 2; y chn, nguyn dng ; x 1, nguyn => x =1 v y = 4 CTPT A. V d 3: t chy hon ton 2,64g mt hydrocacbon A thu c 4,032 lt CO2 (ktc). Tm CTPT A? GII * Tm thnh phn cc nguyn t : mC (trong A) = mC (trong CO2) = (4,032/ 22,4)*12 = 2,16g mH = mA mC = 2,64 2,16 = 0,48g => CTN : C3H8 => CTTN : (C3H8)n Bin lun : S H < 2 s C +2 => 8n < 6n + 2 => n < 1 m n nguyn dng =>n = 1 CTPT A : C3H8 7.1.2) Phng php da vo phn ng chy: Du hiu nhn bit bi ton dng ny : bi t chy mt cht hu c c cp n khi lng cht em t hoc khi lng cc cht sn phm (CO2, H2O) mt cch trc tip hoc gin tip (tc tm c khi lng CO2, H2O sau mt s phn ng trung gian).

1) Phng php gii: Bc 1 : Tnh MA ( phn II.2.1.1) Bc 2 : t A : CxHy * Vit phng trnh phn ng chy. MA(g) 44x 9y mA(g) mCO2 mH2O * Lp t l tnh x,y hoc * T suy ra CTPT A Mt s lu : 1) Nu bi cho: oxi ha han tan mt cht hu c A th c ngha l t chy han tan cht hu c A thnh CO2 v H2O 2) Oxi ha cht hu c A bng CuO th khi lng oxy tham gia phn ng ng bng gim khi lng a(g)ca bnh ng CuO sau phn ng oxi ha. Thng thng trong bi ton cho lng oxi tham gia phn ng chy, tm khi lng cht hu c A nn ch n nh lut bo ton khi lng mA + a = mCO2 + mH2O 3) Sn phm chy (CO2, H2O) thng c cho qua cc bnh cc cht hp th chng. 4) Bnh ng CaCl2 (khan), CuSO4 (khan), H2SO4 c, P2O5, dung dch kim, hp th nc. Bnh ng cc dung dch kimhp th CO2. Bnh ng P trng hp th O2. 5) tng khi lng cc bnh chnh l khi lng cc cht m bnh hp th. 6) Nu bi ton cho CO2 phn ng vi dung dch kim th nn ch n mui to thnh xc nh chnh xc lng CO2. 7) Vit phng trnh phn ng chy ca hp cht hu c vi oxy nn oxy li cn bng sau t v sau n v trc. Cc nguyn t cn li nn cn bng trc, t v trc ra v sau phng trnh phn ng. 2) Bi tp v d : V d 1 : t hon ton 0,58g mt hydrocacbon A c 1,76g CO2 v 0,9g H2O. Bit A c khi lng ring DA @ 2,59g/l. Tm CTPT A Tm tt : 0,58g X + O2 => (1,76g CO2; 0,9 g H2O) DA @ 2,59g/l. Tm CTPT A? GII : * Tm MA : Bit DA => MA = 22,4.2,59 @ 58 * Vit phng trnh phn ng chy, lp t l tm x,y MA(g) 44x 9y mA(g) mCO2 mH2O = => x = 4 y =10 Vy CTPT A : C4H10 V d 2 : Khi t chy han tan 0,42 g mt Hydrocacbon X thu tan b sn phm qua bnh 1 ng H2SO4 c, bnh 2 ng KOH d. Kt qu, bnh 1 tng 0,54 g; bnh 2 tng 1,32 g. Bit rng khi ha hi 0,42 g X chim th tch bng th tch ca 1,192 g O2 cng iu kin. Tm CTPT ca X

Tm tt : Tm CTPT X? GII * Tnh MX : 0,42g X c VX = VO2 ca 0,192g O2 (cng iu kin) => nX = nO2 => => * Gi X : CxHy MX 44x 9y (g) 0,42 mCO2 mH2O (g) Ta c : (1) bi cho khi lng CO2, H2O gin tip qua cc phn ng trung gian ta phi tm khi lng CO2, H2O * Tm mCO2, mH2O : - Bnh 1 ng dd H2SO4 s hp th H2O do tng khi lng bnh 1 chnh l khi lng ca H2O : rm1 = mH2O=0,54g (2) - Bnh 2 ng dd KOH d s hp th CO2 do tng khi lng bnh 2 chnh l khi lng ca CO2 : rm2 = mCO2 =1,32g (3) (1), (2), (3) => => x = 5 y = 10 Vy CTPT X : C5H10 (M = 70vC) 7.1.3 Phng php th tch (phng php kh nhin k): v Phm vi ng dng : Dng xc nh CTPT ca cc cht hu c th kh hay th lng d bay hi. v C s khoa hc ca phng php : Trong mt phng trnh phn ng c cc cht kh tham gia v to thnh ( cng iu kin nhit , p sut) h s t trc cng thc ca cc cht khng nhng cho bit t l s mol m cn cho bit t l th tch ca chng. 1) Phng php gii Bc 1 : Tnh th tch cc kh VA, VO2, VCO2, VH2O (hi) Bc 2 : Vit v cn bng cc phng trnh phn ng chy ca hydrocacbon A di dng CTTQ CxHy Bc 3 : Lp cc t l th tch tnh x,y 1(l) (l) x(l) (l) VA(l) VO2 (l) VCO2 (l) VH2O (hi)(l) hay Cch khc : Sau khi thc hin bc 1 c th lm theo cch khc: - Lp t l th tch VA: VB : VCO2 : VH2O ri a v t l s nguyn ti gin m:n:p:q. - Vit phng trnh phn ng chy ca hp cht hu c A di dng: mCxHy + nO2pCO2 + qH2O - Dng nh lut bo ton nguyn t cn bng phng trnh phn ng chy s tm c x v y =>CTPT A

* Mt s lu : - Nu VCO2 : VH2O = 1:1 => C : H = nC : nH = 1: 2 - Nu tan cho oxy ban u d th sau khi bt tia la in v lm lnh (ngng t hi nc) th trong kh nhin k c CO2 v O2 cn d. Bi tan l lun theo CxHy - Nu tan cho VCxHy = VO2 th sau khi bt tia la in v lm lnh th trong kh nhin k c CO2 v CxHy d. Bi tan l lun theo oxy. - Khi t chy hay oxi ha han ton mt hydrocacbon m gi thit khng xc nh r sn phm, th cc nguyn t trong hydrocacbon s chuyn thnh oxit bn tng ng tr: N2 => kh N2 Halogen => kh X2 hay HX (ty bi) 2. Bi tp v d V d 1: Trn 0,5 l hn hp C gm hydrocacbon A v CO2 vi 2,5 l O2 ri cho vo kh nhin k t chy th thu c 3,4 l kh, lm lnh ch cn 1,8 l. Cho hn hp qua tip dung dch KOH (c) ch cn 0,5 l kh. Cc V kh o cng iu kin. Tm CTPT ca hydrocacbon A. Tm tt : CxHy : a (l) Gi 0,5 l hn hp CO2 : b (l) 0,5l hn hp + 2,5l O2 CO2 ,O2 d,H2O CO2,O2d O2 d GII : * O2 d , bi tan l lun theo Hydrocacbon A a a ax a (lt) CO2 => CO2 b b (lt) Ta c Vhh = a + b = 0,5 (1) VCO2 = ax + b = 1,8 0,5 = 1,3 (2) VH2O = a = 3,4 1,8 = 1,6 (3) VO2 d = 2,5 - a = 0,5 => ax + a = 2 (4) => ax + 3,2/4 = 2 => ax = 1,2 (5) (2), (3) VCO2 = b = 0,1 Vhh = a + b = 0,5 => a = 0,4 => x = ax /a = 3 => y = ay/a = 8 Vy CTPT ca A l C3H8 V d 2 : Trn 12 cm3 mt hydrocacbon A th kh vi 60 cm3 oxi (ly d) ri t chy. Sau khi lm lnh nc ngng t ri a v iu kin ban u th th tch kh cn li l 48 cm3, trong c 24cm3 b hp th bi KOH, phn cn li b hp th bi P. Tm CTPT ca A (cc th tch kh o trong cng iu kin nhit v p sut) Tm tt : GII : * Tnh cc V: VCO2 = 24cm3 VO2 d = 48 24 = 24cm3 => VO2 p = 60 24 = 36 cm3

* Tm CTPT : Cch 1: Tnh trc tip t phng trnh phn ng t chy: 12 => 12 => 12x (cm3) VCO2 =12x = 24 => x = 2 VO2 d = 60 12 = 24 => y = 4 => CTPT ca A: C2H4 Cch 2: Lp t l th tch 1 x (cm3) 12 36 24 (cm3) => x = 2 v y = 4 => CTPT ca A: C2H4 Cch 3: Nhn xt: t 12 cm3 A dng 36 cm3 oxy v to ra 24 cm3 CO2 Suy ra LBT (O): => LBT (C): 12x = 24 => x = 2 LBT (H) :12y = 48 => y = 4 Vy CTPT ca A l C2H4 V d 3 : Trong mt bnh kn th tch 1dm3 c mt hn hp ng th tch gm hydrocacbon A v O2 133,5 oC, 1 atm. Sau khi bt tia la in v a v nhit ban u (133,5 oC) th p sut trong bnh tng ln 10% so vi ban u v khi lng nc to ra l 0,216 g. Tm CTPT A Tm tt : GII : Tm CTPT A? V hn hp ng th tch nn nA = nO2 = 0,03/2 = 0,015 mol => CxHy d, bin lun theo O2 Sau khi a v nhit ban u, cc kh to p sut c trong bnh gm H2O, CO2, CxHy d c s mol l : n2 = n1 . P2/P1 = 0,03.110/100 = 0,033 mol nH2O = 0,216/18 = 0,012 mol LBT khi lng (O) : nO2 = n CO2 + 1/2n H2O => n CO2 = nO2 1/2nH2O = 0,015-0,012/2 = 0,009mol nCxHyd = n2 - nCO2 - nH2O = 0,033-0,012-0,009 =0,012mol =>nCxHyphn ng = 0,015-0,012 = 0,003 mol 1 x (mol) 0,003 0,015 0,009 0,012 (mol) Ta c : => x = 3 y=8 Vy CTPT A : C3H8 7.1.4 Phng php gi tr trung bnh (xc nh CTPT ca hai hay nhiu cht hu c trong hn hp): L phng php chuyn hn hp nhiu gi tr v mt gi tr tng ng, nhiu cht v mt cht tng ng

c im Phng php gi tr trung bnh c dng nhiu trong ha hu c khi gii bi tan v cc cht cng dy ng ng. Mt phn bn cht ca gi tr trung bnh c cp n vic tnh phn trm n v v khi lng hn hp kh trong bi tan t khi hi chng u lp 10. Do , hc sinh d dng lnh hi phng php ny xc nh CTPT ca hai hay nhiu cht hu c trong hn hp. Phng php khi lng phn t trung bnh ca hn hp () Cht tng ng c khi lng mol phn t l khi lng mol phn t trung bnh ca hn hp. Cc bc gii : Bc c bn : Xc nh CTTQ ca hai cht hu c A,B Bc 1 : Xc nh CTTB ca hai cht hu c A, B trong hn hp Bc 2 : Tm qua cc cng thc sau : Hoc Gi s MA< MB => MA CTPT ng ca A v B Phm vi ng dng: s dng c li nhiu i vi hn hp cc cht cng dy ng ng 1) Phng php CTPT trung bnh ca hn hp: v Phm vi p dng : Khi c hn hp gm nhiu cht, cng tc dng vi mt cht khc m phng trnh phn ng tng t nhau (sn phm, t l mol gia nguyn liu v sn phm, hiu sut, phn ng tng t nhau), c th thay th hn hp bng mt cht tng ng, c s mol bng tng s mol ca hn hp. Cng thc ca cht tng ng gi l CTPT trung bnh. v Phng php gii : Bc 1 : t CTPT ca hai cht hu c cn tm ri suy ra CTPT trung bnh ca chng : t A : CxHy ; B : CxHy => CTPTTB : Bc 2 : Vit phng trnh phn ng tng qut v d liu bi cho tnh Bc 3 : bin lun Nu x x < < x y y< < y Da vo iu kin x, x, y, y tha mn bin lun suy ra gi tr hp l ca chng => CTPT A, B. v Phm vi ng dng : Phng php gii ny ngn gn i vi cc bi tan hu c thuc loi hn hp cc ng ng nht l cc ng ng lin tip. Tuy nhin c th dng phng php ny gii cc bi ton hn hp cc cht hu c khng ng ng cng rt hiu qu. Ngoi phng php trn cn c phng php s C, s H, s lin kt p trung bnh (). Phng php gii tng t nh hai phng php trn v Mt s lu : 1) Nu bi cho 2 cht hu c A, B l ng dng lin tip th : m = n + 1 ( y n, m l s C trong phn t A, B) 2) Nu bi cho 2 cht hu c A, B hn km nhau k nguyn t C th m = n + k. 3) Nu bi cho 2 cht hu c A, B cch nhau k nguyn t C th : m = n + (k +1) 4) Nu bi cho anken, ankin th n, m 2. 5) Nu bi ton cho A, B l hydrocacbon th kh trong iu kin thng (hay iu kin tiu chun) th n, m < 4 v Bi tp v d : Bi 1: t chy han tan 19,2 g hn hp 2 ankan lin tip thu c 14,56 l CO2 ( OoC, 2 atm). Tm CTPT 2 ankan. GII : Gi CTPT trung bnh ca hai ankan : v

=> . Cch 1: phng php s C trung bnh () S mol hn hp S mol CO2 : nCO2 = . = 1,3 Hn hp gm 2 ankan lin tip CnH2n+2 CmH2m+2 ; n = mhh / nhh = 19,2/0,5 = 38,4 MA < 38,4 < MB = MA + 14 A CH4 C2H6 C3H8 C4H10 MA 16 30 44 58 38,4 38,4 38,4 38,4 MB 30 44 58 72 Vy A : C2H6 B : C3H8 7.1.5 - Phng php bin lun 1. Da vo gii hn xc nh CTPT ca mt hydrocacbon: - Khi s phng trnh i s thit lp c t hn s n cn tm, c th bin lun da vo gii hn : A : CxHy th : y < 2x + 2; y chn, nguyn dng ; x 1, nguyn. - Nu khng bin lun c hay bin lun kh khn c th dng bng tr s tm kt qu. - iu kin bin lun ch yu ca loi ton ny l : ha tr cc nguyn t. Phng php bin lun

trnh by trn ch c th p dng xc nh CTPT ca mt cht hoc nu nm trong 1 hn hp th phi bit CTPT ca cht kia. 2. Bin lun theo phng php ghp n s xc nh CTPT ca mt hydrocacbon : a) Cc bc c bn : Bc 1 : t s mol cc cht trong hn hp l n s. Bc 2 : ng vi mi d kin ca bi ton ta lp mt phng trnh ton hc. Bc 3 : Sau ghp cc n s li rt ra h phng trnh ton hc. Chng hn : a + b = P (vi a, b l s mol 2 cht thnh phn) an + bm = Q (vi n, m l s C ca 2 hydrocacbon thnh phn) Bc 4 : c th xc nh m, n ri suy ra CTPT cc cht hu c thnh phn, c th p dng tnh cht bt ng thc : Gi s : n < m th n(x + y) < nx + my < m(x + y) Hoc t mi lin h n,m lp bng tr s bin lun - Nu A, B thuc hai dy ng ng khc nhau ta phi tm x, y ri th vo phng trnh nx + my = Q xc nh m, n => CTPT. 3. Mt s phng php bin lun xc nh dy ng ng v CTPT hydrocacbon : v Cch 1 : Da vo phn ng chy ca hydrocacbon, so snh s mol CO2 v s mol H2O. Nu t 1 hydrocacbon (A) m tm c : * nH2O > nCO2 (A) thuc dy ng ng ankan ptp : * nH2O = nCO2 => (A) thuc dy ng ng anken hay olefin hoc (A) l xicloankan ptp : * nH2O < nCO2 => (A) thuc dy ng ng ankadien, ankin hoc benzen ptp : ( ng ng ankin hoc ankadien) ( ng ng benzen) v Cch 2 : Da vo CTTQ ca hydrocacbon A : * Bc 1 : t CTTQ ca hydrocacbon l : CnH2n+2-2k ( y k l s lin kt p hoc dng mch vng hoc c 2 trong CTCT A) iu kin k 0, nguyn. Nu xc nh c k th xc nh c dy ng ng ca A. - k = 0 => A thuc dy ng ng ankan - k = 1 => A thuc dy ng ng anken - k = 2 => A thuc dy ng ng ankin hay ankadien - k = 4 => A thuc dy ng ng benzen. chng minh hai ankan A, B thuc cng dy ng ng, ta t A : CnH2n+2-2k ; B : CmH2m+2-2k. Nu tm c k = k th A,B cng dy ng ng. * Bc 2 : Sau khi bit c A,B thuc cng dy ng ng, ta t CTTQ ca A l CxHy. V B l ng ng ca A, B hn A n nhm CH2- th CTTQ ca B :CxHy (CH2)n hay Cx+nHy+2n. * Bc 3 : Da vo phng trnh phn ng chy ca A, B, da vo lng CO2, H2O, O2 hoc s mol hn hp thit lp h phng trnh ton hc, ri gii suy ra x, y, n Xc nh c CTPT A, B. v Cch 3 : da vo khi nim dy ng ng rt ra nhn xt : Cc cht ng ng k tip nhau c khi lng phn t lp thnh mt cp s cng cng sai d = 14. C mt dy n s hng M1, M2, ,Mn lp thnh mt cp s cng cng sai d th ta c : + S hng cui Mn = M1 + (n-1)d

+ Tng s hng S = .n + Tm M1, , Mn suy ra cc cht Trong mt bi ton thng phi kt hp nhiu phng php. V d : t chy mt hn hp gm 2 hydrocacbon A, B (c M hn km nhau 28g) th thu c 0,3mol CO2 v 0,5 mol H2O. Tm CTPT & tn A, B GII : Hydrocacbon A, B c M hn km nhau 28g => A, B thuc cng dy ng ng. Cch 1 : A, B + O2 => CO2 + H2O >1 => A, B thuc dy ng ng ankan. t CTTB A, B : : a mol a => a => a(+1) (mol) Ta c => = 1,5 t CTTQ A, B : CnH2n+2 v CmH2m+2 Gi s n< m => n< 1,5 => n = 1 => CTPT A : CH4 (M = 16) => MB = 16 + 28 = 44 => CTPT B : C3H8. Cch 2 : t CTTQ A, B : CnH2n+2 : a mol v CmH2m+2 : b mol Cc ptp chy : a an a(n+1-k) (mol) b bm b(m+1-k) (mol) Ta c : => (a+b)(1-k) = 0,2 => k = 0 v ch c k = 0 th phng trnh mi c ngha. => a + b = 0,2 v an + bm = 0,3 Gi s n < m => n(a+b) < m (a+b) => n n < < m Bin lun tng t cch trn suy ra CTPT A : CH4 v B : C3H8. 7.2 PHNG PHP GII MT S BI TON LP CTPT HYDROCACBON Bi 1 : Cracking ankan A, ngi ta thu c mt hn hp kh B gm 2 ankan v 2 anken. T khi hi ca B so vi H2 dB/H2 = 14,5. Khi dn hn hp kh B qua dung dch Br2 d, khi lng hn hp kh gim i 55,52%. a) Tm CTPT ca A v cc cht trong B. b) Tnh % th tch cc cht kh trong B. GII bi ny da vo tnh cht phn ng cracking v p dng nh lut bo ton khi lng tm MA kt hp vi phng php ghp n s gii. =14,5.2 = 29 Theo LBT khi lng : khi lng A em cracking = khi lng hn hp B => mAtham gia p = mB (1) Phn ng cracking lm tng gp i s mol hydrocacbon nn nB = 2nA tham gia p (2) (1) chia (2) => = MA => MA = 29.2 = 58 MA = 14n + 2 = 58 => n= 4 v CTPT A l C4H10

Cc ptpu cracking A : C4H10 => CH4 + C3H6 a => a a (mol) C4H10 => C2H6 + C2H4 b => b b (mol) Gi A, B ln lt l s mol A b cracking theo 2 phn ng trn. hh B gm : CH4 : a (mol) C2H6 : b (mol) C3H6 : a (mol) C2H4 : b (mol) Khi dn hh qua dd Br2 th 2 anken b hp th. => m2anken = 55,52%mB = 55,52%mA => mC3H6 + mC2H4 = 55,52%.58 (a+b) 42a + 28b = 32,2016 (a+b) 9,7984a = 4,2016b b @ 2,3a (mol) nB = 2(a + b) = 2 (a + 2,3a) = 6,6a (mol) cng iu kin, t l s mol cng chnh l t l v th tch => %CH4 = %C3H6 = = 15% %C2H6 = %C2H4 = Bi 2 : Hydrocacbon (X) dn xut t aren. Ha hi (X) trn vi oxi va trong mt kh nhin k, t hon ton hn hp ri a v nhit ban u, p sut trong bnh bng 12 ln p xut ca (X) ban u. a v 0oC p sut kh gim cn 2/3. 5,2 gam (X) lm mt mu dung dch cha 8 gam Brm. Xc nh CTCT ca (X) GII Bi ny c 2 cch gii : v cng iu kin T, V khng i th t l s mol bng t l p sut. PV = nRT v Khi a v OoC, hi nc b ngng t. Kh cn li l CO2 P3 = 2/3P2 nCO2 = 2/3 (nCO2 + nH2O) Cch 1 : Dng phng php thng thng gii. v Gi s s mol ca X l 1mol 1 => x => y/2 (mol) n2 = x + y/2 => n2 = 12n1 hay x + y/2 = 12 (1) x = 2/3 (x + y/2) 3x = 2x + y x = y (2) (1), (2) => x + x/2 = 12 => x = 8 = y => CTPT X : C8H8 Cch 2 : Dng phng php bin lun gii Gi CxHy : n1 (mol) => n2 = 12n1 (1) Khi a v OoC : n3 = 2/3n2 = 2/3.12n1 = 8n1 = nCO2 => nH2O = 12n1 8n1 = 4n1 n1 => 8n1 => 4n1

x : y/2 = 8n1 : 4n1 => x= y. v CTN ca X (CH)n hay CnHn CnHn + kBr2 => CnHnBrk 13n => 160k 5,2 => 8 X l dn xut ca Benzen => n 6 => k < 8/6 = 1,33 => k = 1 v n = 8. v Vy CTPT A : C8H8 r = (8.2 + 2 8) = 5 A l dn xut ca benzen, A li lm mt mu dd Br2 => c ni i C=C nhnh. v CTCT A : A l Styren. Bi 3 : t chy 19,2 g hn hp 2 ankan k tip th thu c V lt CO2 (0oC, 2 atm). Cho V lt CO2 trn qua dd Ca(OH)2 th thu c 30g kt ta. Nu tip tc cho dd Ca(OH)2 vo n d th thu c thm 100g kt ta na. a) Xc nh CTPT 2 ankan. b) Tnh thnh phn % theo khi lng 2 hydrocacbon. GII bi ny, t chy hn hp gm 2 ankan lin tip nn dng phng php trung bnh gii. a) Xc nh CTPT 2 ankan : t CTTQ 2 ankan X : CnH2n+2 : a (mol) Y : CmH2m+2 : b (mol) CTPT trung bnh 2 ankan Gi s n < m => n< < m = n + 1. CO2 + Ca(OH)2 => CaCO3 + H2O 2CO2 + Ca(OH)2 => Ca(HCO3)2 Khi cho thm dd Ca(OH)2 vo n d : Ca(HCO3)2 + Ca(OH)2 => 2CaCO3 + 2H2O p dng LBT khi lng th mCO2 = mCO2(trong ) => nCO2 = nCaCO3 = (mol) => mCO2 = 1,3.44 = 57,2 (g) 44 19,2 57,2 Ta c t l : = 2,6 Ta c n < = 2,6 < m = n+1 => n = 2 v m =3 v Vy CTPT 2 ankan l C2H6 v C3H8 b) Tnh % cc hydrocacbon trn : C2H6 + 7/2O2 => 2CO2 + 3H2O a => 2a (mol) C3H8 + 5O2 => 3CO2 + 4H2O b => 3b (mol) nCO2 = 2a + 3b = 1,3 (1) mhh = 30a + 44b = 19,2 (2) (1) , (2) => a = 0,2 ;b = 0,3 (mol)

%C2H6 = % C3H8 = Bi 4 : Mt hn hp X gm 2 hydrocacbon lin tip nhau thuc cng mt dy ng ng v u th kh ktc. t chy X vi 6,4g O2 (ly d) v cho hn hp CO2, H2O, O2 d i qua bnh ng dung dch Ca(OH)2 d th c 100g kt ta v cn li 1 kh thot ra c V= 11,2l(0,4atm,OoC). a) Xc nh dy ng ng A,B b) Xc nh CTPT ca A, B Tm tt : nO2b = 64/32 = 2 mol GII bi ny, t chy 2 hydrocacbon lin tip thuc cng mt dy ng ng nn s dng phng php trung bnh gii. a) Xc nh dy ng ng ca A, B : nO2 d = PV/RT = 11,2.0,4/(0,082.273) = 0,2 mol nO2 pu = 2-0,2 = 1,8 (mol) kh CO2, H2O b hp th vo dd Ca(OH)2 d nCO2 = nCaCO3 = 100/100 = 1 mol p dng LBT nguyn t (O) cho ptpu chy : nO2 p = nCO2 + nH2O => nH2O = 2(nO2 p nCO2) = 2(1,8-1) = 1,6 mol. Ta thy nH2O > nCO2 => hai hydrocacbon trn thuc dy ng ng ankan. CTPT trung bnh 2 ankan l : x => (3+1)/2x => x => x (+1) (mol) nCO2 = x = 1 nH2O = x(+1) = 1,6 x = 0,6 = 1,67 1 < =1,67 < m= n + 1 => n= 1 v m = 2 => CTPT 2 ankan l CH4 v C2H6 Bi 5 : t chy 560cm3 hn hp kh (ktc) gm 2 hydrocacbon c cng s nguyn t cacbon ta thu c 4,4g CO2 v 1,9125g hi nc. a) Xc nh CTPT cc cht hu c. b) Tnh %khi lng cc cht. c) Nu cho lng CO2 trn vo 100 ml dd KOH 1,3M; Tnh CM mui to thnh. GII bi ny, ta dng phng php s nguyn t H trung bnh kt hp vi phng php bin lun gii. a) Xc nh CTPT cc hydrocacbon : t CTPT 2 hydrocacbon trn : CTPT trung bnh 2 hydrocacbon trn : Gi s y < y => y < < y S mol hn hp kh nhh = mol nCO2 = 4,4/44 = 0,1 (mol) nH2O = 1,9125/18 = 0,10625 (mol)

0,025 => 0,025x => 0,025/2 CTPT A, B c dng : A : C4Hy v B : C4Hy Ta c y < < y hay y < 8,5 y =10 => CTPT B : C4H10 Tng t bin lun tm CTPT A : y < 8,5 y chn y 2 4 6 8 A C4H2 C4H4 C4H6 C4H8 Vy c 4 cp nghim : v v v c) Tnh CM cc mui to thnh : nKOH = V.CM = 0,1.1,3 = 0,13 (mol) Ta c : = => To thnh 2 mui. CO2 + 2KOH => K2CO3 + H2O a 2a a (mol) CO2 + KOH => KHCO3 b b b (mol) Ta c : (mol) CM(K2CO3 ) = (M) CM(KHCO3) = (M) Bi 6 : t chy hon ton hn hp gm ankin (A) v ankan (B) c V = 5,6 lt (kc) c 30,8g CO2 v 11,7g H2O Xc nh CTPT A,B. Tnh % A,B. Bit B nhiu hn A mt C GII bi ny, t chy hn hp 2 hydrocacbon khng phi l ng ng ca nhau nn khng dng phng php trung bnh c m s dng phng php ghp n s v bin lun gii. Gi 5,6 l hh :(mol) (n 2; m 1) a an a(n-1) (mol) b bm bm (mol) n hn hp = a+ b = (mol) (1) n CO2 = an + bm =(mol) (2) n H2O = a(n-1) + bm =(mol) (3) (2), (3) => an - a + bm = 0,65 0,7 - a = 0,65 a = 0,05 mol (1) b = 0,25 a = 0,25-0,05 = 0,2 mol (2) an + bm = 0,05n +0,2m = 0,7 n + 4m = 14 => m < 3,5 n = 14 4m m = n +1 v B nhiu hn A mt C Bin lun :

m 1 2 3 n 10 6 2 Vy m = 3 n =2 Vy CTPT A, B: Bi 7 : Mt hn hp kh (X) gm 1 ankan, 1 anken v 1 ankin c V =1,792 lt ( ktc) c chia thnh 2 phn bng nhau: - Phn 1: Cho qua dung dch AgNO3/NH3 d to 0,735 g kt ta v th tch hn hp gim 12,5% - Phn 2 : t chy hon ton ri hp th sn phm chy vo 0,2 lt dung dch Ca(OH)2 0,0125M thy c 11g kt ta Xc nh CTPT ca cc hydrocacbon. GII : bi ny, c nhiu th nghim vi nhiu d kin, ta nn dng phng php ghp n s gii. nhh = mol gi a, b, c ln lt l s mol ca ankan, anken, ankin trong mi phn => a + b + c = 0,04 mol (1) v Phn 1 + dd AgNO3/NH3 d => 0,735g (kt ta) Vhh gim 12,5% => Vankin = 12,5%(1/2Vhh) => nankin = c = 12,5%*0,04 = 0,005 mol (2) => M(kt ta)= => Trong phn t kt ta ch c mt nguyn t bc. Vy ankin ban u l ankin-1 t CTPT kt ta CnH2n-3Ag M(kt ta) = 14n +105 = 107 => n= 3 Vy CTPT ankin l C3H4. Ta c a + b = 0,04- c =0,04 -0,005 = 0,035 mol (3) v Phn 2 : C3H4 + 4O2 => 3CO2 + 2H2O 0,005 => 0,015 (mol) CmH2m + 3m/2O2 => mCO2 + mH2O b => mb (mol) CnH2n+2 + (3n+1)/2O2 => nCO2 + (n+1)H2O a => na (mol) nCO2 = 0,015 + mb + na (mol) (4) nCa(OH)2 = 0,2 * 0,0125 = 0,115 mol nCaCO3 = 11/100 = 0,11(mol) Khi cho CO2 vo dd Ca(OH)2 c th xy ra cc phn ng sau : Ca(OH)2 + CO2 => CaCO3 + H2O (5) Ca(OH)2 + 2CO2 => Ca(HCO3 )2 (6) v TH 1 : S mol CO2 thiu so vi dd Ca(OH)2, ch xy ra phng trnh phn ng s (5) nCO2 = 0,015 + mb + na = nCaCO3 = 0,11 mol => mb + na = 0,095 mol (7) Cch 1 : Dng phng php bin lun da vo gii hn Gi s n < m => na + nb < na + mb < ma + mb

n(a + b) < na + mb < m(a + b) n Ca(HCO3 )2 (6) (0,115-0,11) 2.0,005 nCO2 = 0,11 + 2.0,005 = 0,12 (mol) => na + mb = 0,12 0,015 = 0,105 (mol) Gi s n 1< n < => CTPT v *n=3 Gi s m < n => 2< m < BaCO3 (kt ta) + H2O nCO2 = nBaCO3 = mol

=> mH2O = 46,5 0,75.44 = 13,5(gam) => nH2O = (mol) Cch 1 : Ta thy khi t hai hydrocacbon cho s mol nCO2 = nH2O = 0,75 mol Cch 2 : bin lun theo phng php s trung bnh t CTPT trung bnh ca hai hydrocacbon trn l : 1 => => () (mol) 0,3 => 0,3 => 0,3() (mol) nCO2 = 0,3 = 0,75 (mol) => = 2,5 nH2O = 0,3() = 0,75 (mol) thay = 2,5 vo phng trnh trn => =1 => c hai trng hp : * A, B u l anken * A l ankan, B l ankin (hoc ngc li) v TH 1 : A, B l anken. t CTPT (mol) t CTPT trung bnh 2 anken 0,3 => 0,3 nCO2 = 0,3= 0,75 =>= 2,5 Gi s n< m => n= 2. => CTPT A l C2H4. A l anken nh nht nn ta ch c t l : (loi) v TH 2 : A l ankan, B l ankin (mol) a => an => a(n+1) (mol) b => bm => b(m-1) (mol) ta c : nCO2 = an + bm = 0,75 (1) nhh = a + b = 0,3 (2) nH2O = a(n+1) + b(m-1) = an + bm + (a-b) = 0,75 (3) Thay (1) vo (3) : => a- b = 0 hay a= b (2) => a = b = 0,15 (mol) (1) => n + m = Xt t l phn t lng gia A v B ta c hai trng hp : MA : MB = 22 : 13 => n = 3 (A : C3H8) v n =2 (B : C2H2) MB : MA = 22 : 13 => => n = 1,7 v m = 3,28 (loi) Vy hai hydrocacbon l : (mol) Tnh khi lng cc cht trong hn hp : mC3H8 = 0,15.44 = 6,6 (g) mC2H2 = 0,15.26 = 3,9 (g) b) Xc nh tn v tnh khi lng sn phm :

nBr2 = 0,5.0,2 = 0,1(mol) Dung dch Br2 b mt mu han ton chng t 0,1 mol Br2 trong dd phn ng ht. S mol kh thot ra khi dd Br2 l 5,04/22,4 = 0,225 (mol) trong c 0,15 mol C3H8. => nC2H2 p = 0,225 0,15 = 0,075 (mol) Hai phn ng c th xy ra : C2H2 + Br2 => C2H2Br2 (lng) a => a => a (mol) C2H2 + 2Br2 => C2H2Br4 (lng) b => 2b => b (mol) Ta c h phng trnh : => => Tn 2 sn phm : C2H2Br2 : 1,2- Dibrometen; C2H2Br4 : 1,1,2,2-Tetrabrometan. Bi 9 : Mt hn hp gm mt s hydrocacbon lin tip trong dy ng ng c khi lng phn t trung bnh () = 64. 100oC th hn hp ny th kh, lm lnh n nhit phng th mt s cht b ngng t. cc cht kh c khi lng phn t trung bnh (= 54). Cc cht lng c (=74). Tng khi lng cc cht trong hn hp u l 252. Bit khi lng phn t cht nng nht gp i cht nh nht. Tm CTPT cc cht v % th tch cc cht trong hn hp. GII : bi ny, p dng tnh cht ng ng trong ton hc gii Gi a1, a2, , an l khi lng phn t ca cc hydrocacbon trn. * p dng tnh cht ton hc : Cc hydrocacbon lin tip thuc cng mt dy ng ng s to nn mt cp s cng c cng sai d = 14 an = a1 + (n-1)d S= Vi an = 2a1 => 2a1 = a1 + (n-1).14 => a1 = 14(n-1) => S = 1,5na1 = 252 Hay 15,5.14n(n-1) = 252 => 21n12 - 21n1 - 252 = 0 n = 4(nhn) hay n = -3 (loi) a1 = 14(4-1) = 42 t hydrocacbon u l A1 : CxHy M1 = 12x + y = 42 m : y chn y < 2x +2 x 1 2 3 4 y 30 18 6 b = 6a (2) => c = 2,5d(3) Thay (2), (3) vo (1) : => => d = 2a (4) => c= 2,5.2a = 5a (3) nhh = a + b + c + d = a + 6a + 5a + 2a = 14a cng iu kin, t l v s mol bng t l v th tch %C3H6 = %C4H8 = %C5H10 = %C6H12 =

8 BI TON XC NH THNH PHN HN HP CC HIDROCACBON BIT CTPT

v

8.1 MT S LU KHI GII BI TON HN HP : Khai thc tnh cht ha hc khc nhau ca tng loi hydrocacbon, vit cc phng trnh phn ng. t a, b, c, ln lt l th tch (hoc s mol) kh trong hn hp. Lp cc phng trnh i s : bao nhiu d kin l by nhiu phng trnh. Cc th nghim thng gp trong ton hn hp : + t chy hn hp trong O2 : thng dng lng d O2 (hoc ) phn ng xy ra hon ton, nu thiu oxi bi ton s tr nn phc tp v sn phm c th l C, CO, CO2, H2O, hoc sn phm ch gm CO2, H2O ng thi d hydrocacbon. + Phn ng cng vi H2 : cho hn hp gm hydrocacbon cha no v H2 qua Ni, toC (hoc Pd,to) s c phn ng cng. gim th tch hn hp bng th tch H2 tham gia phn ng. Ta lun c : S mol hn hp trc phn ng ln hn s mol hn hp sau phn ng. T> S Khi lng hn hp trc v sau phn ng bng nhau (LBTKL). mhh T = mhhS => < => dT < dS + Phn ng vi dd brm v thuc tm d, tng khi lng ca dd chnh l khi lng ca hydrocacbon cha no. CnH2n+2-2k + kBr2 => CnH2n+2-2kBr2k + Phn ng c trng ca ankin-1 : 2R(CCH)n + nAg2O => 2R(CCAg)n (kt ta) + nH2O Khi lm ton hn hp do s mol cc cht lun thay i qua mi th nghim do khi qua th

nghim mi ta nn lit k s mol ca hn hp sau v trc mi th nghim. Lu : trong cng thc tnh PV = nRT th V l Vbnh. V d : Mt bnh kn c dung tch 17,92 lt ng hn hp gm kh hidro v axetilen ( OoC v 1 atm) v mt t bt Ni xc tc. Nung nng bnh mt thi gian sau lm lnh n 0oC. a) Nu cho lng kh trong bnh qua dd AgNO3/NH3 s sinh ra 1,2 gam kt ta vng nht. Tm s gam axetilen cn li trong bnh. b) Cho lng kh cn li qua dd Brom thy khi lng dung dch tng ln 0,41 gam. Tnh s gam etilen to thnh trong bnh. c) Tnh th tch etan sinh ra v th tch H2 cn li sau phn ng. Bit t khi hn hp u (H2 + C2H2 trc phn ng) so vi H2 = 4. Bt Ni c th tch khng ng k. GII a) Tnh lng axetilen cn d : v Phn 1 : Sn phm chy to kt ta vng nht vi ddAgNO3/NH3 chng t hn hp cn axetilen d Cc ptp : C2H2 + Ag2O C2Ag2 (kt ta) + H2O nC2Ag2 = (mol) Lng axetilen cn li trong bnh : nC2H2 d = 2nC2H2 p = 2nC2Ag2 = 2.0,005 = 0,01 (mol) b) Tnh s gam etilen to thnh trong bnh : v Phn 2 : Cc ptp : C2H4 + Br2 => C2H4Br2 b => b => b (mol) C2H2 + 2Br2 => C2H2Br4 0,005 => 2.0,005 (mol) p dng LBT khi lng : mbnh tng = mC2H4 + mC2H2 => mC2H4 = mbnh tng mC2H2 = 2(0,41- 0,005.26) = 0,56 (g) nC2H4 = (mol) c) Th tch etan sinh ra v th tch H2 cn li : v Phn 3 : nhh = (mol) Gi x (mol) l s mol H2 trong 0,8 mol hn hp ban u. hh = 4.2 = 8 hh = => x = 0,6 (mol) nC2H2 b = 0,2 (mol) Cc ptp : C2H2 + H2 C2H4 0,02 0,02 (mol) C2H2 + 2H2 C2H6 y => 2y => y (mol) Gi y l s mol etan to thnh. nC2H2 p to etan = y = 0,2 (0,01 + 0,02) = 0,17 (mol)

=> nEtan = 0,17 (mol) nH2 cn li = 0,6 (0,02 + 2.0,17) = 0,24 mol Ghi ch : ta nn t cc n s ngay t u v phi cng n v. Qua mi th nghim s gip ta tm mt n s. Lu lng hn hp mang phn ng trong mi th nghim c th khc nhau nhng t l thnh phn cc cht trong hn hp khng i. 8.2 PHNG PHP GII MT S BI TON XC NH THNH PHN HN HP CC HYDROCAC BIT CTPT Bi 1 : t chy hon ton 100cm3 hn hp A gm : C2H6, C2H4, C2H2 v H2 th thu c 90cm3 CO2. Nung nng 100cm3 A c s hin din ca Pd th thu c 80cm3 hn hp kh B. Nu cho B tip tc qua Ni, to th thu c cht duy nht. Tm % cc cht trong hn hp. GII : t 100cm3 hh A gm : C2H6 : a C2H4 : b C2H2 : c H2 : d (cm3) => a + b + c + d = 100 (cm3) v TN1 : Cc ptp : C2H6 + 7/2O2 => 2CO2 + 3H2O a => 2a (mol) C2H4 + 3O2 => 2CO2 + 2H2O b => 2b (mol) C2H2 + 5/2O2 => 2CO2 + H2O c => 2c (mol) H2 + 1/2O2 => H2O d => d (mol) Lu : H2 cng chy trong Oxi, sn phm l H2O. VCO2 = 2(a + b + c) = 90 (cm3) => a + b + c = 45 (cm3) => d = 100 45 = 55(cm3) (1) v TN2 : xc tc Pd,toC th mt lin kt b t, sn phm cng l anken. C2H2 + H2 C2H4 c => c (cm3) Th tch hn hp gim : Vkh gim = 2c c = c = 100 80 = 20 (cm3) (2) Hn hp kh B gm : C2H6 : a C2H4 : b + c (cm3) H2 : d c = 55 20 = 35 v TN3 : C2H4 + H2 C2H6 b+c b+c (cm3) V ch thu c mt kh duy nht => C2H4 v H2 u ht. => b + c = 35 =>b = 35 c = 35 20 = 15 (cm3) a = 100 (b + c + d) = 100 (15 + 20 + 55) = 10 (cm3) % th tch cc cht trong hn hp :

%VC2H6 = =10% %VC2H4 = = 15% %VC2H2 = = 20% %VH2 = = 55% Bi 2 : Cho 11 gam hn hp gm 6,72 lt hydrocacbon mch h A v 2,24 lt mt ankin. t chy hn hp ny th tiu th 25,76 lt Oxi. Cc th tch o ktc. a) Xc nh loi hydrocacbon. b) Cho 5,5 gam hn hp trn cng 1,5 gam hidro vo mt bnh kn cha sn mt t bt Ni ( ktc) un nng bnh phn ng xy ra hon ton ri a v OoC. Tnh thnh phn % hn hp cui cng v p sut trong bnh. GII : Da vo ptp chy, t s mol cc cht v gii h phng trnh tm cc gi tr x, n. a) Xc nh loi hydrocacbon : S mol cc cht : nA =(mol) nankin = (mol) nO2 = (mol) Gi 11g hn hp (mol) Cc ptp : 0,3=> 0,3(x + y/4) (mol)

0,1 => 0,1(3n-1)/2 (mol) nO2 = 0,3(x + ) + 0,1() = 1,15 mhh = (12x + y)0,3 + (14n - 2)0,1 = 11 36x + 3y + 14n = 112 (1) 4x + y + 2n = 16 (2) (1) 7.(2) => y = 2x Thay y = 2x vo (1), (2) : 36x + 6x +14n = 112 4x + 2x + 2n = 16 3x + n = 8 => x < => x = 2 C2H4 n = 2 C2H2 b) Tnh thnh phn % hn hp cui cng v p sut trong bnh : nH2 = 1,5/2 = 0,75 (mol) Hn hp mi gm C2H4 : 0,015 mol C2H2 : 0,05 mol H2 : 0,75 mol Cc ptp : C2H4 + H2 C2H6 0,15 => 0,15 => 0,15 C2H2 + 2H2 C2H6 0,05 => 0,1 => 0,05 Sn phm thu c gm : C2H6 : 0,2 mol H2 d : 0,5 mol T l %th tch chnh l t l % s mol : %VC2H6 =

%VH2 = Tnh p sut : PVbnh = nRT Trc phn ng n1 = nC2H4 + nC2H2 + nH2 = 0,15 + 0,05 + 0,75 = 0,95 (mol) Sau phn ng n2 = nC2H6 + nH2d = 0,7 (mol) cng iu kin Vbnh, T = const => P2 = (atm) (P1 = 1atm ) Bi 3 : Mt hn hp kh A gm C2H2 v H2 c t khi hi so vi khng kh bng 0,4. un nng A vi xc tc Ni mt thi gian thu c hn hp kh B, t khi ca B so vi khng kh bng . Nu cho ton b lng B qua dung dch KMnO4 d th cn li kh D thot ra ngoi, t khi ca D so vi H2 bng 4,5. Cc th tch o ktc. a) Tnh thnh phn % th tch ca hn hp A. b) Tnh t s th tch ca A so vi th tch B. Gii thch s thay i th tch . c) Tnh thnh phn % th tch ca hn hp kh D d) Bit VB = 3,136 lt, hi nu hp thu ht lng B ny trong dd Brom d th khi lng cc sn phm thu c l bao nhiu? Tm tt : GII : a. % th tch ca hn hp A : Trong A, t C2H2 : a (mol) H2 : b (mol) = 0,4.29 = 11,6 => = 11,6 => b = 1,5a %C2H2 = = = 40% %H2 = 100% - 40% = 60% b. T l th tch ca A so vi B : Phn ng c th xy ra : C2H2 + H2 C2H4 C2H2 + 2H2 C2H6 Phn ng cng H2 lm gim s mol kh nhng khng lm thay i khi lng kh => nB < nA => VB < VA. Ta c : mA = mB => .nA = .nB => = Vi = 29.(gam/mol) v = 11,6 => V t l th tch bng t l s mol kh nn VA : VB = 10 : 7 c. % th tch hn hp kh D : Hn hp B gm C2H4, C2H6, C2H2 d v H2 d khi cho qua dung dch KMnO4 th C2H4 v C2H2 d b oxi ha v gi li trong dung dch : C2H4 + [O] + H2O => C2H4(OH)2 C2H2 + 4[O] => HOOCCOOH Hn hp kh D thot ra gm C2H6 v H2 d. Trong hn hp D gi C2H6 : x(mol) v H2 d : y (mol). Ta c = 4,5.2 = 9 => = 9 => y = 3x => %C2H6 = .100% = = 25% %H2 = 100% - 25% = 75% d. Khi lng cc sn phm : C2H2 + H2 C2H4 u u u (mol)

C2H2 + 2H2 C2H6 x 2x x (mol) Gi u (mol) l s mol C2H4 thu c => B gm : C2H4 : u (mol) C2H6 : x (mol) C2H2 d : a (u + x) (mol) H2 d : 3x (mol) nB = 0,14 (mol) => u + x + a (u + x) + 3x = 0,14 => a + 3x = 0,14 (1) nH2 ban u = 1,5a(mol) => u + 2x +3x = 1,5 =>1,5a = u + 5x (2) = => a = 0,8 (mol) T (1) => x = 0,02 (mol) T (2) => u = 0,02 (mol) Trong B ch c C2H4 : 0,02 (mol) v C2H2 : 0,04 (mol) cho phn ng cng vi dung dch Br2 : C2H4 + Br2 => C2H4Br2 0,02 => 0,02 (mol) C2H2 + 2Br2 => C2H2Br4 0,04 => 0,04 (mol) => mC2H4Br2 = 0,02.188 = 3,76 (gam) mC2H2Br4 = 0,04.346 = 13,84 (gam) Bi 4 : Mt bnh kn 2 lt 27,3oC cha 0,03 mol C2H2 ; 0,015 mol C2H4 v 0,04 mol H2 c p sut P1. Tnh P1 - Nu trong bnh c mt t bt Ni lm xc tc (th tch khng ng k) nung bnh n nhit cao phn ng xy ra hon ton, sau a v nhit ban u c hn hp kh A c p sut P2. - Cho hn hp A tc dng vi lng d dd AgNO3/NH3 thu c 3,6 gam kt ta. Tnh P2. Tnh s mol mi cht trong A.. GII : Tnh p sut P1 : Tng s mol cc cht trc phn ng : n1 = nC2H2 + nC2H4 + nH2 = 0,03 + 0,015 + 0,04 = 0,085 (mol) p dng phng trnh trng thi kh l tng : atm Tnh p sut P2 v s mol cc cht : Cc ptp : C2H2 + H2 C2H4 a => a => a (mol) C2H4 + H2 C2H6 b => b => b (mol) V s mol H2 = 0,04 < nC2H2 + nC2H4 = 0,045 (mol) nn phn ng ht H2 t a, b l s mol H2 tham gia hai phn ng trn => a + b = 0,04 (1)(mol) C2H2 cn d sau phn ng trn tc dng vi lng d dd AgNO3/NH3 : HCCH + Ag2O C2Ag2 (kt ta) + H2O 0,015 0,015 (mol) nC2Ag2 = (mol) => nC2H2 d = 0,015 mol => nC2H2 phn ng = a = 0,03 0,015 = 0,015 (mol)

b = 0,04 a = 0,04 0,015 = 0,025 (mol) nC2H4 = nC2H4 b + a = 0,015 + 0,015 = 0,03 (mol) nC2H4 d = 0,03 b = 0,03 0,025 = 0,005 (mol) p sut P2 : n2 = nC2H2 d + nC2H4 d + nC2H6 = 0,015 + 0,005 + 0,025 = 0,045 (mol) (atm) Bi 5 : Cho a gam CaC2 cha b% tp cht tr tc dng vi nc th thu c V lt C2H2 (ktc) 1) Lp biu thc tnh b theo a v V 2) Nu cho V lt trn vo bnh kn c than hot tnh nung nng lm xc tc,to trong bnh toC p sut P1. Sau phn ng thu c hn hp kh, trong sn phm phn ng chim 60%V, nhit khng i, p sut P2 Tnh hiu sut ca phn ng. 3) Gi s dung tch bnh khng i, th tch cht rn khng ng k. Hy a) Lp biu thc tnh p sut P2 theo P1 v hiu sut h. b) Tnh khong gi tr ca P2 theo P1 GI I : 1) Lp biu thc tnh b theo a v V : Cch 1 : CaC2 + 2H2O => Ca(OH)2 + C2H2 (mol) Gi V l th tch C2H2 sinh ra. nC2H2 = mCaC2 = 64. mtp cht = (a-) b% = Cch 2 : CaC2 + 2H2O => Ca(OH)2 + C2H2 64(g) 22,4(lt) a 0,01b(g) V(lt) b(%) Ta lp c t l : =>b = => b%= 2) Tnh hiu sut phn ng : Xt phn ng : 3C2H2 C6H6 x => x/3 (lt) Gi x l th tch C2H2 tham gia phn ng trn Tng s mol cc cht sau phn ng : (lt) n = V x + x/3 = V 2/3x (l) VC6H6 = 60%Vhh Cch 1 : Tnh theo cht tham gia : Hiu sut phn ng h = Cch 2 : Tnh hiu sut phn ng theo sn phm : Theo phn ng : 3C2H2 C6H6 V => V/3 (theo l thuyt)

h% = = 81,81% 3) a) Lp biu thc tnh p sut P2 theo P1 v hiu sut h : Ta c pt TTKLT : PVbnh = nRT Vbnh, T = const => V2, V1 : S mol cc cht trong bnh. V V2 = V 2/3x V1 = V => => b) Tnh khong gi tr ca P2 theo P1 : Ta c 0 < h < 100 h = 0 => = 1 h = 100 => = => P1 < P2 < P1 Bi 6 : Khi sn xut t n ta thu c hn hp rn gm CaC2, Ca v CaO (hh A). Cho hn hp A tc dng ht vi nc th thu c 2,5 lt hn hp kh kh X 27,0oC v 0,9856atm. T khi ca X so vi metan bng 0,725. a) Tnh % khi lng mi cht trong A b) un nng hn hp kh X vi bt Ni xc tc mt thi gian th thu c hn hp kh Y, chia Y lm hai phn bng nhau. - Phn th nht cho li t t qua bnh nc Brom d thy cn li 448 ml hn hp kh X (ktc) v t khi so vi Hidro l 4,5. Hi khi lng bnh nuc Brom tng bao nhiu gam? - Phn th hai em trn vi 1,68 lt O2 (ktc) trong bnh kn dung tch 4 lt. Sau khi bt tia la in t chy, gi nhit bnh 109,2oC. Tnh p sut bnh nhit bit dung tch bnh khng i. GII : a) Tnh % khi lng mi cht trong A : Gi 5,52g hh A CaC2 : a Ca : b CaO : c (mol) mhh X = 64a + 40b + 56c = 5,52 (1) Lu : hn hp A tc dng vi nc, c Ca v CaO cng c phn ng. CaC2 + 2H2O => Ca(OH)2 + C2H2 a a (mol) Ca + H2O => Ca(OH)2 + H2 b b (mol) CaO + H2O => Ca(OH)2 c (mol) 2,5 lt kh thu c gm : C2H2 : a H2 : b = 0,725.16 = 11,6 nhhX = a + b = = (mol) (2) = 0,725.16 = 11,6 (gam/mol) => mX = .nX = 11,6.0,1 = 1,16 (gam)

=> 26a + 2b = 1,16 (3) (2), (3) => (mol) Theo cc phn ng trn : nCaC2 = nC2H2 = 0,04 mol => %CaC2 = nCa = nH2 = 0,06 mol => %Ca = % CaO = 100% - (46,38 + 43,48)% = 10,14% b) tng khi lng bnh Brom : * Khi nung nng hn hp X vi xc tc Ni, c th xy ra 2 phn ng : C2H2 + H2 => C2H4 C2H2 + 2H2 => C2H6 => hn hp kh Y c th gm C2H4, C2H6, C2H2 d, H2 d. * Khi cho hn hp Y qua bnh ng dd Br2 d th C2H2, C2H4 b hp thu : C2H2 + 2Br2 => C2H2Br4 (lng) C2H4 + Br2 => C2H4Br2 (lng) => hn hp kh Z thot ra gm C2H6 v H2. nZ = 0,02 mol v = 4,5.2 = 9 => mZ = 9.0,02 = 0,18 gam * p dng LBT khi lng, ta c : m1/2hh Y = m1/2hh X = 1,16/2 = 0,58gam So snh hn hp Y v Z, ta thy tng khi lng bnh ng dung dch Br2 l tng khi lng ca C2H2 v C2H4 tc l mY - mZ Vy tng khi lng bnh Brom = 0,58 0,18 = 0,4 gam v Tnh p sut bnh sau phn ng chy : So snh hn hp X vi Y v p dng LBT nguyn t, ta c : trong hn hp Y = trong hn hp X = 2.0,02 = 0,04 (mol) trong hn hp Y = trong hn hp X = 2.0,02 + 2.0,03 = 0,1 (mol) * Sn phm chy gm : nCO2 = nC = 0,04 (mol); nH2O = nH = 0,05 (mol) Mt khc, trong CO2 v trong H2O = 0,04.2 + 0,05 = 0,13 (mol) nO ban u l 0,075.2 = 0,15 (mol) => nO d = 0,15 0,13 = 0,02 (mol) => nO2 d = 0,02/2 = 0,01 (mol) cc kh trong bnh sau khi t = 0,04 + 0,05 + 0,01 = 0,1 (mol) Vy p sut bnh l : Patm

9.BI TP TNG HP V HYDROCACBONv Mt s lu khi gii bi tp tng hp v hydrocacbon : Bi tp tng hp l mt dng bi tp trong c c phn tnh ton km theo cu hi l thuyt hoc cu hi th nghim. Bi tp hn hp thng c cc dng sau : Tm CTPT ca mt hay nhiu hydrocacbon sau yu cu : Xc nh CTCT ng ca cc cht qua th nghim cho cht tc dng vi mt cht no thu c sn phm c th. Xc nh CTCT ri vit phng trnh phn ng iu ch mt cht hydrocacbon khc hoc iu ch cht t nguyn liu chnh ban u l g. a ra phng php phn bit cc hydrocacbon mi tm c hoc nu cch tch ring, tinh ch tng cht trong hn hp cc cht mi tm c.

V phng php lm bi tp loi ny, chng ta vn dng cc phng php hng dn trong phn bi tp l thuyt v bi tp tm CTPT, bi tp hn hp gii. Sau y l mt s bi tp v d : Dng 1 : bi yu cu xc nh CTPT ca sn phm th, t gi thit cho xc nh ng CTCT ca hydrocacbon ban u. Bi 1 : Khi tin hnh phn ng th gia ankan B vi hi Br2 c chiu sng ngi ta thu c hn hp X ch gm 2 sn phm phn ng (mt cht v c v mt cht hu c) th hi. T khi hi ca X so vi khng kh bng 4. a) Lp CTPT ca B v chn cho M mt CTCT thch hp. b) Nu tin hnh phn ng th 3 nguyn t hidro trong phn t B bng Clo th c th thu c my ng phn? GII : bi cho t khi hi ca sn phm th nn ta tm CTPT sn phm ri suy ra CTCT B a. Lp CTPT ca B v chn CTCT ng ca B. Gi k l s nguyn t Brom th vo phn t B : CnH2n+2 + kBr2 => CnH2n+2-kBrk + kHBr a => a ak (mol) Gi a (mol) l s mol B tham gia phn ng Sn phm phn ng gm : CnH2n+2-kBrk : a mol v HBr : ak mol = 29.4 = 116 => => 14n + 44k = 114 n= k 1 2 3 n 5 1,8 CTPT B : C5H12 v dn xut ca B : C5H11Br k = 1 : phn ng xy ra theo t l mol 1:1 v thu c duy nht mt sn phm C5H11Br => B phi c cu to i xng. => CTCT B : Neopentan hay 2,2 dimetylpropan b. Ta thu c 3 ng phn ca dn xut 3 clo ca B : Dng 2 : Sau khi tm c CTPT, CTCT ca cc hydrocacbon bi yu cu vit ptp iu ch cc cht Bi 2 : Hn hp kh X gm 4 hydrocacbon A, B, C, D iiu kin chun. Trn X vi O2 va t chy ht X trong mt bnh kn nhit T1 > 100oC v p sut 0,8amt. Bt tia la in t chy hon ton hn hp ri a bnh v nhit T1, o li p sut trong bnh vn c tr s 0,8atm. Lm li th nghim vi cc hn hp X c thnh phn A, B, C, D khc nhau vn thu c kt qu nh c. a) Lp CTPT A, B, C, D bit rng MA < MB < MC < MD. b) Vit ptp iu ch D t A v B t C GII : Nhit sau khi t T1 > 100oC => H2O th hi cng iu kin nhit , th tch p sut bnh trc v sau khi t khng i => s mol kh trong bnh trc v sau phn ng bng nhau.

Khi thay i thnh phn ca hn hp X m kt qu khng thay i => khi t chy tng cht th tng s mol trc v sau phn ng cng bng nhau. t cng thc ca mt cht trong hn hp l : CxHy a => a(x + y/4) => ax => ay/2 (mol) Ta c : nT = nS => a + a(x + 0,25y) = ax + 0,5ay =>1 + x + 0,25y = x + 0,5y => 0,25y = 1 => y = 4 => Vy c 4 hydrocacbon trn u c 4 nguyn t H trong phn t. Mt khc do A, B, C, D u th kh nn x < 4 => Vy 4 hydrocacbon trong X l CH4, C2H4, C3H4, C4H4 Theo th t MA < MB < MC < MD th A : CH4, B: C2H4, C: C3H4, D: C4H4. b. Vit cc ptp iu ch : iu ch D t A : 2CH4 C2H2 + 3H2 2C2H2 C4H4 (vinylaxetilen) iu ch B t C : C3H4 + 2H2 C3H8 C3H8CH4 + C2H4 Dng 3 : Tm CTPT ca cc hydrocacbon sau nu cch nhn bit hoc tinh ch hoc tch cc cht trong hn hp hydrocacbon . Bi 3 : t chy mt s mol nh nhau ca 3 hydrocacbon L, L, M ta thu c lng CO2 nh nhau v t l s mol H2O v CO2 i vi K., L, M tng ng bng 0,5; 1, 1,5. a) Xc nh CTPT K, L, M b) Nu cch nhn bit 3 kh trn ng trong 3 l mt nhn c) Hy tch ring 3 cht trong hn hp trn. GII t cng thc chung cho 3 hydrocacbon l CnH2n +2-2k vi k l s lin kt trong phn t cc hydrocacbon trn. a an => a(n+1-k) (mol) 3 hydrocacbon t vi s mol nh nhau thu c lng CO2 nh nhau nn K, L, M c cng s C trong phn t. T= K th T = 0,5 => 0,5n = n + 1 k => n = 2(k 1) 0 < n < 4 v k 0 => n = 2, k = 2 => K : C2H2 L th T = 1 => n = 2 v k = 1 => CTPT L : C2H4 M th T = 1,5 => n = 2 v k = 0 => CTPT M : C2H6 b) Nhn bit 3 kh trn ng trong 3 l mt nhn : - Ly mi kh mt t lm mu th. - Dn ln lt 3 kh vo dd AgNO3/NH3, kh no to c kt ta vng nht l C2H2. - Hai kh cn li khng c hin tng g c dn tip qua ddBr2 d, kh no lm mt mu nu ca dd Br2 l C2H4, kh cn li khng c hin tng g thot ra ngoi l C2H6 H2C=CH2 + Br2 => BrH2CCH2Br

c) Cch tch 3 cht trn ra khi hn hp ca chng : - Cng thc hin qua cc th nghim nh trn ta thu c kh C2H6 thot ra ngoi. - Tinh ch li C2H2 bng cch cho dd axt HCl vo kt ta bc axetilua, kh axetilen c hon nguyn s bay ra ngoi : C2Ag2 + 2HCl => C2H2 + 2AgCl(kt ta) - Tinh ch li C2H4Br2 bng cch cho thm ddKOHc/ancol vo dd Br2 b mt mu th kh C2H4 c hon nguyn s bay ra ngoi : Hoc : Ghi ch : Trn y ch l mt s bi tp v d nh, nu cc em lm tt bi tp phn II.1& II.2 th s lm c bi tp phn ny.

IV.MT S CU HI TRC NGHIM B SUNG PHN BI TP V HYDROCACBON TRONG CHNG TRNH HA HC THPTCu 1 : c im hay c tnh no sau y gip ta thy c cu to ho hc l yu t quyt nh tnh cht c bn ca hp cht hu c? A. S phn cc ca lin kt cng ho tr. B. S lng nguyn t ca mi nguyn t. C. Hin tng ng ng v hin tng ng phn D. Tt c u sai. Cu 2: I Cc cht ng phn th c cng CTPT II - Nhng cht c cng khi lng phn t th l ng phn ca nhau. A. I & II u ng B. I ng, II sai C. I sai, II ng D. I & II u sai Cu 3: ng phn l nhng cht c : A. Cng thnh phn nguyn t v c khi lng phn t (M) bng nhau. B. C cng CTPT nhng CTCT khc nhau. C. Cng tnh cht ho hc D. a, b, c u ng Cu 4: I Nhng cht ng phn l nhng cht hn km nhau k nhm CH2 II - Nhng cht c tnh cht ho hc tng t nhau l ng ng ca nhau. A. I & II u ng B. I ng, II sai C. I sai, II ng D. I & II u sai Cu 5: S ng phn ca cht c CTPT C4H8 (ng phn phng v ng phn hnh hc) l : A. 3 B. 4 C. 5 D. 6 Cu 6: Khi cho isopentan tc dng vi Cl2 (1:1) ta thu c s sn phm : A. 1 sn phm duy nht B. 2 C. 3 D. 4 Cu 7: Trong cc hp cht : Propen (I); 2-metylbuten-2 (II); 3,4-dimetylhexen-3(III); allyl clorua (IV); 1,2-

diCloeten (V). Cht no c ng phn hnh hc? A. III, V B. II,IV C. I, II, III, IV D. I, V Cu 8: Cho bit tn ca hp cht sau theo IUPAC ? A. 1-Clo-4-Etylpenten-4 B. 1-clo-4-metylenhexan C. 2-etyl-5-Clopenten-1 D. 5- Clo-2-etylpenten-1 Cu 9: Chn tn ng ca cht c CTCT sau : A. 5-Clo-1,3,4-trimetylpentin-1 B. 6-Clo-4,5-Dimetylhexin-2 C. 1-Clo-2,3-Dimetylhexin-4 D. Tt c u sai Cu 10: Nu hidro ha C6H10 ta thu c isohexan th CTCT ca C6H10 l : Cu 11: Quy tc Maccopnhicop ch p dng cho : A. Anken i xng v tc nhn i xng. B. Anken bt i v tc nhn bt i C. Anken bt i v tc nhn i xng D. Hydrocacbon khng no bt i v tc nhn bt i. Cu 12 : I-- Xicloankan v ankan u l nhng hydrocacbon no nn chng l ng ng ca nhau. II -- Tt c nhng hydrocacbon khng no u c tnh cht ha hc nh nhau. A. I v II u sai B. I ng, II sai C. I sai, II sai D. I sai, II ng Cu 13: Nhng hp cht no sau y c th c ng phn hnh hc (cis-trans) CH3CH = CH2 (I) ; CH3CH = CHCl (II) ; CH3CH = C(CH3)2 (III) (IV) A. (I), (IV), (V) B. (II), (IV), (V) C. (III), (IV) D. (II), III, (IV), (V) Cu 14: (V)

Ankan A c 16,28%H trong phn t (v khi lng). vy CTPT v s ng phn tng ng ca A l : A. C6H14 v 4 ng phn B. C6H14 v 5 ng phn C.C5H12 v 3 ng phn D.C6H14 v 6 ng phn Cu 15: Cho propen, propin, divinyl tc dng vi HCl(t l 1:1), s sn phm thu c l : A. 2,2,3 B. 2,3,2 C. 2,3,1 D. Tt c u sai. Cu 16: Nhng loi hydrocacbon no hc tham gia c phn ng th? A. Ankan B. ankin C. benzen D. Tt c cc hydrocacbon trn. Cu 17 : Chn cu tr li ng : A. Anken l nhng hydrocacbon m phn t c cha mt lin kt i C=C B. Anken l nhng hydrocacbon m CTPT c dng CnH2n, n 2, nguyn. C. Anken l nhng hydrocacbon khng no c CTPT CnH2n, n 2, nguyn. D. Anken l nhng hydrocacbon mch h m phn t c cha mt lin kt i C=C Cu 18: Nhng hp cht no sau y khng th cha vng benzen? a. C8H6Cl2 b. C10H16 c. C9H14BrCl d. C10H12(NO2)2. A. a, b B. b,c C. c, d D. a, c, d Cu 19 : Cho xicloankan c CTCT thu gn sau : 1/ (CH2)4CHCH3 2/ CH3CH(CH2)2CHCH3 3/ (CH2)2CHCH2CH3 4/ CH3CH(CH2)4CHCH2CH3 Xicloankan bn nht l : A. (1) B. (2) C. (3) D. (4) Cu 20 : Phng php iu ch no sau y gip ta thu c 2-Clobutan tinh khit hn ht ? A. n-Butan tc dng vi Cl2, chiu sng, t l 1:1. B. Buten-2 tc dng vi hidroclorua C. Buten-1 tc dng vi hidroclorua D. Butadien-1,3 tc dng vi hidroclorua Cu 21:

Th t nhn bit cc l mt nhn N2(I), H2(II), CH4(III), C2H4(IV), C2H2(V) A.5-4-1-3-2 B.5-4-2-1-3 C.5-4-3-2-1 D. Tt c u ng Cu 22 : Xc nh X, Y, Z, T trong chui phn ng sau : A. X : butan, Y: Buten-2, Z : Propen, T : Metan B. X : Butan, Y: Etan, Z : CloEtan, T : iCloEtan C. X : Butan, Y: Propan, Z : Etan, T : Metan D. Cc p trn u sai. Cu 23: T CTPTTQ ca hydrocacbon CnH2n+2-2k (k0), ta c th suy ra cc trng hp no sau y? A. k = 1 => X l anken CnH2n, (n2), n nguyn B. k = 2 => X l ankin CnH2n-2, (n2), n nguyn B. k = 4 => X l aren CnH2n-6, (n6), n nguyn D. Tt c u ng Cu 24 : Khi t chy mt hydrocacbon X ta thu c (s mol CO2/ s mol H2O =2) . Vy X c th l : A. C2H2 B. C12H12 C. C3H6 D. A,B u ng Cu 25 : t chy mt hn hp gm nhiu hydrocacbon trong cng mt dy ng ng nu ta thu c s mol H2O > s mol CO2 th CTPT tng ng ca dy : A. CnHn, n 2 B. CnH2n+2, n 1 (cc gi tr n u nguyn) C. CnH2n-2, n 2 D. Tt c u sai Cu 26 : t chy hon ton hn hp hai hydrocacbon ng ng c khi lng phn t hn km nhau 28vC, ta thu c 4,48 l CO2 (ktc) v 5,4g H2O. CTPT ca 2 hydrocacbon trn l : A. C2H4 v C4H8 B. C2H2 v C4H6 C. C3H4 v C5H8 D. CH4 v C3H8 Cu 27: Th t nhn bit cc l mt nhn ng cc kh : C2H6 (I), C2H4 (II), C2H2 (III), CO2 (IV), H2(V) A. III, II, IV, I, V B. IV, III, II, I, V C. III, IV, II, I, V D. Tt c u ng Cu 28: Cng thc thc nghim ca mt ng ng ca benzen c dng (C3H4)n th CTPT ca ng ng l : A. C12H16 B. C9H12 C. A, C ng

D. A, C sai. Cu 29: Khi t chy metan trong kh Cl2 sinh ra mui en v mt cht kh lm qu tm ha . Vy sn phm phn ng l : A. CH3Cl v HCl B. CH2Cl2 v HCl C. C v HCl D. CCl4 v HCl Cu 30 : t chy 2 hydrocacbon ng ng lin tip ta thu c 6,43g nc v 9,8gam CO2. vy CTPT 2 hydrocacbon l : A. C2H4 v C3H6 B. CH4 v C2H6 C. C2H6 v C3H8 D. Tt c u sai. Cu 31 : Trong mt bnh kn cha hn hp A gm hydrocacbon X v H2 vi xt Ni. Nung nng bnh mt thi gian ta thu c mt kh B duy nht. t chy B ta thu c 8,8g CO2 v 5,4g H2O. Bit VA=3VB. Cng thc ca X l : A. C3H4 B. C3H8 C. C2H2 D. C2H4 Cu 32 : Mt hn hp kh X gm ankin B v H2 c t khi hi so vi CH4 l 0,6. Nung nng hn hp X vi Ni xt phn ng xy ra hon ton thu c hn hp kh Y c t khi hi so vi CH4 l 1. Cho hn hp Y qua dd Brom d th bnh cha Brom c khi lng tng ln l : A. 8g B. 16g C. 0 D. Tt c u sai. Cu 33 : t chy mt hn hp hydrocacbon ta thu c 2,24l CO2 (ktc) v 2,7 gam H2O th th tch O2 tham gia phn ng chy (ktc) l : A. 5,6 lt B. 2,8 lt C. 4,48 lt D. 3,92 lt Cu 34 : Khi t chy mt hydrocacbon A, thu c 0,108g nc v 0,396g CO2. Cng thc n gin nht ca A l : A. C2H3 B. C3H4 C. C4H6 D. Tt c u sai Cu 35 : Hn hp A gm mt ankan v mt anken. t chy hn hp A th thu c a (mol) H2O v b

(mol) CO2. Hi t s T = a/b c gi tr trong khong no? A. 1,2< T B => TNT (thuc n) A. A l Toluen, B l n-heptan B. A l benzen, B l Toluen C. A l n-hexan, B l Toluen D. Tt c u sai Cu 37 : Khi cng HBr vo 2-metylbuten-2 theo t l 1:1, ta thu c s sn phm. A. 2 B. 3 C. 4 D. Tt c u sai Cu 38 : Anken thch hp iu ch : A. 3-etylpenten-2 B. 3-etylpenten-3 C. 3-etylpenten-1 D. 3,3-Dimetylpenten-1 Cu 39 : Khi cho Br2 tc dng vi mt hydrocacbon thu c mt dn xut brom ha duy nht c t khi hi so vi khng kh bng 5,207. CTPT ca hydrocacbon l : A. C5H12 B. C5H10 C. C4H10 D. Khng xc nh c. Cu 40 : I- t chy mt ankin thu c s mol CO2 > s mol H2O II- Khi t chy mt hydrocacbon X m thu c s mol CO2> s mol H2O th X l ankin? A. I & II u ng B. I ng, II sai C. I sai, II ng D. I & II u sai Cu 41: Cho 1,12gam mt anken tc dng va vi dd Br2 ta thu c 4,32 gam sn phm cng. Vy CTPT ca anken c th l : A. C2H4 B. C3H6 C. C2H2 D. p s khc Cu 42 : t chy mt th tch hydrocacbon A cn nm th tch oxi. Vy CTPT ca A l :

A. C3H6 B. C2H12 C. C3H8 D.B v C u ng Cu 43: Hn hp 2 ankan lin tip c dhh/H2 = 24,8. CTPT ca 2 ankan l : A.CH4 ; C2H6 B.C2H6 C3H8 C.C3H8 v C4H10 D. Tt c u sai Cu 44 : t chy mt s mol nh nhau ca 3 hydrocacbon K, L, M ta thu c lng CO2 nh nhau v t l s mol H2O v CO2 i vi K, L, M tng ng bng 0,5 : 1 : 1,5. CTPT ca K, L, M ln lt l : A. C3H8, C3H4, C2H4 B. C2H2, C2H4, C2H6 C. C12H12, C3H6, C2H6 D. B v C ng Cu 45 : Hai xicloankan M, N u c t khi hi so vi metan bng 5,25. khi monoclo ha (c chiu sng) th M cho 4 hp cht, N ch cho mt hp cht duy nht. Tn ca M v N l : A. metyl xiclopentan v dimetyl xiclobuan B. xiclohexan v metyl xiclopentan C. xiclohexan v isopropan xiclopropyl D. A, B, C u ng Cu 46 : t chy hon ton mt hydrocacbon X vi mt lng va oxi. Dn hn hp sn phm chy qua H2SO4 th th tch kh gim hn mt na. Dy ng ng ca X l : A. ankan B. anken C. ankin D. ankadien E. aren Cu 47 : t chy V(lt) hn hp kh X ktc gm 2 hydrocacbon to thnh 4,4gam CO2 v 1,8gam H2O. Cho bit 2 hydrocacbon trn cng hay khc dy ng ng v thuc dy ng ng no (ch xt cc dy ng ng hc trong chng trnh) A. Cng dy ng ng anken hoc xicloankan B. Khc dy ng ng : ankan v ankin (s mol bng nhau) C. Khc dy ng ng : ankan v ankadien (s mol bng nhau) D. Tt c u ng. Cu 48 : Cho 0,896 lt ( ktc) hn hp kh A gm 2 hydrocacbon mch h. Chia A thnh 2 phn bng nhau. Phn 1 : Cho qua dd Br2 d, lng Br2 nguyn cht phn ng l 5,6 gam Phn 2 : t chy hon ton to ra 2,2 gam CO2. Tm CTPT 2 hydrocacbon. A. C4H8 v C2H2 B. CH4 v mt hydrocacbon khng no. C. C2H2 v C2H4 D. Tt c u sai. Cu 49 : Hn hp kh A gm Etan v Propan. t chy hn hp A thu c