Phuong phap viet_cong_thuc_dong_phan_hidrocacbon

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  1. 1. Trang 1PHNG PHP VIT CNG THC CU TO CAHIDROCACBONI. Dat vn d:Ho hc l mot mn khoa hc thc nghiem nghin cu nhiu hien tngtrong t nhin v d i sng. L mot mn hc thit thc ph c vdac lc chod i sng con ng i.Nham gip hc sinh mot kin thc vng vng, bit phn tch v nhan dnhcc s vat, hien tng, t tin khi hc ho hc. Th vn d rn luyen ky nangcho hc sinh c mot phng php t duy thc hnh tt l mot vn d rt cnthit v cp bch.Ring bn thn ti qua nhiu nam ging dy ti cc tr ng PTTH, ti nhanthy rang trong khi vit cng thc cu to ca cc cht, dac biet l dng cngcng thc phn t nhng c nhiu cng thc cu to khc nhau th hc sinhlp 11, tham chlp 12 cn nhiu b ng v cc em vit khng dy d cccng thc cu to khc nhau hoac sai v th t lin kt.V le d ti xin trnh by mot s vn d c bn nham gip hc sinh vitdc cng thc cu to dng v t c s trng lap cng thc. Kin thc ny sert cn cho cc em khi hc ln chng trnh ha lp 12 v l nn tng rt ttde cc em hc hacc cp cao hn.II. NOI DUNG D TIA. Phng php tin hnh:- Phn loi hp cht hu c: Hidrocacbon (Hidrocacbon no, hidrocacbonkhng no hay hidrocacbon thm) .- Cch vit cng thc cu to ca mot s hp cht hu c.
  2. 2. Trang 2B. Noi dung:1. C s l thuyt:M c tiu:- Hc sinh cn hc v nam vng cc dnh nghia dng phn, ankan, xicloankan, anken, ankadien, ankin, benzen.- Hc sinh nam dc cch vit cng thc cu to ca t$ng hp cht hu c.- Xc dnh nhanh chng s lng dng phn ca t$ng cht.Noi dung:* Cc dnh nghia:Cc hp cht hu c c cng cng thc phn t nhng c cu to hahc khc nhau, dan ti tnh cht ha hc khc nhau dc gi l cc cht dngphn ca nhau.Ankan l nhng hidrocacbon no mch h c cng thc chungCnH2n+2(n 1).Xicloankan l nhng hidrocacbon no mch vng c cng thc chungCnH2n (n3).Anken l hidrocacbon khng no, mch h cha mot ni di C = C trongphn t, c cng thc chung l CnH2n(n2)Ankadien: l hidrocacbon mch h, cha 2 ni di trong phn t, ccng thc chung l CnH2n-2(n 3)Ankin l l hidrocacbon mch h, cha mot ni ba C C trong phn t,c cng thc chung l CnH2n-2(n 3).Hidrocacbon thm ( aren) lah loi hidrocacbon trong cng thc phn tc mottj hay nhiu nhn benzen, di dien cho dy dng dang aren lphn t benzen c cng thc tong qut l: CnH2n-6(n 6 )
  3. 3. * Cc loi dng phn thng gap trong chng trnh ha hc pho thng:+ Dng phn cu to:- Dng phn mch cc bon ( Mch khng nhnh, mch c nhnh, mch vng)- Dng phn nhm chc- Dng phn v tr ( v tr lin kt boi, v tr nhm chc)+ Dng phn hnh hc (cis trans)* Phng php vit cng thc cu to hp cht hu c.n 1 nu ( n 1) l sTrang 3Ankan:Vd 1: Vit cc dng phn c the c ca ankan c cng thc C7H16Gi i:B!c 1:Vit mch C di dng mch thang n nguyn t C. Dc dng phnth nht.CCCCCCC (1)B!c 2.1:B) 1 nguyn t Cmch chnh n nguyn t Ctrn lm mch nhnh.Mch chnh by gigm m = n 1 nguyn t C.Di chuyen mch nhnh t$ v tr C2 dn v tr C2chan, dn v tr Cn nu (n -1) l s l) de dc cc dng phn tip theo.2CCCCCC (2)CCCCCCC (3)CB!c 2.2:Tip t c b) 2 nguyn t C de lm nhnh. Mch chnh by gigm a = n 2 nguyn t C.- Vit cc dng phn gm hai nhnh moi nhnh gm 1 nguyn t Cbang cch c dnh nhm ny di chuyen nhm kia t$ v tr C2 dn v tr Ca-1
  4. 4. a +1 nu a lTrang 4CCCCCC (4)CCCCCC (5)C CCCCCC (6)C CDi chuyen dng th i hai nhnh cng lc cng lin kt cng 1 nguynt C t$ v tr C2 ln lt dn v tr Ca nu a l s chan, dn v tr C22s l). CCCCCC (7)CVit cc dng phn ch, gm 1 nhnh gm 2 nguyn t C lm nhnh batdu t$ v tr C3 dn v tr Ca-2 th d$ng li de trnh trng lap.CCCCC (8)CCDe thc hien dc bc vit cc dng phn ch, gm 1 nhnh gm p =2, 3, 4 nguyn t C lm nhnh bat du t$ v tr Cp+1 dn v tr Ca-p-1nydi h-i phn t ban du phi c ti thieu l 3p + 1 s nguyn t C trong phnt.B!c 2.3:B) 3 nguyn t C de lm nhnh. Mch chnh by gigm b = n 3nguyn t C.- V s nguyn t C trong phn t C7H16 l 73 . 3 + 1 nn khng thevit cc dng phn ch, gm 1 nhnh gm 3 nguyn t C lm nhnh.Vit cc dng phn gm 3 nhnh moi nhnh gm 1 nguyn t C bangcch c dnh nhm ny di chuyen nhm kia t$ v tr C2 dn v tr Ca-1.
  5. 5. De thc hien dc bc vit cc dng phn gm q = 2, 3, 4.. nhnhmoi nhnh gm 1 nguyn t C di h-i phn t ban du phi c ti thieu l 2q+ 1 s nguyn t C trong phn t.n (nu n chan) v dn v trTrang 5CCCCC (9)C CDe thc hien dc bc vit cc dng phn gm q = 2, 3, 4 nhnhlin kt vi q nguyn t Cmch chnh m moi nhnh gm 1 nguyn t Cdi h-i phn t ban du phi c ti thieu l 2q + 2 s nguyn t C trong phnt. Phn t C7H16 khng tha mn diu kien ny nn n ch$ c 9 dngphn.B!c 3: Din H vo mch C sao cho dng ha tr ca cc nguyn t ta sedc tt c cc dng phn cn tm.AnkenGi i:B!c 1: Xc dnh do bt b o ha ( s lin kt . hoac s vng ca phn t ccng thc CxHy) theo cng thc:2x 2 y2a+ =Nu a = 1, 2th c dng phn l xicloankan v dng phn cis - trans. Vitcc dng phn xicloankan v dng phn cis - trans d. Phn ny cc em tvit ly.B!c 2:Vit mch C di dng mch thang n nguyn t C v vit lin kt . v tr C1. Dc dng phn th nht.C % CCCCCC (1)Di chuyen lin kt . t$ v t C1 dn v tr C2Cn 1 ( nu n l s l)) se dc cc dng phn tip theo.2
  6. 6. n 1 nu ( n 1) l s chan, dnTrang 6CC % CCCCC (2)CCC % CCCC (3)B!c 3.1:B) 1 nguyn t Cmch chnh n nguyn t Ctrn lm mch nhnh.Mch chnh by gigm m = n 1 nguyn t C.Di chuyen lin kt . t$ v tr C1 dn v tr Cm-1 de dc cc dng phntip theo.C % CCCCC (4)CCC % CCCC (5)CCCC % CCC (6)CCCCC % CC (7)CCCCCC % C (8)CDi chuyen mch nhnh t$ v tr C2 dn v tr C2v tr Cn nu (n -1) l s le de dc cc dng phn tip theo.2C % CCCCC (9)CCC % CCCC (10)CCCC % CCC (11)CCCCC %CC (12)C
  7. 7. Trang 7CCCCC % C (13)CB!c 3.2:Tip t c b) 2 nguyn t C de lm nhnh. Mch chnh by gigm a = n 2 nguyn t C.- Vit cc dng phn gm hai nhnh moi nhnh gm 1 nguyn t C bang cchc dnh nhm ny di chuyen nhm kia t$ v tr C2 dn v tr Ca-1CC % CCCC (14)CCCC % CCC (15)CCCCC % CC (16)CCCCCC % C (17)Cng v!i moi cng thc thu dc Di chuyen lin kt . t$ v tr C1 dn v trCa-1 de dc cc dng phn tip theo.C % CCCC (18)C CCC % CCC (19)C CCCC % CC (20)C CCCCC % C (21)C C
  8. 8. a 1 nu a le. Nu phn t c mch chnha +1 nu a l s le.Trang 8C % CCCC (22)C CCC % CCC (23)C CCCC % CC (24)C CNu phn t c mch chnh di xng th di chuyen lin kt . t$ v trC1 dn v tra C nu a chan, dn C22khng di xng th di chuyen lin kt . t$ v tr C1 dn v tr Ca-1CCCC % C (25)C C(phn t c mch C di xng nn CTCT (25) trng vi (22) v (24) trng vi (23)Di chuyen dng th i hai nhnh cng lc cng lin kt cng 1 nguyn t C t$v tr C2 ln lt dn v tr Ca nu a l s chan, dn v tr C22CC % CCCC (26)CCCC % CCC (27)CVit cc dng phn ch, gm 1 nhnh gm 2 nguyn t C lm nhnh batdu t$ v tr C3 dn v tr Ca-2 th d$ng li de trnh trng lap.C % CCCC (28)CCDe thc hien dc bc vit cc dng phn ch, gm 1 nhnh gm p =2, 3, 4 nguyn t C lm nhnh bat du t$ v tr Cp+1 dn v tr Ca-p-1 ny dih-i phn t ban du phi c ti thieu l 3p + 1 s nguyn t C trong phn t.
  9. 9. ng v!i moi cng thc thu dc Di chuyen lin kt . t$ v tr C1 dn v trCa-1 ( nu mch chnh di xng th d$ng liv tr Cde dc cc dngn 3 Trang 92phn tip theo).CC % CCC (29)CCB!c 3.3:B) 3 nguyn t C de lm nhnh. Mch chnh by gigm b = n 3nguyn t C.- V s nguyn t C trong phn t C7H14 l 73 . 3 + 1 nn khng thevit cc dng phn ch, gm 1 nhnh gm 3 nguyn t C lm nhnh.Vit cc dng phn gm 3 nhnh moi nhnh gm 1 nguyn t C bangcch c dnh nhm ny di chuyen nhm kia t$ v tr C2 dn v tr Ca-1.De thc hien dc bc vit cc dng phn gm q = 2, 3, 4nhnh moi nhnh gm 1 nguyn t C di h-i phn t ban du phi c ti thieul 2q + 1 s nguyn t C trong phn t.CC % CCC (30)C CDe thc hien dc bc vit cc dng phn gm q = 2, 3, 4nhnh linkt vi q nguyn t Cmch chnh m moi nhnh gm 1 nguyn t C di h-iphn t ban du phi c ti thieu l 2q + 2 s nguyn t C trong phn t.CCC % CC (31)C CCCCC % C (32)C C
  10. 10. Trang 10* Ch :V c bn vit cc dng phn ca anken, ankin ging vi ankan. T$khung cacbon ca ankan ta di chuyen v tr lin kt di de dc cc dng phnca anken hoac ankin v thm bc vit dng phn xicloankan v cis trans.Di vi ankin th c thm dng phn vtr lin kt .: he lin kt . lin hpv khng lin hp.Khi di chuyen lin kt . phi ch tr ng hp mch cacbon di xngde loi b- mot s dng phn trng lap.Xc dnh do bt bo ha ( s lin kt ( hoac s vng ca phn t) ccng thc CxHyOzNtXv) theo cng thc:- Do bt bo ha D ca mot hp cht hu c l tong s lin kt p v svng trong mot hp cht hu c.2 2 ( )x + y + v + t2D =Ch : - Cng thc tnhtrn ch, p d ng cho hp cht cong ha tr.- Cc nguyn t ha tr II nh oxi, lu huynh khng nh hng ti dobt bo ha.- 1 lin kt di ( = )Do bt bo ha D = 1- 1 lin kt ba ( )Do bt bo ha D = 2- 1 vng noDo bt bo ha D =1VD: - Benzen: C6H6 c2.6 2 64+ D = =2 Phn t c 3 lin kt p + 1 vng = 4.- Stiren: C7H8 c2.7 2 65+ D = =2 Phn t c 4 lin kt p + 1 vng = 5.H
  11. 11. Trang 112. Phng php :Phng php chung: Cc bc th ng dng de vit cng thc cu tohay xc dnh cc dng phnBc 1: Tnh do bt bo ha (s lin kt p v s vng).Bc 2: Vit cu trc mch cacbon (khng phn nhnh, c nhnh, vng)v da lin kt boi (di, ba) vo mch cacbon nu c.Bc 3: Da nhm chc vo mch cacbon (thng th ng cc nhm chccha cacbon th ng dc da lun vo mchbc 3). Lu dn tr nghp km bn hoac khng tn ti ca nhm chc (v dnhm OH khng bnv se b chuyen v khi gan vi cacbon c lin kt boi).Bc 4: Din s H vo de dm bo d ha tr ca cc nguyn t, sau dxt dng phn hnh hc nu c. Ch vi cc bi tap trac nghiem c thekhng cn din s nguyn t H.Ti tin hnh ging dy cho hai nhm di tng hc sinh ca lp 11A3vit cc cng thc cu to khc nhau ca cng cng thc phn t hp chthu c theo hng dan sch gio khoa:+ Nhm 1: Nhm di tng hc sinh kh, gi-i.+ Nhm 2: Nhm di tng hc sinh trung bnh, yu.p dng phng php vit cng thc cu to cho hp cht h u c sau: Ankan : T$ 4 nguyn t cacbon tr ln c dng phn cu to, d l dngphn mch cacbon. Xicloankan: Du tin vit dng phn c vng ln nht, sau d dn vngc t hn mot C de to mot nhnh, tip theo l vng c t hn 2C de to hainhnh CH3 hoac mot nhnh C2H5, gi mot nhnh CH3 v di chuyen nhnhCH3 cn li, lm tng t dn vng c 3C. Anken : C hai loi dng phn- Dng phn cu to: Dng phn mch cacbon ( mch thang , mch nhnh)v dng phn v tr lin kt di.- Dng phn hnh hc: Diu kien de c dng phn hnh hc
  12. 12. + Phi c ni di C=C trong phn t.+ Moi nguyn t mang ni di phi mang hai nhm th (hay nguyn t)khc nhau. Nu mch chnh nam cng pha ta c dng phn cis, nu mchchnh nam khc pha ta c dng phn trans.Trang 12R1 R3C = CR2 R4Diu kien : R10 R2 , R30 R4 Ankadien: Tng t anken c hai loi dng phn: cu to v hnh hc Ankin: T$ 4 nguyn t cacbon tr ln c dng phn v tr nhm chc, t$C5 tr di c dng phn mch cacbon. Aren: T$ C8H10 tr di c cc dng phn v v tr tng di ca cc nhmankyl xung quanh vng benzen v v cu to mch cacbon ca mch nhnh.Kt qu
  13. 13. thu dc nh sau:+ Nhm 1: Dt dc kt