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8/8/2019 Sang Kien Kinh Nghiem Mon Hoa
1/17
Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 1
PHN I: M UI. L Do Chn Ti: ng Phn l vn kh, rng v suyn sut ch ng trnh ha hu c v n lin
quan n cu to ha hc, tnh cht ca cht,s bin i cht ny thnh cht khc lnhng vn then cht ca b mn ha hc.
Vn ng phn lun c quan tm nhiu ch ng trnh hc, kim tra, thituyn sinh i hc-Cao ng trong cc nm qua. Mu thun gia bi lm ca hc sinh vi p n ca thi. S lng tng ca hc sinh khng bit m nh xc nh ng, s l ng cccht hay ch a, lm th no xc nh ng, cc cht mt cch nhanh chng.
p ng yu cu chy ua thi gian vi h nh thc thi trc nghim hin nay, vinhng tr n tr ca hc sinh nh nu, cng vi s yu mn ngh nghip ti xin c nggp mt phn nh kinh nghim ca m nh gp phn gii quyt cc tn tinh trnh by,ng thi hn ch ti a nhng sai lm v kin thc khi ging dy b mn hahc camnh. Ti chon ti: Gii Quyt Vn ng Phn Trong Ha HuC
II. PhmVi D Ti:Ch nghin cu cc vn ng b mn ha hu c ca ch ng trnh THPT.i
t ng l cc vn ng phnd b thiu st, nhm ln,gy kh kh n i vi hc sinh lp11, lp 12, hc sinh luyn thi i hc.
III. nh Ngh a Mt S T Vit Tc Trong Ti:- NTL : Nguyn t l !(l nguyn t khc C v H )- CTPT : Cng thc phn t.- CTCT : Cng thc cu to.
- p : ng phn.- mp : Mt phng.- # hay : Khc.- hchc: Hp cht h c- h/s Hc sinh.- lk: Lin kt.- TH: Trng hp.Ni
tro PDF
Trial
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8/8/2019 Sang Kien Kinh Nghiem Mon Hoa
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 2
PHN II: NI DUNGI. C S L ThuytLin Quan n Vn ng Phn:I.1. Khi nim ng ng v ngh a:I.1.a. Khi nim cht ng ng ng ng l nhng cht c thnh phn phn t h n km nhau mt hay nhiu nhm
CH2 (nhm metilen) nh ng c tnh cht ha hc t ng t nhau. Cc cht ng ng hpthnh mt dy gi l dy ng ng ca cc cht .
L u : Cc cht cng mt dy ng ng th phi t ng t v cu to ha hc v tnhcht ha hc.Phn t khicc cht ng ng lin tip nhau lp thnh mt cp s cng vicng sai d = 14.
V d : Cc chtCH3OH, CH3CH2OH CH3(CH2)xOH (x 0) u thuc mt dy ng ng.I.1.b. ngh a ng ng: Nu bit c 1 cht c thtrong dyng ng th ta s t m c cng thc chung
cho dyng ng ca chng.V d1: Tm cng thc chung cho dyng ng ca ancol metylic c cng thc CH3OHGi : Ta c: CH3OH=> cc cht ng ng c dngCH3(CH2)xOHhay C1+xH3+2xOH t n =1+x th => CnH2n+1OH v x 0 => n 1. VyCnH2n+1OH , n 1 l cng thc chung
cho dyng ng ca ancol metylic. Nhng cht trong cng mt dy ng ng c tnh cht t ng t nhau, do bi
ton hn hp cc cht ng ng c gii quyt bi cng thc chung.V d2: Cho 13,4 (g) hn hp hai ancol X, Y cng mt dy ng ng phn ng ht vi
Na thu c 2,24 lt H2 ( ktc). Hai ancol X, Y lA. CH3OH v C2H5OH B. C2H5OH v C3H7OHC. CH3OH v C3H7OH D. C2H5OH v C4H9OHGi : ( khl khng cho bit thng tin v 2 ancol, nh ng thng tin l ra p n)
Nhn p n ta t cng thc chung cho X, Y l OH H C nn 12 hay OH R , a molBo ton nguyn t hidro linhng ta c: a.1 = nH2.2 = 0,2 mol=>
hh
hh
n
m M =
2,0
4,13 = 67 => p n ng phi c ancol c M > 67(C5) => chnD
32 46 74 8860CnH2n+1OH , n > 1)
CH4O C2 C3 C4 C5( e y : a thuoc phan t kho i:
I.2. Khi nim cht ng phn ng phn l nhng cht khc nhau nhng c cng cng thc phn t.
L u : Nhng cht l ng phn ca nhau tuy c cng cng thc phn t nhng ccu to ha hc khc nhau.I.3. Phn loi ng phn (khng xtng phn quang hc, ng phn cu dng)
ong Phan(cung CTPT)
p cau tao
p lap the
khacCTCT
khacCa u truc khong gianCung CTCT
p mach C ( do mach C thayoi: (khongnhanh, conhanh, machvong)pnhom chc( dos thayoi ve ban cha t nhom chc)p v tr (do s thayoi v tr nhom chc, vtr (so lng) lienke t pi tren mach C)
phnh hoc
(cac nhon the ln namkhac phao i vi mplien ket pi)
(cac nhon the ln namcungphao i vi mp lien ket pi)p cis-
p trans-C = C
A B
a b
C = CA
Ba
b
A # aB # b
A # aB # b
? Khong xet
L u : iu kin c ng phn h nh hc:
Nitro PD
F Trial
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8/8/2019 Sang Kien Kinh Nghiem Mon Hoa
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 3
+ Phi c lin kt i gia hai nguyn t C ( C = C )+ Mi nguyn t C mang lin kt i phi lin kt vi hai nhm th khc nhau
V d: s tr n c ng phn h nh hc th A a v B bI.4. Khi nim ha tr ca nhm chcNhm chc ha tr mt ( hai, ba ) l nhm chc c kh nng to c mt (hai, ba) lin
kt cng ha tr vi nguyn t hay nhm nguyn t khc. Nhm chc ha tr mt gm:-Cl; -OH; -CH=O; -COOH ; -NH2 ; -C CH ; RCOO- Nhm chc ha tr hai gm:-O- ; -CO- ; -NH- Nhm chc ha tr ba gm:
l N (amin bc ba)
I.5. bt bo ha (tng s lin kt trong mt phn t cht hu c ) k hiu: aI.5.a. Cch thnh lp cng thc tnh a trong cht hu c (A) CxHyOzNtClvTheo thuyt cu to ha hc ta rt ra mt s nhn xt sau:+ S electron ha tr ca nguyn t C, N, O, H, halogen ln l t l 4, 3, 2, 1, 1
( : trn N vn cn 1 cp e ha tr ch a lin kt) => s e ha tr = 4x+y+ 2z+ 3t +1v+ Mi lin kt cng ha tr(lin kt , ) c thc hin bng 2 electron ha tr+ S lin kt gia cc nguyn t trong phn t = s nguyn t - 1
=> s e ha tr to lin kt (trongA) = (x + y + z + t + v 1).2+ Gi a l tng s lin kt trong phn t=> s e ha tr to lin kt = 2.aV: s e ha tr ban u = s e ha tr to lin kt + s e ha trto lin kt nn tac: 4x + y + 2z + 3t + v = (x + y + z + t + v 1).2 + 2.a
Hay: a =2
)()22( v yt x ( )
Vy l biu thc tnh s lin kt trong phn t CxHyOzNtClv , vi Nc ha tr 3Nu N s dng 5 e ha tr th N phi dng hp cht ion (NH4 , R-NH3 , NO3 ) th
s lin kt trong hp cht ion = a + 1 hay a (cht ion)= )()2()22( v yt x
: Mt lin kt t ng ng vi mt vng no(s khp vng cng cn 2 e ha tr)Vy c th xem a = s lin kt + s vng no.
I.5.b. ngh a ca a (tng s lin kt ) Bit a , ta c th suy n c cu to ha hcca hp cht hu c (lin kt n,
lin kt i , lin kt ba, mch h , mch vng, vng thm ..)V d 1: C3H6 (a =1) => C3H6 c 1 hay 1 vng no. Vy C3H6 c 2 ng phn cu to
l propen v xiclopropan y hc sinh dqun tr ng hp mch v ng ! Bit a , ta c th vit ng , s lng cc ng phn cu to, v d
(2x +2) +t - (y+v)2
ne u a =0 phant A: no ( ch co lien ket n)neua =1 phan t A:
co 1lien ket pi (C=C,C=O )co 1vongno. (xiclo)
ne u a =2 phan t A:
co1 lkba(C C ,C N)co 2 lk oi (C=C=C, C=C- C=O )co 1 lkoi +1 vong no( , C =C ...)
co 2 vong no(
a =
, , )
Tongso lk pi + So vongno
Bit a , bits nguyn t l ta c th suy n loi nhm chc trong cht AV d2: Xt chtAmch h: CxHyOz , a=
2
22 y x , nguyn t l y l nguyn t oxi=> y = 2x+2- 2a, nu t x = n th Al CnH2n+2-2aOz.
Nitro PD
F Trial
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 4
Nu a = 0, z = 1 (0 + 1 oxi) => A(CnH2n+2O) c 2 trng hp l:(1) Ancol no, n chc, mch h (s C 1) v d: CH3-OH(2) Ete no, n chc, mch h (s C 2) v d: CH3-O-CH3V d3: C4H10O (0 +1oxi) c 7ng phn cu to mch h: 4 ancol v 3 ete.C - C - C - C
C - C - CC
- OHC C C C
C C CC
-O -
"N hom chc hoa tr II" y hc sinh th ng qu n 3 ng phn ete !
Nu a = 1, z = 1 (1 + 1 oxi) => A(CnH2nO) c 4 tr ng hp l:(1) An ehit no, n chc, mch h (s C 1) v d: HCH=O(2) Xeton no, n chc, mch h (s C 3) v d: CH3-CO-CH3(3) Ancol khng no (1 ), n chc, mch h (s C 3) v d: HO-CH2-CH=CH2(4) Ete khng no (1 ), n chc, mch h (s C 3) v d:CH3-O-CH=CH2V d 4: C3H6O (1 +1oxi) c 4ng phn cu to mch h: 1 an ehit,1 xeton,1ancol
khng no(1 ), 1 ete khng no (1 ).An ehit: CH3-CH2-CH=O ; Xeton: CH3-CO-CH3Ancol khng no: HO-CH2-CH=CH2; Ete khng no: CH3-O-CH=CH2 y hc sinh th ng qu n hai tr ng ancol, ete khng no !
Nu a = 1, z = 2(1 + 2 oxi) => A(CnH2nO2) c 9 tr ng hp l:(1) Axit no, n chc, mch h (s C 1) v d: HCOOH(2) Este no, n chc, mch h (s C 2) v d: HCOO-CH3(3) Tp chc 1 anehit + 1 ancol no, mch h (s C 2) v d:HO-CH2-CH=O(4) Tp chc 1 anehit + 1ete no, mch h (s C 3) v d:CH3-O-CH2-CH=O(5) Tp chc 1 xeton + 1 ancol no, mch h (s C 3) v d:HO-CH2-CO-CH3(6) Tp chc 1 xeton + 1 ete no, mch h (s C 4) v d: CH3-O-CH2-CO-CH3(7) Tp chc 1 ancol +1 ete khng no(1 ) (s C 3) v d: HO-CH2-O-CH=CH2
(8) Ancol 2 chc khng no (1 ) (s C 4) v d: HO-CH2-CH=CH-CH2-OH(9) Ete 2 chc khng no (1 ) (s C 4) v d: CH3-O-CH2-O-CH=CH2V d5: C3H6O2 (1 +2oxi) c 8ng phn cu to mch h:Axit: (1) CH3-CH2-COOHEste: (2) HCOO-CH2-CH3 , CH3-COO-CH3Tp chc 1 anehit +1 ancol no, mch h (2) HO-CH2-CH2-CH=O, CH3-CH(OH)-CH=OTp chc 1 anehit + 1 ete no, mch h (1) CH3-O-CH2-CH=OTp chc 1 xeton + 1 ancol no, mch h (1) HO-CH2-CO-CH3Tp chc 1 ancol +1 ete khng no, mch h (1) HO-CH2-O-CH=CH2Ancol 2 chc ; ete hai chc ; tp chc xeton + ete khng tn ti v s C < 4 y hc sinh th ng qu n bn tr ng hp tp chc ! Bit a ta c th suy ra iu kin tn ti cht.V d6: (A) CxHy hay CxHyOz ta c a =
2
22 y x hay 2a = 2x +2-y ( M=12x+y+16z )
V a 0 nn => iu kin l: y 2x+2 v y, M lun chn (v 2a v 2x+2 chn)V d7: (A) CxHyNt hay CxHyOzNt ta c a =
2
22 yt x hay 2a = 2x +2 + t yV a 0 nn => iu kin l: y 2x+2+t v y v t cng chn (l). T ng t hp cht l
CxHyCly hay CxHyOzClv c iu kin l: y+v 2x+2 v y v v cng chn (l)V d8: Hp cht (A) c CTPT l (C4H9Cl)n c my ng phn cu toA. 2 B. 3 C. 4 D. 5Gi : A:(C4H9Cl)n hay C4nH9nCln , iu kin (n+9n) 2.4n+2 => n 1 => chn n = 1Vy A l C4H9Cl (0 +1 Cl) => p n ng lC v A c 4ng phn lCl-CH2-CH2-CH2-CH3 , CH3-CHCl-CH2-CH3 , Cl-CH2-CH(CH3)-CH3 , CH3-CCl(CH3)-CH3. y hc sinh th ng khng bit t m gi tr n !
Nitro PD
F Trial
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 5
Bit a ta c th t m c cng thc chung ca mt dy ng ng no hay ta cth xc nhCTPTca mt cht t cng thc chung.
V d9: Tm cng thc chung ca dy ng ng cc axit n chc khng no mt linkt i, mch h?
Gi : Theo => cc axit cn t m phi c 2 lin kt (1 trng gc + 1 trong chc) t cng thc cc axit l : CxHyOz ( z = 2 v axit cho l n chc)
Vi a = 2 222 y x
= 2 => y = 2x-2. Vy cng thcchung ca dy axit cn t m lCxH2x-2O2 hay CnH2n-2O2 , vi n 3. ( 2C mang lin kt i + 1C trong nhm chc)
V d10: Axit cacboxylic (A) no, mch h c cng thc (C3H4O3)n. Tm CTPT ca AGi : (A) : (C3H4O3)n hay C3nH4nO3n . A c 3n Oxi => s nhm axit l
2
3n => a =2
3n (Ano, mch h nn s lin kt ch nhm chc axit)
Vy ta c:a =2
423.2 nn =2
3n => n = 2 => CTPT ca A l C6H8O6 hay C3H5(COOH)3I.6. Nhm chcNhm chc l nhm nguyn t gy ra nhng phn ng c tr ng ca phn t hp
cht hu c . n chc : Hp cht ch c1 nhm chc trong phn t.V d1: A l hp cht n chc c cng thc (CH2O)n. SCTCT ca A lA. 1 B. 2 C. 3 D. 4Gi : V A n chc => s nguyn t oxi= 1(chc cha 1 oxi) hoc = 2.Vi n =1 => CTPT CH2O c 1ng phn l: H-CH=OVi n = 2 => CTPT C2H4O2 c 2 ng phn ( 1 axit: CH3COOH , 1 este: HCOOCH3) y hc sinh nhm: n chc l 1 oxi ch khng ngh n chc l 1 nhm chc. a chc : Hp cht c t2 hay nhiu nhm chc ging nhau trong phn t.V d 2:SCTCT ca ancol a chc ng vi cng thc C3H8Ox l
A. 2 B. 3 C. 4 D. 5Gi : Theo iu kin tn ti ancol, theo bi th => 2 x 3x = 2 => CTPT C3H8O2 c 2 ng phn ancol a chcx = 3 => CTPT C3H8O3 c 1 ng phn ancol a chc => ChnB. Tp chc : Hp cht c 2 hay nhiu loi nhm chc khc nhau. Mt loinhm chc : Hp cht c th l n chc hoc a chc.V d 3: (A) l hp cht c cng thc C3H8Ox. Bit A tc dng c vi Na v ch cha
mt loi nhm chc trong phn t. SCTCTca A lA. 2 B. 3 C. 4 D. 5Gi : Theo bi th A l ancol vi 1 x 3 => ChnD.HO-CH2-CH2-CH3 , CH3-CH(OH)-CH3HO-CH2- CH(OH)-CH3 , HO-CH2-CH2-CH2-OH , CH2(OH)-CH(OH)-CH2(OH)I.7. Cc b c vit ng phn cu to B c 1: Phitnh nhm nhanh s lin kt (lin t ng ti s vng no) B c 2: Vit dng mch C , in ng s lin kt vo mch C B c 3: B tr nhm chc vo mch C B c 4: in H vo sao cho ng ha tr ca CVi h nh thc thi trc nghim hin nay, khng nht thit phi tin hnh theo trnh t cc
bc nh trn, m ta cn mt s vn kinh nghim sau:Vi mch C ta bt u mch thng mch nhnh (1 nhnh 2 nhnh .)Vi lin kt trong gc hirocacbon ta dng m i tn cho n chy vo cc lin kt t
u mch n v tr i xng th dng li.Vi mch vng ta bt u t vng to vng nh, khng nhnh c nhnh.Vi nhm chc ha tr I ta cho n chy vo cc nguyn t sao cho ng ha tr.Vi nhm chc ha trII ta cho n chy vo gia cc lin kt C-C trn mch C.
Nitro PD
F Trial
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 6
Vi nhm chc ha trIII ta vit ba C , cc C cn li cho chy vo mch C.Nu c 2 nhm chc th c nh mt nhm tr n mi mch v cho nhm kia chy.
V d 1:C bao nhiung phn cu to ng vi cng thc C5H10Gi : C5H10 c a = 1 => C5H10 c 1 mch h hoc c 1 vng noNu mch h (anken) ta lm nh sau:
C - C - C - C - C C - C - C - CC C - C - C
C
C
a u hoa tr CPi
=> c 5ng phn mch hNu mch vng: (xicloankan):
C CC
=> c 5ng phn mch vngI.8. Tnh cht ha hc c tr ng ca nhm chc.Cn nm k tnh cht cc nhm chc sau: OH; O ; CH=O; CO ; COOH ;
COO-R ; NH2 ; -Cl ; C CH ; CONH (peptit) ; R-NH3 (hp cht ion). .
1. hchc (A)+ Na H2
Muo i=> Aphai coH linh ong. VayAla: - OH(ancol ; phenol )
- COOH (axit , tap chc ...)
2. hchc (A) => A phai co cac nhom chc sau:+dd.NaOH(to , to )
San pham++
- OH (phenol); R-X (dan xuat halogen)- COOH (axit , tap chc ...)- COO-R' (este) , ( to)-CO-NH-(hp cha t co lk peptit ) ( to)- NH4 / R-NH3 (hp chat ion)( t
o)
3.hchc (A) =>A phai co nhom-CH=O. VayAla:+ddAgNO3 /NH3
toSan pham
- CH=O (anehit)- HCOOH(axit fomic)- HCOO-R' (este cua axitfomic)- HCOONH4, HCOONa(muoicua HCOOH)
- glucoz , fructoz, mantoz
Ag
+ddAgNO3 /NH3 San phamey:ne uhchc (A) Aco nhom-CH=O nh (3)Ala ankinau mach (HC CH; R-(C CH)x=>
4. hchc (A) => A phai co nhom -CH=O nh (3)+Cu(OH)2 / OH-
toSan pham
Cu2O
5. hchc (A) => A phai co:+Cu(OH)2 San pham
ey: - neu A co nhieu nhom -OH canh nhau => se taophc mau xanh lam
(ngay kt)
- nhieu nhom OH canh nhau: (C2H4(OH)2, C3H5(OH)3,, glucoz, fructoz, mantz,sacaroz... )- co nhom axit (-COOH)- co cha t 2 lien ket peptit tr len (p mau biure)
- ne u A co nhom -COOH se hoa tan Cu(OH)2 tao dd mau xanh( do ion Cu2+ hirat hoa )
- ne u A co t 2 lk peptit tr len se tao phc mau tm (tripeptit, polipeptit , protein ...)
6.hchc (A)
=>Aphai co:
+ddBr2(nc brom) co nhat mauddBr2(kt)- lien ke t pi goc HCkemben (anken, ankaien,ankin, h/c khong no:CH2=CH-COOH...- co nhom chc -CH=O nh (3) (tr fructoz)- co cau tao ac biet: NH2OH ,,
: mt s cht bn c th lm mt mu Br2 khan khi c xt thch hp (CxHy,benzen, hp
chtcacbonyl c H , axit c H
7.hchc (A) =>A phai co lien ke t pi (tr lkpi trong -COOH; -COO-R'+H2(Ni, to) Sanphamcong
Nitro PD
F Trial
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 7
II. Gii Quyt Vn ng Phn:II.1. Phn tch gii quyt mt s bi tp ng phn c bnII.1.a. S l ng cht ng phn v ha tr ca cc nguy n t lV d a.1: Xc nh s ng phn cu to ca cc cht ng viCTPTln lt l C4H10 ;
C4H9Cl ; C4H10O ; C4H11N v rt ra nhn xt ?H ng gii quyt:Vi: C4H10 (0 ) => ch c ng phn mch C (mch h)(c 2ng phn)
C - C - C - C va C - C - CC
Vi: C4H9Cl (0 +1Cl) => cng phn mch C v v tr nhm chc (4 p)C - C - C - C
C - C - CC
- Cl
"Cach tm so ong phan hp chat cha nhom chc hoa tr I"Vi: C4H10O (0 + 1 O) => cng phn mch C, v tr nhm chc, loi chc (7p)
C - C - C - CC - C - C
C
- OH C C C CC C C
C
-O -
"Nhom chc hoa tr II"Vi: C4H11N (0 + 1 N) => cng phn mch C, v tr nhm chc v bc amin
C - C - C - C
C - C - CC
- NH2C C C C
C C CC
-NH- C - N - CC
C
Nhn xt: Vi cng s C, cng s th s lng ng phn t l thun vi ha tr canguyn t l ! Tc l s ng phn hp cht cha N > O > Cl
V d a.2: C14 M253 HA 2010: Trong s cc cht: C3H8, C3H7Cl, C3H8O v C3H9N; chtc nhiu ng phn cu to nht l
A. C3H9N. B. C3H7Cl. C. C3H8O. D. C3H8.V d a.3: Cho cc cng thc phn t sau : C3H7Cl , C3H8O v C3H9N. Hy cho bit s sp
xp no sau y theo chiu tng dn s lng ng phn ng vi cc cng thcphnt A. C3H7Cl < C3H8O < C3H9N B. C3H8O < C3H9N < C3H7ClC. C3H8O < C3H7Cl p n C, hc sinh lng tng khi vit p mch CV d a.5: (A) l hp cht n chc c cng thc C5H10O2. C bao nhiung phn
cu toca Atham gia c phn ng trng g ng ?A. 3 B. 4 C. 5 D. 6
H ng gii quyt: C5H10O2 c a = 1, 2 oxi , m A n chc => A l axit hoc este, v Ac phn ng trang g ng => A l este ca axit formit(HCOO-R)
C - C - C - CC - C - C
C
HCOO-
=> Chn p n B, hc sinh lng tng khi vit p mch C
Nitro PD
F Trial
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 8
V d a.6: (A) l hi rocacbon c cng thc C5H10. C bao nhiung phn cu to caA lm nht mu dung dch brom iu kin th ng ?
A. 5 B. 6 C. 7 D. 8H ng gii quyt: C5H10 c a = 1 => A c 1 hay 1 vng no. M A lm mt mu dung
dch brom => A l anken hay dn xut ca xiclopropan (vng ba cnh).Anken:
C - C - C - C - C C - C - C - CC
C - C - CC
C
a u hoa tr CPi
=> c 5 ankenVng ba cnh:
CC
=> c 3ng phn chc vng ba cnh => Chn p nD y hc sinh qu n tr ng hp v ng 3 cnh ! => chn p n AV d a.7: (A) l dn xut ca benzen c cng thc C8H10O. Gi a l tng sng phn
cu to ca A ; b l tng s ng phn cu to ca A tc dng c vi Nanh ng khng tcdng c vi NaOH; c l tng s ng phn cu to ca A phn ng c vi NaOH ; d ltng s ng phn cu to ca A khng tc dng c vi Na v NaOH. Vy tng a+b+c+d l
A. 10 B. 20 C. 30 D. 40H ng gii quyt: C8H10O c a = 4, 1oxi. A l dn xut ca benzen (3 +1vng = 4)=> Oxi khng th mang lin kt pi => n nhm OH hoc ete (-O-)(A) tc dng c vi NaOH => A l phenol ( c = 3 hay 10 ? )
C - C
- OHv
CC C C C
C- OH
h/s chn 3 l sai lm ! v cn vng c 2 nhnh. => c 10ng phn + NaOH(A) tc dng c vi Nam khng phn ng vi NaOH => A l ancol thm
C - C
- OHV
- CH2-O H
C
=> b = 5 . h/s thng chn 2(A) khng tc dng c vi Na v khng phn ng vi NaOH => A l ete
- O-
C - C
v
CC C C C
C
- O-
=> d = 5Tm li A c 5 ete , 5 ancol thm , 10 phenol. Vy chn p nDV d a.8: C bao nhiung phn cu to ng vi cng thc phn t C4H8OA. 3 B. 11 C. 15 D. 19H ng gii quyt: C4H8O c a =1 => c mch h hay c 1 vng no.
Nu C4H8O dng mch h, 1 + 1oxi c cc tr ng hp sau:TH1:An ehit no, n chc mch h.(2 ng phn) (c phn ng trng g ng)TH2:Xeton no, n chc mch h. (1 ng phn)TH3:Ancol khng no 1 pi n chc. (4 ng phn) (c phn ng vi Na)TH4:Ete khng no 1 pi n chc. (4 ng phn)TH5:Ancol mch vng no n chc. (4 ng phn) (c phn ng vi Na)TH6:Ete mch vng no n chc. (4 ng phn)Cc ng phn tr n c xc nh theo cc s d i y.
Nitro PD
F Trial
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C - C - C - CH=OC C C - C -
O
C - C - C - CC - C - C
C
C - C - C - C
- OH C C C C
C C CC
-O-
C C C C
C - OH C - O-
II.1.b.Tng hp gy nh cho hc sinh v Phn t khi, S C, S ng phnSau khi nghin cu cch vit ng phn cc cht cha nhm chc ha trI ; cha
nhm chc ha tr II , ha tr III ta c th tng hp s ng phnca cc dy ng ng nhsau: ( quy lut bin i phn t khi , s ng phnC1 C4 v mt s ngoi lC5)
CnH2n+2O(n >1)
R-OHSo pAncol no ,n chc, mach h
CH 4O 1 p32
46
60
74
88
C2H6O 1 p
C3H8O 2 p
C4H10O 4 p
C 5H 12O 8 p
Rat t thay
e thi
CnH2n+3N(n >1)
-NH2 , -NH - ,So pAmin n o,n chc, mach h
C H5N 1 p31
45
59
73
87
C2H7N 2 p
C3H9N 4 p
C4H11N 8 p
C 5H 13N 17 p
R a t t tha y
e thi
-N -
CnH2nO (n >1)
R-CH=OSo pAnehit no,n chc, m ach h
CH2O 1 p30
4 4
58
72
86
C2H4O 1 p
C3H6O 1 p
C4H8O 2 p
C 5H 10O 4 p
R a t t tha y
e thi
CnH2nO2 (n >1)
R-COOHSo pAxit no ,n chc, mach h
CH2O 1 p46
60
74
88
10 2
C2H4O 1 p
C3H6O 1 p
C4H8O 2 p
C 5H 10O 4 p
R a t t tha y
e thi
CnH2nO2(n > 2)R-COO-R'
So pEste no,n chc, mach h
C1
6074
88
102
C2H4O2 1 pC3H6O2 2 p
C4H8O2 4 p
C5H10O2 9 p
Rat t thay e th i
CnH2n+1O2N (n > 2)H2N-R-COOH
So paminoaxit (1pi)no, mach h
C1
7589
103
C2H5O2N 1 pC3H7O2N 2 p
C4H9O2N 5 p
Ra t t tha y e thi
CH5O2N 1 p63
7791
105
C2H7O2N 2 pC3H9O2N 4 p
C4H11O2N 9 p
Ra t t tha y e thi
CH6O3N2 1 p46
6074
88
C2H8O3N2 2 pC3H10O3N2 4 p
C4H8O3N2 8 p
Ra t t tha y e thi
CnH2n+3O2N (n > 1)R-COONH3-R '
So php cha t ion,no, mach h
CnH2n+4O3N2 (n > 1)R-NH3NO3
So php cha t ion(3O,2N) no , m.h
C5H11O2N 12 p
Nhn xt:Ly M (CH4O) = 16+16 = 32 l chun => cc ng ng lin tip th + 14 vCM (amin n, no, h) = M(ancol) - 1M (an ehit / xeton n, no h) = M (ancol) - 2 (v c 1 pi nn mt 2H )M (axit / este n, no h Cn ) = M (ancol n, no, h Cn+1)
Nitro PD
F Trial
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 10
i vi hp cht cha N CxHyO2Nth ly M = 75 l chun (Glyxin)S ng phn cc cht ng ng dng nh bin i c tnh qui lut t C1 C5 (ch
c mt s t ngoi l). Qui lut l: Khi c xut hin ng phn th ta ly s ng phn nhn 2 th c s ng phn ca cht tip theo
Ngoi l:(c 3 ngoi l)S p amin C5 = s pamin C4 x 2 1 = 17S p este C5= s p este C4 x 2 + 1 = 9Dy aminoaxit v dy hp cht ion(O2N)bt u C4 c ngoi l v c nhiu nhm
chc nn phc tp h n. S p C4= s p C3 x 2 + 1 (khng xt C5 , thi cng n)II.2. Phn tch giiquyt mt s bi tp ng phn trong thi tuyn sinh
V d 1: C6 M263 HA 2008: S ng phn este ng vi cng thc phn t C4H8O2 lA. 5. B. 2. C. 4. D. 6.Gi : Este C4H8O2 c a = 1 => c 4 p ( thuc bng tng hp)=> chn C
V d 2: C29 M231 C A 2007S hp cht n chc, ng phn cu to ca nhau c cng cngthc phn t C4H8O2, u tc dng c vi dung dch NaOH l
A. 5. B. 3. C. 6. D. 4.
Gi : C4H8O2 c a = 1, n chc, mch h , phn ng vi NaOH => n l axit hoceste: Vy c 2 axit C4 + 4 este C4 => chnCV d 3: C30 M182 C A 2009: S hp cht l ng phn cu to, c cng cng thc phn tC4H8O2, tc dng c vi dung dch NaOH nh ng khng tc dng c vi Na l
A. 3. B. 2. C. 4. D. 1.Gi : C4H8O2 c a = 1 , 2O , khng tc dng vi Na, nhng tc dng c vi NaOH
=> n khng l axit, ancol m C4H8O2 l este n chc no, mch h (khng th c mchvng). Vy c 4 este C4 => chnC.V d 4: C32 M174 HB 2010: Tng s hp cht hu c no, n chc, mch h, c cng cngthc phn t C5H10O2, phn ng c vi dung dch NaOH nh ng khng c phn ng trng
bc l A. 4. B. 5. C. 8. D. 9.Gi : Theo th => C5H10O2 ( a = 1, 2O) l axit hoc este tr este ca axit formic.Vy c 4 axtit C4 + 9 este C5 4 este HCOO-R ( nh V d a.5. / II.1a)= 9=> ChnD
V d 5: C18 M253 HA 2010: Tng s cht hu c mch h, c cng cng thc phn tC2H4O2 l
A. 1. B. 2. C. 4. D. 3.Gi : C2H4O2 c a = 1, 2O, C2. Nh V d 5 / I.5.b.=> cc trng hp sauC2H4O2 l axit C2 c 1 pC2H4O2 l este C2 c 1 pC2H4O2 l an ehit + ancol c 1 p. Hc sinh quyn tr ng hp ny => chn B (sai)Khng c cc trng hp khc. Vy chnD
V d 6: C20 M268 C A 2010:ng vi cng thc phn t C3H6O c bao nhiu hp cht mchh bn khi tc dng vi kh H2 (xc tc Ni, to) sinh ra ancol ?
A. 3. B. 4. C. 2. D. 1.Gi : Nh V d 4 / I.5.b. Ta d dng => C3H6O l an ehit, xeton v ancol khng no
=> chn AV d 7: C18 M175 HA 2009: Cho 10 gam amin n chc X phn ng hon ton vi HCl (d ),thu c 15 gam mui.S ng phn cu to ca Xl
A. 4. B. 8. C. 5. D. 7.
Gi : Ta c Mamin n =
5,36
101510 = 73. Amin = 73 l C4H11N c 8 p => chnB
Nitro PD
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V d 8: C6 M182 C A 2009: S ng phn cu to ca amin bc mt c cng cng thc phnt C4H11N l
A. 4. B. 2. C. 5. D. 3.Gi : C4H11N c a =0, C4 => c 4 ng phn amin bc 1 t ng t nh ancol C4 v
nhn NH2 t ng t nhm OH. Vy chnANhn xt. S ngphn amin bc 1 = s ng phn ancol (u no, n chc, h)
S ng phn amin bc 2(-NH-)= s ng phn ete (-O-)(nhm ha tr 2)V d 9: C11 M285 HB 2007: S cht ng vi cng thc phn t C7H8O (l dn xut cabenzen) u tc dng c vidung dch NaOH l
A. 2. B. 4. C. 3. D. 1.Gi : C7H8O c a = 4 = 3pi+ 1 vng => O ch c th nhm OH , -O-, theo, t gi
V d a.7 / II.1.a ta c 3ng phn thuc hp cht phenol lC
- O H=> chn C
V d 10: C30 M231 C A 2007: Este X khng no, mch h, c t khi h i so vi oxi bng 3,125v khi tham gia phn ng x phng ho to ra mt an ehit v mt mui ca axit hu c . Cbao nhiu cng thc cu to ph hp vi X?A. 2. B. 5. C. 3. D. 4.
Gi : Meste = 3,125x32 = 100. Este c M = 100 l C5H8O2 v bit Meste = 102 lC5H10O2 (bng tng hpII.1b.)
C5H8O2 c a = 2. theo th X c dng R-COO-CH=CH-R.C - COO-CH= C
CH-COO-CH=C - C=> chn p n D
Nhn xt: thira i ra linhng cht quen thc t C1 C5 vi nhiu gc nh n khc nhau,nu ra C6, C7 th lun c hn ch s ng phn t gi ca bi hoc l dn xut ca
benzen.My nm gn y ra bi tp ng phn lun gn lin vi tnh cht hahc c
tr ng, do hc sinh cn nm k phn l thuyt mcI.8. s c cch suy lun rt nhanhchng ti p s.
Cc n m tip n c l i thi s khai thc vn ng phn nhiu hp chtCxHyO2Nhoc CxHyO2N3 !
II.3. Phn tch gii quyt mt s bi tp ng phn tng hp, nng caoBi II.3.1:Khi cng Cl2 vo axetilen th thu c hai sn phm ng phn. Xc nh
CTCT ca hai sn phm .Gi : Cc phn ng c th xy ra lCH CH + Cl2 1:1 CHCl=CHCl (A)CH CH + 2Cl2 2:1 CHCl2 CHCl2 (B)Nu xt v ng phn cu to th c sn phm A, B u c 1 ng phn cu to duy
nht. Vy theo yu cu bi th 2 sn phm ng phn phi l ng phn h nh hc canhau. Vy Atha mn iu kin c ng phn h nh hc (L u : I.3.)v 2 ng phn l
C C
H
Cl
H
Clong phan Cis-
C C
H
Cl
Cl
Hong phanTrans- (Cl l nhm ln h n H v ZCl = 17, ZH =1)
Nitro PD
F Trial
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Sang Kien Kinh Nghiem Trang 12
Bi II.3.2:Ancol (A) khi tch H2O th thu c mt hn hp gm 3 anken th kh iu kin thng. Xc nh CTCT ca ancol A, vit CTCT ca 3 sn phm tr n ?
Gi :A O H 2 anken (th kh) => A l ancol no, n chc, mch h, c s C 4.V C2H4 v C3H6 khng cng phn cu to.Vy A l C4H9-OHAncol O H 2 nhiu anken => A l ancol bc cao khng i xng,v trong 3 anken
ng phn phi c 2 anken l ng phn h nh hc. Vy A v cc sn phm l :
C CH
CH3
H
CH3Cis- But-2-en
C CH
CH3
CH3
HTrans- But-2-en
CH3-CH-CH2-CH3OH
CH2 = CH-CH2-CH3 (khong co ong phan hnh hoc)
CH3-CH = CH-CH3
But-1-en- H2Oxt,t0
(A)
Bi II.3.3:Hp cht hu c A(C, H, O) c phn t khi M = 74. Gi a l tng s ngphn cu tomch hca A v b l tng s ng phn cu tomch hca A c kh nngtham gia phn ng trang g ng. Tng a+b l :
A. 14 B. 16 C. 23 D. 25Gi : t A: CxHyOz => 12x + y +16z = 74 (1), ( s 74 rt quen th c ! )(1) => 16z < 17 => z chn z = 1, 2, 3, 4.
Vi z = 1 => 12x+y = 58 =>22)1258(
83,412
58
x x y
x => chon x = 4, y = 10
Vy viz = 1=> CTPTl C4H10O (0 +1Oxi)Vy viz = 2=> CTPTl C3H6O2 (1 +2Oxi) , : T ng 1O => gim1CH4Vy viz = 3=> CTPTl C2H2O3 (2 +3Oxi)D thyz = 3 loi. Tm li ta c:
A(C,H,O) = 74mach h
C4H10O
C3H6O2
C2H2O3
(0 pi+1O)4 Ancol nh:V du a.1 / II.1.a
3 Ete nh:V du a.1 / II.1.a
1 Aaxit nh:V du 5 / I.5.b.2 Este nh:V du 5 / I.5.b.2 HO-R-CH=O nh:V du 5 / I.5.b.1 R-O-R'-CH=O nh:V du 5 / I.5.b.1 R-O-R'-CO-R" nh:V du 5 / I.5.b.1 HO-R'-O-R(pi)" nh:V du 5 / I.5.b.
(1 pi+1O)
1 HOOC - CH=O (axit + anehit)1 H-COO- CH=O (este + anehit)(2 pi+1O)
A(C,H,O) = 88 ?
A(C,H,O) = 60 ?
+_
+_ 14
14
Vy a =17 , b = 6 => chn p n C.Bi II.3.4:Hp cht thm ng vi cng thc phn t C8H8O2 c a ng phn khng tc
dng vi Na nh ng tc dng c vi dd NaOH v nc brom; c b ng phn khng tcdng vi Na nh ng tc dng c vi dd KOH theo t l mol t ng ng 1:2. Tng a+b l
A.7. B.5. C. 6. D. 8.Gi : Nh V d a.7 / II.1.a => p n D. (este ca HCOOH v este ca phenol)
Nitro PD
F Trial
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II.4. Bi Tp Tho lun , Tham KhoBi II.4.0:Hp cht hu c A c cng thc C3H7O2N. C bao nhiung phn cu to
mch h ca Atc dng c vi dung dch NaOH.?Gi : Bn c cng suy ngm.(A) CxHyO2N1 NaOH Sn phm => A c th l :
Phenol cha N !
Axit cha N !Este cha N !Hp cht ion R-COONH3-R (bc 1, 2, 3) !Hp cht c lin kt peptit (nhm CO NH )Hp cht cha N cht c bit !
Ngoi ccng phn tc dng c vi dung dch NaOH cn c cc ng phn nokhc khng ?
Bi II.4.1:ng phn no ca C2H3Cl3 khi tc dng vi dung dch NaOH d thu chp cht hu c c phn t khi M = 82. Xc nh CTCT ca ng phn v vit phn ngxy ra ?
Gi : Bn c cng suy ngm.Bi II.4.2: C32 M231 C A 2007: Hp cht hu c X (phn t c vng benzen) c cng thc
phn t l C7H8O2, tc dng c vi Na v vi NaOH.Bit rng khi cho X tc dng vi Nad , s mol H2 thu c bng s mol X tham gia phn ng v X ch tc dng c vi NaOHtheo t l s mol 1:1. Cng thc cu to thu gn ca X l
A. C6H5CH(OH)2. B. HOC6H4CH2OH.C. CH3C6H3(OH)2. D. CH3OC6H4OH.Bi II.4.3: C47 M231 C A 2007:Khi thc hin phn ng tch nc i vi r u (ancol) X,
ch thu c mt anken duy nht. Oxi ho hon ton mt l ng cht X thu c 5,6 lt CO2 ( ktc) v 5,4 gam nc. C bao nhiu cng thc cu to ph hp vi X?
A. 5. B. 4. C. 3. D. 2.Bi II.4.4: C40 M253 HA 2010:C bao nhiu tripeptit (mch h) khc loi m khi thy phnhon ton u thu c 3 aminoaxit: glyxin, alanin v phenylalanin ?A. 6. B. 9. C. 4. D. 3.Bi II.4.5: C40 M175 HA 2009: X phng ho mt hp cht c cng thcphn t C10H14O6
trong dung dch NaOH (d ), thu c glixerol v hn hp gm ba mui (khng cng phnhnh hc). Cng thc ca ba mui l:
A. CH2=CH-COONa, CH3-CH2-COONa v HCOONa.B. HCOONa, CH C-COONa v CH3-CH2-COONa.C. CH2=CH-COONa, HCOONa v CH C-COONa.D. CH3-COONa, HCOONa v CH3-CH=CH-COONa.
Bi II.4.6: C45 M175
HA 2009: Cho cc hp cht hu c
: C2H2; C2H4; CH2O; CH2O2 (mchh); C3H4O2 (mch h, n chc). Bit C3H4O2 khng lm chuyn mu qu tm m. S chttc dng c vi dung dch AgNO3 trong NH3 to ra kt ta l
A. 3. B. 4. C. 5. D. 2.Bi II.4.7: C21 M263 HA 2008: Khi phn tch thnh phn mt r u (ancol) n chc X th
thu c kt qu: tng khi l ng ca cacbon v hi ro gp 3,625 ln khi lng oxi. Sngphn r u (ancol)ng vi cng thc phn t ca X l
A. 3. B. 4. C. 2. D. 1.Bi II.4.8: C43 M174 HB 2010: C bao nhiu cht hu c mch h dng iu ch 4-
metylpentan-2-ol ch bng phn ng cng H2 (xc tc Ni, to) ?A. 3. B. 5. C. 2. D. 4.Bi II.4.9: C18 M148 HB 2009S ipeptit ti a c th to ra t mt hn hp gm alanin
v glyxin lA. 3. B. 1. C. 2. D. 4.
Nitro PD
F Trial
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Bi II.4.10: C2 M285 HB 2007: Cc ng phn ng vi cng thc phn t C8H10O (u ldn xut ca benzen) c tnh cht: tch n c thu c sn phm c th tr ng hp to polime,khng tc dng c vi NaOH. S lng ng phn ng vi cng thc phn t C8H10O,tho mn tnh cht tr n l
A. 1. B. 4. C. 3. D. 2.Bi II.4.11: C9 M285 HB 2007: Cho tt c cc ng phn n chc, mch h, c cng
cng thc phn t C2H4O2 ln l t tc dng vi: Na, NaOH, NaHCO3. S phn ng xy ra lA. 2 B. 5 C. 4. D. 3Bi II.4.12: C31 M268 C A 2010:ng vi cng thc phn t C2H7O2N c bao nhiu cht
va phn ng c vi dung dch NaOH, va phn ng c vi dung dch HCl ?A. 4. B. 2. C. 1. D. 3.Bi II.4.13: C57 M268 C A 2010: S amin thm bc mt ng vi cng thc phn t C7H9N
lA. 3. B. 2. C. 5. D. 4.Bi II.4.14:Cho cc ng phn ng vi cng thc phn t C7H7Cl tc dng vi dd NaOH
long, un nng th tng s ancol thm v phenol thu c ti a lA. 2. B. 1. C. 4. D. 3.Bi II.4.15:S ng phn ng vi cng thc phn t C4H8Br2 khi thy phn trong ddkim cho sn phm ancol a chc l:A. 7. B. 6. C. 8. D.9Bi II.4.16:Cho 6 cht: CH3CH=Ch2; CHBr=CHBr; CH2=CH-CH=CH2; CH3-CH=CH-CH-
CH=CH2; CH3-C(CH3)=CH-CH3; CH3-[CH2]7CH=CH[CH2]7-COOH. S cht c ng phn h nhhc l:
A. 4. B. 2. C. 3. D.1.
Nitro PD
F Trial
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Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 15
PHN III: KT LUNI. Thnh cng ca ti.Xut pht t tnh cp thit ca ti ti tin hnh nghin cu v gii quyt c cc
vn c bn ca ng phn ch ng trnh THPT,ng thiti pht hin v sng tomt sni dung sau:
Nu v xy dng h thng l thuyt vng chc, cn thit nht cho vn ng phnm hc sinh ang cn. c bit a ra cc khi nim mi nh : Ha tr ca nhm chc, Bng tng hp tnh cht c tr ng ca nhm chc , Bng tng hp s ng phn ca ccng nglin tip , Cch tnh bt bo ha v cc ngh a qi him ca n , cng viS mi tn t m s ng phn mt cch nhanh chng.
nu ln c nh ng ch kin thc m hc sinh rt hay nhm ln, thiu st cngvi gi v h ng gii quytchng.
tng hp c mi lin h: Yu t ng ng , Khi lng phn t, S C, , Slin kt pi , S nguyn t l v ha tr ca n. c bitthng qua phn tch cc v d phthin ra hai quy lut mi l: S ng phn t l vi ha tr canguyn t l , Khi c xut hinng phn th ta ly s ng phn nhn 2 th c s ng phn ca cht tip theo trongmt dy ng ng(C1 C5 vi vi ngoi l)
a ra 2 bi tp tho lun hay,xy dng h thng bi tp nng cao v c bn phn loi hc sinh, h thng thi tham kho
C s l thuyt v bt bo ha ngh a ca n v bng tng hp tnh cht ctr ng ca nhm chc s l cng c c lc cho victm chtv tnh ton trong mn ha h uc .
II. Hn ch, xut.Hn ch v thi gian nghin cu cng nh khun kh ca ti nn h thng bi
tp xy dng y ch a c a dng v phong ph , m rng v nng cao h n.Cc tit luyn tp tr n lp qu t cng nh c s l thuyt a ra trong sch gio
khoa ch a c y nn ti c mt s xut nh sau:Mt l: Nn a cch tnh bt bo ha v ngh a ca n vo trong mt bi c
thm no sau bi ng ng- ng phn.Hai l: Gio vin nn tranh th thi gian ch racc qui lutmang tnh bn cht v cch
nh phn t khi, s ng phn cc Em cm thy gn gi vi cc cht hu c h n.
III. Li Kt.Lm th no cc em hc sinh yu mn, quan tm nhiu n mn ha hc, hiu, lm
c bi tp ha, c th gii quyt c cc vn trong cuc sng sau ny trn gc nhnha hc l quan im ging dy ca ti-Mt gio vin mi vo ngh. D c gng ht m nhsong n ng lc ca bn thn vn cn hn ch, trong qu tr nh th hin chc chnc ch cnch a cht ch. Rt mong s pht hin, gp chn t nh ti sm l cm nan cho hc sinhl ti liu hay cho bn c.
Xin Chan Thanh Cam n !
------------------------- het ----------------------
Nitro PD
F Trial
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8/8/2019 Sang Kien Kinh Nghiem Mon Hoa
16/17
Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
Sang Kien Kinh Nghiem Trang 16
MC LCTrang
Phn I. M u------------------------------------------------------------------ 1I. L do chn ti --------------------------------------------------------------- 1
II. Phm vi ti ------------------------------------------------------------------ 1III. nh ngha mt s t vit tc----------------------------------------------- 1Phn II. Ni Dung.-------------------------------------------------------------- 2
I. C S L Thuyt Lin Quan n Vn ng Phn:------------------ 2I.1. Khi nim ng ng v ngh a: ------------------------------------ 2I.1.a. Khi nim cht ng ng---------------------------------------------- 2I.1.b. ngh a ng ng:------------------------------------------------------ 2I.2. Khi nim cht ng phn------------------------------------------------- 2I.3. Phn loi ng phn------------------------------------------------------- 2
I.4. Khi nim ha tr ca nhm chc---------------------------------------- 3I.5. bt bo ha -------------------------------------------------------------- 3I.5.a. Cch thnh lp cng thc tnh a trong cht hu c --------------- 3I.5.b. ngh a ca a(tng s lin k t ) -------------------------------------- 3I.6. Nhm chc ------------------------------------------------------------------- 5I.7. Cc b c vit ng phn cu to------------------------------------- 5I.8. Tnh cht ha hc c tr ng ca nhm chc.-------------------------- 6
II. Gii Quyt Vn ng Phn:---------------------------------------------- 7II.1. Phn tch gii quyt mt s bi tp ng phn c bn-------------- 7II.1.a. S l ng cht ng phn v ha tr ca cc nguyn t l ------- 7II.1.b. Tng hp gy nh cho hc sinh v Phn t khi,S C, S ng phn----------------------------------------------------- 9II.2. Phn tch gii quyt mt s bi tp ng phn
trong thi tuyn sinh-------------------------------------------------- 10II.3. Phn tch gii quyt mt s bi tp ng phn
tng hp, nng cao----------------------------------------------------- 11II.4. Bi Tp Tho lun , Tham Kho-------------------------------------- 13
PHN III: KT LUN------------------------------------------------------ 15I. Thnh cng ca ti. --------------------------------------------------------- 15
II. Hn ch, xut.--------------------------------------------------------------- 15III. Li Kt.-------------------------------------------------------------------------- 15
Nitro PD
F Trial
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8/8/2019 Sang Kien Kinh Nghiem Mon Hoa
17/17
Trng THPT So 2 c Pho Giao Vien: Nguyen Ch Thanh
TI LI U THAM KHO[1]. Ph ng php gii bi tp Ho hc Hu c . PGS.TS Nguyn Thanh
Khuyn NXB
HQG H Ni, nm2006[2]. Ph ng php gii bi tp Ho hc 11, Tp 2. TS. Cao C Gic - NXB HQG H Ni 2008
[3]. Chuyn bi dng Ho hc 11. Nguyn nh - NXB Nng 2006[4]. Sch bi tp Ho hc lp 11- NXBGD H Ni, nm 2007[5]. Sch gio khoa Ho hc lp 11- NXBGD H Ni, nm 2007[6]. thi tuyn sinh H, C cc nm 2007, 2008, 2009, 2010.
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Nitro PD
F Trial
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