Ñeà oân 10:
DESCRIPTION
Ñeà oân 10:. Toång keát Nhoùm nguyeân toá (C, H, O). - Caùc loaïi chöùc. - Coâng thöùc toång quaùt. - Giaùo khoa cô baûn röôïu. CAÙC Baøi taäp traéc nghieäm oân thi ñai hoïc - 2007. NGUYEÃN TAÁN TRUNG ( TTLT CLC VÓNH VIEÃN). Phaàn 1:. Nhoùm C, H, O. - Caùc loaïi chöùc. - PowerPoint PPT PresentationTRANSCRIPT
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Ñeà oân 10:
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CAÙC Baøi taäp traéc nghieäm oân thi ñai hoïc - 2007
NGUYEÃN TAÁN TRUNG ( TTLT CLC VÓNH VIEÃN)
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Phaàn 1:
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Caùc loaïi chöùc:
C, H, O
Röôïu; EÂte; Phenol
Andehyt; Xeâtoân
Axit; Este Gluxit
Röôïu
: (-OH)
EÂte
: (-O-)
Phenol
: Nhaân benzen, nhoùm (-OH)
CH3-
OH CH3-O-
CH3 C6H5-OH
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Caùc loaïi chöùc:
C, H, O
Röôïu; EÂte; Phenol
Andehyt; Xeâtoân
Axit; Este Gluxit
EÂtePhenol
AndehytXeâtoân
- C - HO
- C - O
CH3 - CHO
CH3-CO -
CH3
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Caùc loaïi chöùc:
C, H, O
Röôïu; EÂte; Phenol
Andehyt; Xeâtoân
Axit; Este Gluxit
EÂtePhenol
AxitEste
- C -OHO
- C-O-O
CH3 - COOH
CH3-COO -
CH3
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Caùc loaïi chöùc:
C, H, O
Röôïu; EÂte; Phenol
Andehyt; Xeâtoân
Axit; Este GluxitGluxit
Cn(H2O)m
Monosacarit
Ñisaccarit
Polisaccarit
(Gucuzô,Fructozô)
(Saccaroâzô, Mantozô)(Tinhboät, Xenlulozô)
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Ví duï 1:
Moät soá hôïp chaát höõu cô chöùa C, H, O coù M = 74 ñvC1.CTPT cuûa caùc hôïp chaát höõu cô treân2.Caùc CTCT coù theå coù:
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Böôùc 1: Ñaët CTTQ
Böôùc 2: Laäp phöông trình ñaïi soá (*) Böôùc 3: Giaûi phöông trình (*)Gôïi yù:-Neáu phöông trình (*) coù 3
aån daïng:
B1: Cho cz < d Mieàn giaù trò cuûa z.
B2: Xeùt töøng z ñeå x,y CTPT
ax + by + cz = d.
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Giaûi yù 1:
- Ñaët CTTQ (A): CxHyOz
- Theo ñeà coù: MA = 74 ñ.v.
12x + y + 16z = 74 ( *)
16z < 74 z < 4,625
z = 1; 2; 3; 4.Vôùi z = 1:
(*) 12x + y = 58
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y = 58 – 12xx
y
Ñieàu kieän:
z = 1; 2; 3; 4.
- (A): CxHyOzCoù 12x + y + 16z = 74 ( *)
Vôùi z = 1:(*) 12x + y = 58
Giaûi yù 1:
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ÑK: 0 < y 2.x + 2 + t – u
y + t + u = soá chaün.
Ñieàu kieän hoaù trò:
Vôùi: CxHyOzNtXu; X: Cl ; Br
Vôùi: CxHyOz
ÑK: 0 < y 2.x + 2
y = soá chaün.
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y = 58 – 12xx
y
z = 1; 2; 3; 4.
- (A): CxHyOzCoù 12x + y + 16z = 74 ( *)
Vôùi z = 1:(*) 12x + y = 58
Giaûi yù 1:
ÑK: 0 < y 2.x + 2
y = soá chaün.
1
46
2
34
3
22
4
10
5
aâm
Vôùi z = 1: C4H10O
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y = 42 – 12x
z = 1; 2; 3; 4.
- (A): CxHyOzCoù 12x + y + 16z = 74 ( *)
Vôùi z = 2:(*) 12x + y = 42
Giaûi yù 1:
ÑK: 0 < y 2.x + 2
y = soá chaün.
Vôùi z = 2: C3H6O2
0 < y 2.x + 20 < 42 - 12x 2.x + 22,8 x < 3,5
x=3 y=6
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z = 1; 2; 3; 4.
- (A): CxHyOzCoù 12x + y + 16z = 74 ( *)
Vôùi z = 3:(*) 12x + y = 26
Giaûi yù 1:
Vôùi z = 3: C2H2O3
26 =
12. +
x = ?y= ?R P
Q dö k R=P.Q + K
2 2
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M(c,h,o) =74
C4H10O
C3H6O2
C2H2O3
(Röôïu, EÂte)
( Axit, Este, . . .)
( Axit - Andehyt)
Toùm laïi:
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CTPTTQCTPTTQ:
a*: lk trong phaân töû.
COÂng thöùc toånG quaùt (CTTQ)
CnH2n+2–2a*Om
CTCTTQ:CTCTTQ:
a: lk trong
goác hydrocacbon
CnH2n+2–2a - m(Chöùc )m
CTTQ cuûa ñeà
a*=?m =?
CTTQ cuûa ñeà
a =?m =?
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D. CnH2n-
2(CHO)2
Ví duï 2:Coâng thöùc toång quaùt cuûa Andehit no, 2 chöùc laø: A. CnH2n+1(CHO)2
C. CnH2n(CHO)2B. CnH2n-1(CHO)2
CnH2n+2-2a-
m(chöùc)mAndehit no, 2 chöùc
° a = 0
° m =2 CnH2n(CHO)
2
CC
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D. CnH2n-
2(OH)2
Ví duï 3:Coâng thöùc toång quaùt cuûa röôïu no, 2 chöùc laø:
A. CnH2n+1(OH)2
C. CnH2n(OH)2B. CnH2n-1(OH)2
CnH2n+2-2a-
m(chöùc)mRöôïu no, 2 chöùc
° a = 0
° m =2
CnH2n(OH)2
CC
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D. CnH2n-
1CHO
Ví duï 4:Ñoàng ñaúng cuûa andehyt CH2= CH-CHO laø: A. (C2H3-CHO)n
C. CnH2n+1CHOB. CnH2n-3CHO
CnH2n+2-2a-
m(chöùc)mAdehyt ñeà cho laø chöa no coù 1lk C=C, 1 chöùc
° a = 1 ° m =1
CnH2n-1CHO
DD
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Vieát CTTQ cuûa :a.Röôïu ñôn
chöùc, no.b.Röôïu no.c.Röôïu 3 chöùc,
no. CnH2n+2–2a*Om
CTTQ:
a*=?m =?
Röôïu ñôn chöùc, no.Röôïu ñôn chöùc, no.
= 0= 0= 1= 1
CCnnHH2n+22n+2OO
Ví duï 5:
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Ví duï 5Ví duï 5::
Vieát CTTQ cuûa :a.Röôïu ñôn chöùc, no:
CCnnHH2n+22n+2OO
b.Röôïu no.c.Röôïu 3 chöùc, no.
CnH2n+2–2a*Om
CTTQ:
a*=?m =?
Röôïu no.Röôïu no.
= 0= 0= khoâng xaùc ñònh= khoâng xaùc ñònh
CCnnHH2n+22n+2OOmm
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Ví duï 6Ví duï 6::
A laø röôïu ñôn coù moät lieân keát C=Ccoù KLPT khoâng quaù 60 ñvC. Tìm A
PP tìm CTPTDöïa treân BTÑS Ñaët CTTQ
Laäp pt (töø M) Giaûi pt treân.
KLPT khoâng quaù 60 ñvCKLPT khoâng quaù 60 ñvCröôïu ñôn coù moät lieân keát C=Cröôïu ñôn coù moät lieân keát C=C
CnH2n+2–2a-m (OH)m
Ñaët CTTQÑaët CTTQ
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CnH2n+2-2a-m(OH)mÑeà:Ñeà: A laø röôïu chöa no coù 1lk C=C, 1 chöùc
a = 1
m =1 CnH2n-1OH
A laø röôïu ñôn coù moät lieân keát C=Ccoù KLPT khoâng quaù 60 ñvC. Tìm A
Ví duï 6Ví duï 6::
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A laø röôïu ñôn coù moät lieân keát C=Ccoù KLPT khoâng quaù 60 ñvC. Tìm A
Ví duï 6Ví duï 6::
A: CnH2n-1OH
GiaûiGiaûi:
KLPT khoâng quaù 60 ñvCKLPT khoâng quaù 60 ñvC
14n + 16 60
n 3,14 n 3,14
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Ñieàu kieän toàn taïi röôïuÑieàu kieän toàn taïi röôïu
Moãi C chæ gaén toái ña 1 nhoùm(-OH)
Nhoùm (-OH) chæ gaén treân C no
Soá nhoùm OH Soá C no
( Moái lieân heä soá C, soá O )
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A laø röôïu ñôn coù moät lieân keát C=Ccoù KLPT khoâng quaù 60 ñvC. Tìm A
Ví duï 6Ví duï 6::
A: CnH2n-1OH
GiaûiGiaûi:
14n + 16 60 n 3,14 n 3,14
A: CH2=CH- CH2-OH
Maø: n 3
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Ví duï 7Ví duï 7::
A laø röôïu ñôn coù moät lieân keát C=Ccoù KLPT khoâng quaù 60 ñvC. Tìm A
PP tìm CTPTDöïa treân BTÑS Ñaët CTTQ
Laäp pt (töø M) Giaûi pt treân.
KLPT khoâng quaù 60 ñvCKLPT khoâng quaù 60 ñvCröôïu ñôn coù moät lieân keát C=Cröôïu ñôn coù moät lieân keát C=C
Ñaët CTTQÑaët CTTQ
CnH2n+2–2a-m (OH)m
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CnH2n+2-2a-
m(chöùc)mÑeà:Ñeà: A laø röôïu chöa no coù 1 lk C=C
a = 1
m =m CnH2n-
m(OH)m
A laø röôïu ñôn coù moät lieân keát C=Ccoù KLPT khoâng quaù 60 ñvC. Tìm A
Ví duï 7Ví duï 7::
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A laø röôïu ñôn coù moät lieân keát C=Ccoù KLPT khoâng quaù 60 ñvC. Tìm A
Ví duï 7Ví duï 7::
A: CnH2n-m(OH)m
GiaûiGiaûi:
KLPT khoâng quaù 60 ñvCKLPT khoâng quaù 60 ñvC
14n + 16 m 60
n 3,14 n 3,14
CnH2nOm
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Ví duï 8Ví duï 8::
A laø röôïu ñôn chöùc, coù 1 lieân keát C=C, maïch hôû vaø coù %O= 27,58Tìm CTPT cuûa A.
ñôn chöùc, coù 1 lieân keát C=C ñôn chöùc, coù 1 lieân keát C=C %O= 27,58%O= 27,58
PP tìm CTPTDöïa treân BTÑS Ñaët CTTQ
Laäp pt (töø %) Giaûi pt treân.
Ñaùp aùn:
CH2=CH-CH2-OH
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Phaàn 2:
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Ñònh nghóaÑònh nghóa(Röôïu laø gì?):
Röôïu laø hchc coù chöùa nhoùm(-OH) gaén treân Cno cuûa goác hydrocacbon.
Ví duï:CH3-OH; CH3 –CH2-OHCH2= CH-CH2-OH
CH2-OH
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CTPTTQCTPTTQ:
a*: lk trong phaân töû.
COÂng thöùc toånG quaùt (CTTQ)
CnH2n+2–2a*Om
CTCTTQ:CTCTTQ:
a: lk trong
goác hydrocacbon
CnH2n+2–2a -m(OH)m
CTTQ cuûa ñeà
a*=?m =?
CTTQ cuûa ñeà
a =?m =?
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Ñieàu kieän toàn taïi röôïuÑieàu kieän toàn taïi röôïu
Moãi C chæ gaén toái ña 1 nhoùm(-OH)
Nhoùm (-OH) chæ gaén treân C no
Soá nhoùm OH Soá C no
( Moái lieân heä soá C, soá O )
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Ví duï 8Ví duï 8:
Ñem oxi hoaù hoaøn toaøn 1 mol röôïu no (A) Caàn 2,5 mol oxi . Tìm CTPT-CTCT (A)?
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Caùc baøi toaùn tìm CTPT höõu cô
Tìm CTPTDöïa treân phaûn öùng chaùy Caùch giaûi: B1. Ñaët CTTQ
B2. Vieát phaûn öùng chaùy B3. Laäp phöông trình
B4. Giaûi phöông trình
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Ví duï 8Ví duï 8:Ñem oxi hoaù hoaøn` toaøn 1 mol röôïu no (A) Caàn 2,5 mol oxi . Tìm CTPT-CTCT (A)?
PP 4 böôùc
(A)?
Toùm taét:(A):Röôïu no
1 mol 2,5 mol O2
CO2 (Ñoát)
H2O
B1. Ñaët CTTQB2. Vieát pöù chaùyB3. Laäp phöông trình B4. Giaûi phöông trình
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(A)?
Toùm taét :(A):Röôïu no
1 mol 2,5 mol O2
CO2 (Ñoát)
H2O
Ñaët CTTQ A: CnH2n+2Om
CnH2n+2Om+ O2CO2+ H2On (n+1)3n+1-m
2
1 mol 2,5 mol
1 3n+1-m2
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Ñaët CTTQ AÑaët CTTQ A: CnH2n+2Om
CnH2n+2Om+ O2CO2+ H2On (n+1)3n+1-m
2
1 mol 2,5 mol
1 3n+1-m2
Theo treân ta coù:
1 2,5
1 3n+1-m2=
3n+1-m = 5
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Ñaët CTTQ AÑaët CTTQ A: CnH2n+2Om Ta coù: 3n+1-m = 5
n = 4 + m3
m
n
Vôùi: n,m +
;n m
m
m 2
1 2
5/3
1
2
Choïn: m = 2 n = 2
Vaäy: A laø C2H6O2
CH2 CH2
OH OH
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Ví duï 9Ví duï 9:
Ñem oxi hoaù hoaøn toaøn 1 mol röôïu no (A) Caàn 3,5 mol oxi . Vaäy (A) coù theå laø:
A. C3H7OH B. C2H5OHCH
2
CH2OH
OH
C. D
OHOH
CH2CH CH2
OH
Ñoát 1mol röôïu noCaàn x,5 mol O2
SoáC = SoáO = x
DD
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Caùc tröôøng hôïpï hoã bieán thöôøng gaëp:ï hoã bieán thöôøng gaëp:TH1: Nhieàu nhoùm (-OH) gaén treân 1 C no
Nguyeân taéc: Töï taùch nöôùc bieán thaønh saûn phaåm môùi
Hai nhoùm (-OH) gaén treân Cno baäc 1:
R- CH OHO H
H2OR- CO
H +
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Caùc tröôøng hôïpï hoã bieán thöôøng gaëp:ï hoã bieán thöôøng gaëp:TH1: Nhieàu nhoùm (-OH) gaén treân 1 C no
Nguyeân taéc: Töï taùch nöôùc bieán thaønh saûn phaåm môùi
Hai nhoùm (-OH) gaén treân Cno baäc 2:
R- C OHO H
H2OR- CO
R’ +
R’
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Caùc tröôøng hôïpï hoã bieán thöôøng gaëp:ï hoã bieán thöôøng gaëp:TH1: Nhieàu nhoùm (-OH) gaén treân 1 C no
Nguyeân taéc: Töï taùch nöôùc bieán thaønh saûn phaåm môùi
Ba nhoùm (-OH) gaén treân Cno baäc 1:
R- C OHO H
H2OR- CO
OH +
OH
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TH2: Nhoùm (-OH) gaén treân 1 C khoâng no
Caùc tröôøng hôïpï hoã bieán thöôøng gaëp:ï hoã bieán thöôøng gaëp:TH1: Nhieàu nhoùm (-OH) gaén treân 1 C no
Nguyeân taéc: Coù söï chuyeån vò H linh ñoäng vaø lieân keát bieán thaønh saûn phaåm môùi
Nhoùm (-OH) gaén treân C khoâng no baäc 1:
R CH CHOH
R-CH2- CO
H
R C CH2OH
R- CO
CH3
Nhoùm (-OH) gaén treân C khoâng no baäc 2:
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Ví duï 10:Ví duï 10:
OH
Vieát caùc phaûn öùng sau :
a. CH2 CH2OH
H2SO4, ñaëc170oC
b. CH3-COO-CH=CH2 + NaOHto
XH2SO4, ñaëc
170oC Y °Laø pöù taùch nöôùc taïo lK C=C °X : Coù nhoùm OH vaø Soá C 2
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Ví duï 10Ví duï 10::
OH
Vieát caùc phaûn öùng sau :
a. CH2 CH2OH
H2SO4, ñaëc170oC
b. CH3-COO-CH=CH2 + NaOHto
CH3CHO + H2O
R–COO–R’ + NaOH to
R–COONa + R’–OH
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Ví duï 11::Vieát caùc phaûn öùng sau:
a. CH2Cl-COO-CHCl2 + NaOH
b. HCOO-CHCl2-CH2Cl + NaOH to
to
°R–COO–R’ + NaOH
R–COONa + R’–OH°RCln + nNaOH
R-(OH)n + nNaCl
to
to
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° Caùc phaûn öùng cuûa röôïu etylic ° Caùc sô doà
phaûn öùng cuûa röôïu etylic
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C2H5OHCH2=CH2
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Phaûn öùng ñieàu cheá röôïuPhaûn öùng ñieàu cheá röôïu etylicetylic
+H2
O
+NaOH +H2 +HCl
+NaOH
Men röôïu
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C2H5OHC2H4
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Pöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylicPöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylic
CH2=CH-CH=CH2
CH3COOH
(1)
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Caùc phaûn öùng cô baûn cuûa röôïu etylic(1): Phaûn öùng taùch nöôùc taïo olefin
CnH2n+1O
H
H2SO4, ñaëc
170oCCnH2n +H2O
C2H5OHH2SO4, ñaëc
170o
C
C2H4 +H2O
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C2H5OHC2H4
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Pöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylicPöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylic
CH2=CH-CH=CH2
CH3COOH
(2)
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Caùc phaûn öùng cô baûn cuûa röôïu etylic(2): Phaûn öùng taïo este voâ cô
R(OH)n + n
HX
RXn + nH2O
C2H5OH +
HCl
C2H5Cl +
H2O
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C2H5OHC2H4
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Pöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylicPöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylic
CH2=CH-CH=CH2
CH3COOH
(3)
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Caùc phaûn öùng cô baûn cuûa röôïu etylic(3): Phaûn öùng taïo Andehit
C2H5OH +
CuO
CH3CHO + Cu
+H2O
R(CH2OH)n+nC
uO
R(CHO)n+nCu+n
H2O
to
to
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C2H5OHC2H4
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Pöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylicPöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylic
CH2=CH-CH=CH2
CH3COOH
(4)
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Caùc phaûn öùng cô baûn cuûa röôïu etylic(4): Phaûn öùng cuûa röôïu vôùi Na, K.
C2H5OH +
Na
C2H5ONa +1/2 H2
R(ONa)n+ n/2
H2R(OH)n + nNa
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C2H5OHC2H4
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Pöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylicPöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylic
CH2=CH-CH=CH2
CH3COOH
(5)
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Caùc phaûn öùng cô baûn cuûa röôïu etylic(5): Phaûn öùng taïo este höõu cô (ñôn chöùc)
R C OH + HO R’ O
H2SO4ñaëc,tO
H2O R C OR’O
+
CH3
O
C OH + C2H5OHH2SO4ñaëc,tO
CH3
O
C O C2H5 + H2O
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C2H5OHC2H4
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Pöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylicPöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylic
CH2=CH-CH=CH2
CH3COOH
(6)
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Caùc phaûn öùng cô baûn cuûa röôïu etylic(6): Phaûn öùng leân men giaám.
CH3COOH + H2 O C2H5OH+ O2
men giaám
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C2H5OHC2H4
C2H5 X C2H5 ONa CH3 CHO
Este cuûa etylic
C6H12O6(Glucoâzô)
Pöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylicPöù ñieàu cheá vaø tính chaát hoaù hoïc röôïu etylic
CH2=CH-CH=CH2
CH3COOH
(7)
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Caùc phaûn öùng cô baûn cuûa röôïu etylic(7): Phaûn öùng taùch nöôùc vaø taùch H2.
CH2=CHCH=CH2
+ + H2 O
+ H2
C2H5OHAl2O3
400 – 500OC2
2
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Aùp duïng 1Aùp duïng 1: : (CÑ KT- TAØI CHÍNH THAÙI (CÑ KT- TAØI CHÍNH THAÙI NGHUYEÂN - 2004)NGHUYEÂN - 2004) Vieát caùc phaûn öùng theo sô ñoà:
XenluloâzôGlucoâzôRöôïu etylic
Ñietyl eâte
Butadien-1,3
(1) (2)
(3)
(4)
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ví duï 11ví duï 11: : (CÑ KT- TAØI CHÍNH THAÙI (CÑ KT- TAØI CHÍNH THAÙI NGHUYEÂN - 2004)NGHUYEÂN - 2004) Sô ñoà ñöôïc vieát laïi:
XenluloâzôGlucoâzôRöôïu etylic
Ñietyl eâte
Butadien-1,3
(1) (2)
(3)
(4)
(C6H10O5)n C6H12O6
C2H5OH
C2H5 O C2H5
CH2=CHCH=CH2
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ví duï 12ví duï 12: : (ÑH,CÑ – khoái B- 2003)(ÑH,CÑ – khoái B- 2003) Vieát caùc phaûn öùng theo sô ñoà:
A BtoC
D
EF
C
G
H
+X
+Y+X
+Y
Bieát: -E laø röôïu etylic.-G, H laø polime