Giải chi tiết các câu liên quan đến anđêhit, xeton, axit cacboxylic trong đề thi đại học

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)H/I) J)=H/:I) )K/I L)K/<I)
S mol HCHO=s mol CO
2
=0,35 (mol)
HCHOH
2
O
0,35 0,35 (mol) Suy ra s mol H
2
=(11,7:18)-0,35=0,3(mol)
% th tớch H
2
=(0,3:0,65).100%=46,15%
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K
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V:
?A)
S mol Ag = 2 ln s mol anehit chng t anehit n chc
S mol H

2
=2 ln s mol anehit chng t anehit cha no cha 1 pi
ỏp ỏn A
=:17113&415



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=
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K

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K
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&M.8&&G>
)K) J)= ) L)
CH=CH, HCHO, HCOOH, , 4 cht ny u tỏc dng vi AgNO
3
/NH
3
ỏp ỏn B
=H,&11$,->1/M1T7)6 &17-&/K>/
&O01::/>&

?@&1A)'8&.[/K>&91CO\>OOP1':Q),&(>
)/U

U) J)/
K
)
)/



L)/U)
0,3 mol axit trung ho 0,5 mol NaOH thỡ phi cú 1 axit n chc v mt axit hai chc
n trung bỡnh l 5/3 nờn chn D
ỏp ỏn D
J
=6 &17-&%&13&415/&01/K:

/=KH<>&

?@&1A)J8&1(
+"N?A

&.T&.0]^()3&>
)
K

K
) J)_U_) )

_U

U) L)



)
-nCO
2
= nH
2
O = 0,197. X tỏc dng vi Cu(OH)
2,
t
0
andehit no n chc
ỏp ỏn D

,&251"1,&`1"1?2/`1(1\! -a&b11$A)&
C$c)C%&&71OZ8&N'/!.=/=<>&

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!.H/=

)T&"113&M&dC&.D^ >01G`&.>C>0&>
)U

U;/<;I) J)UH/I)
)U

U=/<<I) L)U=/<HI)
- RCOOH + Na RCOONa + ẵ H
2
; R(COOH)
2
+ 2Na R(COONa)
2
+ H
2
X x/2 y y
nH
2
= x/2+y = 0,02(1) , nCO
2
= nx

+ ny = 0,06 (2) 1,5 <n < 3. Vy n = 2 (CH
3

COOH v HOOC-COOH)
Thay vo (1), (2) x = 0,02 , y = 0,01
0,1.90
% .100 42,86(%)
0,1.90 0,2.60
HOOC COOH- = =
+
ỏp ỏn D
:Le-17113&01!f,8&E1^&DOCg&%!T&h&.7!+>
)
K
/



//
K
)
J)
K
//



/
K
)
)/
K
/

/
K

L)
K
/



//
K
)
t
0
sụi
: CH
3
CHO < C
2
H
5
OH < HCOOH < CH
3
COOH.
ỏp ỏn A
Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013

Biên soạn GV Nguyễn Ngọc Sản Mobile: 01694043249
:17113&!
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U

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K
U?AU


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K
U

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K
UU

K
7113&^&71OZ01N'/?A

>
)?1A/?OA/?iA J)?A/?$A/?1A )?A/?1A/?OA L)?1A/?OA/?EA
Tỏc dng vi Na v Cu(OH)
2
Cú ớt nht 2 Nhúm OH cnh nhau (hoc COOH).
ỏp ỏn C
K13&415/51"1&71OZhGN:>OOP1/=Q/&
01%& KKH>5%&1>?@&1A)'8 &17-&>0&.a/!(3&Z8&
!+Y17-$9#OOP1?A

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13&415&.>
)



J)
K

K




)

K
L)
K

;
n
X
= n
NaOH
= 0,04; n
ancol
= 0,015< n
X
. 1 axit v 1 este ca axit ú. C
n
H
2
n
O
2

n
CO

2
+
n
H
2
O, 44x+18x
=6,82.
x= 0,11.
n
= nCO
2
/nX = 2,75 ; n.(0,04-0,015) +m.0,015 = 2,75.0,04 ; n= 2 (CH
3
COOH) v m =
4(CH
3
COOC
2
H
5
)
K:17113&415
?:AW ?A1>/51"1/M1@W
?KA,1>W ?=AE&E/51"1/M1@W
?AEW ?HA1>T?1(%&>a8&T_A/M1@
?;AW ?<AE&/51"1/M1@W
?A,&/51"1/M1@ ?:A,&T?1(%&>a8&T_A/51"1
Le-17113& &17-&^1! >

$c! >

>
)?KA/?A/?HA/?<A/?A J)?KA/?=A/?HA/?;A/?:A
)?A/?KA/?A/?;A/?A L)?:A/?KA/?A/?HA/?<A
nCO
2
= NH
2
O CT chung hp cht C
n
H
2n
O
x
Cú (3),(5),(6),(8),(9) tho món.
ỏp ỏn A
K13&4152>j8&8/^&71OZN'1(+"&.7$M1)J8&C
&.D >0,&./2>C>0&>K/KKI=K/=I)T&"113&M1G2&05">
)U

U

UU

U

U

U
J)U?

K
AUU

U
)U

UU

U

U
L)
K


U
K
Th: loi B( khỏc dóy ), loi D(khụng tỏc dng vi Na). Cỏc cht trờn u cú CT chung C
n
H
2n
O
2
32 53,33 32 43,24
% 2;% 3
14 32 100 14 32 100
X Y
O n O m
n m
= = ị = = = ị =

+ +
ỏp ỏn C
K1>%&13&415?1"//A+"&N'R1N'
K
&9^
!.>)3&>
)E&->E>1> J),&1
),&KU.,.1 L)1>o-.,$Ek->1
Tỏc dng vi NaHCO
3
l axit: R(COOH)
n
+ nNaHCO
3
+ nCO
2
+ Cú nX = nCO
2
1nhúm COOH.
Tỏc dng vi Na nX = nH
2
cú thờm 1 nhúm OH. (axit 3-hidroxipropanoic: HO-CH
2
-CH
2
-COOH tho mn)
A.HOCH
2
-CH
2

OH v D.o- HO-C
6
H
4
-CH
2
OH khụng td NaHCO
3
; B.HOOC-(CH
2
)
4
-COOH td vi NaHCO
3
-2a mol CO
2
.
ỏp ỏn C
6 &17-&:>13&415/&01=>

)3&&71OZ01N'/&
+"&.7$M1+"1%J.

&E&l>g>::)T&"113&M1G>
)U_U J)U

U

U_U
)U

U

U

U L)U

U_U
nCO
2
= 4 cú 4C; T/dng vi Na cú -OH hoc -COOH; cú phn ng trỏng Ag cú -CH=O; cng Br
2
(1:1) cú 1liờn
kt C=C
CTCT l HO-CH
2
-CH=CH-CH=O
Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013

Biªn so¹n GV NguyÔn Ngäc S¶n Mobile: 01694043249
Đáp án D
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)K=/ J);/KH )/< L):</=
Số C = n
CO2
/n
hh
= 3 vậy ancol là C
3
H
7
OH → 4H
2
O . Vì n
Nước
< n
CO2
nên axit không no.
Axit có 3C có 2TH: CH
2
=CH-COOH → 2H
2
O ; x + y = 0,5 và 4x + 2y = 1,4. Ta có x= 0,2 và y = 0,3 (nhận)
CH≡C-COOH → 1H
2
O ; x + y = 0,5 và 4x + y = 1,4. Ta có x= 0,3 và y = 0,2 (loại n
Y
< n

X
)
Este là CH
2
=CH-COOC
3
H
7
. Với m
CH2=CH-COOC3H7
= 0,2*0,8*114 = 18,24 (g)
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Sơ đồ : C
6
H
5
CH(CH
3
)
2






O
H O
+
+
→
(CH
3
)
2
CO.

  

xeton cumen
n mol n
= = =

  

m g
= =
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Axit RCOOH, muối RCOOM: 0,1(R+45) + 0,1(R + M + 44) = 15,8
⇒
0,2R + 0,1M = 6,9.
R = 1
⇒

M = 67 (loại); R = 15
⇒
M = 39 (K); R = 29
⇒
M = 11(loại) . Vậy axit là CH
3
COOH (axit etanoic)
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
K
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;
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

;

=


)
N
hh
= 2n
H2
= 0,6 (mol). Do axit tác dụng đủ với CH
3
OH nên n
2 axit
= n
CH3OH
. = 0,3 (mol).
  
R COOH CH OH R COOCH H O
− + → − +
.

R COOCH

M
−
= 25/0,3 = 83,33
⇒
R = 24,33 (CH
3
- và C
2
H
5
-)
Vậy 2 axit là CH
3
COOH và C
2
H
5
COOH.
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A. 0,015. =%% :%% >%%
J+
 -.
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N%O`% =%j]% :%O`V%L)y+;/K<)
Hướng dẫn:
71:
Gọi công thức chung của các chất là C
n
H
2n-2
O
2
C
n
H
2n-2
O
2
+ O
2
→ nCO
2
+ (n-1)H
2
O
→ Từ pt cháy ta thấy nC

n
H
2n-2
O
2
= nCO
2
– nH
2
O
nCO
2
= nCaCO
3
= 18/100 = 0,18
gọi nH
2
O

=a → n C
n
H
2n-2
O
2
= 0,18 – a.; gọi số mol O
2
là y
Chuyªn ®Ò «n thi ®¹i häc Ho¸ h÷u c¬ 2007-2013


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+ Bo ton nguyờn t O : (0,18 a)2 + 2y = 0,18.2 + a -3a + 2y = 0 (1)
+ Bo ton khi lng : 3,42 + 32y = 0,18.44 + 18a -18a + 32y = 4,5 (2)
(1)v(2) a = 0,15
m(CO
2
+ H
2
O) = 0,18.44 + 0,15.18 = 10,62
Khi lng dung dch gim = 18 10,62 = 7,38gam.
71
C
n
H
2n-2
O
2
nCO
2
0,18/n 0,18
M = 3,42n/0,18 = 19n 14n + 30 = 19n n = 6
nH
2
O = 0,18(n-1)/n = 0,15 m (CO
2
+ H
2
O) = 0,18.44 + 0,15.18 = 10,62 gam.
Thu vo 10,62 gam, mt i 18 gam , vy sau p khi lng dung dch gim = 18 10,62 = 7,38
gam.

:$#6#oA4 $C_$]E!L4 (486##6#
oA:

: A:
@


A: +Z#"(5] %&']E$"$"!L
6##6#*!K4<k#9+(+,g./%j2 ,<k#"
)/;) =%% :%V@% >%%
Hng dn:
n axitsalixylic = 43,2/180 = 0,24
oUCH
3
COO-C
6
H
4
-COOH + 3NaOH CH
3
COONa + o-NaO-C
6
H
4
-COONa + 2H
2
O.
0,24 0,72
V NaOH = 0,72/1 = 0,72 lớt
;&526$"$"$#C$6#l(486$#:.

!"_$#

.!L_?6m%
:$$#l2+J!L-:.

+8(486$#:.

%OQ<l#"
N% 6#% J),&,>1) :%4%>%n$%
Hng dn:
S C trung bỡnh = nCO
2
/naxit = y/x
S nhúm COOH = nCO
2
/naxit = y/x
Cht cú s C = s nhúm chc ( loi A, C)
Axit fomic chỏy cú nCO
2
= nH
2
O (loi) chn B
:;&526$"$"C$6#ERS!"4a(HM
#Q134T:?: $[\(48k#919:.

41!"6$#

.%='(E#QoY2
2 ,6!"k#"

N%k?



x y
% =%k?

@
V
x y
)p_



x y+
) >%k?

@
V
x y+
%
Hng dn :
Cụng thc chung ca cỏc axit trờn l : C
n
H
2n-4
O
4
C
n

H
2n-4
O
4
nCO
2
+ (n-2)H
2
O
T phng trỡnh ta thy : naxit = (nCO
2
nH
2
O)/2
naxit = (V/22,4 y)/2
Khi lng axit = xgam = mC/axit + mH/axit + mO/axit
x = 12V/22,4 + 2y + 64(V/22,4 y)/2
x = 44V/22,4 30y V = 28/55(x + 30y)
Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013

Biên soạn GV Nguyễn Ngọc Sản Mobile: 01694043249
Hn$!"$#%g$2+J!L-:.

+87(48@#919:.

41%/0124526$"$"*V@#919.

41(48:.

!"6$#

.%j2 ,<6#"
N%% =%% :%% L)/H)
Hng dn:
Cụng thc chung ca cỏc axit l R(COOH)
X
R(COOH)
x
+ xNaHCO
3
R(COONa)
x
+ xCO
2
+ xH
2
O
0,7/x 0,7
n O/axit = 0,7.2 = 1,4
Axit + O
2
CO
2
+ H
2
O
0,7 0,4 0,8 y
Bo ton nguyờn t O : nO/axit + nO/O
2
= nO/CO

2
+ nO/H
2
O 1,4 + 0,4.2 = 0,8.2 + y
y = 0,6
<O ()C$6#$4DERSCp++-.T
R$"CM++(]E(48(51%-3(4526$"$"7'
9$41*+Z#"
N%#9% J)K/KH>&) :%#9% >%#9%
Hng dn:
n axit = (m mui m axit)/22 = (5,2 3,88)/22 = 0,06
M axit = 3,88/0,06 = 194/3
C
n
H
2n
O
2
= 194/3 n = 7/3
C
7/3
H
14/3
O
2
+ 5/2 O
2
7/3CO
2
+ 7/3H

2
O.
0,06 0,15
V O
2
= 0,15.22,4 = 3,36 lớt.
K&526$"$"4(48'919:.

Cp'9D8L $Z4a(
1oo4M2(X%g$$#q+!LM#8+8++N

$-

7(48
$#N%#"
)E&i1) =%4$RSE%
:%4% >%41T$RSE%
Hng dn:
Andehit t chỏy cú n CO
2
= n H
2
O andehit no, n
nAg to ra / n andehit = 0,04/0,01 = 4 HCHO.
=HHDM$4DE!"M$4EF5$#
#LD5$#F(48M'9DCp'9<@-

4$Z $4a(1oo
4M2(X%-3(4526$"CM Q7(48#9:.

41%:O:O<
F#*#8#"^
N%:

A:

A: !" :A: =%:

A: !" :A:

A:

A:
:%A: !" :A: L)
K
UU

U
Hng dn:
nN
2
= nX = 5,6/28 = 0,2
nCO
2
= 0,48
C trung bỡnh = 0,48/0,2 = 2,4 (loi C)
Dựng quy tc ng chộo da vo s C v C trung bỡnh v d kin s mol X ln hn s mol Y X l
CH
3
COOH

Da vo d kin s mol mi cht theo quy tc ng chộo v khi lng hn hp l 15,52 gam ta tỡm
c Y l HOOC-CH
2
-COOH
Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013
@