- Cho Tuidc brom vao hai mSu thuf con lai, mau thijf nao cho ket tua t r ^ n g \k anilin^ mSu thuf k h o n g c6 h i e n tLfomg g i x a y r a 1^ metanol.
b) Tri ch m6i chat mot it l ^m mku thuf
- D i j n g Cu( 0 H ) 2 de n h a n biet glixerol v i tao dung dich mku xanh.
- D u n g Cu( 0 H ) 2 dun nong de n h a n biet C H 3 C H O v i tao k e t t u a do gach.
- D u n g niidc b r o m di nhkn biet CeHgNHa v i tao k e t t u a t r ^ n g . (phan ling cdc em xem d phan li thuyet)
Bdi DI/ONG HHA H n r 17
p a i 5. Cho 4 chat t r e n tac dung vdri Cu(OH)2 trong m o i trUcfng k i e m v a dun n6ng, t a n h a n thay ong nghiem chtfa ho t i n h bot k h o n g phan ufng, ong nghiem chufa glixerol cho dung dich mau xanh l a m , 6ng nghi§m chufa glucozcr cho k e t tua CU2O mau do gach, ong nghiem chufa long trSng trufng
CO mau t i m dac trttog. (HS tU viet cdc phuang trinh hoa hoc).
Litu y: Vdi long trdng tntng, Cu(OH)2 đ phan ling vdi cdc nhdm peptit (-CO-NH-) cho sdn phdm cd mau tim.
Bat 6 .
- N h o v a i giot dung dich axit n i t r i c dac vao o n g n g h i e m dt/ng cac dung dich t r e n , c h i c6 m o t chat t r o n g o n g n g h i e m c6 k e t tiia v a n g la long t r a n g trufng.
Goc - C 6 H 4- O H ciia m o t so n h o m amino axit t r o n g p r o t e i n da p h a n dng vdri H N O 3 cho gdc mcJi m a n g n h o m - N O 2 c6 m a u vang.
- C 6 H 4- O H + 2 H N O 3 > - C 6 H 2( N 0 2 ) 2 0 H i + 2 H 2 O
(mau vang)
- Cho 3 chat con l a i tac dung vori Cu( 0 H ) 2 chat nao t r o n g o n g n g h i e m hoa t a n Cu( 0 H ) 2 cho dung dich m a u x a n h l a m l a glixerol, 2 chat con
l a i k h o n g tac dung. (HS tU viet phuang trinh hoa hoc).
- De p h a n biet xa phong v a ho t i n h bot, cho dung dich i o t vao 2 o n g n g h i e m de n h a n r a ho t i n h bot (dung dich m a u xanh), chat t r o n g o n g n g h i e m c5n l a i k h o n g p h a n tfng la xa phong.
Bai 7.
C 2 H 5 N H 2 > H - N - H > < ^ ^ h Q H ,
T i n h bazcr g i a m d a n Bai 8.
- V 6 n g benzen h u t electron m a n h hcfn nguyen tuf H n e n cac a m i n thcfm CO t i n h bazof yeu hcfn N H 3 .
- Goc m e t y l - C H 3 day electron m a n h hcfn nguyen tuf H n e n c^c a m i n
CO n h o m - C H 3 c6 t i n h bazcf m a n h hcfn N H g .
- Trong cac a m i n t h o m , nhom nitro -NO2 c6 lien k e t doi \k n h o m the hut
electron n e n l a m giam k h a nang k e t hcfp H^ cua cap electron t i i do ciia
- N H 2 , do do p - n i t r o a n i l i n c6 t i n h bazcf yeu nhat. {Hoc sinh tU sdp xep)
Bai 9 . T r a t ttr sap xep l a : H < H I < I V < I
c a c n h 6 m h u t electron l a m g i a m t i n h bazcf cua a n i l i n . N h o m - N O 2 h u t
electron m a n h han clo r a t nhieụ Cac n h o m day electron ( - C H 3 ) l ^ m t a n g t i n h baza cua a n i l i n .
B a i 1 0 . V i X c h i chdfa cac nguyen to C, H , N n e n no la m o t a m i n , v i tac
d u n g v d i R C l t h e o t i le m o l 1 : 1, tijfc l a p h a n tuf c h i chiJa m o t n h 6 m chufc a m i n , n g h i a la c6 1 n g u y e n tuf N t r o n g p h a n tuf:
1 4 x 1 0 0
= - ^ 3 ^ =
G o i cong thufc p h a n tuf cua X l a C x H y N H g t a c6: 12x + y + 16 = 59 => 12x + y = 43
L a p b a n g :
X 1 2 3
y 3 1 (loai) 19 (loai) 7 (hcfp li)
V a y cong thufc p h a n tuf cua X 1^ C3H9N. C o n g thufc cau tao cua X :
C H 3 - C H 2 - C H 2 - N H 2 C H 3- C H- N H 2 CH3 C H 3 - C H 2- N H- C H 3 CH3 - N - CH3 C H ,
: propylamin (amin bgc mot) : isopropylamin (amin bgc mot)
: etylmetylamin (amin bgc hai) : trimetylamin (amin bgc ba)
B a i 11. G o i cong thuTc t o n g quat cua Z la H O O C - R- N H 2 .
H O O C - R- N H 2 + H C l H O O C - R- N H 3 C I (gam) R + 6 1 , R + 97,5 (gam) 15,1 18,75 <:> 18,75R + 1143,75 = 1 5 , I R + 1472,25 o 3,65R = 328,5 =i> R = 90 => 12x + y = 90 ^ n g h i e m hgfp l y : x = 8 va y = 9. Bai 12. +; Cach 1: a) P h ^ n ufng: C n H 2 „ . i N H 2 + H C l (mol) a C„ H 2 „ > . i N H 2 + H C l (mol) b ->
P h a n tuf Itfong cua h a i a m i n :
C„ H 2„ , i + N H 3 C I a
C„ , H 2„ . . i + N H 3 C l
b M l = 14n + 17 nihenhgp = ( 1 4 n + 17)a + ( 1 4 m + 17)ab = 1,52 <=> 14(na + m b ) + 17(a + b) = 1,52 (1) M 2 = 1 4 m + 17 B6\G HdA HOC 12 K h o i lUOng m u o i : ( 1 4 n + 5 3 , 5) a + (14m + 5 3 , 5) b = 2,98 <=> 14(na + m b ) + 5 3 , 5( a + b) = 2,98 (2) G i a i he phtfcfng t r i n h (1) v a (2) => a + b = 0,04 Só m o l h a i a m i n c u n g l a só m o l cua H C l : C „ , „ , „ = i x l 0 0 0M(HCi) Y 2 0 0 = M i j ^ . 0 , 2 M I b) K h i a = b => n a + m b = ăn + m) = 0,06 . 0,04 0,06 _ a = b = — — = 0,02 =i> m + n = — — = 3 2 0,02 [ C o n g thufc 2 a m i n : C H 3 N H 2 v a C 2 H 5 N H 2 . +) Cdch 2: