September 19, 2022 | Author: Anonymous | Category: N/A
ABC-1, 2, 3 & 4
(A) PHENOL Preparation of phenol (5-Methods) 1.
Cumene hydroperoxide method
CH3 CH3 CH3 CH CH3 –C –O –O –H
This is an industrial process to convert cumene
O2/h
H3O+
into phenol and acetone in the presence of oxygen from air followed by hydrolysis.
2.
Dow's process.
+ CH3 –C –CH3 acetone
OH
(1) NaOH 350ºC High pressure (2) HCl
In this process chlorobenzene is heated at 350°C (under high pressure) with sodium hydroxide
3.
O
Cumene hydroperoxide
Cl
+NaCl+ H2O Phenol
Chlorobenzene
which yields phenol. phenol.
OH
Fusion of benzene sulphonic acid with concentrated NaOH
OH
SO3H
(1) NaOH/high temp. high pressure
This is commercial process for synthesizing phenol. Benzene sulphonic acid is melted (fused) with sodium hydroxide at (300°C – –320°C) 320°C) followed
(2) H Benzene Sulphonic acid
Phenol
by hydrolysis which yields phenol.
4.
Decarboxylation of salicylic acid
OH
OH
OH COONa
Distillation of salicyclic acid with soda-lime
COOH
or
Soda lime
(NaOH + CaO) produces phenol.
5.
Hydrolysis of benzene diazonium salt
NH2
temperature.Then this aqueous solution is heated to get phenol.
NaNO2 , HCl
Benzene diazonium salt is prepared by reacting aromatic primary amine with NaNO2, HCl at low
+ – N N Cl
H2O, 5º C romatic Primary amine
Aryl diazonium salt
+ – NN Cl
OH BoilingH BoilingH 2O
Phenol
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ABC1TO4 - 18
ABC-1, 2, 3 & 4
Chemical reactions of phenol [5-Reactions] 1
(a) Reaction with Br 2 /H2O is added to in presence of it forms white 2,4,6-tribromo-
(b) Reaction with Br 2 /CS2
Br White ppt.
OH
Reimer Tiemann carboxy carboxylation lation
Br
+
or CH3COOH < 5ºC
Br OH
OH
OH CHO
(1) (1) CHCl CHCl3 , NaOH NaOH, ( 2) 2) H2O
+ CHO
Major OH
OH
Phenol when heated with CCl4 and sodium hydroxide followed by hydrolysis forms salicylic acid. Note: Salicyclic acid can be used Note: in formation of aspirin. (Aspirin is used as painkiller)
OH
Br 2, CS2
Phenol when heated with chloroform and NaOH followed by H2O forms salicyladehyde.
3.
+ HBr
OH
Reimer Tiemann formylation
Br
Br 2 EtOH tOH / H2O
In presence of non-polar solvent (like CS2) or acids like CH 3COOH at low temperature, only monobromo product is obtained.
2.
OH
Br
OH
When bromine solution of phenol ethanol or H2O, precipitate of phenol.
CC CCll , Na NaOH OH,
OH
COOH
+
4
H
COOH
Major
O
OH
O –C –CH3 COOH
COOH AcCl or Ac2O,
pyridine
Aspirin
4.
Kolbe's Schmidt reaction
OH
Phenol when reacted with hydroxide ion in presence of CO 2 forms a complex which on acidification forms salicyclic acid.
5.
Bakelite is made by condensation
COO –
(i)
OH
(ii)
CO2
(iii) HCl
COOH
CH2O
OH
OH COOH
+
OH
Bakelite formation
OH
O –
Major
OH CH2OH
reaction between formaldehyde.
phenol
and
NaOH
CH2OH Polymerises
OH
OH
OH
OH
Bakelite
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ABC1TO4 - 19
ABC-1, 2, 3 & 4
Lab test for phenol 1.
Litmus test : Phenol turns blue litmus red. Phenols behave as weak acid because of presence of polar O – –H H group, they ionise in aqueous solution to give H + ions.
O
OH
+ H2O
H3O+
Phenoxide
Phenol
ion
2.
Neutral FeCl3 : Phenols give a violet-coloured water soluble complex with neutral ferric chloride. 3– + 6Ph – –OH OH + FeCl3 Fe(OPh) 6 + 3H + 3HCl
Violet complex NH2
(B) ANILINE Preparation of aniline [3-Methods] 1. Reduction by Metals
3. Hofmann bromamide degradation reaction In this reaction an unsubstituted amides (only 1°) treated with NaOH/KOH and bromine to give a
2. Reduction by H2
NH2
Metal / Acid
primary amine that amide. has one carbon lesser than starting NO2
NH2
NO2
Sn / HCl
NH2
O Ni / H2 or Pd / H2
R –C –NH2 + Br 2 + 4NaOH
+ 6H
R –NH2 + Na2CO3 + 2NaBr + 2H2O
NH2
R can be : Alkyl or phenyl This method is used to prepar 1° aliphatic or aromatic amines.
Fe/ HCl + 6H
Special : NH2
NO2 (NH4 )2 S
Selective reduction
(various other sulphides can be used.) NO2
NO2
Chemical reactions of aniline : 1. Preparation of diazonium salt: salt: Aniline reacts with NaNO2/H+ to form diazonium salt. +
N2 Cl
NH2
–
NaNO2/HCl (0-5ºC) Benzene diazonium chloride
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ABC-1, 2, 3 & 4
Note : 1. 1. Primary aliphatic amines amines react with nitrous acid to form aliphatic diazonium salt which being unstable, liberate nitrogen gas. 2. 2. Primary aromatic amines form arene diazonium salts which are stable for a short short time in solution at low temperatures (273-278 K).Due to its instability, the diazonium salt is not generally stored and is used immediately after its preparation. 2.
Chemical reactions of diazonium salt Diazonium salt opens the way to prepare many other compounds, see following chart.
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ABC1TO4 - 21
ABC-1, 2, 3 & 4
OH
Boiling H2O +
N N HSO4 –
N=N
Ph-OH dil. NaOH
OH Orange-dye
Benzene diazonium salt
N=N
Ph-NH2
NH2 Yellow-dye
dil. HCl
1. 1.
Complete the following reactions.
NNCl
NNCl CuCl/HCl or Cu2Cl2/HCl
i.
CuBr/HBr
ii.
CH3
CH3
NNCl
NNCl Cu/HBr
Cu/HCl
iv.
iii. CH3
CH3
NNCl
NNCl C2H5OH
H3PO2
vi. vi.
v.
CH3
CH3
NNCl
NNCl H2O,
HBF4
vii.
viii.
–HCl
CH3
CH3
2. 2.
Conversion. Ph –Cl
Ph –NO2 H2/Pd
Ph –NH2
Sandmayer reaction
NaNO2/HCl
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ABC1TO4 - 22
ABC-1, 2, 3 & 4
Lab test of aniline : : 1.
Carbylamine reaction reaction Primary amines (aliphatic as well as aromatic) react with chloroform (CHCl 3) on heating in the presence of ethanolic solution of KOH to form isocyanides (also called carbylamines) which are foul smelling substances. Secondary and tertiary amines do not undergo this reaction, therefore this reaction is used as a test for primary amines (aliphatic as well as aromatic). R –NH2 + CHCl3 + 3KOH Heat
Ex.
CH3 –NH2 + CHCl3 + 3KOH
R –NC + 3KCl + 3H2O
Heat
CH3 –NC + 3KCl + 3H 2O
NH2
Ex.
2.
NC
+ CHCl3 + 3KOH
Heat
+ 3KCl + 3H 2O
Azo dye test Azo compounds are usually intensely colored because of the azo linkage ( –N=N –N=N –). –). Azo compounds, because of their intense color’s and because they can be synthesized from relatively inexpensive compounds, are used extensively as dyes. Synthesis of orange-red dye from 2-naphthol [ -naphthol] and aniline. N=N
OH NH2
3.
NaNO2/H2SO4 (0 –5°C)
+ N2
OH NaOH pH = 8-10
Orange-red Dye
Bromine water test (Br 2 + H2O): Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline. Aniline also gives test with Br 2 + CS2 NH2
Br
NH2
Br 2 / H2O
+ 3HBr
+ 3Br 2 Aniline
Br
Br
2,4,6- Tribromoaniline
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ABC1TO4 - 23
ABC-1, 2, 3 & 4
ONLY ONE OPTION CORRECT TYPE 1.
The process of conversion of cumene in the presence of oxygen and light followed by hydrolysis lead to the formation of _______ . (A) Phenol (B) Aniline (C) Anisole (D) Benzene
2. 2.
Identify the product of following reaction. CH3 –CH –CH3 (i) (i) O /h
2
(ii) H3O
CH3 OH
(A)
(B)
(C)
CH3
3. 3.
(D) None of these
CH3
Identify the product of the following reaction. Cl (i) NaOH (ii)HCl
350 º C High pressure
NaCl H O 2
Cl
OH
OH
OH Cl
(A)
(C)
(B)
(D) Cl
Cl
Cl SO3H (i) NaOH/ ? (ii) H
4. 4. CH3
SO3H
SO3H
CH3
OH
OH
(D)
(C)
(B)
(A)
OH CH3
CH3
5. 5.
OH CH3
Give the product for following reaction. OH COONa Soda lime
C2H5 OH
OH COOH
(A)
(B) C2H5
(D) None of these
(C) C2H5
C2H5
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ABC1TO4 - 24
ABC-1, 2, 3 & 4
6. 6.
Identify the X and Y in the following reaction. NH2 NaNO2 , HCl
H2O, 5ºC
Y H2O
+
N2Cl
OH
(A) X = +
(B) X =
OH
–
(C) X =
OH
–
Y= N2Cl
7. 7.
X
Y= N2Cl
Y=
(D) X =
Y=
What will be the product of following reaction? OH Cl2 EtOH / H2O
OH
OH
OH
OH Cl
(A)
(B)
(C)
Cl
Cl
(D)
Cl Cl
8.
Cl
Cl
Find the product for the following reaction. OH Br 2
Cs2 , 5º C
CH3 OH
OH
Br Br
(A) Br
9. 9.
(D) None of these
(C)
(B) CH3
Give the major product for the following reaction. OH CHCl / NaOH
3
Me OH
OH CHO
(A)
(B) CHO
10. 10.
OH
OH
Me
COOH
Me
(C)
(D) Me
Me
The reagent used for Reimer Tiemann carboxylation in order to form salicylic acid is _______ . (A) CHCl3 / NaOH / H+ (B) CCl4 / NaOH / H+ (C) OH – / CO2 / HCl (D) None of these
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ABC1TO4 - 25
ABC-1, 2, 3 & 4
11. 11.
Aspirin can be formed from following reaction using which reagent? O O –C –CH3
OH
COOH
COOH
O
O +
(B) (C2H5 –C –)2O / H+
(A) (CH3 –C –)2O / H 12. 12.
(C) CHCl3 / NaOH
(D) None of these
Give product for following reaction. OH CHCl3 KOH
CHO OH
CHO
(A) 13. 13.
COOH
(B)
OH
(C)
(D) None of these
The reactant used in the reaction is : OH COOH
Reactant
– OH / CO2 HCl
OH OH
OH COOH
(A)
(C)
(B)
(D)
NO2 Sn/HCl + 6H
14. 14.
R is :
R,
CH3 NO2
(A)
(B)
NH2
NH2
NH2
(C)
(D) CH3
CH3 Cl
NH2
15. 15.
Fe/HCl
R
+ 6H
, R is :
NO2
(A)
(B)
NO2
NO2
(C)
NO2
(D)
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ABC1TO4 - 26
ABC-1, 2, 3 & 4 NH2
16. 16.
Ni/H2
R
, R is : CH3 NO2
NO2
(A)
NO2
NO2
(B)
(C)
(D)
CH3
17. 17.
Br 2/NaOH
R
C2H5 – –NH NH2, R is : : CH3
(A)
(B) C2H5 –CH –C –NH2
C2H5 –CH2 –C –NH2 O
(C)
O CH3
(D) C2H5 –C C –NH2
C2H5 –C –NH2
CH3 O
O +
N2Cl –
CH3
HC 18. 18.
3
NaNO2/HCl (0 –5°C)
R
R, isCH3
NO2
NH2 CH3
H3C
(A)
(B) CH3
CH3
CH3
CH3
H3C
(C)
(D) CH3
19. 19.
CH3
H3C
CH3
Which NH of them produces diazonium cation at 0°C ? NH –CH 3
2
CH3
(A)
(B)
(C) CH3 –CH –CH2 –NH –NH2
(D) CH3 –C –NH2 CH3
+
N2Cl –
P
20. 20.
, P is : C2H5
(A) Cu2Cl2/HCl
C2H5
(B) Cu/HCl
(C) C2H5OH
(D) HBF4
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ABC1TO4 - 27
ABC-1, 2, 3 & 4 NC
21. 21.
CHCl3 + 3KOH/
R
R is C2H5
NO2
N2Cl –
(B)
(A)
C2H5
C2H5 NH2
(C)
CN
(D) C2H5
C2H5
OH
NH2 NaNO2/HCl (0-5°C)
22. 22.
(X) NaOH (pH=8-10)
(Y)
SO3H Y is : +
N2Cl
O
(A) HO3S (C) HO
–
(B) HO
N=N
HO
2
SO3Na
+
+ 2
–
–
N Cl
(D)
N2Cl
SO
1.
(A)
2. 2.
(A)
3. 3.
(C)
4. 4.
(C)
5. 5.
(B)
6. 6.
(C)
7. 7.
(D)
8.
(B)
9. 9.
(B)
10. 10.
(B)
11. 11.
(A)
12. 12.
(B)
13. 13.
(C)
14. 14.
(C)
15. 15.
(C)
16. 16.
(B)
17. 17.
(C)
18. 18.
(B)
19. 19.
(A)
20. 20.
(C)
21. 21.
(C)
22. 22.
(B)
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ABC1TO4 - 28
