1. An aromatic primary amine A with molecular formula C6H7N undergoes diazotisation to give B. B when treated with hypo phosphorous acid gives C. Identify A, B and C.
Since (A) undergoes diazotisation, (A) may be a Nuclear aromatic amine. Comparing with molecular formula (A) may be Aniline.
C6H5NH2 + NaNO2 + HCl → C6H5N = N - Cl (Diazotisation)
(A) Aniline 273K (B) Benzene diazonium chloride
C6H5N = N - Cl + H2 → C6H6 + N2 + HCl (Cuprous salts catalyse this reaction)
(B) H3PO2 / Cu+ (C) Benzene
Compound A
|
Compound B
|
Compound C
|
C6H5NH2
|
C6H5N = N - Cl
|
C6H6
|
Aniline
|
Benzene diazonium chloride
|
Benzene
|
The simplest aromatic nitro compound is Nitro benzene (A). Since (B) undergoes Carbylamine reaction (B) may be a Primary amine.
Sn/HCl
C6H5NO2 → C6H5NH2 + 2H2O
(A) Nitro benzene 6(H) (B) Aniline
C6H5NH2 + CHCl3 + 3KOH → C6H5NC + 3KCl + 3H2O (Carbylamine reaction)
(B) Aniline
Compound A
|
Compound B
|
C6H5NO2
|
C6H5NH2
|
Nitro benzene
|
Aniline
|
Uses of compound A - Nitro benzene
4. An organic compound (A) having molecular formula C2H7N is treated with nitrous acid to give (B) of molecular formula C2H6O which on mild oxidation gives compound (C) of molecular formula C2H4O which answers Tollen’s reagent test. Identify A, B, C.
5. An organic compound (A) with molecular formula C6H7N gives (B) with HNO2 / HCl at 273K. The aqueous solution of (B) on heating gives (C) which gives violet colour with neutral FeCl3. Identify the compounds A, B and C.
6. An organic compound A of molecular formula C2H5NO on treatment with Na / C2H5OH gives B (C2H7N) and with Br2 / KOH gives C (CH5N). Identify A, B. C.
7. An organic compound A of molecular formula C2H5ON treated with bromine and KOH -gives B of molecular formula CH5N. Identify A and B. Write the equation involved.
8.
HNO2
9. Compound A is yellow coloured liquid and it is called oil of mirbane. A on reduction with tin and HCl gives B. B answers carbylamine test. Identify A and B.
10. How will you convert acetamide to methyl amine? Give equation.
14.
17. An yellow coloured liquid (A) called as ‘Oil of Mirbane’ is reduced with Sn / HCl to give compound (B). Identify A and B write the equation
ONE MARKS FIVE MARKS
1. To prepare explosives like TNT, 1, 3, 5-trinitro benzene.
2. Used in making dye stuffs and pharmaceuticals.
Nitro benzene has the smell of ‘bitter almonds’ and is called ‘the oil of Mirbane’.
3. An organic compound (A) having molecular formula C2H7N is treated with nitrous acid to give (B) of molecular formula C2H6O which answers iodoform test. Identify (A) and (B) and explain the reaction.
Since compound (B) undergoes Iodoform test, (B) may be Ethyl alcohol. So (A) may be a Primary amine.
CH3CH2NH2 + O = N- OH → CH3CH2OH + N2 + H2O
(A) Ethyl amine (B) Ethyl alcohol
I2 I2 NaOH
CH3CH2OH → CH3CHO → CI3CHO → CHI3 + HCOONa (Iodoform test)
(B) Ethyl alcohol Acetaldehyde Iodoform
Compound A
|
Compound B
|
CH3CH2NH2
|
CH3CH2OH
|
Ethyl amine
|
Ethyl alcohol
|
Since compound (C) answers Tollen’s reagent test, (C) may be an aliphatic aldehyde. So (B) may be a Primary alcohol and (A) may be a Primary amine.
CH3CH2NH2 + O = N - OH → CH3CH2OH + N2 + H2O
(A) Ethyl amine (B) Ethyl alcohol
CH3CH2OH → CH3CHO
(B) Ethyl alcohol (C) Acetaldehyde
CH3CHO + 2Ag+ + 3OH– → CH3COO– + 2Ag + 2H2O (Tollen’s reagent test)
(C) Acetaldehyde Acetate ion (Silver mirror)
Compound A
|
Compound B
|
Compound C
|
CH3CH2NH2
|
CH3CH2OH
|
CH3CHO
|
Ethyl amine
|
Ethyl alcohol
|
Acetaldehyde
|
Since (A) undergoes diazotisation, (A) may be a Nuclear aromatic amine. Comparing with molecular formula (A) may be Aniline.
Since (C) gives violet colour with neutral FeCl3 (C) is Phenol
C6H5NH2 + NaNO2 + HCl → C6H5N = N - Cl (Diazotisation)
(A) Aniline 273K (B) Benzene diazonium chloride
C6H5N = N - Cl + H2O → C6H5OH + N2 + HCl
(B) (C) Phenol
Compound A
|
Compound B
|
Compound C
|
C6H5NH2
|
C6H5N = N - Cl
|
C6H5OH
|
Aniline
|
Benzene diazonium chloride
|
Phenol
|
Na / C2H5OH
CH3CONH2 + 4 [H] → CH3CH2NH2 + H2O
(A) Acetamide (B) Ethylamine
Br2 / KOH
CH3CONH2 → CH3NH2 + CO2 (Hoffman’s bromamide reaction)
(A) Acetamide (C) Methyl amine
Compound A
|
Compound B
|
Compound C
|
CH3CONH2
|
CH3CH2NH2
|
CH3NH2
|
Acetamide
|
Ethylamine
|
Methyl amine
|
Since an amide (A) is treated with bromine and alkali, the amide is converted into primary amine (B) containing one carbon less than that of amide.
Br2 / KOH
CH3CONH2 → CH3NH2 + CO2 (Hoffman’s bromamide reaction)
(A) Acetamide (B) Methyl amine
Compound A
|
Compound B
|
CH3CONH2
|
CH3NH2
|
Acetamide
|
Methyl amine
|
C6H5CH2NH2 → C6H5CH2OH
Benzylamine (A) Benzyl alcohol
[O]
C6H5CH2OH → C6H5CHO
(A) Benzyl alcohol (B) Benzaldehyde
Zn / Hg
C6H5CHO → C6H5CH3 (Clemmenson reduction)
(B) Benzaldehyde HCl (C) Toluene
Compound A
|
Compound B
|
Compound C
|
C6H5CH2OH
|
C6H5CHO
|
C6H5CH3
|
Benzyl alcohol
|
Benzaldehyde
|
Toluene
|
Yellow coloured liquid and it is called oil of mirbane is - Nitro benzene (A).
Sn/HCl
C6H5NO2 → C6H5NH2 + 2H2O
(A) Nitro benzene 6(H) (B) Aniline
Since (B) undergoes Carbylamine reaction (B) may be a Primary amine.
C6H5NH2 + CHCl3 + 3KOH → C6H5NC + 3KCl + 3H2O (Carbylamine reaction)
(B) Aniline
Compound A
|
Compound B
|
C6H5NO2
|
C6H5NH2
|
Nitro benzene
|
Aniline
|
Acetamide reacts with Br2 / KOH to give methyl amine.
Br2 / KOH
CH3CONH2 → CH3NH2 + CO2
Hoffman’s hypobromite reaction or Hoffman’s bromamide reaction or Hoffman reaction
11. What is diazotisation? Give an example.
A cold solution of sodium nitrite reacts with aniline dissolved in hydrochloric acid, a clear solution is obtained. This solution contains ‘benzene diazonium chloride’. This reaction is known as ‘diazotisation’.
HCl
C6H5NH2 + O=N–OH → C6H5– N = N – Cl + H2O
273K
12. What is Gabriel phthalimide synthesis? Or Explain / Write Gabriel’s phthalimide synthesis.
Phthalimide reacts with KOH to form Potassium phthalimide. This reacts with an alkyl halide to give N-alkyl phthalimide, which in turn reacts with KOH to form a pure aliphatic primary amine and potassium phthalate.
13. When benzamide is treated with bromine and alkali gives compound A. Also when benzamide is reduced by LiAlH4 compound B is formed. Find A and B. Write the equations.
C6H5CONH2 → C6H5NH2
Benzamide Br2/KOH (A) Aniline
C6H5CONH2 → C6H5CH2NH2
Benzamide LiAIH4 (B) Benzylamine
Compound A
|
Compound B
|
C6H5NH2
|
C6H5CH2NH2
|
Aniline
|
Benzylamine
|
Sn / HCl
CH3NO2 + 6 [H] → CH3NH2 + 2H2O
Nitro methane (A) Methyl amine
CH3NH2 + CHCl3 + 3KOH → CH3– N º C + 3KCl + 3H2O (Carbylamine reaction)
(A) Chloroform (B) Methyl isocyanide
H2 / Pt
CH3– N º C → CH3NHCH3
(B) Methyl isocyanide (C) Dimethylamine
Compound A
|
Compound B
|
Compound C
|
CH3NH2
|
CH3– N º C
|
CH3NHCH3
|
Methyl amine
|
Methyl isocyanide
|
Dimethylamine
|
15. How
is chloropicrin prepared? What is its use?
Nitromethane reacts
with Chlorine in presence of alkali, NaOH to form Chloropicrin, CCl3NO2
NaOH
CH3NO2 + 3Cl2 → CCl3NO2 + 3HCl
Use
Chloropicrin is used as Soil sterilizing agent
16. How is methyl cyanide
obtained from acetamide?
Acetamide on dehydration
by heating with P2O5 forms Methyl cyanide.
P2O5
CH3CONH2 → CH3C
≡ N
– H2O 17. An yellow coloured liquid (A) called as ‘Oil of Mirbane’ is reduced with Sn / HCl to give compound (B). Identify A and B write the equation
Nitro benzene, C6H5NO2 (A) is called ‘Oil of Mirbane’.
Sn /
Conc. HCl
C6H5NO2 + 6 [H] → C6H5NH2 + 2H2O
(A) Nitro benzene (B) Aniline / Amino benzene
Compound
A
|
Compound
B
|
C6H5NO2
|
C6H5NH2
|
Nitro benzene
|
Aniline / Amino benzene
|
Unit – 21 Bio molecules
1. What are
carbohydrates? Give two examples
2. Give the structure
of sucrose.
3. What is starch? What
are the ultimate hydrolysis products?
4. What is the action
of Conc. HI on glucose?
5. What is Saponification?
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