April 06, 2012

Organic Nitrogen Compounds Three Marks

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
2.  An aromatic simplest nitro compound A on reduction using Sn and HCl gives B. B undergoes carbylamine reaction. Identify A and B. Give any one use of compound A.
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
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
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.
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
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.
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  + H2OC6H5OH + N2 + HCl
(B)                                           (C) Phenol

Compound A
Compound B
Compound C
C6H5NH2
C6H5N = N - Cl
C6H5OH
Aniline
Benzene diazonium chloride
Phenol
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.
                                    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
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.
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
8.
                                     HNO2
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
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.
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
10.  How will you convert acetamide to methyl amine? Give equation.
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
14.  
                                      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

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