Organic Nitrogen Compounds Five Marks

1. Distinguish between primary, secondary and tertiary amines.

S.No	Primary amine RNH2	Secondary amine R2NH	Tertiary amine R3N
1	With HNO2 forms alcohol.	Forms N-nitroso amine.	Forms salt.
2	With CHCl3 / KOH forms carbylamines	No reaction.	No reaction.
3	With acetyl chloride forms N-alkyl acetamide.	Forms N, N-dialkyl acetamide.	No reaction.
4	With CS2 and HgCl2 alkyl isothiocyanate is formed.	No reaction.	No reaction.
5	With Diethyl oxalate         COOC2H5               CONHR |             + 2RNH2 → | COOC2H5               CONHR	Forms N,N-dialkyl oxamic ester, a liquid CONR2 | COOC2H5	No reaction.
6	With three molar proportion of alkyl halide, quarternary ammonium salt is formed. RNH2 + 3RX →R4N+X–	With two molar proportion of alkyl halide, quarternary ammonium salt is formed. R2NH + 2RX→ R4N+X–	With only one molar proportion of alkyl halide, quarternary ammonium salt is formed. R3N + RX → R4N+X–

 2. Explain / Write a note on the reduction of nitrobenzene under different conditions.

a) Strongly acidic medium:

When reduced with Tin and Hydrochloric acid, Nitrobenzene is converted to Aniline.

                                          Sn/Conc.HCl

C6H5NO2 + 6 [H]      →      C6H5NH2 + 2H2O

                                    or Fe/Conc.HCl

b) Neutral medium:

When reduced with a neutral reducing agent like Zinc dust and aqueous Ammonium chloride, Nitrobenzene forms Phenyl hydroxylamine.

                                        Zn/NH4Cl

C6H5NO2 + 4 [H]  → C6H5NHOH + H2O

c) Alkaline medium:

In alkaline medium, Nitro benzene on reduction forms the intermediate products Nitrosobenzene (C6H5NO) and Phenyl hydroxylamine (C6H5NHOH). These undergo bimolecular condensation reaction.

d) Catalytic reduction:

Lithium aluminium hydride is a powerful hydride ion donor. So LiAlH4 reduces Nitro benzene to Aniline. This reduction can also be carried out by H2/Ni - Catalytic Hydrogenation

                   LiAlH4

C6H5NO2  →  C6H5NH2

                               H2/Ni

C6H5NO2       →      C6H5NH2

                      Catalytic

                         Hydrogenation

e) Electrolytic Reduction:

When Nitro benzene is reduced electrolytically in presence of Concentrated Sulphuric acid, Phenyl hydroxylamine is first produced which rearranges to give p-Amino phenol.

3. Explain / Write the following reactions: i) Carbylamine reaction ii) Gabriel's phthalimide synthesis.

i) Carbylamine reaction:

1. Primary amines on heating with Chloroform and alcoholic Potash / KOH forms a foul smelling substance called Carbylamine or Alkyl isocyanide.

                                                                 Δ

CH3NH2 + CHCl3 + 3KOH → CH3NC + 3KCl + 3H2O

Or

2. Aniline reacts with Chloroform and alcoholic KOH to give an offensive smelling liquid, Phenyl isocyanide.

                                                                   Δ

C6H5NH2 + CHCl3 + 3KOH → C6H5NC + 3KCl + 3H2O

This reaction is characteristic of primary amines. Secondary and tertiary amines do not undergo this reaction.

ii) Gabriel’s phthalimide synthesis of primary amine.

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.

4. Explain the following reactions in aniline: a) Coupling reaction b) Schotten-Baumann reaction c) Formation of Schiff’s base

a) Diazonium coupling reaction:

Diazonium salt reacts with Aromatic amine and Phenols to give Azo compounds of the general formula Ar – N = N – Ar'

This reaction is known as Coupling reaction since all these compounds are intensely coloured and used as dyes, thousands of azo dyes have been synthesised by this procedure.

The aromatic compound with which it couples should have any one of the groups like –OH, –NH2, –OR, –NHR, –NR2 etc. coupling usually occurs at para - position.

b) Schotten-Baumann reaction:

C6H5NH2 reacts with Benzoyl chloride in presence of Sodium hydroxide to give Benzanilide.

                                                    NaOH

C6H5NH2 + ClCOC6H5 → C6H5NHCOC6H5

c) Formation of Schiff’s base:

1. Primary amines condense with Aromatic aldehydes forming Schiff’s bases.

C6H5CHO + H2N CH3 → C6H5CH = NCH3

                                                Schiff’s base - Benzal-N-methyl amine

Or

2. Aniline is Primary aromatic amine. It reacts with Aldehydes to form Aldimines or Schiff’s bases.

C6H5NH2 + O = CHR → C6H5N = CHR

                                               Aldimine or Schiff’s base

5. How are i) phenol, ii) chlorobenzene, iii) biphenyl prepared by using benzene diazonium chloride?

i) When the aqueous solution of C6H5 N2 Cl is boiled, Phenol is obtained.

C6H5 N2 Cl → C6H5OH + N2 + HCl

       HO ––– H

ii) Sand Meyer Reaction: When aqueous solution of Diazonium chloride is warmed with Cu2Cl2 in Halogen acid, Halobenzene is formed.

                          HCl

C6H5N2Cl  →  C6H5 – Cl + N2

                        Cu2Cl2

Or

Gattermann reaction: When the Diazonium chloride solution is warmed with Copper powder and the Hydrogen halide, the corresponding Halobenzene is obtained.

                          Cu

C6H5N2Cl → C6H5Cl + N2

                         HCl

iii) Gomberg Bachmann Reaction: Decomposition of Benzene diazonium chloride in presence of Sodium hydroxide and Benzene forms Biphenyl.

                                        NaOH

C6H5N2Cl + C6H6 → C6H5– C6H5 + N2 + HCl

6. How are the following conversions carried out? a) Nitromethane to methylamine b) methylamine to methyl isocyanide.

a) Reduction: Sn/HCl or Fe/HCl or H2/Raney Nickel reduces Nitromethane to Methylamine.

                             Sn/HCl

CH3– NO2 + 6 [H] → CH3NH2 + 2H2O

b) Carbylamine reaction: Methylamine on heating with Chloroform and Alcoholic Potash / KOH forms a foul smelling Methyl isocyanide.

CH3NH2 + CHCl3 + 3KOH → CH3NC + 3KCl + 3H2O

7. How are the following conversions carried out? i) Nitrobenzene to phenyl hydroxylamine ii) Aniline to phenyl isocyanide

i) Nitrobenzene reduced with Zinc dust and aqueous Ammonium chloride (Neutral reducing agent) forms Phenyl hydroxylamine.

                                    Zn/NH4Cl

C6H5NO2 + 4 [H]  → C6H5NHOH + H2O

ii) Aniline reacts with Chloroform and alcoholic KOH to give Phenyl isocyanide.

                                                  Δ

C6H5NH2 + CHCl3 + 3KOH → C6H5NC + 3KCl + 3H2O

8. How can the following conversions be effected / carried out? i. Nitrobenzene to anisole ii. Chlorobenzene to phenyl hydrazine iii. Aniline to Benzoic acid

i. Nitrobenzene to anisole:

             Sn/Conc.HCl                     NaNO2                                        CH3OH

C6H5NO2    →    C6H5NH2  →  C6H5– N = N – Cl  →  C6H5OCH3

                                                 HCl, 0oC

Stage 1. When C6H5NO2 is reduced with Tin and Conc.HCl Aniline or Amino benzene is formed.

                           Sn / Conc.HCl

C6H5NO2 + 6 [H]     →      C6H5NH2 + 2H2O

Stage 2. A cold solution of Sodium nitrite reacts with Aniline dissolved in Hydrochloric acid, Benzene diazonium chloride is obtained. Diazotisation

                                        HCl

C6H5NH2 + O=N–OH → C6H5– N = N – Cl + H2O

Stage 3. By warming Benzene diazonium chloride with alcohol Anisole is obtained.

C6H5    N2   Cl → C6H5OCH3 + N2 + HCl

CH3O––––––H               

ii. Chlorobenzene to phenyl hydrazine

          NH3                   HNO2                  Sn/Conc.HCl

C6H5Cl → C6H5NH2 → C6H5N2Cl → C6H5NHNH2

                 CuCl2                     HCl

Stage 1. By the ammonolysis of Chloro benzene at high temperature and pressure in presence of Copper salts, Aniline is prepared.

                         CuCl2

C6H5Cl + 2NH3 → C6H5NH2 + NH4Cl

Stage 2. A cold solution of Sodium nitrite reacts with Aniline dissolved in Hydrochloric acid, Benzene diazonium chloride is obtained. Diazotisation

                                       HCl

C6H5NH2 + O=N–OH → C6H5– N = N – Cl + H2O

Stage 3. Diazonium chloride is reduced to phenyl hydrazine on treatment with SnCl2/HCl or Zn/HCl or NaHSO3.

                      4[H]

C6H5N2Cl → C6H5NHNH2 + HCl

iii. Aniline to Benzoic acid

              HNO2                     KCN                  H3O+

C6H5NH2 → C6H5N2Cl → C6H5CN → C6H5COOH

                      HCl               Cu2(CN)2 or Cu

Stage 1. A cold solution of Sodium nitrite reacts with Aniline dissolved in Hydrochloric acid, Benzene diazonium chloride is obtained. Diazotisation

                                       HCl

C6H5NH2 + O=N–OH → C6H5– N = N – Cl + H2O

Stage 2. Benzene diazonium chloride when treated with Cuprous cyanide / Potassium cyanide mixture, Phenyl cyanide is formed.

                         Cu2(CN)2

C6H5   N2  Cl → C6H5CN + N2 + HCl

      NC––––––K

Or

Treatment of Benzene diazonium chloride with KCN solution in presence of Copper gives Phenyl cyanide.

                               Cu

C6H5N2Cl + KCN →C6H5CN + N2 + KCl

Stage 3. Phenyl cyanide is hydrolysed with aqueous acid to give Benzoic acid

                      H+                                  H+

C6H5CN → (C6H5CONH2) → C6H5COOH

                    H2O                                H2O

9. How does nitrous acid react with primary, secondary and tertiary amines?

Reaction with Nitrous acid:

a) Primary amines react with Nitrous acid to form Alcohols and Nitrogen gas.

CH3NH2 + O = N – OH → [CH3–N = N – OH] → CH3OH + N2

Primary amine                               Unstable

Aliphatic Diazonium compound is unstable because of absence of resonance stabilisation.

b) Secondary amines react with Nitrous acid to form N-nitroso amines which are water insoluble yellow oils.

(CH3)2 N H + HO – N = O → (CH3)2 N – N = O

Secondary amine                         N-nitroso dimethyl amine - Yellow oil

                                                     (Insoluble in water)

c) Tertiary amines react with Nitrous acid to form Trialkyl ammonium nitrite salts which are soluble in water.

(CH3)3 N + HONO → (CH3)3 NH+ NO2–

Tertiary amine                Trimethyl ammonium nitrite

                                        (Salt soluble in water)

10. Write notes on the following i) Mustard oil reaction ii) Diazotisation reaction iii) Gomberg reaction.

i)  Mustard oil reaction:

When primary amines are warmed with Carbon disulphide and Mercuric chloride, Alkyl isothiocyanate, having a pungent mustard like odour is obtained.

CH3NH2         +         S = C = S      →     CH3– N = C = S            +          H2S

Carbondisulphide                                            Methyl isothiocyanate

This reaction is characteristic of primary amines. Secondary and tertiary amines do not undergo this reaction.

ii) Diazotisation - Reaction with Nitrous acid:

A cold solution of Sodium nitrite reacts with Aniline dissolved in Hydrochloric acid, a clear solution of ‘Benzene diazonium chloride’ is obtained. This reaction is known as ‘Diazotisation’.

                                           HCl

C6H5NH2 + O = N–OH    →   C6H5– N = N – Cl

                                                   273K

iii) Gomberg Bachmann reaction:

Decomposition of diazonium salts in presence of Sodium hydroxide and Benzene, results in the formation of Biphenyl.

                                      NaOH

C6H5N2Cl + C6H6 → C6H5– C6H5 + N2 + HCl

 11. Write any three methods of preparing benzyl amine?

1. Reduction of benzonitrile : Benzyl amine is prepared by the reduction of Benzonitrile catalytically Or with Lithium Aluminium Hydride.

2. Reduction of benzamide : Benzyl amine is prepared by the reduction of Benzamide catalytically Or with Lithium Aluminium Hydride.

3. From benzylbromide : Benzyl amine is prepared by the action of alcoholic ammonia on Benzyl bromide

C6H5CH2 – Br + H – NH2 → C6H5CH2NH2 + HBr

 12. Explain the reduction of Nitrobenzene in alkaline medium
 Nitro benzene on reduction forms the intermediate products Nitrosobenzene (C6H5NO) and Phenyl hydroxylamine (C6H5NHOH). These undergo bimolecular condensation reaction.
 

13. Srarting from benzene diazonium chloride how will you obtain the following i) Phenol   ii) Anisol  iii) p-hydroxy azobenzene

i) Phenol : When the aqueous solution of C6H5 N2 Cl is boiled, Phenol is obtained.

C6H5 N2 Cl → C6H5OH + N2 + HCl

       HO ––– H

ii) Anisol : By warming Benzene diazonium chloride with alcohol Anisole is obtained.

C6H5    N2   Cl → C6H5OCH3 + N2 + HCl

CH3O––––––H  

iii) p-hydroxy azobenzene :

Benzene Diazonium chloride reacts with Phenol to give p - hydroxy azobenzene
 .
 

14. Srarting from Benzene diazonium chloride how will you obtain the following

      i) Anisol                                         ii) Nitrobenzene                                   iii) Phenyl hydrazine
ONE MARKS  THREE MARKS