Short answer Questions

Organic Nitrogen Compounds | Chemistry

Short answer Questions

1. Write down the possible isomers of the C4 H9NO2 give their IUPAC names

2. There are two isomers with the formula CH3NO2 . How will you distinguish between them?

Nitro form

1. Less acidic

2. Dissolves in NaOH slowly

3. Decolourises FeCl₃ solution

4. Electrical conductivity is low

Aci – form

1. More acidic and also called pseudoacids (or) nitronic acids

2. Dissolves in NaOH instantly

3. With FeCl₃ gives reddish brown colour

4. Electrical conductivity is high

3. What happens when

i. 2 – Nitropropane boiled with HCl

ii. Nitrobenzene undergo electrolytic-reduction in strongly acidic medium.

iii. Oxidation of tert – butylamine with KMnO4

iv. Oxidation of acetoneoxime with trifluoroperoxy acetic acid.

i) 2 - Nitropropane boiled with HCl

When 2 - Nitropropane boiled with HCl gives acetone.

ii. Nitrobenzene electrolytic reduction in strongly acidic medium.

iii. Oxidation of tert - butylamine with KMnO₄

tert - butyl amine is oxidised with aqueous KMnO₄ give tert - nitro alkanes.

iv. Oxidation of acetoneoxime with trifluoroperoxy acetic acid.

Oxidation of acetoneoxime with trifluoroperoxy acetic acid gives 2 - nitropropane.

4. How will you convert nitrobenzene into

i. 1,3,5 - trinitrobenzene

ii. o and p- nitrophenol

iii. m – nitro aniline

iv. azoxybenzene

v. hydrozobenzene

vi. N – phenylhydroxylamine

vii. Aniline

Answer:

5. Identify compounds A,B and C in the following sequence of reactions.

Answer:

6. Write short notes on the following

i. Hofmann’s bromide reaction

ii. Ammonolysis

iii. Gabriel phthalimide synthesis

iv. Schotten – Baumann reaction

v. Carbylamine reaction

vi. Mustard oil reaction

vii. Coupling reaction

viii. Diazotisation

ix. Gomberg reaction

i) Hofmann's bromide reaction

When amides are treated with bromine in the presence of aqueous or ethanolic solution of KOH, primary amines with one carbon atom less than the parent amides are obtained.

RCONH₂ —- ^Br2/KOH→ R NH₂ + K₂ CO₃ + 2KBr + H₂O

ii) Ammonolysis

When Alkyl halides (or) benzylhalides are heated with alcoholic ammonia, mixtures of 1°, 2° and 3° amines and quaternary ammonium salts are obtained. 

iii) Gabriel phthalimide synthesis

Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine.

iv) Schotten - Baumann reaction

Aniline reacts with benzoyl chloride in the presence of NaOH to give N -phenyl benzamide.

v) Carbylamine reaction

Aliphatic (or) aromatic primary amines react with chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This test used to identify the primary amines.

 C₂H₅ - NH₂ [Ethylamine] + CHCl₃ [Chloroform] + 3K0H → C₂H₅ - NC [Ethyl isocyanide] + 3KCl + 3H₂O

vi) Mustard oil reaction

When primary amines are treated with carbon disulphide (CS₂), N - alkyldithio carbonic acid is formed which on subsequent treatment with HgCl₂, give an alkyl isothiocyanate.

vii) Coupling reaction

Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds. Coupling generally occurs at the para position. If para position is occupied then coupling occurs at the ortho position. Coupling tendency is enhanced if an electron donating group is present at the para - position to N₂⁺ C𝑙⁻ group.

viii) Diazotisation

Aniline reacts with sodium nitrite and hydrochloric acid at low temperature (273 - 278 K) to give benzene diazonium chloride which is stable for a short time and slowly decompose even at low temperatures.

ix) Gomberg reaction

Benzene diazonium chloride reacts with benzene in the presence of sodium hydroxide to give biphenyl.

7. How will you distinguish between primary secondary and tertiary alphatic amines.

8. Account for the following

i. Aniline does not undergo Friedel – Crafts reaction

ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines

iii. pKb of aniline is more than that of methylamine

iv. Gabriel phthalimide synthesis is preferred for synthesising primary amines.

v. Ethylamine is soluble in water whereas aniline is not

vi. Amines are more basic than amides

vii. Although amino group is o – and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m – nitroaniline.

i. Aniline does not undergo Friedel - Crafts reaction.

Aniline does not undergo Friedel - Crafts reaction (alkylation and acetylation). Aniline is basic in nature and it donates its lone pair to the lewis acid AlCl₃ to form an adduct which inhibits further the electrophilic substitution reaction

ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines

Diazonium salts of aromatic amines are more stable. This is due to the dispersal of the positive charge over the benzene ring

iii) pK_b of aniline is more than that of methylamine

The lone pair of electrons on the N atom present in the aniline is involved delocalized over the benzene ring. Therefore lone pair of electron on the nitrogen atom is not available for donation.

Whereas in methylamine due to +1 effect, the lone pair of electrons are available for donation and shows higher basicity than aniline.

Smaller the value of pK_b, stronger is the base. Methylamine is strong base than aniline. Hence pK_b of aniline is more than that of methylamine.

iv) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Gabriel phthalimide synthesis is preferred for synthesizing primary amines. It involves S_N² nucleophilic substitution of alkyl halide by the anion formed by the phthalimide.

v) Ethylamine is soluble in water whereas aniline is not

When Ethylamine was added to water, it forms intermolecular Hydrogen bonding with water and therefore soluble water. Aniline does not form H-bond with water to a very large extent due to the presence of aromatic ring. Hence aniline is insoluble.

vi) Amines are more basic than amides

The carbonyl group present in the amide draws electron from the -NH₂ group results the lone pair of electron on the nitrogen atom of -NH₂ group is less available for donation and the basicity decreased. The lone pair of electrons on the amine is easily available to act as a base. Hence amines are more basic than amides.

vii) Although amino group is o - and p - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m - nitroaniline.

Direct nitration of aniline gives o and p - nitro aniline. In a strong acid medium aniline is protonated to form anilinium ion which is m - directing and hence m - nitro aniline is also formed

9. Arrange the following

i. In increasing order of solubility in water, C6 H5NH2 ,(C2H5 )2NH,C2 H5NH2

ii. In increasing order of basic strength

a) aniline, p- toludine and p – nitroaniline

b) C6 H5 NH2 ,C6 H5 NHCH3 ,C6 H5 NH2 ,p-Cl-C6 H4 -NH2

iii. In decreasing order of basic strength in gas phase

(C2 H5 )NH2 ,(C2H5 )NH, ( C2H5 )3 N and NH3

iv. In increasing order of boiling point

C6H5OH, (CH3)2NH, C2H5NH2

v. In decreasing order of the pKb values

C2 H5NH2 , C6H5 NHCH3 ,(C2 H5 )2 NH and CH3NH2

vi. Increasing order of basic strength

C6 H5NH2 ,C6H5 N(CH3 )2 ,(C2H5 )2 NH and CH3 NH2

vii. In decreasing order of basic strength

i. In increasing order of solubility in water, C₆H₅NH₂ ,(C₂H₅)₂NH,C₂H₅NH₂

C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

ii. In increasing order of basic strength

a) aniline, p- toluidine and p - nitroaniline

p - nitroaniline < aniline< p- toluidine

b) C₆H₅NH₂,C₆H₅NHCH₃ ,p-Cl-C₆H₄ -NH₂

p-Cl-C₆H₄ -NH₂ < C₆H₅NH₂ < C₆H₅NHCH₃ ,

iii. In decreasing order of basic strength in gas phase

C₂H₅NH₂, (C₂H₅)₂NH,( C₂H₅)₃ N and NH₃

(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃

iv. In increasing order of boiling point

C₆H₅OH, (CH₃)₂NH, C₂H₅NH₂

(CH₃)₂NH < C₂H₅NH₂ < C₆H₅OH 

v. In decreasing order of the pK_b values C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and CH₃NH₂

C₆H₅NHCH₃ < CH₃NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

vi. Increasing order of basic strength

C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂ NH and CH₃NH₂

C₆H₅NH₂ < C₆H₅N(CH₃)₂ <CH₃NH₂<(C₂H₅)₂ NH

10. How will you prepare propan – 1- amine from

i) butane nitrile ii) propanamide ii) 1- nitropropane

11. Identify A,B,and C

12. How will you convert diethylamine into

i) N, N – diethylacetamide ii) N – nitrosodiethylamine

13. Identify A,B and C

14. Identify A,B,C and D

15. Complete the following reaction

16. Predict A,B,C and D for the following reaction

17. A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO2 . (B) on heating with liquid ammonia followed by treating with Br2 /KOH gives (c) which on treating with NaNO2 and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.

Solution:

Answer:

A - 1, 2 - dibromobutane

B - Butanoic acid

C - 1-amino propane

D - Propionic acid [Molecular mass 74]

18. Identify A to E in the following sequence of reactions.

EVALUATE YOURSELF:

1. Write all possible isomers for the following compounds.

i) C₂H₅NO₂ ii) C₃H₇-NO₂

2. Find out the product of the following reactions.

3. Predict the major product that would be obtained on nitration of the following compunds.

4. Draw the structure of the following compounds

i. Neopentylamine

ii. Tert - butylamine

iii. α- amino propionaldehyde

iv. Tribenzylamine

v. N - ethyl - N - methylhexan - 3- amine.

Answer:

5. Give the correct IUPAC names for the following amines

Answer:

(i) pentan-2-amine

(ii) N-methylbutan-2-amine

(iii) cyclohexanamine

(iv) 3-aminophenol

(v) N, N, N - triphenyl amine

NITRO COMPOUNDS