C₂H₅NH₂ , C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂ 

_{For the above compounds, decreasing order of the pKb values (in gaseous phase) is -}

1. C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

2. C₂H₅NH₂ > C₆H₅NHCH₃ > C₆H₅NH₂ > (C₂H₅)₂NH

3. C₆H₅NHCH₃ > C₆H₅NH₂ > C₂H₅NH₂ > (C₂H₅)₂NH

4. (C₂H₅)₂NH > C₆H₅NHCH₃ > C₂H₅NH₂ > C₆H₅NH₂

The increasing order of boiling point of :

 C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂ is-

1. (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

2. (CH₃)₂NH > C₂H₅NH₂ > C₂H₅OH

3. (CH₃)₂NH < C₂H₅NH₂ > C₂H₅OH

4. (CH₃)₂NH <  C₂H₅OH< C₂H₅NH₂ 

The increasing sequence of solubility of the following molecules ;C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂  

in water is -

1. C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

2. C₆H₅NH₂ > (C₂H₅)₂NH < C₂H₅NH₂

3. C₆H₅NH₂ > (C₂H₅)₂NH > C₂H₅NH₂

4. C₆H₅NH₂ < (C₂H₅)₂NH > C₂H₅NH₂

$\left(i\right) Aromatic amine + Nitrous acid \to A$
$\left(ii\right) Aliphatic amine + Nitrous acid \to B$

The A and B in the above reactions are -

Amines are less acidic than alcohols of comparable molecular masses because :

1. N is more electronegative than O.

2. O is more electronegative than N.

3. N can accomodate negative charge easily than O

4. Amides are more stable than alcohols

An aromatic compound ‘A’ on treatment with ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. The Substrate A is -

1. Aniline

2. Benzoic acid 

3. Benzamide

4. Benzamine

A method for the identification of primary, secondary and tertiary amines is:

1. Azo-dye test

2. Carbylamine test

3. Hinsberg’s test

4. None of these

Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because :

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2\mathrm{Cl}+{\mathrm{H}}_3{\mathrm{PO}}_2+{\mathrm{H}}_2\mathrm{O}$:

The product from the above reaction would be -

$1.{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{NC}$

$2. {\mathrm{C}}_6{\mathrm{H}}_6{\mathrm{CH}}_3$

$3.{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CN}$

$4. {\mathrm{C}}_6{\mathrm{H}}_6$