The incorrect statement among the following is :

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 among the given compounds is:

C₂H₅OH, (CH₃)₂NH, and C₂H₅NH₂

1. (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

2. (CH₃)₂NH > C₂H₅NH₂ > C₂H₅OH

3. (CH₃)₂NH < C₂H₅OH < C₂H₅NH₂ 

4. (CH₃)₂NH > C₂H₅OH > C₂H₅NH₂

The increasing sequence of solubility of the following molecules in water is: 
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂  

1. C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

2. C₆H₅NH₂ > (C₂H₅)₂NH < C₂H₅NH₂

3. C₆H₅NH₂ > (C₂H₅)₂NH > C₂H₅NH₂

4. C₆H₅NH₂ < (C₂H₅)₂NH > C₂H₅NH₂

Consider the following statements related to ammonolysis.

The true statement(s) are:
1. a, b
2. a, b, c
3. a, c
4. b

X, Y, and Z in the above sequence are:

1. X= Fe/HCl, Y= ${\mathrm{Br}}_2/{\mathrm{FeCl}}_3$, and Z = ${\mathrm{H}}_2\mathrm{O}$/ warm

2. X= ${\mathrm{Br}}_2/{\mathrm{FeCl}}_3$, Y= Fe/HCl, and Z = ${\mathrm{H}}_2\mathrm{O}$/ warm

3.  X= $\mathrm{HBr}/<mspace\; width="1em"></mspace>\mathrm{hv}$, Y= Fe/HCl, and Z = ${\mathrm{H}}_3{\mathrm{PO}}_2$

4. X = $\mathrm{HBr}/<mspace\; width="1em"></mspace>\mathrm{hv}$ Y = ${\mathrm{Br}}_2/{\mathrm{FeCl}}_3$, and Z = ${\mathrm{H}}_3{\mathrm{PO}}_2$

Benzyl chloride $\stackrel{\mathrm{KCN}}{\to }$ X $\stackrel{LiAl{H}_4}{\to }$ Y

The product X and Y are -

1. 2-Phenylethanenitrile and Benzoic acid

2. Benzonitrile and Benzoic acid

3. 2-Phenylethanenitrile and 2-Phenylethanamine

4. Benzonitrile and 2-Phenylethanamine

The formula of A and B in the following reaction is -

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2\mathrm{Cl}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\stackrel{\mathrm{CuCN}}{\to }<mspace\; width="1em"></mspace>{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CN}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\stackrel{{\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>/{\mathrm{H}}^{+}}{\to }<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\mathrm{A}<mspace\; width="1em"></mspace>\stackrel{{\mathrm{NH}}_3<mspace\; width="1em"></mspace>,∆}{\to }\mathrm{B}$
1. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH},<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CONH}}_2$
2. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH},<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2$
3. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{NH}}_2,<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{N}=\mathrm{NH}$
4. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{NH}}_2,<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{N}=\mathrm{NH}$

 Predict the product formed when benzenediazonium chloride reacts with hypophosphorous acid and water.

C₆H₅N₂Cl + H₃PO₂ + H₂O → Product

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