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

1.	Alkyl halides do not undergo electrophilic substitution with the anion formed by the phthalimide
2.	Alkyl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide
3.	Aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide
4.	Aryl halides do not undergo electrophilic substitution with the anion formed by the phthalimide

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

The product from the above reaction would be -

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

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

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

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

The primary amine is -

1. ${\left({\mathrm{CH}}_3\right)}_2{\mathrm{CHNH}}_2$

2. ${{\mathrm{CH}}_3\left({\mathrm{CH}}_2\right)}_2\mathrm{NH}-{\mathrm{CH}}_3$

3. ${{\mathrm{CH}}_3\left({\mathrm{CH}}_2\right)}_2\mathrm{N}{\left({\mathrm{CH}}_3\right)}_2$

4. None of the above

Which of the following is a secondary amine is -

$1. {\mathrm{CH}}_3{\left({\mathrm{CH}}_2\right)}_2{\mathrm{NH}}_2$
$2. {\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_2{\mathrm{NCH}}_3$
$3. {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NHCH}}_3$
$4. {\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}_2$

The tertiary amine is-

$1. {\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_2{\mathrm{NCH}}_3$
$2. {\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}_2$
$3. \mathrm{m}-{\mathrm{BrC}}_6{\mathrm{H}}_4{\mathrm{NH}}_2$
$4. \mathrm{None} \mathrm{of} \mathrm{these}$

The chemical test used to distinguish between aniline and benzylamine is-

1. Nitrous acid.

2. Reduction Reaction

3. Hinsberg reagent

4. None of above

The chemical test used to distinguish between aniline and N-methyl aniline is-

1. Carbylamine test 

2. Hoffmann bromamide reaction

3. Acetylation

4. None of above

The chemical test used to distinguish between methylamine and dimethylamine is-

1. Hoffmann bromamide reaction

2. Hinsberg's reagent 

3. Carbylamine test

4. Both 2 and 3

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= $\mathrm{HBr}/ \mathrm{hv}$, Y= Fe/HCl, and Z = ${\mathrm{H}}_3{\mathrm{PO}}_2$

4. X = $\mathrm{HBr}/ \mathrm{hv}$ Y = ${\mathrm{Br}}_2/{\mathrm{FeCl}}_3$, and Z = ${\mathrm{H}}_3{\mathrm{PO}}_2$

X, and Y in the following sequence are :

1.	X =  Br2 / CS2 , and Y =
2.	X =  Br2 / H2O  and Y =
3.	X =  Br2 / CS2 , and Y =
4.	X =  Br2 / H2O  and Y =