Nitration of aniline in strong acidic medium also gives m-nitroaniline because

1. In spite of substituents nitro group always goes to only m-position

2. In electrophilic substitution reactions amino group is meta directive

3. In absence of substituents nitro group always goes to only m-position

4. In acidic (strong) medium aniline is present as anilinium ion

An aromatic amine (A) was treated with alcoholic potash and another compound (Y) when a foul smelling gas was formed with formula C₆H₅NC. (Y) was formed by reacting a compound (Z) with Cl₂ in the presence of slaked lime. Compound (Z) is:

1. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2$

2. ${\mathrm{CH}}_3\mathrm{OH}$

3. ${\mathrm{CH}}_3{\mathrm{COCH}}_3$

4. ${\mathrm{CHCl}}_3$

The correct order of basicities of the following compounds is:

$\underset{1}{{\mathrm{CH}}_3-\underset{{\mathrm{NH}}_2}{\overset{\mathrm{NH}}{\underset{\left|\right|}{\overset{\left|\right|}{\mathrm{C}}}}}}, \underset{2}{C{H}_{3}C{H}_{2}N{H}_2} , \underset{3}{{\left(C{H}_3\right)}_{2}NH} , \underset{4}{ C{H}_3-\stackrel{O}{\stackrel{\left|\right|}{C}}-N{H}_2}$

1. 2 > 1 > 3 > 4

2. 1> 3 > 2 > 4

3. 3 > 1 > 2 > 4

4. 1 > 2 > 3 > 4

Which reagent transforms nitromethane into methylamine?

1. Zn/HCl

2. Zn/NaOH

3. Zn/C₂H₅OH

4. Ni/H₂

Ethylamine undergoes oxidation in the presence of KMnO₄ followed by hydrolysis to form:

1. An acid

2. An alcohol

3. An aldehyde

4. A N-oxide

In the above reaction the product X will be

1.
2.
3.
4.

The major product of the following reaction would be

1.
2.
3.
4.

The product (C) in the below mentioned reaction is:
 \(\xrightarrow[]{{(NH_4)}_2S}\ A\ \xrightarrow[0-5\ ^\circ C]{NaNO_2/HCl}\ B \xrightarrow[\Delta]{C_2H_5OH}\ C\)

1. Nitrobenzene

2. 1,3-Diethoxybenzene

3. Ethoxybenzene

4. Benzene