$$LiAl{H_4}$$  in ether reduces aryl nitro compounds to azo compounds.
\[\underset{\text{Nitrobenzene}}{\mathop{2{{C}_{6}}{{H}_{5}}N{{O}_{2}}}}\,\xrightarrow{\frac{LiAl{{H}_{4}}}{\text{ether}}}\]     \[\underset{\text{ Diazobenzene}}{\mathop{{{C}_{6}}{{H}_{5}}-N=N-{{C}_{6}}{{H}_{5}}}}\,\]

\[C{{H}_{3}}CN\xrightarrow{Na/{{C}_{2}}{{H}_{5}}OH}\underset{\left( A \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}}}\,\]        \[\xrightarrow{HN{{O}_{2}}}\underset{\left( B \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,\xrightarrow[573\,K]{Cu}\underset{\left( C \right)}{\mathop{C{{H}_{3}}CHO}}\,\]

The order of basicity is I > III > II > IV.
The lone pair of electrons on $$N$$  is more readily available for protonation in I and III then in II. III contains an oxygen atom which has $$– I$$  effect due to which it is less basic than I. In compound IV lone pair of $${e^ - }s$$  on $$N–$$  atom is contributed towards the aromatic sextet formation and hence is not at all available for protonation. Hence option (B) is correct.

When acetamide is heated with $$aq.$$ $$NaOH$$  it forms $$N{H_3}$$  gas but ethylamine cannot form $$N{H_3}.$$
\[C{{H}_{3}}CON{{H}_{2}}+{{H}_{2}}O\xrightarrow{NaOH\Delta }\]       \[C{{H}_{3}}COONa+N{{H}_{3}}\]
\[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}+{{H}_{2}}O\xrightarrow{NaOH\Delta }\]       \[\text{No reaction}\]

\[\underset{\left( \text{I} \right)}{\mathop{A\xrightarrow{N{{H}_{3}}}}}\,B\underset{\left( \text{II} \right)}{\mathop{\xrightarrow{\Delta }}}\,C\underset{\left( \text{III} \right)}{\mathop{\xrightarrow[KOH]{B{{r}_{2}}}}}\,\]     \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\]
Reaction (III) is a Hofmann bromamide reaction formation of $$C{H_3}C{H_2}N{H_2}$$    is possible only from a compound $$C{H_3}C{H_2}CON{H_2}$$    which can be obtained from the compound $$C{H_3}C{H_2}CO{O^ - }NH_4^ + \left( B \right)$$      in (II) reaction further propanic acid $$\left( {C{H_3}C{H_2}COOH} \right)$$    on reaction with $$N{H_3}$$  produce $$C{H_3}C{H_2}CO{O^ - }NH_4^ - $$    ( reaction I ) hence the reaction will be

It is a secondary amine.

No explanation is given for this question. Let's discuss the answer together.

More the number of hydrogen atoms linked with nitrogen more will be the intermolecular hydrogen bonding and hence more will be the boiling point. Since $${1^ \circ }$$ amines have two, $${2^ \circ }$$ amines have one while $${3^ \circ }$$ amines have no hydrogen linked to nitrogen, therefore, $${1^ \circ }$$ amines have the highest while $${3^ \circ }$$ amines have the lowest boiling point.

When a primary amide is treated with anqueous solution of potassium hydroxide and bromine, it gives a primary amine which has one carbon atom less than the original amide and the reaction is known as Hoffmann bromamide reaction.