Organic Compounds Containing Nitrogen MCQ Questions & Answers in Organic Chemistry | Chemistry

$$LiAl{H_4}$$  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.

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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.