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

Note the point of difference in the given compounds which here lies at $$\beta $$ - carbon. In I, II, III, the $$\beta $$ - carbon atoms are $$s{p^3},s{p^2}$$   and $$sp$$  hybridised respectively which in turn cause the difference in their $$s$$ - character. We know that more is the $$s$$  character of an atom, greater will be its electron-withdrawing nature. Thus $$sp$$  ( $$50\% \,\,s$$   character ) hybridised carbon is most electron-withdrawing, while $$s{p^3}$$  ( $$25\% $$  $$s$$ - character ) is least electron-withdrawing. Further, we know that presence of an electron-withdrawing group decreases basicity of an amine. Thus

Quaternary ammonium salt, as it carries a positive charge over nitrogen. e.g., $${\left( {C{H_3}} \right)_4}\mathop N\limits^ + C{l^ - }$$

Diazonium cation is a weak electrophile and hence reacts with electron rich compounds containing electron donating groups such as $$ - OH, - N{H_2}$$   and $$ - OC{H_3}$$  groups and not with compounds containing electron withdrawing groups such as $$ - N{O_2},$$  etc.

When benzene sulphuric acid and $$p$$ - nitrophenol are treated with $$NaHC{O_3},$$   the gases released, respectively are $$C{O_2},C{O_2}.$$
The balanced chemical equation for the reaction of benzene sulphonic acid with sodium bicarbonate is as shown below.
$$PhS{O_3}H + NaHC{O_3} \to $$     $$PhS{O_3}Na + C{O_2} + {H_2}O$$
The balanced chemical equation for the reaction of $$p$$ - nitro phenol with sodium bicarbonate is as show below.
$$p - {O_2}N - {C_6}{H_4} - OH + NaHC{O_3} \to $$        $$p - N{O_2} - {C_6}{H_4} - ONa + C{O_2} + {H_2}O$$

The given reaction follows $${S_N}2$$  mechanism and $${S_N}2$$  reactions are favoured in polar aprotic medium like $$DMSO,DMF...$$    etc.
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br+NaCN\xrightarrow{DMF}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CN+NaBr\]
So, the correct option is (C).

\[-\ddot{N}{{H}_{2}}\]  group is acetylated by acetic anhydride in methylene chloride (solvent).

Note that \[-CON{{H}_{2}}\]   group does not undergo acetylation because here lone pair of electrons is delocalised.

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