Chemistry MCQs for NEET — Practice Questions with Answers

(c) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2 \underset{0°\mathrm{c}}{\overset{{\mathrm{NaNO}}_2+\mathrm{HCl}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2\mathrm{Cl}$

 Order of basic nature in a aqueous solution on the basis of K_b.

$({\mathrm{CH}}_3{)}_2\mathrm{NH}>{\mathrm{CH}}_3{\mathrm{NH}}_2>({\mathrm{CH}}_3{)}_3\mathrm{N}>{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2$

(a) Trinitrobenzene is an explosive compound formed during nitration of C₆H₆ with fuming HNO₃.

(a) ${\mathrm{CH}}_3-\mathrm{N}\equiv \mathrm{C} +\hspace{0.17em}4\mathrm{H} \stackrel{{\mathrm{LiAlH}}_4}{\to }{\mathrm{CH}}_3{\mathrm{NHCH}}_3$

                                         Dimethylamine

On catalytic reduction or with lithium aluminium hydride (LiAlH₄) or with nascent hydrogen, alkyl isocyanide yield $2°$ amine whereas cyanide gives $1°$ amine on reduction.

When an organometallic compound (RMgX) reacts with cyanogen chloride (ClCN), the Grignard reagent undergoes nucleophilic addition followed by elimination, leading to the formation of an organo-nitrile (RCN). Hence, the correct option is RCN.

$$\begin{gathered} \underset{1°}{{\mathrm{R}}_1{\mathrm{CH}}_2{\mathrm{NO}}_2}\stackrel{HN{O}_2}{\to }{R}_1-\underset{NOH}{\underset{\left|\right|}{C}}-N{O}_2\stackrel{NaOH}{\to }\underset{\mathrm{Red}}{R_1-\underset{NONa}{\underset{\left|\right|}{C}}-N{O}_2} \\ \underset{2°}{{\mathrm{R}}_2{\mathrm{CHNO}}_2}\stackrel{HN{O}_2}{\to }{R}_2-\underset{NO}{\underset{|}{C}}-N{O}_2\stackrel{NaOH}{\to }\underset{Blue}{R_1-\underset{NO}{\underset{|}{C}}-N{O}_2} \\ \underset{\mathit{3}\mathit{°}}{{\mathit{R}}_{\mathit{3}}{\mathit{CNO}}_{\mathit{2}}}\stackrel{\mathit{H}\mathit{N}{\mathit{O}}_{\mathit{2}}}{\mathit{\to }}No\mathit{ }reaction \end{gathered}$$

 $$\begin{gathered} {\mathrm{CH}}_3\mathrm{C}\equiv \mathrm{N}\underset{{\mathrm{LiAlH}}_4}{\overset{\mathrm{Reduction}}{\to }}{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{NH}}_2\stackrel{{\mathrm{HNO}}_3}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH} \\ \mathrm{B} \mathrm{is} 1° \mathrm{amine} \mathrm{i}.\mathrm{e}. \mathrm{Ethanamine} \\ \therefore \mathrm{A} \mathrm{is} {\mathrm{CH}}_3\mathrm{CN} \end{gathered}$$

(c) Key Idea

(i) Presence of electron withdrawing substituent decreases the basicity while the presence of electron releasing substituent like, -CH₃, -C₂H₅, etc, increases the acidity.

(ii) HNO₂ converts -NH₂ group of aliphatic amine into -OH while that of aromatic amines into -N=NCl.

Since, phenyl group is a electron withdrawing group, it decreases the basicity. Alkyl group, on the other hand, being electron releasing, increases the basicity. Thus, alkyl amines are more basic as compared to aryl amines as well as ammonia.

$R-{\mathrm{NH}}_2\stackrel{{\mathrm{HNO}}_2}{\to }R\mathit{ }-\mathrm{OH}$ 

Thus, HNO₂ (nitrous) converts alkyl amines to alcohols.

But aryl amines react with nitrous acid to form diazonium salt.

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\underset{0-5°\mathrm{c} (273-278\mathrm{k})}{\overset{{\mathrm{HNO}}_2}{\to }}{\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{N}}_2^{+}\underset{·}{\mathrm{Cl}}$

at $\mathrm{o}-5°\mathrm{c}$ temperature ${\mathrm{NaNO}}_2 + \mathrm{HCl} \stackrel{ }{\to }{\mathrm{HNO}}_2 + \mathrm{NaCl}$

Thus, HNO₂ does not convert aryl amines into phenol.

(b) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\stackrel{\mathrm{Diazotisation}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2\mathrm{Cl} + {\mathrm{H}}_2\mathrm{O}$