Amines MCQs for NEET — Chemistry Questions with Answers

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

(a)      $R{\mathrm{NH}}_2 + {\mathrm{CHCl}}_3 + 3\mathrm{KOH}(\mathrm{alc}.) \stackrel{ }{\to } R\mathrm{NC} + 3\mathrm{KCl} + 3{\mathrm{H}}_2\mathrm{O}$

                                                                       Alkyl isocyanide

This reaction is known as carbylamone test.

(only $1°$ amine gives this reaction)

(d) Aliphatic amines are more basic than aromatic amines as the later are more stabilised due to resonance.

(d) 2,4-dinitro fluorobenzene is called Sanger's reagent. When this reagent reacts with amino group of peptide chain. It form 2,4-dinitrophenyl derivatives which on hydrolysis form DNP derivatives of amino acids.

In the Hofmann's reaction, an alkanamide undergoes rearrangement to form a primary amine with one less carbon atom. If the alkanamide gives 1-phenylethylamine on Hofmann's reaction, the starting alkanamide must be 2-phenylpropanamide.

(b) $R\mathrm{Br} + \mathrm{KCN} \stackrel{ }{\to }R\mathrm{CN} \stackrel{\mathrm{Reduction}}{\to } {\mathrm{RCH}}_2{\mathrm{NH}}_2$

(b) ${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2\stackrel{{\mathrm{HNO}}_2}{\to }{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{{\mathrm{PCl}}_5}{\to }{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}\stackrel{{\mathrm{NH}}_3}{\to }{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2$

(b) C₃H₈O is alcohol and C₃H₆O₂ is acid. Thus, C₃H₉N is primary amine.