Chemistry MCQs for NEET — Practice Questions with Answers

Tertiary nitro compounds cannot show tautomerism because they do not have a labile hydrogen atom attached to the carbon atom. Tautomerism involves the shifting of a labile hydrogen atom from one position to another in the same molecule.

A secondary amine is a compound in which two of the hydrogen atoms of ammonia (NH3) have been replaced by alkyl or aryl groups. It has the general formula R2NH, where R can be an alkyl or aryl group.

(c) When amides react with bromine in the presence of caustic alkali to form a primary amine carrying one carbon atom less than the parent amide, then the reaction is known as Hofmann bromide reaction.

$R{\mathrm{CONH}}_2 + {\mathrm{Br}}_2 + 4\mathrm{KOH} \stackrel{\mathrm{Heat}}{\to } R{\mathrm{NH}}_2 + {\mathrm{K}}_2{\mathrm{CO}}_3 +2\mathrm{KBr} + 2{\mathrm{H}}_2\mathrm{O}$

(a) The reagent which can convert -CONH₂ group is used for this reaction. Among the given reagents only NaOH/Br₂ converts -CONH2 group to -NH₂ group, thus it is used for converting acetamide to methyl amine. This reaction is called Hoffmann bromide reaction, in which primary amides on treatement with Br₂/NaOH from primary amines.

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CONH}}_2 + \mathrm{NaOH} + {\mathrm{Br}}_2 \stackrel{ }{\to }{\mathrm{CH}}_3{\mathrm{NH}}_2 + \mathrm{NaBr} + {\mathrm{Na}}_2{\mathrm{CO}}_3 + {\mathrm{H}}_2\mathrm{O} \\ \mathrm{Acetamide} \mathrm{Methyl} \mathrm{amide} \end{gathered}$$

 ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Cl}$ + KCN $\to {\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CN}$ + KCl

                                                          Butanonitrile 

(c) $2R{\mathrm{NH}}_2 + 2\mathrm{HCl} + {\mathrm{PtCl}}_4 \stackrel{ }{\to }\left(R{\mathrm{NH}}_3\right)2{\mathrm{PtCl}}_6; R{\mathrm{NH}}_3{\mathrm{AuCl}}_4$

its a test of 2 degree of amine

$\mathrm{R}-\mathrm{NC} + 2{\mathrm{H}}_2\mathrm{O} \stackrel{{\mathrm{H}}^{+}}{\to } \mathrm{R}-{\mathrm{NH}}_2 + \mathrm{HCOOH}$

                                   $1° \mathrm{Amine}$

Deamination of n-Bu-NH2 using NaNO2/HCl gives a mixture of 1-butene, cis-2-butene, and trans-2-butene. The formation of isobutene is not possible as it requires a rearrangement which does not occur under these conditions.