Chemistry MCQs for NEET — Practice Questions with Answers

The process of rapid equilibrium between a carbonyl compound with a hydrogen atom on its alpha-carbon and its corresponding enol form is known as keto-enol tautomerism. This tautomerism is an important phenomenon in organic chemistry, as it affects the reactivity and properties of the compounds involved.

Reaction of carbonyl compounds with ammonia derivatives give addition product followed by the elimination reaction. Slightly acidic medium generate a nucleophilic centre for the attack of weak base like ammonia derivatives.

Clemmensen reduction of a ketone is carried out in the presence of zinc-mercury amalgam (Zn-Hg) and hydrochloric acid (HCl). This reducing mixture converts the carbonyl group of the ketone into a methylene group, effectively reducing the ketone to an alkane.

The Cannizzaro reaction involves the disproportionation of a non-enolizable aldehyde, forming an alcohol and a carboxylate salt. It does not result in the formation of a carbon-carbon bond.

$$\begin{gathered} >\mathrm{C}=\mathrm{O}+\mathrm{HCN} \stackrel{}{\to } >\mathrm{C}<\begin{array}{c}\mathrm{OH}\\ \mathrm{CN}\end{array}\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to } >\mathrm{C}< \begin{array}{c}\mathrm{OH}\\ \mathrm{COOH}\end{array} \left(\mathrm{It} \mathrm{is} \mathrm{\alpha }-\mathrm{hydroxy} \mathrm{acid}\right) \\ \mathrm{If} \mathrm{it} \mathrm{is} \mathrm{racemic} \mathrm{mixture}, \mathrm{therefore} \mathrm{such} \text{'}\mathrm{c}\text{'} \mathrm{atom} \mathrm{must} \mathrm{be} \mathrm{asymmetric} \mathrm{carbon} \mathrm{atom}. \\ \begin{array}{c}{\mathrm{CH}}_3\\ \mathrm{H}\end{array}>\mathrm{C}=\mathrm{O}+\mathrm{HCN} \stackrel{}{\to } \begin{array}{c}{\mathrm{CH}}_3\\ \mathrm{H}\end{array}>\mathrm{C}<\begin{array}{c}\mathrm{OH}\\ \mathrm{CN}\end{array} \stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }\begin{array}{c}{\mathrm{CH}}_3\\ \mathrm{H}-\mathrm{C}-\mathrm{OH}\\ \begin{array}{c}\mathrm{COOH}\\ \mathrm{d}-\end{array}\end{array}+\begin{array}{c}{\mathrm{CH}}_3\\ \mathrm{HO}-\mathrm{C}-\mathrm{H}\\ \begin{array}{c}\mathrm{COOH}\\ \mathrm{I}-\end{array}\end{array} \\ \mathrm{Acetaldehyde} \mathrm{Racemic} \mathrm{mixture} \\ \begin{array}{c}{\mathrm{CH}}_3\\ {\mathrm{CH}}_3\end{array}>\mathrm{C}=\mathrm{O}+\mathrm{HCN} \to \begin{array}{c}{\mathrm{CH}}_3\\ {\mathrm{CH}}_3\end{array}>\mathrm{C}<\begin{array}{c}\mathrm{OH}\\ \mathrm{CN}\end{array}\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to } \begin{array}{c}{\mathrm{CH}}_3\\ {\mathrm{CH}}_3\end{array}>\mathrm{C}< \begin{array}{c}\mathrm{OH}\\ \mathrm{COOH}\end{array} \\ \mathrm{Acetone} \mathrm{It} \mathrm{is} \mathrm{not} \mathrm{optical} \mathrm{active} \\ \begin{array}{c}{\mathrm{C}}_2{\mathrm{H}}_5\\ {\mathrm{C}}_2{\mathrm{H}}_5\end{array}>\mathrm{C}=\mathrm{O}+\mathrm{HCN} \stackrel{}{\to } \begin{array}{c}{\mathrm{C}}_2{\mathrm{H}}_5\\ {\mathrm{C}}_2{\mathrm{H}}_5\end{array}>\mathrm{C}< \begin{array}{c}\mathrm{OH}\\ \mathrm{CN}\end{array}\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }\begin{array}{c}{\mathrm{C}}_2{\mathrm{H}}_5\\ {\mathrm{C}}_2{\mathrm{H}}_5\end{array}>\mathrm{C}< \mathrm{COOH} \\ \mathrm{Diethyl} \mathrm{ketone} \mathrm{It} \mathrm{is} \mathrm{not} \mathrm{optical} \mathrm{active} \\ \begin{array}{c}\mathrm{H}\\ \mathrm{H}\end{array}>\mathrm{C}=\mathrm{O}+\mathrm{HCN} \stackrel{}{\to } \begin{array}{c}\mathrm{H}\\ \mathrm{H}\end{array}>\mathrm{C}<\begin{array}{c}\mathrm{OH}\\ \mathrm{CN}\end{array} \stackrel{{\mathrm{H}}_2\mathrm{O}}{\to } \begin{array}{c}\mathrm{H}\\ \mathrm{H}\end{array}>\mathrm{C}<\begin{array}{c}\mathrm{OH}\\ \mathrm{COOH}\end{array} \\ \mathrm{Formaldehyde} \mathrm{It} \mathrm{is} \mathrm{not} \mathrm{optical} \mathrm{active} \end{gathered}$$

On condensation of two moles of ethyl acetate in the presence of sodium ethoxide, gives ethyl aceto-acetate (ester). This condensation is an example of Claisen condensation because it is possible in that ester which have a-hydrogen atom.

$$\begin{gathered} \mathrm{O} \mathrm{O} \\ \parallel \parallel \\ {\mathrm{CH}}_3-\mathrm{C}-\mathrm{O}-{\mathrm{C}}_2{\mathrm{H}}_5 + \mathrm{H}-{\mathrm{CH}}_2{\mathrm{COOC}}_2{\mathrm{H}}_5 \stackrel{{\mathrm{NaOC}}_2{\mathrm{H}}_5}{\to } {\mathrm{CH}}_3{\mathrm{CCH}}_2{\mathrm{COOC}}_2{\mathrm{H}}_5 + {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH} \end{gathered}$$

(b) When acetamide is heated with aq. NaOH it forms NH₃ gas but ethylamine cannot form NH₃.

 CH₃CONH₂ + H₂O$\stackrel{\mathrm{NaOH}}{\to }$CH₃COONa + NH₃

CH₃CH₂NH₂ + H₂O $\stackrel{\mathrm{NaOH}}{\to }$No reaction

Basically, all Azo dyes are derivatives of aniline.