JEE Main 2023 — p Block Elements (Group 15, 16, 17 & 18) Question with Solution

(A) Hell-Volhard-Zelinsky reaction:

Carboxylic acid react with chlorine or bromine in the presence of phosphorus to give $\mathrm{\alpha }-$halogenated carboxylic acids. The halogenation continues till all the hydrogen atom in the $\mathrm{\alpha }-$position are replaced by halogen atoms. This reaction is called HVZ (Hell-Volhard-Zelinsky) reaction.

$\underset{\mathrm{Acetic} \mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH} }\underset{-\mathrm{HCl}}{\overset{{\mathrm{Cl}}_2, {\mathrm{P}}_4}{\to }} \underset{\mathrm{Chloroacetic} \mathrm{acid}}{{\mathrm{ClCH}}_2\mathrm{COOH}}\underset{-\mathrm{HCl}}{\overset{{\mathrm{Cl}}_2, {\mathrm{P}}_4}{\to }}\underset{\mathrm{Dichloroacetic} \mathrm{acid}}{{\mathrm{Cl}}_2\mathrm{CHCOOH}}\underset{-\mathrm{HCl}}{\overset{{\mathrm{Cl}}_2, {\mathrm{P}}_4}{\to }}\underset{\mathrm{Trichloroacetic} \mathrm{acid}}{{\mathrm{Cl}}_3\mathrm{CCOOH}}$

(B) Iodoform reaction:

The Iodoform reaction is a chemical reaction where a iodoform$\left({\mathrm{CHI}}_3\right)$  is produced by the full iodonation of a methyl ketone (${\mathrm{RCOCH}}_3$, where $\mathrm{R}$ can be either a hydrogen atom, an alkyl or an aryl group) in the presence of a base.

$$\begin{gathered} 2\mathrm{NaOH}+{\mathrm{I}}_2\to \mathrm{NaOI}+\mathrm{NaI} \\ {\mathrm{CH}}_3{\mathrm{COCH}}_3+3\mathrm{NaOI}\to {\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{CHI}}_3+2\mathrm{NaOH} \end{gathered}$$

(C) Etard reaction:

 In Etard reaction, chromyl chloride oxidises methyl group to a chromium complex, which on hydrolysis gives corresponding Benzaldehyde. The equation for the reaction is:

${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_3 + {\mathrm{CrO}}_2{\mathrm{Cl}}_2 \stackrel{{\mathrm{CS}}_2}{\to } {\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CH}\left({\mathrm{OCrOHCl}}_2\right) \stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to } {\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CHO}$

 

(D) Gatterman-Koch reaction

When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives Benzaldehyde or substituted Benzaldehyde. The equation for the reaction is:

${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_3 \underset{\mathrm{Anh}.{\mathrm{AlCl}}_3/\mathrm{CuCl}}{\overset{\mathrm{CO} + \mathrm{HCl}}{\to }} {\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CHO}$