Question

Match list-I and list-II

Column 1	Column 2
$\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\mathrm{Cl}\to \mathrm{R}-\mathrm{CHO}$	${\mathrm{Br}}_2/\mathrm{NaOH}$
$\mathrm{R}-{\mathrm{CH}}_2-\mathrm{COOH}\to \mathrm{R}-\underset{\underset{\mathrm{Cl}}{|}}{\mathrm{CH}}-\mathrm{COOH}$	${\mathrm{H}}_2/\mathrm{Pd}-{\mathrm{BaSO}}_4$
$\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{NH}}_2\to \mathrm{R}-{\mathrm{NH}}_2$	$\mathrm{Zn}\left(\mathrm{Hg}\right)/\mathrm{Conc}.\mathrm{HCl}$
$\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_3\to \mathrm{R}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$	${\mathrm{Cl}}_2/\mathrm{Red}\mathrm{P},{\mathrm{H}}_2\mathrm{O}$

Answer 1

(A) $\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\mathrm{Cl}\to \mathrm{R}-\mathrm{CHO}$- (II) ${\mathrm{H}}_2/\mathrm{Pd}-{\mathrm{BaSO}}_4$

1. The Rosenmund Reaction uses palladium-barium sulphate as a catalyst to hydrogenate acyl chloride and reduce it to aldehyde.
2. It is a reduction reaction in which hydrogen is added.
3. Rosenmund catalyst is palladium on barium sulphate. Palladium is used to initiate the reduction process, but barium sulphate is used to inhibit the activity of palladium because it has a small surface area and so prevents over-reduction.
4. $\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\mathrm{Cl}\stackrel{{\mathrm{H}}_2/\mathrm{Pd}-{\mathrm{BaSO}}_4}{\to }\mathrm{R}-\mathrm{CHO}$

(B) $\mathrm{R}-{\mathrm{CH}}_2-\mathrm{COOH}\to \mathrm{R}-\underset{\underset{\mathrm{Cl}}{|}}{\mathrm{CH}}-\mathrm{COOH}$- (IV) ${\mathrm{Cl}}_2/\mathrm{Red}\mathrm{P},{\mathrm{H}}_2\mathrm{O}$

1. Carboxylic acids are halogenated at the alpha carbon via the Hell Volhard Zelinsky reaction.
2. Using red phosphorus, halogens, and water, the Hell-Volhard-Zelinsky reaction transforms carboxylic acid with alpha hydrogen into alpha -halo carboxylic acid.
3. $\mathrm{R}-{\mathrm{CH}}_2-\mathrm{COOH}\stackrel{{\mathrm{Cl}}_2/\mathrm{Red}\mathrm{P},{\mathrm{H}}_2\mathrm{O}}{\to }\mathrm{R}-\mathrm{C}\underset{\underset{\mathrm{Cl}}{|}}{\mathrm{H}}-\mathrm{COOH}$

(C) $\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{NH}}_2\to \mathrm{R}-{\mathrm{NH}}_2$- (I) ${\mathrm{Br}}_2/\mathrm{NaOH}$

1. A primary amide is changed into a primary amine in a process known as the Hoffmann bromamide reaction. The primary amine has one less carbon atom upon conversion.
2. In this reaction, an amide is introduced to bromine in an aqueous or ethanolic sodium hydroxide solution, which causes the amide to degrade and produce primary amine.
3. $\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{NH}}_2\stackrel{{\mathrm{Br}}_2/\mathrm{NaOH}}{\to }\mathrm{R}-{\mathrm{NH}}_2$

(D) $\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_3\to \mathrm{R}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$- (III) $\mathrm{Zn}\left(\mathrm{Hg}\right)/\mathrm{Conc}.\mathrm{HCl}$

1. Aldehydes or ketones can be converted to alkanes through a reaction called the Clemmensen reduction, which involves zinc amalgam and hydrochloric acid.
2. The reduction of aryl-alkyl ketones produced by Friedel-Crafts acylation benefits particularly from this process. The reduction of cyclic ketones or aliphatic and zinc metal makes the reaction more efficient.
3. $\mathrm{R}-\stackrel{\underset{\left|\right|}{\mathrm{O}}}{\mathrm{C}}-{\mathrm{CH}}_3\stackrel{\mathrm{Zn}\left(\mathrm{Hg}\right)/\mathrm{conc}.\mathrm{HCl}}{\to }\mathrm{R}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

Final answer:

Reaction	Reagent
(A) $\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-\mathrm{Cl}\to \mathrm{R}-\mathrm{CHO}$	${\mathrm{H}}_2/\mathrm{Pd}-{\mathrm{BaSO}}_4$
(B) $\mathrm{R}-{\mathrm{CH}}_2-\mathrm{COOH}\to \mathrm{R}-\underset{\underset{\mathrm{Cl}}{|}}{\mathrm{CH}}-\mathrm{COOH}$	${\mathrm{Cl}}_2/\mathrm{Red}\mathrm{P},{\mathrm{H}}_2\mathrm{O}$
(C) $\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{NH}}_2\to \mathrm{R}-{\mathrm{NH}}_2$	)${\mathrm{Br}}_2/\mathrm{NaOH}$
(D )$\mathrm{R}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_3\to \mathrm{R}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$	$\mathrm{Zn}\left(\mathrm{Hg}\right)/\mathrm{Conc}.\mathrm{HCl}$