Match List I with List II
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A-IV, B-III, C-II, D-I A-IV, B-I, C-II, D-III A-IV, B-I, C-III, D-II A-I, B-II, C-III, D-IV |
A-IV, B-I, C-III, D-II |
The correct answer is option 3. A-IV, B-I, C-III, D-II.
Let us look at each of the given reactions: A. When benzene diazonium chloride reacts with hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)), it undergoes a reduction reaction. The hypophosphorous acid reduces the diazonium group (\(-\text{N}_2^+\)) to a phenyl group (\(-\text{H}\)). The general reaction is: Benzene Diazonium Chloride (\(\text{C}_6\text{H}_5\text{N}_2\text{Cl}\)) is a diazonium salt where the diazonium group (\(-\text{N}_2^+\)) is attached to a benzene ring. It acts as a reducing agent. It reduces the diazonium group to a phenyl group while itself being oxidized to phosphorous acid (\(\text{H}_3\text{PO}_3\)). The benzene diazonium chloride is reduced to benzene. B. When benzene diazonium chloride (\(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\)) reacts with potassium iodide (\(\text{KI}\)), it undergoes a reaction known as the Sandmeyer reaction, specifically for the substitution of the diazonium group with an iodine atom. The reaction is: It is a diazonium salt where the diazonium group (\(-\text{N}_2^+\)) is attached to a benzene ring. Potassium Iodide (\(\text{KI}\)) acts as a source of iodide ions (\(\text{I}^-\)). The diazonium group (\(-\text{N}_2^+\)) is replaced by the iodide ion (\(\text{I}^-\)) in a substitution reaction. The diazonium ion (\(\text{N}_2^+\)) is a good leaving group, and it gets replaced by the iodide ion to form iodobenzene (\(\text{C}_6\text{H}_5\text{I}\)). The by-products are potassium chloride (\(\text{KCl}\)) and nitrogen gas (\(\text{N}_2\)). This reaction is an example of a Sandmeyer reaction where the diazonium group is replaced by an iodine atom, leading to the formation of iodobenzene. C. When benzene diazonium chloride (\(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\)) reacts with water (\(\text{H}_2\text{O}\)), it undergoes hydrolysis. The reaction results in the formation of phenol and the release of nitrogen gas. The reaction is:
D. When benzene diazonium chloride (\(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\)) reacts with aniline (\(\text{C}_6\text{H}_5\text{NH}_2\)), it undergoes a reaction known as the diazo coupling reaction. This reaction results in the formation of a azo compound, specifically phenylazoaniline (also known as aniline yellow). The reaction is:
Benzene Diazonium Chloride (\(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\)) is a diazonium salt where the diazonium group (\(-\text{N}_2^+\)) is attached to a benzene ring. Aniline is an aromatic amine with an amino group (\(-\text{NH}_2\)) attached to a benzene ring. It is a nucleophile in this reaction. The diazonium group (\(-\text{N}_2^+\)) is highly reactive and can couple with nucleophiles like the amino group in aniline. The diazonium ion (\(\text{N}_2^+\)) reacts with the amino group of aniline to form a diazo linkage (\(-\text{N}_2\text{-}\)) between the two benzene rings. This forms the azo compound phenylazoaniline and releases hydrochloric acid (\(\text{HCl}\)) as a by-product. The reaction between benzene diazonium chloride and aniline results in the formation of an azo compound, phenylazoaniline, which is characterized by the presence of an azo group (\(-\text{N}_2\text{-}\)) linking two benzene rings. This type of reaction is widely used in dye chemistry for producing azo dyes. |