Preparation of Haloarenes Questions in English

(A) In the Gattermann reaction,the diazonium group $(-N_2^+Cl^-)$ is replaced by a chlorine atom $(-Cl)$ using copper powder $(Cu)$ in the presence of hydrochloric acid $(HCl)$.
Specifically,the nucleophilic species involved in the substitution is the chloride ion $(Cl^{\ominus})$,and the reagents used are $Cu$ powder and $HCl$.
Therefore,$\underline{X} = Cl^{\ominus}$ and $\underline{Y} = Cu / HCl$.

(B) $1$. Sandmeyer's reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts.
$2$. In this reaction, the diazonium group $(-N_2^+X^-)$ of an aryl diazonium salt is replaced by a halogen atom ($-Cl$, $-Br$, or $-CN$) using copper$(I)$ salts like $CuCl/HCl$ or $CuBr/HBr$.
$3$. Therefore, the correct option is the one describing the replacement of the diazo group to form $Ar-X$.

(C) The reaction of benzene diazonium chloride with potassium iodide $(KI)$ is a standard method to introduce an iodine atom into the benzene ring.
Step $1$: The diazonium salt reacts with $KI$ in an aqueous solution.
Step $2$: The $-N_2^+Cl^-$ group is replaced by an iodine atom to form iodobenzene and nitrogen gas $(N_2)$ is evolved.
Reaction: $C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + N_2 + KCl$.

(A) Step $1$: The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts using copper salts as catalysts.
Step $2$: The reaction involves the replacement of the diazonium group $(-N_2^+)$ with a halogen atom (e.g., $-Cl, -Br, -CN$).
Step $3$: The Mendius reaction involves the reduction of nitriles to primary amines. The Hofmann rearrangement involves the conversion of an amide to a primary amine with one less carbon atom. The Carbylamine reaction is a test for primary amines using chloroform and base.
Step $4$: Therefore, the Sandmeyer reaction is the only one among the options that utilizes a diazonium salt.

(C) Step $1$: Chlorobenzene can be prepared from aniline via the Sandmeyer reaction using $NaNO_2/HCl$ followed by $CuCl/HCl$.
Step $2$: It can be prepared from benzene via electrophilic substitution using $Cl_2/FeCl_3$.
Step $3$: It can be prepared from benzene diazonium chloride via the Sandmeyer reaction.
Step $4$: In phenol, the $C-O$ bond has partial double bond character due to resonance, making it very strong and difficult to break to replace the $-OH$ group with a $-Cl$ atom. Thus, chlorobenzene is not easily prepared from phenol.