Describe the preparation of aryl halides from aromatic hydrocarbons and from amines via the Sandmeyer reaction.

(N/A) Aryl halides cannot be prepared by the method used for alkyl halides because the $C-OH$ bond in phenol is stronger than in alcohols.
$(a)$ Preparation of aryl halides from arenes by electrophilic substitution:
Method: Arenes react with chlorine $(Cl_2)$ or bromine $(Br_2)$ in the presence of a Lewis acid catalyst like iron or iron $(III)$ chloride to form aryl chlorides and aryl bromides,respectively.
Mechanism: This reaction proceeds via electrophilic aromatic substitution. The catalyst generates electrophiles ($Cl^+$ or $Br^+$).
Reaction: $C_6H_6 + X_2 \xrightarrow{Fe/FeX_3, \Delta} C_6H_5X + HX$ (where $X = Cl, Br$).
The ortho and para isomers can be easily separated due to the large difference in their melting points.
Iodination: Requires an oxidizing agent like $HNO_3$ or $HIO_4$ to oxidize $HI$ formed during the reaction.
Fluorination: Cannot be done by this method due to the extremely high reactivity of fluorine.
$(b)$ Preparation of aryl halides from amines via Sandmeyer reaction:
$(i)$ Diazotization: Primary aromatic amines dissolved or suspended in cold aqueous mineral acid are treated with sodium nitrite $(NaNO_2)$ to form diazonium salts $(ArN_2^+ X^-)$.
Reaction: $C_6H_5NH_2 + NaNO_2 + 2HX \xrightarrow{273-278 K} C_6H_5N_2^+ X^- + 2H_2O + NaX$.
$(ii)$ Sandmeyer reaction: The freshly prepared diazonium salt solution is treated with cuprous chloride $(Cu_2Cl_2)$ or cuprous bromide $(Cu_2Br_2)$ to replace the diazonium group $(N_2^+ X^-)$ with $-Cl$ or $-Br$.
Reaction: $C_6H_5N_2^+ X^- \xrightarrow{Cu_2X_2} C_6H_5X + N_2$.