Preparation of Haloalkanes Questions in English

(C) The reaction sequence is as follows:
$1$. Toluene reacts with $Br_2$ in the presence of $UV$ light to undergo free-radical substitution at the benzylic position,forming benzyl bromide $(C_6H_5CH_2Br)$.
$2$. Benzyl bromide reacts with $Mg$ in dry ether to form the Grignard reagent,benzylmagnesium bromide $(C_6H_5CH_2MgBr)$.
$3$. The Grignard reagent reacts with formaldehyde $(CH_2O)$ followed by acidic hydrolysis $(H_3O^+)$ to form $2-$phenylethanol $(C_6H_5CH_2CH_2OH)$.
$4$. Finally,$2$-phenylethanol reacts with $PBr_3$ to replace the hydroxyl group with a bromine atom,yielding ($2$-bromoethyl)benzene $(C_6H_5CH_2CH_2Br)$.

(D) The Swarts reaction is used to replace chlorine atoms with fluorine atoms in alkyl halides using inorganic fluorides like $\text{SbF}_3$ or $\text{Hg}_2\text{F}_2$.
This is the method used for the fluorination of $\text{CCl}_4$ to form $\text{CCl}_2\text{F}_2$ (Freon-$12$).
The reaction is: $\text{CCl}_4 + 2\text{HF} \xrightarrow{\text{SbF}_3} \text{CCl}_2\text{F}_2 + 2\text{HCl}$.

(B) Step $1$: The reaction involves the exchange of a bromine atom in an alkyl bromide with a fluorine atom using a metallic fluoride like $Hg_2F_2$.
Step $2$: This specific method for the synthesis of alkyl fluorides by heating alkyl chlorides or bromides in the presence of metallic fluorides (such as $AgF$, $Hg_2F_2$, $CoF_2$, or $SbF_3$) is called the Swartz reaction.

(C) The preparation of alkyl fluorides is best achieved by heating alkyl chlorides or alkyl bromides in the presence of a metallic fluoride such as $AgF$, $Hg_2F_2$, $CoF_2$, or $SbF_3$. This specific reaction is known as the $Swarts$ reaction.
Step $1$: The general reaction is $R-X + AgF \rightarrow R-F + AgX$ (where $X = Cl, Br$).
Step $2$: Since $Finkelstein$ reaction is used for alkyl iodides and $Sandmeyer$ or $Fittig$ reactions are for aryl halides, the correct choice is the $Swarts$ reaction.

(B) The Finkelstein reaction is a type of nucleophilic substitution reaction $(S_N2)$ used to prepare alkyl iodides from alkyl chlorides or alkyl bromides.
In this reaction, the alkyl halide is treated with sodium iodide $(NaI)$ in the presence of dry acetone.
Acetone acts as a solvent in which $NaI$ is soluble, while the byproduct $NaCl$ or $NaBr$ is insoluble and precipitates out, driving the reaction forward according to Le Chatelier's principle.
Therefore, the correct reagent is $NaI$ in dry acetone.