Preparation of Haloalkanes Questions in English

(C) The industrial preparation of chloroform $(CHCl_3)$ involves the reaction of acetone $(CH_3COCH_3)$ with bleaching powder $(CaOCl_2)$.
Bleaching powder acts as a source of chlorine and provides the necessary alkaline medium for the reaction.
The reaction proceeds in three steps: chlorination,hydrolysis,and formation of chloroform.

(B) The reaction of carbon disulfide $(CS_2)$ with chlorine $(Cl_2)$ in the presence of an anhydrous aluminum chloride $(AlCl_3)$ catalyst is a standard method for the preparation of carbon tetrachloride $(CCl_4)$.
The chemical equation for the reaction is:
$CS_2 + 3Cl_2 \xrightarrow{AlCl_3} CCl_4 + S_2Cl_2$
Thus,the correct product is $CCl_4$.

(B) The conversion of ethylbenzene $(C_6H_5CH_2CH_3)$ to styrene $(C_6H_5CH=CH_2)$ requires a two-step process: halogenation followed by dehydrohalogenation.
Step $1$: Free radical chlorination at the benzylic position using $SO_2Cl_2$ (sulfuryl chloride) yields $1$-chloroethylbenzene $(C_6H_5CHClCH_3)$.
Step $2$: Dehydrohalogenation of $C_6H_5CHClCH_3$ using alcoholic $KOH$ (a strong base) leads to the elimination of $HCl$ to form styrene $(C_6H_5CH=CH_2)$.

(B) Alkyl fluorides are most conveniently prepared by heating suitable chloro- or bromo-alkanes with inorganic fluorides such as $AsF_3$,$SbF_3$,$CoF_2$,$AgF$,$Hg_2F_2$,etc. This reaction is known as the $Swarts$ reaction.
$CH_3Br + AgF \rightarrow CH_3F + AgBr$
$2CH_3CH_2Cl + Hg_2F_2 \rightarrow 2CH_3CH_2F + Hg_2Cl_2$

(C) The reaction of alkyl chlorides or alkyl bromides with $NaI$ in acetone to form alkyl iodides is known as the $Finkelstein$ reaction.
The chemical equation is: $R-X + NaI \xrightarrow{\text{acetone}} R-I + NaX$ (where $X = Cl, Br$).
In this reaction,$NaI$ is soluble in acetone,whereas $NaCl$ and $NaBr$ are insoluble in acetone.
Because $NaCl$ or $NaBr$ precipitate out of the solution,the equilibrium shifts in the forward direction according to $Le$ $Chatelier's$ principle,facilitating the formation of alkyl iodide.
Therefore,the Assertion is correct,but the Reason is incorrect because it states the solubility properties in reverse.

(N/A) $(i)$ $CH_3CH_2CH_2CH_2OH + HI \xrightarrow{ZnCl_2} CH_3CH_2CH_2CH_2I + H_2O$
$(ii)$ $CH_3CH_2CH_2CH_2Cl + NaI \xrightarrow{\text{dry acetone (Finkelstein reaction)}} CH_3CH_2CH_2CH_2I + NaCl$
$(iii)$ $CH_3CH_2CH=CH_2 + HBr$ $\xrightarrow{\text{peroxide (Anti-Markovnikov addition)}} CH_3CH_2CH_2CH_2Br$ $\xrightarrow{NaI/\text{dry acetone}} CH_3CH_2CH_2CH_2I + NaBr$

(N/A) Alcohols $\rightarrow$ Chlorides / Halides: When alcohols react with concentrated halogen acids,the hydroxyl group is replaced by a halogen atom. The reaction of primary and secondary alcohols with $HCl$ requires the presence of $ZnCl_{2}$ as a catalyst. Tertiary alcohols react with concentrated $HCl$ simply by shaking at room temperature.
$(i)$ $R_{3}C-OH + HCl \rightarrow R_{3}C-Cl + H_{2}O$
$(ii)$ $R-OH + HCl \text{ (conc.)} \xrightarrow{\Delta} RCl + H_{2}O$
$(iii)$ $R-OH + HCl \xrightarrow{ZnCl_{2}} RCl + H_{2}O$
$(iv)$ $R-OH + HCl_{(g)} \rightarrow R-Cl + H_{2}O$
The order of reactivity of halogen acids $(HX)$ with alcohols is $3^{0} > 2^{0} > 1^{0}$. Alkyl chlorides can be prepared by passing dry $HCl$ gas through an alcohol solution or by heating a mixture of concentrated aqueous halogen acid and alcohol.
$(b)$ Reaction with phosphorus halides $(PX_{3} / PX_{5})$: The $-OH$ group is replaced by a halogen atom upon reaction with phosphorus halides.
$(i)$ $3R-OH + PX_{3} \rightarrow 3R-X + H_{3}PO_{3} \text{ (where } X = Cl, Br)$
$(ii)$ $R-OH + PCl_{5} \rightarrow R-Cl + POCl_{3} + HCl$
Generally,red phosphorus reacts with bromine and iodine to form $PBr_{3}$ and $PI_{3}$ respectively.
$(c)$ Reaction with thionyl chloride $(SOCl_{2})$: Alcohols react with thionyl chloride to form alkyl chlorides.
$R-OH + SOCl_{2} \rightarrow R-Cl + SO_{2(g)} + HCl_{(g)}$
This method is preferred for preparing alkyl chlorides because the by-products $SO_{2}$ and $HCl$ are gases,leaving behind pure alkyl chloride.
$(d)$ Preparation of bromides: Alcohols are heated with $48\% HBr$ to prepare alkyl bromides.
$ROH + HBr \text{ (48\%)} \xrightarrow{\text{Boiling}} R-Br + H_{2}O$

(N/A) Free radical halogenation of alkanes: Free radical chlorination of alkanes in the presence of $Cl_2$ and $h\nu$ or heat,and bromination in the presence of $Br_2$ and $h\nu$ or heat,results in a mixture of mono- and poly-halogenated compounds. It is difficult to separate these components in pure form. Therefore,this method is not suitable for preparing a single alkyl halide.
$(b)$ Alkyl halides from alkenes:
$(i)$ Hydrohalogenation of alkenes: Addition of hydrogen halides ($HCl$,$HBr$,or $HI$) to alkenes leads to the formation of alkyl halides via electrophilic addition. If two addition products are possible,the major product is formed according to Markovnikov's rule. For example:
$CH_3CH=CH_2 + HI \rightarrow CH_3CHICH_3 \text{ (major)} + CH_3CH_2CH_2I \text{ (minor)}$
$(ii)$ Addition of halogens $(X_2)$ to alkenes: Addition of bromine $(Br_2)$ dissolved in $CCl_4$ to an alkene results in the formation of vicinal dibromides. The reddish-brown color of $Br_2$ in $CCl_4$ disappears as the colorless dibromide is formed,which serves as a laboratory test for unsaturation (double bond).

(N/A) Finkelstein Reaction: Alkyl iodides are often prepared by the reaction of alkyl chlorides or bromides with $NaI$ in dry acetone. This reaction is known as the Finkelstein reaction.
Reaction: $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX_{(s)}$
(where $X = Cl, Br$)
$NaCl$ or $NaBr$ formed precipitates in dry acetone,which facilitates the forward reaction according to Le Chatelier's principle.
$(b)$ Swarts Reaction: Alkyl fluorides are best prepared by heating an alkyl chloride or bromide in the presence of a metallic fluoride such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$. This reaction is known as the Swarts reaction.
Reaction: $H_3C-Br + AgF \rightarrow H_3C-F + AgBr$

(N/A) Finkelstein Reaction (Preparation of Alkyl Iodides from Alkyl Chlorides/Bromides): Alkyl chlorides or bromides are reacted with $NaI$ in dry acetone to form alkyl iodides.
"The reaction of alkyl chlorides or bromides with $NaI$ in dry acetone to form alkyl iodides is known as the Finkelstein reaction."
Reaction: $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX(s)$
(where $X = Cl, Br$).
In this reaction,$NaCl$ or $NaBr$ precipitates out,which shifts the equilibrium forward according to Le Chatelier's principle.
$(b)$ Swarts Reaction (Preparation of Alkyl Fluorides from Alkyl Chlorides/Bromides): Alkyl chlorides or bromides are heated with metallic fluorides such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
"The reaction of alkyl chlorides or bromides with metallic fluorides to form alkyl fluorides is known as the Swarts reaction."
Reaction: $H_3C-Br + AgF \rightarrow H_3C-F + AgBr$

(B) Ethanol can be converted to iodoethane by first converting it into a better leaving group,such as a chloroalkane,and then performing a nucleophilic substitution reaction (Finkelstein reaction) using $NaI$ in acetone.
Step $1$: $C_{2}H_{5}OH + HCl \xrightarrow{ZnCl_{2}} C_{2}H_{5}Cl + H_{2}O$
Step $2$: $C_{2}H_{5}Cl + NaI \xrightarrow{\text{Acetone}} C_{2}H_{5}I + NaCl$

(N/A) The hydroxyl group of an alcohol is replaced by a halogen atom upon reaction with concentrated halogen acids,phosphorus halides,or thionyl chloride.
$(i)$ By reaction of alcohols with halogen acids $(HX)$: When an alcohol is reacted with concentrated halogen acids,the $-OH$ group is replaced by $-X$ to form haloalkanes.
Primary $(1^{\circ})$ and secondary $(2^{\circ})$ alcohols require a catalyst,such as anhydrous $ZnCl_{2}$,when reacted with $HCl$.
Example: $CH_{3}CH_{2}OH + HCl \xrightarrow{ZnCl_{2}} CH_{3}CH_{2}Cl + H_{2}O$
The reaction of tertiary $(3^{\circ})$ alcohols with concentrated $HCl$ is carried out simply by shaking at room temperature.
Example: $(CH_{3})_{3}C-OH + HCl \rightarrow (CH_{3})_{3}C-Cl + H_{2}O$
Bromoalkanes are prepared by constant boiling of the suitable alcohol with concentrated $HBr$ $(48\%)$ or by the reaction of alcohol with $NaBr$ in the presence of $H_{2}SO_{4}$.
$CH_{3}CH_{2}OH + HBr \rightarrow CH_{3}CH_{2}Br + H_{2}O$
$CH_{3}CH_{2}OH + NaBr + H_{2}SO_{4} \rightarrow CH_{3}CH_{2}Br + NaHSO_{4} + H_{2}O$

(N/A) $(i)$ Finkelstein Reaction: Alkyl iodides are often prepared by the reaction of alkyl chlorides or alkyl bromides with $NaI$ in dry acetone.
$CH_3CH_2Br + NaI \xrightarrow{\text{Acetone}} CH_3CH_2I + NaBr$
$\text{Bromoethane} \rightarrow \text{Iodoethane}$
$CH_3CH_2Cl + NaI \xrightarrow{\text{Acetone}} CH_3CH_2I + NaCl$
$NaCl$ or $NaBr$ thus formed is precipitated in dry acetone. It facilitates the forward reaction according to Le Chatelier's Principle.
$(ii)$ Swarts Reaction: The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride or alkyl bromide in the presence of a metallic fluoride such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
$CH_3Br + AgF \rightarrow CH_3F + AgBr$

(C) The given reactions represent the preparation of haloalkanes from alcohols using phosphorus halides.
$1$. The reaction of alcohol with phosphorus trichloride is:
$3 ROH + PCl_3 \rightarrow 3 RCl + H_3 PO_3$ $(A)$
$2$. The reaction of alcohol with phosphorus pentachloride is:
$ROH + PCl_5 \rightarrow RCl + HCl + POCl_3$ $(B)$
Thus,$A$ is $H_3 PO_3$ and $B$ is $POCl_3$.

(B) The reaction sequence is a Wurtz-Fittig type coupling reaction.
Step $1$: $A \xrightarrow{SOCl_2, \Delta} B$. This is the conversion of an alcohol to an alkyl chloride. Propan$-2-$ol $(CH_3CH(OH)CH_3)$ reacts with $SOCl_2$ to form $2-$chloropropane $(CH_3CHClCH_3)$ as $B$.
Step $2$: $B + C_2H_5Cl \xrightarrow{Na/\text{ether}} \text{2-Methylbutane}$. This is a Wurtz reaction where $2-$chloropropane $(CH_3CHClCH_3)$ reacts with chloroethane $(C_2H_5Cl)$ in the presence of sodium and dry ether to form $2-$methylbutane $(CH_3CH(CH_3)CH_2CH_3)$.
Therefore,substrate $A$ is Propan$-2-$ol.

(D) The reactivity of alcohols with halo acids depends on the stability of the carbocation intermediate formed during the reaction.
Since the stability of carbocations follows the order $3^{\circ} > 2^{\circ} > 1^{\circ}$,the reactivity of alcohols towards halo acids also follows the order $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Therefore,the statement $D$ is false.

(A) Step $1$: Ethanol $(CH_3CH_2OH)$ reacts with $NaBr$ in the presence of $H_2SO_4$ and heat to undergo nucleophilic substitution,forming ethyl bromide $(CH_3CH_2Br)$ as product '$A$'.
$CH_3CH_2OH + NaBr + H_2SO_4 \xrightarrow{\Delta} CH_3CH_2Br + NaHSO_4 + H_2O$
Step $2$: Ethyl bromide $(CH_3CH_2Br)$ reacts with magnesium metal in the presence of dry ether to form a Grignard reagent,ethyl magnesium bromide $(CH_3CH_2MgBr)$,which is product '$B$'.
$CH_3CH_2Br + Mg \xrightarrow{\text{Dry ether}} CH_3CH_2MgBr$

(D) The Swartz reaction is a method used to prepare alkyl fluorides by heating alkyl chlorides or alkyl bromides in the presence of metallic fluorides such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
The general reaction is: $R-X + AgF \rightarrow R-F + AgX$ (where $X = Cl, Br$).
Thus,the product obtained is an alkyl fluoride.

(B) Alkyl fluorides are prepared by heating alkyl chlorides or bromides with metal fluorides such as $AgF$,$Hg_2F_2$,$AsF_3$,$SbF_3$,etc. This reaction is known as the Swartz reaction.
$R-Cl + AgF \longrightarrow R-F + AgCl \downarrow$

(D) The Finkelstein reaction is a type of halogen exchange reaction used to prepare alkyl iodides from alkyl chlorides or alkyl bromides.
The general reaction is: $R-X + NaI \longrightarrow R-I + NaX$ (where $X = Cl, Br$).
Thus,alkyl iodides are obtained.

(A) The reaction of alcohol with halogen in the presence of sunlight is a radical substitution reaction typically used for alkanes,not for the preparation of alkyl halides from alcohols.
Alcohol reacts with $HCl$ in the presence of anhydrous $ZnCl_2$ (Lucas reagent) to form alkyl chlorides.
Alcohol reacts with $HBr$ in the presence of $NaBr$ and $H_2SO_4$ to form alkyl bromides.
Alcohol reacts with $HI$ in the presence of $NaI$ and $H_3PO_4$ to form alkyl iodides.
Therefore,the reaction of alcohol with halogen in the presence of sunlight does not produce alkyl halides.

(A) Swartz Reaction: It involves the heating of alkyl chlorides or bromides with metallic fluorides like $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$ to produce alkyl fluorides.
General reaction: $R-X + AgF \rightarrow R-F + AgX$ (where $X = -Cl, -Br$).

(A) The $Swartz$ reaction is a method used for the preparation of alkyl fluorides by heating alkyl chlorides or bromides with metallic fluorides like $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
The reaction is represented as: $R-Cl + AgF \xrightarrow{\Delta} R-F + AgCl$.
Option $A$ represents the $Swartz$ reaction.

(A) The preparation of Grignard reagent involves the reaction of an alkyl halide $(R-X)$ with magnesium metal in the presence of dry ether as a solvent.
The reaction is represented as: $R-X + Mg \xrightarrow{\text{dry ether}} R-MgX$
Dry ether is essential because Grignard reagents are highly reactive and decompose in the presence of moisture or protic solvents.

(C) The reaction in which alkyl fluorides are prepared by heating alkyl bromides or alkyl chlorides with metallic fluorides (such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$) is known as the $Swarts$ reaction.
The general reaction is: $R-X + AgF \rightarrow R-F + AgX$ (where $X = Cl, Br$).

(A) The given reaction involves the replacement of a chlorine atom in an alkyl chloride $(R-Cl)$ with a fluorine atom using a metallic fluoride $(CoF_{2})$.
This specific method of preparing alkyl fluorides from alkyl chlorides or bromides using metallic fluorides like $AgF$,$Hg_{2}F_{2}$,$CoF_{2}$,or $SbF_{3}$ is known as the $Swarts$ reaction.
Therefore,the correct option is $A$.

(D) The correct answer is $D$.
Chlorination of alkanes in the presence of sunlight (free radical substitution) is not a suitable method for the preparation of alkyl chlorides because it produces a complex mixture of mono-,di-,and polychloroalkanes,which are difficult to separate into individual components.

(A) The reaction of alkyl halides $(RX)$ with silver nitrite $(AgNO_2)$ is a standard laboratory method for the preparation of nitroalkanes.
The reaction proceeds as follows:
$RX + AgNO_2 \rightarrow RNO_2 (\text{Nitroalkane}) + AgX$
Here,$AgNO_2$ is a covalent compound,which allows the nitrogen atom to act as the nucleophile,leading to the formation of the $C-N$ bond.

(A) The given reaction is a Wurtz reaction: $2R-Cl + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaCl$.
To find '$A$',we split the product $3,4-$diethyl$-3,4-$dimethylhexane at the bond between $C3$ and $C4$.
The product is $CH_3CH_2-C(CH_3)(CH_2CH_3)-C(CH_3)(CH_2CH_3)-CH_2CH_3$.
Splitting this at the central $C-C$ bond gives two identical fragments: $CH_3CH_2-C(CH_3)(CH_2CH_3)-$.
Replacing the radical with a chlorine atom gives $CH_3CH_2-C(Cl)(CH_3)-CH_2CH_3$,which is $3-$chloro$-3-$methylpentane.

(A) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal to form a symmetrical alkane.
To obtain $2,2,3,3-$tetramethylbutane,we must identify the alkyl group that,when doubled,forms this structure.
The structure of $2,2,3,3-$tetramethylbutane is $(CH_3)_3C-C(CH_3)_3$.
Splitting this molecule at the central $C-C$ bond gives two $(CH_3)_3C-$ groups.
Therefore,the required alkyl halide is $tert-$butyl bromide,$(CH_3)_3C-Br$.

(C) The preparation of Freon-$12$ $(CCl_2F_2)$ from carbon tetrachloride $(CCl_4)$ is achieved by the Swarts reaction.
In this reaction,$CCl_4$ is treated with antimony trifluoride $(SbF_3)$ in the presence of antimony pentachloride $(SbCl_5)$ as a catalyst.
The chemical equation is: $3CCl_4 + 2SbF_3 \xrightarrow{SbCl_5} 3CCl_2F_2 + 2SbCl_3$.

(A) The Swarts reaction is a method used to prepare alkyl fluorides by heating alkyl chlorides or alkyl bromides in the presence of metallic fluorides.
Commonly used metallic fluorides include $AgF$,$Hg_2F_2$,$CoF_2$,and $SbF_3$.
$HF$ (hydrogen fluoride) is not a metallic fluoride and is not used as a reagent in the Swarts reaction.
Therefore,the correct option is $A$.

(A) To prepare $2-$methylpropane by the Wurtz reaction,we need to couple a methyl group with an isopropyl group. Therefore,the haloalkanes required are chloromethane $(CH_3Cl)$ and $2-$chloropropane $(CH_3CHClCH_3)$.
The reaction is as follows:
$CH_3Cl + CH_3CHClCH_3 + 2Na \xrightarrow{\text{dry ether}} CH_3CH(CH_3)CH_3 + 2NaCl$
Thus,the correct option is $A$.

(C) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal in dry ether to form a symmetrical alkane.
To prepare $2,3-$dimethylbutane,which has a symmetric structure,we need two molecules of $2-$bromopropane:
$2 CH_3-CH(Br)-CH_3 + 2 Na \xrightarrow{\text{dry ether}} CH_3-CH(CH_3)-CH(CH_3)-CH_3 + 2 NaBr$
Thus,the required alkyl halide is $2-$bromopropane.

(D) The reaction of ethanol with phosphorus pentachloride $(PCl_5)$ is a standard method for the preparation of alkyl halides.
The chemical equation is:
$CH_3CH_2OH + PCl_5 \rightarrow CH_3CH_2Cl + HCl + POCl_3$
Here,the products formed are chloroethane $(CH_3CH_2Cl)$,hydrogen chloride $(HCl)$,and phosphorus oxychloride $(POCl_3)$.

(D) The Swarts reaction is a method used for the synthesis of alkyl fluorides.
It involves the heating of an alkyl chloride or alkyl bromide in the presence of a metallic fluoride such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
In the given options,$2 R-CH_2-Br + CoF_2 \longrightarrow 2 R-CH_2-F + CoBr_2$ represents the Swarts reaction as it involves the exchange of a halogen atom with fluorine using a metallic fluoride.

(B) The reaction of allylbenzene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ proceeds via the peroxide effect (Kharasch effect),which follows Anti-Markovnikov addition.
In this reaction,the $Br$ atom attaches to the less substituted carbon atom of the double bond.
The product formed is $C_6H_5-CH_2-CH_2-CH_2Br$.
Since the bromine atom is attached to a primary carbon atom,the product is a primary alkyl halide.

(B) The reactions are analyzed as follows:
$A$: Propene reacts with $HI$ to form $2-$iodopropane via Markovnikov addition.
$B$: Propan-$1-$ol reacts with $NaI$ and $H_3PO_4$ to form $1-$iodopropane.
$C$: $1-$chloropropane reacts with $NaI$ in dry acetone (Finkelstein reaction) to form $1-$iodopropane.
$D$: Propane reacts with $I_2$ in the presence of $hv$ to form $2-$iodopropane as the major product due to the stability of the secondary free radical.
Thus,$B$ and $C$ are the correct methods for the preparation of $1-$iodopropane.

(B) Alkyl iodides are prepared by the reaction of alkyl chlorides or alkyl bromides with $NaI$ in dry acetone. This reaction is known as the Finkelstein reaction.
$R-X + NaI \xrightarrow{\text{Acetone}} R-I + NaX$ (where $X = Cl, Br$)
$NaCl$ or $NaBr$ formed in this reaction is precipitated in dry acetone. This precipitation facilitates the forward reaction according to Le Chatelier's Principle.

(A) The $Finkelstein$ reaction is a classic example of a halogen exchange reaction.
It is an $S_{N}2$ reaction where an alkyl halide (usually chloride or bromide) is converted into an alkyl iodide by treatment with sodium iodide $(NaI)$ in acetone.
The general reaction is: $R-X + NaI \rightarrow R-I + NaX$ (where $X = Cl, Br$).
Other options like $Sandmeyer$,$Fittig$,and $Wurtz-Fittig$ reactions are not halogen exchange reactions.

(D) The correct matches are:
$(A)$ Sandmeyer reaction: Used to prepare aryl halides like $Ar-Cl$ or $Ar-Br$ from diazonium salts. Thus,$(A-III)$.
$(B)$ Finkelstein reaction: $R-X + NaI \longrightarrow R-I + NaX$ (where $X = Cl, Br$). Thus,$(B-I)$.
$(C)$ Swarts reaction: $R-X + AgF \longrightarrow R-F + AgX$. Thus,$(C-II)$.
Therefore,the correct sequence is $A-III, B-I, C-II$.

(A) The reaction proceeds in two steps:
Step $1$: Electrophilic addition of $HBr$ to ethene $(CH_2=CH_2)$ gives bromoethane $(CH_3CH_2Br)$. Thus,$X = HBr$.
Step $2$: Bromoethane reacts with $NaI$ in dry acetone to form iodoethane $(CH_3CH_2I)$ via the Finkelstein reaction. Thus,$Y = NaI / \text{dry } CH_3COCH_3$.
Therefore,the correct reagents are $X = HBr$ and $Y = NaI / \text{dry } CH_3COCH_3$.

(A) The Swarts reaction is a method used for the preparation of alkyl fluorides from alkyl chlorides or alkyl bromides by heating them in the presence of metallic fluorides such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
The reaction is represented as:
$CH_3-CH_2-Br \xrightarrow{CoF_2} CH_3-CH_2-F$
Therefore,option $A$ represents the Swarts reaction.

(B) The Swarts reaction is a method used for the synthesis of alkyl fluorides 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 is a classic example of a halogen exchange reaction.
The general reaction is: $H_3C-Br + AgF \rightarrow H_3C-F + AgBr$.

(B) The Swarts reaction is used to prepare alkyl fluorides from alkyl chlorides or alkyl bromides.
This is achieved by heating the alkyl chloride or bromide in the presence of a metallic fluoride,such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
For example:
$CH_3-Br + AgF \rightarrow CH_3F + AgBr$
Thus,the Swarts reaction yields fluorohydrocarbons.

(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 $(C_6H_5CH_2CH_2Br)$.