Preparation of Haloalkanes Questions in English

(B) The reaction of carbon disulfide $(CS_2)$ with chlorine $(Cl_2)$ in the presence of an iodine $(I_2)$ catalyst is a standard method for the preparation of carbon tetrachloride $(CCl_4)$.
The balanced chemical equation is:
$CS_2 + 3Cl_2 \xrightarrow{I_2} CCl_4 + S_2Cl_2$

(A) The structure of $2, 3-$dimethylbutane is $CH_3-CH(CH_3)-CH(CH_3)-CH_3$.
There are two types of equivalent hydrogens in this molecule:
$1.$ Primary hydrogens: Monochlorination at a primary carbon gives $ClCH_2-CH(CH_3)-CH(CH_3)-CH_3$ ($1-$chloro$-2, 3-$dimethylbutane). This molecule has one chiral center at $C-2$,so it exists as one pair of enantiomers ($d$ and $l$ forms).
$2.$ Tertiary hydrogens: Monochlorination at a tertiary carbon gives $CH_3-CCl(CH_3)-CH(CH_3)-CH_3$ ($2-$chloro$-2, 3-$dimethylbutane). This molecule is achiral because it does not have a chiral center.
Thus,only $1$ pair of enantiomers is obtained.

(A) The metal used in the preparation of Grignard reagent is Magnesium $(Mg)$.
Haloalkanes,aryl halides,and vinyl halides react with magnesium metal in the presence of a dry ether solvent to yield organomagnesium halides,which are known as Grignard reagents.
The general reaction is: $R-X + Mg \xrightarrow{\text{dry ether}} R-Mg-X$.

(B) The reaction of $1-$butene $(CH_3CH_2CH=CH_2)$ with $HBr$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom with fewer hydrogen atoms.
Therefore,$CH_3CH_2CH=CH_2 + HBr \rightarrow CH_3CH_2CH(Br)CH_3$ ($2-$bromobutane).
Reaction $(1)$ gives a mixture of $2-$bromopentane and $3-$bromopentane.
Reaction $(3)$ gives $2,3-$dibromobutane.
Reaction $(4)$ follows anti-Markovnikov addition,yielding $1-$bromobutane.

(D) The reaction of toluene with chlorine in the presence of light (ultraviolet light) proceeds via a free radical mechanism.
This reaction occurs at the side chain (methyl group) rather than the benzene ring.
The reaction is as follows:
$C_6H_5CH_3 + Cl_2 \xrightarrow{h\nu} C_6H_5CH_2Cl + HCl$
The product formed is $Benzyl \ chloride$.
Side-chain chlorination is favored by high temperature,light,and the absence of a halogen carrier (like $FeCl_3$ or $AlCl_3$).

(D) The preparation of alkyl halides is least preferred by direct halogenation of alkanes because it results in a mixture of mono-,di-,tri-,and poly-substituted alkyl halide derivatives,which are difficult to separate. The reaction proceeds as follows:
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$ (chloromethane)
$CH_3Cl + Cl_2 \rightarrow CH_2Cl_2 + HCl$ (dichloromethane)
$CH_2Cl_2 + Cl_2 \rightarrow CHCl_3 + HCl$ (trichloromethane)
$CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$ (tetrachloromethane)

(C) The preparation of chloroform $(CHCl_3)$ from ethanol $(C_2H_5OH)$ and bleaching powder $(CaOCl_2)$ involves the following steps:
$1$. Bleaching powder reacts with water to produce calcium hydroxide and chlorine gas: $CaOCl_2 + H_2O \to Ca(OH)_2 + Cl_2$.
$2$. Chlorine acts as an oxidizing agent to oxidize ethanol to acetaldehyde,and then as a chlorinating agent to form chloral $(CCl_3CHO)$.
$3$. Calcium hydroxide then reacts with chloral to produce chloroform: $2CCl_3CHO + Ca(OH)_2 \to 2CHCl_3 + (HCOO)_2Ca$.
Thus,bleaching powder provides both $Ca(OH)_2$ and $Cl_2$ for the reaction.

(C) The formation of chloroform $(CHCl_3)$ from ethanol $(C_2H_5OH)$ and bleaching powder $(CaOCl_2)$ involves the following steps:
$1.$ Hydrolysis of bleaching powder: $CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$
$2.$ Oxidation of ethanol: $CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl$
$3.$ Chlorination of acetaldehyde: $CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO + 3HCl$
$4.$ Hydrolysis of chloral: $2CCl_3CHO + Ca(OH)_2 \rightarrow 2CHCl_3 + (HCOO)_2Ca$
Reduction does not occur in this reaction sequence.

(B) The reaction of ethanol $(C_2H_5OH)$ with thionyl chloride $(SOCl_2)$ in the presence of pyridine is the preferred method for preparing ethyl chloride $(C_2H_5Cl)$.
This is because the by-products,$SO_2$ and $HCl$,are gases and escape,leaving behind pure ethyl chloride.
The reaction is: $C_2H_5OH + SOCl_2 \xrightarrow{\text{Pyridine}} C_2H_5Cl + SO_2 \uparrow + HCl \uparrow$

(A) $CH_3I$ can be prepared by the reaction of methanol $(CH_3OH)$ with hydrogen iodide $(HI)$ in the presence of a catalyst like $ZnCl_2$ or by heating.
The reaction is: $CH_3OH + HI \rightarrow CH_3I + H_2O$.

(C) Grignard reagents are prepared by the reaction of an alkyl halide with magnesium metal in the presence of dry ether as a solvent.
$C_2H_5I + Mg \xrightarrow{\text{Dry ether}} C_2H_5MgI$
Here,$C_2H_5I$ (Ethyl iodide) is an alkyl halide,which reacts with $Mg$ to form the Grignard reagent,Ethyl magnesium iodide.

(B) The reaction of an alcohol with $NaCl$ does not occur because the chloride ion $(Cl^-)$ is a poor nucleophile and the hydroxyl group $(-OH)$ is a poor leaving group.
To convert an alcohol to an alkyl chloride,stronger reagents like $HCl$ in the presence of $ZnCl_2$ (Lucas reagent),$PCl_5$,or $SOCl_2$ are required.

(A) The reaction of alcohols with $SOCl_2$ (thionyl chloride) is considered the best method for preparing alkyl chlorides.
This is because the by-products formed,$SO_2$ and $HCl$,are both gases.
Since these by-products escape into the atmosphere,the resulting alkyl chloride is obtained in a highly pure state.
The reaction is: $R-OH + SOCl_2 \xrightarrow{Pyridine} R-Cl + SO_2 \uparrow + HCl \uparrow$

(B) Grignard reagents are prepared by the reaction of alkyl halides with magnesium metal in the presence of dry ether as a solvent.
The general reaction is: $RX + Mg \xrightarrow{\text{Dry ether}} RMgX$.

(B) is the correct answer.
Grignard reagent $(RMgX)$ is prepared by the reaction of an alkyl halide $(RX)$ with magnesium metal in the presence of dry ether.
The reaction is: $RX + Mg \xrightarrow{\text{Dry ether}} RMgX$.

(A) The reaction between $CHCl_3$ and $HF$ is a nucleophilic substitution reaction known as the Swarts reaction.
The chemical equation is: $CHCl_3 + 3HF \to CHF_3 + 3HCl$.
The product formed is $CHF_3$,which is known as fluoroform.
The molecular weight of $CHF_3$ is calculated as: $12.01 + 1.01 + 3 \times 19.00 = 70.02 \approx 70 \ g/mol$.

(B) The preparation of Grignard reagent $(RMgX)$ involves the reaction of an alkyl halide $(RX)$ with magnesium metal in the presence of dry ether.
$I_2$ (iodine) powder is often used as a catalyst or initiator to remove the oxide layer from the surface of the magnesium metal,thereby facilitating the reaction.
Therefore,the correct option is $B$.

(D) The conversion of an alcohol to an alkyl bromide is best achieved using phosphorus tribromide $(PBr_3)$.
The reaction is as follows:
$3CH_3CH_2CH_2CH_2OH + PBr_3 \rightarrow 3CH_3CH_2CH_2CH_2Br + H_3PO_3$
Thus,option $(d)$ is the correct reagent.

(B) The preparation of a Grignard reagent involves the reaction of an alkyl halide with magnesium metal in the presence of dry ether as a solvent/catalyst.
$R-X + Mg \xrightarrow{\text{Dry ether}} R-MgX$
For example: $CH_3I + Mg \xrightarrow{\text{Dry ether}} CH_3MgI$
Therefore,the correct option is $B$.

(A) The given reaction is the $Hunsdiecker$ reaction,where a silver salt of a carboxylic acid reacts with bromine in the presence of $CS_2$ to form an alkyl bromide.
The reaction is: $CH_3COOAg + Br_2 \xrightarrow{CS_2} CH_3Br + AgBr + CO_2$.
Therefore,the major product is $CH_3Br$.

(B) The reaction of ethanol with thionyl chloride $(SOCl_2)$ in the presence of pyridine is the most preferred method for the preparation of ethyl chloride because the by-products ($SO_2$ and $HCl$) are gases and escape,leaving pure alkyl halide.
Reaction: $C_2H_5OH + SOCl_2 \xrightarrow{\text{Pyridine}} C_2H_5Cl + SO_2(g) + HCl(g)$

(D) The preparation of butanenitrile $(CH_3CH_2CH_2CN)$ involves the nucleophilic substitution reaction of an alkyl halide with potassium cyanide $(KCN)$.
To obtain a four-carbon nitrile,we need a three-carbon alkyl halide.
$CH_3-CH_2-CH_2-Cl + KCN \xrightarrow{S_N2} CH_3-CH_2-CH_2-CN + KCl$.
Thus,propyl chloride $(CH_3CH_2CH_2Cl)$ reacts with $KCN$ to form butanenitrile.

(B) The reaction of an alcohol with thionyl chloride $(SOCl_2)$ is a standard method for the preparation of alkyl chlorides.
The chemical equation is: $C_2H_5OH + SOCl_2 \rightarrow C_2H_5Cl + SO_2(g) + HCl(g)$.
Since both by-products ($SO_2$ and $HCl$) are gases,they escape the reaction mixture,leaving behind pure alkyl chloride.
Therefore,the correct products are $CH_3CH_2Cl$,$HCl$,and $SO_2$.

(C) The Grignard reagent $(RMgX)$ is prepared by the reaction of an alkyl halide $(RX)$ with magnesium metal in the presence of dry ether $(R-X + Mg \xrightarrow{\text{dry ether}} RMgX)$.
Dry ether is essential because it acts as a solvent and stabilizes the Grignard reagent through coordination,while also preventing the reagent from reacting with moisture or protic solvents.

(B) The reaction of ethanol $(CH_3CH_2OH)$ with thionyl chloride $(SOCl_2)$ in the presence of pyridine is the best method for preparing chloroethane $(CH_3CH_2Cl)$.
The reaction is: $CH_3CH_2OH + SOCl_2 \xrightarrow{\text{pyridine}} CH_3CH_2Cl + SO_2(g) + HCl(g)$.
This method is preferred because the by-products ($SO_2$ and $HCl$) are gases,which escape,leaving behind pure chloroethane.

(A) The addition of $HBr$ to alkenes follows Markovnikov's rule.
For the formation of $2$-bromobutane $(CH_3-CH(Br)-CH_2-CH_3)$,the starting material should be but-$2$-ene $(CH_3-CH=CH-CH_3)$.
When $CH_3-CH=CH-CH_3$ reacts with $HBr$,the proton $(H^+)$ adds to one of the $sp^2$ carbons and the bromide ion $(Br^-)$ adds to the other,resulting in $2$-bromobutane.
Reaction: $CH_3-CH=CH-CH_3 + HBr \rightarrow CH_3-CH(Br)-CH_2-CH_3$.

(B) The preparation of benzyl chloride $(C_6H_5CH_2Cl)$ from toluene $(C_6H_5CH_3)$ involves free radical substitution of a hydrogen atom on the methyl group.
This reaction is typically carried out using chlorine gas $(Cl_2)$ in the presence of ultraviolet $(UV)$ light or heat.
Alternatively,sulfuryl chloride $(SO_2Cl_2)$ can also be used as a chlorinating agent in the presence of a radical initiator like benzoyl peroxide to achieve the same product.
Given the options provided,$SO_2Cl_2$ is a standard reagent used for side-chain chlorination of toluene.

(A) The structure of $2,3-$dimethylbutane is $(CH_3)_2CH-CH(CH_3)_2$.
Monochlorination can occur at two types of hydrogen atoms:
$1$. At the primary carbon atoms (terminal $CH_3$ groups): There are $12$ equivalent primary hydrogens. Replacing one gives $1-$chloro-$2,3-$dimethylbutane. This molecule has no chiral center,so it is achiral.
$2$. At the secondary carbon atoms (the $CH$ groups): There are $2$ equivalent secondary hydrogens. Replacing one gives $2-$chloro-$2,3-$dimethylbutane. This molecule has a chiral center at $C_2$. Thus,it exists as a pair of enantiomers ($R$ and $S$ forms).
Therefore,only $1$ pair of enantiomers is obtained.

(D) Grignard reagents $(RMgX)$ are highly reactive towards compounds containing active hydrogen atoms,such as water $(H_2O)$,alcohols,or acids.
They react violently with water to form alkanes $(RH)$ and magnesium hydroxide halide $(Mg(OH)X)$.
Therefore,the preparation must be carried out in an anhydrous ether medium to prevent decomposition.

(B) Direct halogenation of alkanes is not preferred because it results in a mixture of mono-,di-,tri-,and tetra-halogenated products,which are difficult to separate.
For example: $CH_4$ $\xrightarrow{Cl_2} CH_3Cl$ $\xrightarrow{Cl_2} CH_2Cl_2$ $\xrightarrow{Cl_2} CHCl_3$ $\xrightarrow{Cl_2} CCl_4$

(C) The given reaction is a Swarts reaction,which is used to prepare alkyl fluorides from alkyl chlorides or bromides.
The reaction is driven by the precipitation of the metal halide $(MCl)$ formed as a byproduct.
For the reaction to be most favourable,the metal fluoride $(MF)$ should be soluble in the reaction medium,and the metal chloride $(MCl)$ should be insoluble (precipitate out).
Among the given options,$RbF$ is the most soluble in organic solvents due to its lower lattice energy compared to $LiF$,$NaF$,and $KF$. Consequently,$RbCl$ is more likely to precipitate,driving the reaction forward.

(D) The reaction of alcohols with $HCl$ to form alkyl halides is known as the $Groove's$ process.
$I$. Primary alcohols like $CH_3CH_2OH$ require the catalyst anhydrous $ZnCl_2$ to proceed.
$II$. Primary alcohols do not react with $HCl$ in the absence of $ZnCl_2$.
$III$. Tertiary alcohols like $(CH_3)_3COH$ are highly reactive and form alkyl halides with $HCl$ even at room temperature without $ZnCl_2$ due to the formation of a stable tertiary carbocation.
$IV$. Secondary alcohols like $(CH_3)_2CHOH$ require anhydrous $ZnCl_2$ to react with $HCl$.
Therefore,reactions $I$,$III$,and $IV$ are suitable for the preparation of alkyl halides.

(D) The $Wurtz$ reaction is used for the preparation of higher alkanes from alkyl halides,not for the synthesis of alkyl iodides.
Alkyl iodides are typically prepared by the $Finkelstein$ reaction,which involves the reaction of alkyl chlorides or bromides with $NaI$ in dry acetone.
Therefore,the pair $Alkyl \text{ } iodide : Wurtz \text{ } reaction$ is incorrectly matched.

(A) The reaction described is the Hunsdiecker reaction,which involves the decarboxylation of the silver salt of a carboxylic acid in the presence of bromine.
Step $1$: $CH_3-CH_2-COOH + AgNO_3 \rightarrow CH_3-CH_2-COOAg + HNO_3$
Step $2$: $CH_3-CH_2-COOAg + Br_2 \xrightarrow{\Delta} CH_3-CH_2-Br + CO_2 + AgBr$
Thus,the product $[X]$ is ethyl bromide $(CH_3-CH_2-Br)$.

(B) The reaction of alcohols with thionyl chloride $(SOCl_2)$ is represented as: $R-OH + SOCl_2 \rightarrow R-Cl + SO_2(g) + HCl(g)$.
In this reaction,the by-products formed are $SO_2$ and $HCl$,both of which are gases.
Since these gases escape from the reaction mixture,the equilibrium shifts in the forward direction according to Le Chatelier's principle,resulting in high yields of pure alkyl chlorides.

(C) gem-dichloride (geminal dichloride) has two chlorine atoms attached to the same carbon atom.
$1$. $CH_3CHO + PCl_5 \rightarrow CH_3CHCl_2 + POCl_3$ (This forms $1,1$-dichloroethane,which is a gem-dichloride).
$2$. $CH_3COCH_3 + PCl_5 \rightarrow CH_3CCl_2CH_3 + POCl_3$ (This forms $2,2$-dichloropropane,which is a gem-dichloride).
$3$. $CH_2 = CH_2 + Cl_2 \rightarrow ClCH_2-CH_2Cl$ (This forms $1,2$-dichloroethane,which is a vicinal dichloride,not a gem-dichloride).
$4$. $CH_2 = CHCl + HCl \rightarrow CH_3CHCl_2$ (This follows Markovnikov's rule to form $1,1$-dichloroethane,which is a gem-dichloride).
Therefore,the reaction that does not form a gem-dichloride is $CH_2 = CH_2$ and $Cl_2$.

(B) Bleaching powder $(CaOCl_2)$ acts as a source of chlorine and an oxidizing agent. When it is distilled with acetone $(CH_3COCH_3)$,it undergoes a haloform reaction to produce chloroform $(CHCl_3)$.
The chemical reaction is:
$2 CH_3COCH_3 + 3 Ca(OCl)_2 \rightarrow 2 CHCl_3 + (CH_3COO)_2Ca + 2 Ca(OH)_2 + CaCl_2$
Thus,the product obtained is $CHCl_3$ (chloroform).

(A, B, C, D) All the given reactions are standard chemical transformations for the preparation of haloalkanes.
$1$. $R-Cl + NaCN \rightarrow R-CN$ is a nucleophilic substitution reaction (cyanide displacement).
$2$. $R-OH + NaI \xrightarrow{Conc. H_2SO_4} R-I$ is a method to prepare alkyl iodides,though $H_3PO_4$ is preferred over $H_2SO_4$ to avoid oxidation of $I^-$ to $I_2$.
$3$. $R-I + AgF \rightarrow R-F$ is the Swarts reaction,used for the synthesis of alkyl fluorides.
$4$. $R-OH + SOCl_2 \rightarrow R-Cl + SO_2 + HCl$ is the Darzens process,which is the best method for preparing alkyl chlorides as the by-products are gases.
Since all reactions are correctly represented,the question implies identifying all valid reactions.

(D) Alkyl halides can be prepared from alcohols using various reagents:
$1$. $HCl + ZnCl_2$ (Lucas reagent) is used to convert alcohols to alkyl chlorides.
$2$. $Red \ P + Br_2$ (or $I_2$) is used to convert alcohols to alkyl bromides or iodides.
$3$. $PCl_5$ reacts with alcohols to form alkyl chlorides.
Since all these reagents are standard methods for the preparation of alkyl halides from alcohols,the correct answer is $D$.

(B) The given reaction is the Hunsdiecker reaction,which involves the treatment of a silver salt of a carboxylic acid with bromine in the presence of carbon tetrachloride $(CCl_4)$ under reflux conditions.
In this reaction,the carboxylate group $(-COOAg)$ is replaced by a bromine atom,resulting in the loss of $CO_2$ and the formation of an alkyl bromide with one carbon atom less than the original carboxylic acid.
The reaction is: $C_6H_5-CH_2-COOAg + Br_2 \xrightarrow{CCl_4, \Delta} C_6H_5-CH_2-Br + CO_2 + AgBr$.
Therefore,the major product is benzyl bromide $(C_6H_5-CH_2-Br)$.

(A) The reaction of $1$-phenylethanol with thionyl chloride $(SOCl_2)$ follows the $S_Ni$ mechanism in the absence of pyridine.
In this reaction,the hydroxyl group $(-OH)$ is replaced by a chlorine atom $(-Cl)$ with retention of configuration.
The chemical equation is: $Ph-CH(OH)-CH_3 + SOCl_2 \rightarrow Ph-CH(Cl)-CH_3 + SO_2 + HCl$.

(C) The reaction $C_2H_5OH + NaCl \to C_2H_5Cl + NaOH$ does not occur because the hydroxyl group $(-OH)$ is a poor leaving group and cannot be displaced by a chloride ion $(Cl^-)$ in a simple nucleophilic substitution reaction.
In the other options:
$(A)$ This is the Lucas test reaction,which produces alkyl chloride.
$(B)$ This is free radical chlorination of an alkane.
$(D)$ This is the Hunsdiecker reaction,which produces alkyl bromide.

(B) The given reaction is $C_2H_5Cl + AgF \longrightarrow C_2H_5F + AgCl$.
This is a classic example of a halogen exchange reaction where an alkyl chloride is converted into an alkyl fluoride using a metallic fluoride like $AgF$.
This specific reaction is known as the $Swarts$ reaction.

(A) The reaction of ethyl alcohol with bleaching powder involves three steps: oxidation,chlorination,and hydrolysis.
$1$. Oxidation of ethyl alcohol to acetaldehyde:
$CH_{3}CH_{2}OH + [O] \longrightarrow CH_{3}CHO + H_{2}O$
$2$. Chlorination of acetaldehyde to chloral:
$CH_{3}CHO + 3Cl_{2} \longrightarrow CCl_{3}CHO + 3HCl$
$3$. Hydrolysis of chloral by calcium hydroxide (formed from bleaching powder) to produce chloroform:
$2CCl_{3}CHO + Ca(OH)_{2} \longrightarrow 2CHCl_{3} + (HCOO)_{2}Ca$
Thus,the final product obtained is chloroform.

(A) $1$. The reaction of benzene with acetic anhydride $(CH_3CO)_2O$ in the presence of $AlCl_3$ is a Friedel-Crafts acylation reaction,which yields acetophenone $(X = C_6H_5COCH_3)$.
$2$. The Clemmensen reduction of acetophenone using $Zn-Hg/HCl$ reduces the carbonyl group to a methylene group,yielding ethylbenzene $(Y = C_6H_5CH_2CH_3)$.
$3$. The reaction of ethylbenzene with $NBS$ ($N$-bromosuccinimide) is a radical substitution reaction that occurs at the benzylic position,yielding $1-$bromo$-1-$phenylethane $(Z = C_6H_5CH(Br)CH_3)$.

(B) The reaction $R-X + NaI \xrightarrow{Acetone} R-I + NaX$ (where $X = Cl, Br$) is a halogen exchange reaction used to prepare alkyl iodides.
This specific reaction is known as the Finkelstein reaction.

(D) Alkyl halides can be prepared from alcohols using various reagents:
$1$. $HCl + ZnCl_2$ (Lucas reagent) is used to convert alcohols to alkyl chlorides.
$2$. $Br_2 + \text{Red } P$ is used to convert alcohols to alkyl bromides.
$3$. $KI + H_3PO_4$ is used to convert alcohols to alkyl iodides.
Since all these methods are standard laboratory procedures for the synthesis of alkyl halides from alcohols,the correct option is $D$.

(B) The $Hunsdiecker$ reaction is a name reaction in organic chemistry whereby silver salts of carboxylic acids react with a halogen to produce an organic halide.
It is an example of both a decarboxylation and a halogenation reaction as the product has one fewer carbon atoms than the starting material (lost as $CO_2$) and a halogen atom is introduced in its place.

(B) The $Finkelstein$ reaction is a classic method for the preparation of alkyl iodides from alkyl chlorides or alkyl bromides.
In this reaction,an alkyl halide ($R-X$,where $X = Cl, Br$) is treated with sodium iodide $(NaI)$ in dry acetone.
The reaction proceeds as follows: $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$.
Since $NaCl$ or $NaBr$ is less soluble in dry acetone than $NaI$,the precipitation of the salt drives the reaction forward according to $Le$ $Chatelier's$ principle.

(B) The reaction of toluene with $Br_2$ in the presence of sunlight (or $UV$ light) proceeds via a free radical mechanism.
This is a side-chain halogenation reaction.
$C_6H_5CH_3 + Br_2 \xrightarrow{h\nu} C_6H_5CH_2Br + HBr$.
The product formed is benzyl bromide.