Properties of Haloalkanes Questions in English

(A) compound is chiral if it contains at least one chiral center (a carbon atom bonded to four different groups).
$1$. In $CH_3-CH_2-CH_2Cl$,the carbon atoms are bonded to identical hydrogen atoms or are part of a chain with no asymmetry. Specifically,the $C-2$ carbon is bonded to two hydrogen atoms,making it achiral.
$2$. In $CH_3-CHD-CH_2Cl$,the $C-2$ carbon is bonded to $-CH_3$,$-H$,$-D$,and $-CH_2Cl$. Since all four groups are different,it is chiral.
$3$. In $CH_3-CHCl-CH_2Cl$,the $C-2$ carbon is bonded to $-CH_3$,$-H$,$-Cl$,and $-CH_2Cl$. Since all four groups are different,it is chiral.
$4$. In $CH_3-CH_2-CHDCl$,the $C-1$ carbon is bonded to $-CH_3-CH_2$,$-H$,$-D$,and $-Cl$. Since all four groups are different,it is chiral.
Therefore,$CH_3-CH_2-CH_2Cl$ is the only achiral compound.

(A) The reaction of $1$-chlorobutane $(CH_3CH_2CH_2CH_2Cl)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).
$CH_3CH_2CH_2CH_2Cl + KOH (\text{alc.}) \rightarrow CH_3CH_2CH=CH_2 + KCl + H_2O$.
The product formed is $1$-butene.

(D) The rate of hydrolysis of alkyl halides via the $S_N1$ mechanism depends on the stability of the carbocation intermediate formed.
In the given options,$(C_6H_5)_3CCl$ forms a triphenylmethyl carbocation $(C_6H_5)_3C^+$.
This carbocation is highly stabilized by resonance due to the presence of three phenyl rings.
Since the carbocation formed from $(C_6H_5)_3CCl$ is the most stable,it undergoes hydrolysis most rapidly.

(B) $1$. $C_2H_5I$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form ethene $(X)$: $C_2H_5I \xrightarrow{\text{alc. } KOH} CH_2=CH_2 (X)$.
$2$. Ethene reacts with $Br_2$ (electrophilic addition) to form $1,2$-dibromoethane $(Y)$: $CH_2=CH_2 \xrightarrow{Br_2} CH_2Br-CH_2Br (Y)$.
$3$. $1,2$-dibromoethane reacts with $KCN$ (nucleophilic substitution) to form succinonitrile $(Z)$: $CH_2Br-CH_2Br \xrightarrow{KCN} CN-CH_2-CH_2-CN (Z)$.

(C) Chloroform $(CHCl_3)$ is slowly oxidized by air in the presence of light to form an extremely poisonous gas,carbonyl chloride,also known as phosgene $(COCl_2)$.
The reaction is: $2CHCl_3 + O_2 \xrightarrow{\text{light/air}} 2COCl_2 + 2HCl$.
Therefore,$X$ represents air and light.

(D) The rate of the ${S}_{N}1$ reaction depends on the stability of the carbocation formed and the ease with which the leaving group departs.
In the rate-determining step of the ${S}_{N}1$ reaction,the carbon-halogen bond breaks.
The ease of breaking this bond depends on the strength of the $C-X$ bond.
The bond strength decreases as the size of the halogen atom increases $(F > Cl > Br > I)$.
Therefore,the $C-I$ bond is the weakest and breaks most easily,making the iodide ion the best leaving group.
Thus,$(CH_3)_3C - I$ reacts most rapidly.

(C) The strength of the carbon-halogen $(C-X)$ bond depends on the bond length.
As the size of the halogen atom increases,the bond length increases,and the bond strength decreases.
The order of atomic size is $F < Cl < Br < I$.
Therefore,the order of bond strength is $C-F > C-Cl > C-Br > C-I$.
Thus,the $C-F$ bond in $CH_3F$ is the strongest.

(A) In $CH_2 = CHCl$,the carbon atom attached to the chlorine atom is $sp^2$ hybridized. Due to resonance,the $C-Cl$ bond acquires partial double bond character,making it stronger and shorter,which makes it the least reactive towards nucleophilic substitution.

(A) In the reaction $CH_3CH_2Cl + KOH(aq.) \rightarrow CH_3CH_2OH + KCl$,the hydroxide ion $(OH^-)$ acts as a nucleophile.
It attacks the electrophilic carbon atom bonded to the chlorine atom in the ethyl chloride molecule.
The chloride ion $(Cl^-)$ is displaced as a leaving group.
Since a nucleophile replaces another group,this is a nucleophilic substitution reaction.

(A) The reactivity order for the $S_{N}1$ mechanism depends on the stability of the carbocation intermediate.
$S_{N}1$ reactivity follows the order: $3^{\circ} \text{ alkyl halide} > 2^{\circ} \text{ alkyl halide} > 1^{\circ} \text{ alkyl halide} > \text{methyl halide}$.
$(CH_3)_3C-Br$ is a $3^{\circ}$ alkyl halide,which forms a stable tertiary carbocation.
Therefore,$(CH_3)_3C-Br$ is the most reactive towards the $S_{N}1$ mechanism.

(B) When chloroform $(CHCl_3)$ reacts with concentrated nitric acid $(HNO_3)$,it forms chloropicrin $(CCl_3NO_2)$,which is also known as trichloronitromethane.
The chemical reaction is:
$CHCl_3 + HNO_3 \to CCl_3NO_2 + H_2O$

(A) The $S_N2$ mechanism involves a backside attack by the nucleophile on the carbon atom bonded to the leaving group.
In tertiary alkyl halides,the central carbon atom is surrounded by three bulky alkyl groups.
These bulky groups create significant steric hindrance,which prevents the nucleophile from approaching the carbon atom effectively.
Therefore,tertiary alkyl halides are practically inert to $S_N2$ reactions.

(A) Pure $CHCl_3$ (chloroform) and $CHI_3$ (iodoform) are both haloforms.
$CHCl_3$ is a liquid at room temperature,while $CHI_3$ is a yellow solid.
However,chemically,they can be distinguished by their reaction with aqueous $AgNO_3$.
$CHCl_3$ does not give a precipitate with aqueous $AgNO_3$ because the $C-Cl$ bond is covalent and does not ionize to release $Cl^-$ ions.
Similarly,$CHI_3$ does not give a precipitate with aqueous $AgNO_3$.
Actually,the most common way to distinguish them is by their physical state or by heating with aqueous $KOH$ followed by testing the resulting halide ions.
However,among the given options,the reaction with aqueous $AgNO_3$ is often cited in textbooks to show that neither forms a precipitate,but they can be distinguished by their physical appearance (liquid vs solid).
Given the standard chemistry curriculum,$CHI_3$ is a yellow solid with a characteristic smell,while $CHCl_3$ is a colorless liquid.
If we must choose a reagent,none of the options provided directly distinguish them via a simple precipitate test as both are stable.
However,in many competitive exams,this question is framed to test the knowledge of physical properties or specific chemical tests.
Based on standard practice,the physical state is the primary differentiator.

(A) $S_{N}1$ reactions proceed via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity.
$PhCH_2Cl$ forms a resonance-stabilized benzyl carbocation $(Ph-CH_2^+)$.
$Ph-Cl$ is an aryl halide and is least reactive towards $S_{N}1$.
$CH_3CHClCH_3$ forms a secondary carbocation.
$p-NO_2-C_6H_4CH_2Cl$ forms a benzyl carbocation,but the $NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which destabilizes the carbocation.
Comparing the stability,the unsubstituted benzyl carbocation $(PhCH_2^+)$ is more stable than the secondary carbocation and significantly more stable than the $NO_2$-substituted benzyl carbocation. Thus,$PhCH_2Cl$ is the most reactive.

(D) In an $S_{N}2$ reaction,the rate depends on the ability of the leaving group to depart. The leaving group ability follows the order $I^{-} > Br^{-} > Cl^{-} > F^{-}$. Since the $C-X$ bond strength decreases as the size of the halogen increases $(C-I < C-Br < C-Cl < C-F)$,the bond is easiest to break for the iodide. Therefore,the order of reactivity is $RI > RBr > RCl > RF$.

(A) $C_6H_5I$ (iodobenzene) does not give a precipitate with $AgNO_3$ because the $C-I$ bond has partial double bond character due to resonance,making it resistant to nucleophilic substitution.
$C_6H_5CH_2I$ (benzyl iodide) undergoes hydrolysis with $NaOH$ to form benzyl alcohol and $NaI$. The $I^-$ ions react with $AgNO_3$ to form a yellow precipitate of $AgI$.
Since $B$ gives a yellow precipitate,$B$ must be $C_6H_5CH_2I$ and $A$ must be $C_6H_5I$.

(B) The reactivity order $3^{\circ} \text{ halide} > 2^{\circ} \text{ halide} > 1^{\circ} \text{ halide} > CH_3X$ is characteristic of the $S_N1$ mechanism.
In an $S_N1$ reaction,the rate-determining step is the formation of a carbocation intermediate.
Since $3^{\circ}$ carbocations are more stable than $2^{\circ}$,$1^{\circ}$,and methyl carbocations due to the inductive effect and hyperconjugation,$3^{\circ}$ halides react the fastest.
Conversely,$S_N2$ reactions follow the order $CH_3X > 1^{\circ} > 2^{\circ} > 3^{\circ}$ due to steric hindrance.

(C) The $S_N1$ mechanism proceeds through the formation of a planar carbocation intermediate.
Because the carbocation is planar,the nucleophile can attack from either the front side or the back side with equal probability.
This leads to the formation of both enantiomers in equal amounts,resulting in a racemic mixture.

(B) The boiling point of isomeric haloalkanes decreases with an increase in branching because branching reduces the surface area of the molecule,which in turn decreases the magnitude of van der Waals forces of attraction.
$(a)$ $CH_3CH_2CH_2CH_2Br$ is a straight-chain primary alkyl halide (n-butyl bromide).
$(b)$ $(CH_3)_2CHCH_2Br$ is a branched primary alkyl halide (isobutyl bromide).
$(c)$ $(CH_3)_3CBr$ is a tertiary alkyl halide (tert-butyl bromide),which is the most branched.
Therefore,the order of boiling points is $(c) < (b) < (a)$.

(C) $S_{N}1$ mechanism depends on the stability of the carbocation intermediate formed during the reaction.
Benzyl chloride $(C_6H_5CH_2Cl)$ forms a benzyl carbocation $(C_6H_5CH_2^+)$ upon the loss of a chloride ion.
This carbocation is highly stabilized by resonance with the benzene ring.
Therefore,benzyl chloride readily undergoes $S_{N}1$ reaction compared to the other options provided.

(A) The reaction that shows complete inversion of configuration is the $S_N2$ mechanism.
$S_N2$ reactions are favored by primary alkyl halides due to minimal steric hindrance.
Among the given options,$CH_3Cl$ is a primary halide (methyl chloride) which undergoes $S_N2$ reaction with complete inversion of configuration (Walden inversion).
$(CH_3)_3CCl$ is a tertiary halide and prefers $S_N1$ mechanism,while $(CH_3)_2CHCl$ is a secondary halide.

(B) The reaction is given by: $R-X + Mg \xrightarrow{\text{dry ether}} R-Mg-X$
When an alkyl halide reacts with $Mg$ in the presence of dry ether,it forms an alkyl magnesium halide,which is known as a Grignard reagent.

(C) The reaction of benzyl bromides with ethanol proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate. The rate of the reaction depends on the stability of the carbocation formed. The $p$-methoxy group is an electron-donating group ($+M$ effect),which significantly stabilizes the carbocation through resonance. Therefore,$p$-methoxybenzyl bromide forms the most stable carbocation and reacts most rapidly with ethanol.

(B) In $CH_3COCH_2CH_2Br$,the $\alpha$-hydrogen atoms are acidic due to the electron-withdrawing effect of the carbonyl group.
Upon treatment with alcoholic $KOH$,it forms a carbanion $CH_3CO\bar{C}H-CH_2Br$ which is stabilized by resonance with the carbonyl group.
This stabilization facilitates the elimination reaction,making it the most reactive among the given options.

(A) Isopentane is $2$-methylbutane,which has the structure $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl): $CH_2Cl-CH(CH_3)-CH_2-CH_3$ (achiral).
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$ (chiral,forms a pair of enantiomers).
$3$. At $C_3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$ (chiral,forms a pair of enantiomers).
$4$. At $C_4$ (terminal methyl): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ (achiral).
Thus,there are $2$ chiral centers formed,each resulting in a pair of enantiomers.
Therefore,the number of possible enantiomeric pairs is $2$.

(D) The $S_N2$ mechanism involves a backside attack by the nucleophile on the chiral carbon atom.
This leads to the inversion of configuration,also known as the $Walden$ inversion.
Since the nucleophile attacks from the side opposite to the leaving group,the spatial arrangement of the atoms around the chiral center is inverted.
Therefore,the product formed is a single stereoisomer with an inverted configuration relative to the reactant.

(B) The reaction of chloroform $(CHCl_3)$ with concentrated nitric acid $(HNO_3)$ is a nitration reaction that produces chloropicrin $(CCl_3NO_2)$,which is also known as trichloronitromethane.
The chemical equation is: $CHCl_3 + HNO_3 (\text{conc.}) \rightarrow CCl_3NO_2 + H_2O$.

(B) Alcoholic $KOH$ contains the alkoxide ion $(RO^-)$,which acts as a strong base. It facilitates the removal of a proton from the $\beta$-carbon,leading to the elimination of a hydrogen halide molecule from the haloalkane. This process is known as dehydrohalogenation.

(D) The hydrolysis of the corresponding halide precursor in the presence of acetone typically proceeds via an $S_N1$ mechanism.
In this specific reaction,the carbocation intermediate formed is stabilized by the electron-donating $p$-methoxy group.
The stereochemistry of the product depends on the attack of the nucleophile on the planar carbocation.
Given the structures provided,the reaction leads to the formation of both $K$ and $M$ as diastereomeric products due to the lack of stereospecificity in the $S_N1$ pathway.

(A) The size of the halogen atom increases from $F$ to $I$.
As the size of the halogen atom increases,the overlap between the carbon and halogen orbitals becomes less effective,leading to a longer and weaker bond.
Therefore,the bond dissociation energy decreases in the order: $CH_3-F > CH_3-Cl > CH_3-Br > CH_3-I$.

(D) The compound '$X$' has three carbon atoms and reacts with aqueous $KOH$ to form an aldehyde. This indicates that '$X$' is a geminal dihalide at the terminal position,specifically $1,1$-dichloropropane $(CH_3CH_2CHCl_2)$.
Reaction with aqueous $KOH$: $CH_3CH_2CHCl_2 + 2KOH(aq) \rightarrow CH_3CH_2CHO + 2KCl + H_2O$.
Reaction with alcoholic $KOH$: $CH_3CH_2CHCl_2 + 2KOH(alc) \rightarrow CH_3C \equiv CH + 2KCl + 2H_2O$.
The product $CH_3C \equiv CH$ (propyne) is a terminal alkyne,which reacts with ammoniacal $Cu_2Cl_2$ to form a red precipitate of copper$(I)$ propynide $(CH_3C \equiv CCu)$.

(A) Geminal dihalides (where two halogen atoms are attached to the same carbon atom) on hydrolysis with aqueous $KOH$ followed by dehydration yield carbonyl compounds.
$CH_3-CCl_2-CH_3 + 2KOH(aq) \rightarrow CH_3-C(OH)_2-CH_3 + 2KCl$
$CH_3-C(OH)_2-CH_3 \xrightarrow{-\Delta H_2O} CH_3-CO-CH_3$ (Acetone).
Thus,$2,2$-dichloropropane is the correct isomer.

(B) When chloroform $(CHCl_3)$ is exposed to air and sunlight,it undergoes oxidation to form a highly poisonous gas called carbonyl chloride $(COCl_2)$,also known as phosgene.
The chemical reaction is:
$2CHCl_3 + O_2 \xrightarrow{\text{Light/Air}} 2COCl_2 + 2HCl$

(D) The $E2$ elimination reaction typically follows $Zaitsev's$ rule,which favors the formation of the most substituted alkene.
However,the $Hofmann$ rule is the opposite,where the major product is the least substituted alkene.
This occurs due to steric hindrance or electronic factors that make the removal of a proton from the less substituted carbon more favorable.
Therefore,the $Hofmann$ rule means the double bond is formed towards the less substituted carbon.

(C) In the given reaction,the bromine atom $(-Br)$ is replaced by a hydroxyl group $(-OH)$.
Since the water molecule acts as a nucleophile ($H_2O$ or $OH^-$) attacking the electrophilic carbon center,this is a nucleophilic substitution reaction.
Specifically,this is an $S_N1$ reaction involving the formation of a stable tertiary carbocation intermediate.

(C) Chloroform $(CHCl_3)$ is sensitive to light and air. When exposed to light and oxygen,it undergoes slow oxidation to form a highly poisonous gas called phosgene $(COCl_2)$.
The chemical reaction is: $2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$.
To prevent this oxidation,chloroform is stored in dark-colored bottles to block light and filled up to the brim to minimize contact with air.

(C) The reaction of alkyl halides with sodium metal in the presence of dry ether is known as the Wurtz reaction.
To obtain isobutane $(CH_3-CH(CH_3)-CH_3)$,we need to combine a methyl group $(CH_3-)$ and an isopropyl group $(-CH(CH_3)_2)$.
Therefore,the reactants must be methyl chloride $(CH_3Cl)$ and isopropyl chloride $(CH_3-CHCl-CH_3)$.
The reaction is: $CH_3Cl + CH_3-CHCl-CH_3 + 2Na \xrightarrow{\text{dry ether}} CH_3-CH(CH_3)-CH_3 + 2NaCl$.

(B) When $CH_3CHCl_2$ ($1$,$1$-dichloroethane) is treated with aqueous $KOH$,it undergoes nucleophilic substitution to form a geminal diol,$CH_3CH(OH)_2$.
Since geminal diols are unstable,they immediately lose a water molecule to form acetaldehyde $(CH_3CHO)$.
The reaction is: $CH_3CHCl_2 + 2KOH(aq)$ $\rightarrow CH_3CH(OH)_2 + 2KCl$ $\rightarrow CH_3CHO + H_2O + 2KCl$.

(C) The reactivity towards $S_N1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
In $PhCHCl(CH_3)$,the carbocation formed is $Ph-CH^+-CH_3$,which is a secondary benzylic carbocation.
This carbocation is stabilized by both the phenyl ring (resonance) and the inductive effect of the methyl group.
Comparing the options,$PhCHCl(CH_3)$ forms the most stable carbocation,making it the most reactive towards $S_N1$.

(C) The reaction involves the ionization of $1$-chloro-$1$-phenylethane in the presence of the Lewis acid $SbCl_5$.
$SbCl_5$ acts as a catalyst by abstracting the chloride ion to form a stable planar carbocation intermediate: $C_6H_5-CH^+(CH_3)$.
Since the carbocation is $sp^2$ hybridized and planar,the nucleophile (chloride ion) can attack from either side with equal probability,leading to the formation of a racemic mixture.
Therefore,the racemization occurs due to the formation of a carbocation.

(A) $S_{N^{1}}$ reactions proceed via the formation of a carbocation intermediate. The stability of the carbocation determines the rate of the reaction. Tertiary $(3^{\circ})$ carbocations are the most stable due to the inductive effect and hyperconjugation of the three alkyl groups. In the reaction of $(CH_3)_3C-Cl$ with $KOH$,the rate-determining step involves the formation of a stable tertiary carbocation,$(CH_3)_3C^{\oplus}$. Therefore,$(CH_3)_3C-Cl$ readily undergoes $S_{N^{1}}$ reaction.

(A) The reaction follows Markovnikov's rule,where the electrophile $H^+$ adds to the carbon atom with more hydrogen atoms.
$CH_2=CH-Cl + HCl \rightarrow CH_3-CHCl_2$
Thus,the product formed is $1,1$-dichloroethane (ethylidene chloride).

(A) When toluene $(C_6H_5CH_3)$ reacts with chlorine $(Cl_2)$ in the presence of sunlight (ultraviolet light) and in the absence of a halogen carrier (like $FeCl_3$),the reaction proceeds via a free radical mechanism.
This leads to the substitution of hydrogen atoms on the side-chain methyl group rather than the benzene ring.
The reaction proceeds as follows:
$C_6H_5CH_3 + Cl_2 \xrightarrow{h\nu} C_6H_5CH_2Cl + HCl$ (Benzyl chloride).
Further chlorination can lead to benzal chloride $(C_6H_5CHCl_2)$ and benzotrichloride $(C_6H_5CCl_3)$.
However,the primary product formed under controlled conditions is benzyl chloride $(C_6H_5CH_2Cl)$.

(B) The reaction of ethyl chloride $(CH_3CH_2Cl)$ with alcoholic $KOH$ leads to the formation of ethene $(CH_2=CH_2)$.
This reaction involves the removal of a hydrogen atom from the $\beta$-carbon and a chlorine atom from the $\alpha$-carbon,hence it is called $\beta$-elimination.
Since a hydrogen atom and a halogen atom are removed,it is also known as dehydrohalogenation.
Therefore,both $(a)$ and $(d)$ are correct descriptions,but given the options,$(b)$ is the most appropriate choice.

(A) The reaction of a dihaloalkane with sodium in ether is an intramolecular Wurtz reaction.
In $1$-bromo-$3$-chlorocyclobutane,the halogen atoms are at positions $1$ and $3$.
When treated with $2$ equivalents of sodium,the sodium atoms remove the halogen atoms,leading to the formation of a new carbon-carbon bond between the $1$ and $3$ positions.
This results in the formation of bicyclo[$1.1.0$]butane.
Thus,the correct product is bicyclobutane.

(D) Step $1$: The reaction of propan$-1-$ol with $PCl_5$ is a nucleophilic substitution reaction where the hydroxyl group is replaced by a chlorine atom.
$CH_3CH_2CH_2OH + PCl_5 \rightarrow CH_3CH_2CH_2Cl (A) + POCl_3 + HCl$.
Step $2$: The reaction of $A$ ($1$-chloropropane) with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction),which results in the formation of an alkene.
$CH_3CH_2CH_2Cl + KOH (\text{alc.}) \rightarrow CH_3CH=CH_2 (B) + KCl + H_2O$.
Thus,the final product $B$ is propene.

(A) The reaction is the electrophilic addition of $HCl$ to $1$-phenylpropene $(C_6H_5-CH=CH-CH_3)$.
According to Markovnikov's rule,the $H^+$ ion attaches to the carbon atom with more hydrogen atoms,and the $Cl^-$ ion attaches to the carbon atom with fewer hydrogen atoms.
However,the stability of the intermediate carbocation is the deciding factor.
The addition of $H^+$ to the $C_2$ position (the carbon adjacent to the phenyl ring) forms a resonance-stabilized benzylic carbocation: $C_6H_5-CH^+-CH_2-CH_3$.
The $Cl^-$ ion then attacks this carbocation to form $1$-phenyl-$1$-chloropropane,which is $C_6H_5CHClCH_2CH_3$.

(D) The reaction of ethylbenzene with $Cl_2$ in the presence of heat (or $UV$ light) proceeds via a free radical mechanism.
The hydrogen atom is abstracted from the benzylic position because the resulting benzylic radical is stabilized by resonance with the benzene ring.
The intermediate formed is $Phdot{C}HCH_3$.
Subsequent reaction with $Cl_2$ yields $PhCHClCH_3$ as the major product.
The primary radical $Phdot{C}H_2CH_2$ is less stable,making $PhCH_2CH_2Cl$ the minor product.

(A) The reaction of $C_6H_5CH = CHCH_3$ with $HBr$ follows Markovnikov's rule.
In this reaction,the electrophile $H^+$ attacks the double bond to form the most stable carbocation.
The carbocation formed is $C_6H_5CH^+ - CH_2CH_3$,which is a benzylic carbocation stabilized by resonance with the phenyl ring.
Subsequently,the nucleophile $Br^-$ attacks this carbocation to form the product $C_6H_5CH(Br)CH_2CH_3$.
Thus,the correct option is $A$.

(A) In the presence of light $(h\nu)$,toluene undergoes free radical substitution at the benzylic position to form benzyl chloride $(C_6H_5CH_2Cl)$.