Properties of Haloalkanes Questions in English

(B) In an $S_{N}1$ reaction,the carbocation intermediate formed is planar,allowing the nucleophile to attack from either side,which leads to the formation of a racemic mixture (racemisation).
In an $S_{N}2$ reaction,the nucleophile attacks the carbon atom from the side opposite to the leaving group,resulting in the inversion of configuration (Walden inversion).
Therefore,the correct statement is that $S_{N}1$ leads to racemisation and $S_{N}2$ leads to inversion of configuration,which corresponds to option $(b)$.

(A) In dry acetone,the reaction with iodide ion follows the $S_N2$ mechanism.
In an $S_N2$ reaction,the rate of reaction is inversely proportional to the steric hindrance around the electrophilic carbon atom.
The order of steric hindrance is: $Methyl < Primary < Secondary < Tertiary$.
Therefore,the order of reactivity is: $CH_3Br (IV) > CH_3CH_2Br (I) > (CH_3)_2CHBr (II) > (CH_3)_3CBr (III)$.
Thus,the correct order is $IV > I > II > III$.

(B) The reaction of an alkyl halide $(C_2H_5Cl)$ with $AgCN$ is a nucleophilic substitution reaction.
$AgCN$ is a covalent compound,and the nitrogen atom acts as the nucleophilic center.
Therefore,the reaction proceeds as follows:
$C_2H_5Cl + AgCN \xrightarrow{EtOH / H_2O} C_2H_5-NC + AgCl$
Here,$\underline{X}$ is ethyl isocyanide $(C_2H_5NC)$.
In ethyl isocyanide,the nitrogen atom is directly linked to the ethyl carbon $(C_2H_5-N \equiv C)$.
Thus,statement $(III)$ is correct.

(A) $C_2H_5Cl + \text{moist } Ag_2O \rightarrow C_2H_5OH + AgCl$. Thus,$A-iii$.
$(B)$ $C_2H_5Cl + AgCN \rightarrow C_2H_5NC + AgCl$ (Isocyanide is the major product). Thus,$B-iv$.
$(C)$ $C_2H_5Cl + AgNO_2 \rightarrow C_2H_5ONO + AgCl$ (Alkyl nitrite is the major product). Thus,$C-i$.
$(D)$ $C_2H_5Cl + \text{ethanolic } KOH \rightarrow C_2H_4 + KCl + H_2O$ (Dehydrohalogenation). Thus,$D-ii$.
Therefore,the correct match is $A-iii, B-iv, C-i, D-ii$.

(B) Chloroform $(CHCl_3)$ undergoes aerial oxidation in the presence of light to form a highly poisonous gas called phosgene $(COCl_2)$.
$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$
To prevent this,$1\%$ of ethyl alcohol $(C_2H_5OH)$ is added to chloroform.
Ethyl alcohol acts as a negative catalyst and converts any phosgene formed back into harmless diethyl carbonate.

(C) Only alcoholic $KOH$ undergoes a dehydrohalogenation reaction with alkyl halides. When ethyl chloride $(CH_3CH_2Cl)$ is heated with alcoholic $KOH$,it undergoes elimination to form ethylene $(CH_2=CH_2)$.
The reaction is as follows:
$CH_3CH_2Cl + KOH (alc.) \rightarrow CH_2=CH_2 + KCl + H_2O$

(D) The reaction of haloalkanes with aqueous $NaOH$ or moist $AgOH$ (which acts as $AgOH$ or $Ag_2O + H_2O$) leads to the nucleophilic substitution of the halogen atom with a hydroxyl group $(-OH)$ to form an alcohol.
In the given reaction,$C_2H_5Cl$ reacts with aqueous $NaOH$ to form $C_2H_5OH$ (ethanol).
Similarly,$C_2H_5Cl$ reacts with moist $AgOH$ to form $C_2H_5OH$ (ethanol).
Therefore,both $A$ and $B$ are $C_2H_5Cl$.

(A) The reaction of haloalkanes like $C_2H_5Cl$ with aqueous alkali (e.g.,$\text{aq. } KOH$ or $\text{aq. } NaOH$) undergoes nucleophilic substitution to form alcohols.
Similarly,moist silver oxide $(AgOH)$ also acts as a source of hydroxide ions and converts haloalkanes into alcohols.
Therefore,$A$ can be $\text{aq. } KOH$ and $B$ can be $AgOH$.

(A) In the presence of air and light,chloroform $(CHCl_3)$ undergoes oxidation to form carbonyl chloride,commonly known as phosgene $(COCl_2)$,which is a highly toxic gas.
The chemical reaction is:
$CHCl_3 + \frac{1}{2}O_2 \xrightarrow{\text{Air and light}} COCl_2 + HCl$

(A) The correct answer is $H_3CCH_2CH_2CH_2Cl$ (n-butyl chloride).
For isomeric alkyl halides,the boiling point decreases with an increase in branching.
Branching of the carbon chain makes the molecule more compact,which reduces the surface area.
As the surface area decreases,the magnitude of intermolecular van der Waals forces of attraction also decreases.
Therefore,straight-chain isomers have higher boiling points compared to their branched-chain counterparts with the same molecular formula.
Among the given options,$H_3CCH_2CH_2CH_2Cl$ is a straight-chain isomer,while the others are branched,making it the one with the highest boiling point.

(A) $1$. The reaction of acetone $(CH_3)_2C=O$ with $CH_3MgBr$ followed by acidic hydrolysis gives tert-butyl alcohol $(A)$,which is $(CH_3)_3C-OH$.
$2$. Treatment of $(A)$ with $SOCl_2$ leads to the formation of tert-butyl chloride $(B)$,which is $(CH_3)_3C-Cl$.
$3$. The reaction of $(B)$ with $CH_3ONa$ (a strong base) under heating conditions leads to dehydrohalogenation via an $E2$ mechanism,resulting in the formation of isobutylene ($2$-methylpropene) as the major product $(C)$,which is $(CH_3)_2C=CH_2$.

(A, C) Ethyl chloride undergoes nucleophilic substitution with ethanolic $KCN$ to form ethyl cyanide:
$C_2H_5Cl + KCN \xrightarrow{\text{Ethanol}} C_2H_5CN + KCl$
$(B)$ Chlorobenzene does not undergo nucleophilic substitution with $KCN$ under normal conditions due to the partial double bond character of the $C-Cl$ bond.
$(C)$ Benzaldehyde reacts with $HCN$ (generated from $KCN$ in the presence of an acid or in alcoholic medium) to form a cyanohydrin:
$C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN$
$(D)$ Salicylic acid does not react with $KCN$.

(C) The preparation of $Me_{3}CCN$ (pivalonitrile) from $Me_{3}CBr$ via $S_{N}2$ reaction with $NaCN$ is not feasible because $Me_{3}CBr$ is a tertiary alkyl halide,which undergoes elimination rather than substitution.
The reaction of $Me_{3}CMgBr$ (a Grignard reagent) with $ClCN$ (cyanogen chloride) is the most effective method for introducing a cyano group to a tertiary carbon atom.
The reaction is as follows:
$Me_{3}CMgBr + ClCN \rightarrow Me_{3}CCN + Mg(Cl)Br$

(D) The $SbCl_5$ acts as a Lewis acid and abstracts the chloride ion $(Cl^-)$ from the substrate,$(+)-2-$chloro$-2-$phenylethane.
This results in the formation of a planar,$sp^2$ hybridized carbocation intermediate.
Because the carbocation is planar,the chloride ion can attack from either the top or the bottom face with equal probability.
This leads to the formation of both enantiomers,resulting in a racemic mixture.

(A) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
$S_{N}1$ reactions proceed via the formation of a carbocation,and the stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Therefore,the reactivity order for $S_{N}1$ reactions is $3^{\circ} > 2^{\circ} > 1^{\circ}$ alkyl halides.
Among the given options:
$A$. tert-butyl chloride is a $3^{\circ}$ alkyl halide.
$B$. $1-$chlorobutane is a $1^{\circ}$ alkyl halide.
$C$. $2-$methyl$-1-$chloropropane is a $1^{\circ}$ alkyl halide.
$D$. $2-$chlorobutane is a $2^{\circ}$ alkyl halide.
Since tert-butyl chloride forms the most stable $3^{\circ}$ carbocation,it reacts most efficiently via the $S_{N}1$ mechanism.

(A) The reactivity of $S_N 1$ reaction is directly proportional to the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
In option $A$,the carbocation formed is the cyclopentadienyl cation,which is aromatic ($6 \pi$ electrons,planar,cyclic,conjugated). Aromatic compounds are exceptionally stable.
In option $B$,the carbocation is allylic,which is resonance-stabilized but not aromatic.
In option $C$,the carbocation is also allylic but less substituted than $B$.
In option $D$,the carbocation is also allylic but with different substitution.
Since the cyclopentadienyl cation is aromatic,it is the most stable carbocation among the choices,making the corresponding alkyl bromide the most reactive in an $S_N 1$ reaction.

(A) The reactivity of alkyl halides towards $S_N1$ reaction depends on the stability of the carbocation formed in the rate-determining step.
$1$. Allyl chloride $(I)$ forms an allyl carbocation $(CH_2=CH-CH_2^+)$,which is resonance-stabilized.
$2$. Ethyl chloride $(III)$ forms a primary ethyl carbocation $(CH_3-CH_2^+)$,which is less stable than the allyl carbocation.
$3$. Chlorobenzene $(II)$ does not undergo $S_N1$ reaction easily because the $C-Cl$ bond has partial double bond character due to resonance,making the formation of a phenyl carbocation extremely difficult.
Thus,the order of reactivity is $I > III > II$.

(C) The reaction involves the nucleophilic substitution of a primary alkyl halide in the presence of $NaCN$ and $DMF$ (a polar aprotic solvent),which favors the $S_{N}2$ mechanism.
The substrate contains two halogen atoms: an iodine atom directly attached to the benzene ring (aryl iodide) and a chlorine atom attached to a benzylic carbon (benzylic chloride).
$1$. The $C-I$ bond attached to the benzene ring has partial double-bond character due to resonance,making it very resistant to nucleophilic substitution.
$2$. The benzylic $C-Cl$ bond is highly reactive towards $S_{N}2$ substitution because the transition state is stabilized by the adjacent benzene ring.
Therefore,the $CN^-$ nucleophile selectively attacks the benzylic carbon,displacing the $Cl^-$ ion to form $2-(3-iodophenyl)acetonitrile$ as the major product.

(A, B, C) The $S_{N}2$ reaction is a bimolecular nucleophilic substitution reaction.
$1$. The rate of reaction depends on the concentration of both the substrate and the nucleophile,hence it is a second-order reaction.
$2$. The rate increases with the strength of the nucleophile (nucleophilicity).
$3$. Tertiary substrates are sterically hindered,making $S_{N}2$ reactions extremely slow or impossible.
$4$. Optical rotation change is not guaranteed; it depends on the priority of the incoming nucleophile compared to the leaving group according to Cahn-Ingold-Prelog rules.
Therefore,statements $A$,$B$,and $C$ are true.

(D) The reaction sequence involves two $S_N2$ steps,each resulting in an inversion of configuration at the chiral center.
$1$. The starting material is $sec$-butyl chloride. Reaction with aqueous $NaOH$ proceeds via $S_N2$ mechanism,leading to the formation of $sec$-butyl alcohol with inversion of configuration.
$2$. The alcohol then reacts with $p$-toluenesulfonyl chloride $(TsCl)$ in pyridine to form a tosylate intermediate,which is a good leaving group.
$3$. Finally,the tosylate reacts with $NaI$ in acetone via another $S_N2$ reaction,resulting in a second inversion of configuration.
$4$. Since there are two inversions,the final product $C_4H_9I$ will have the same configuration as the starting material $C_4H_9Cl$.
$5$. Comparing the structures,option $D$ correctly represents the intermediate alcohol (inverted) and the final iodide (retained configuration relative to the starting material).

(A) The hydrolysis of alkyl chlorides in aqueous $NaOH$ proceeds via $S_N1$ or $S_N2$ mechanisms.
$1$. Methoxymethyl chloride $(CH_3OCH_2Cl)$ undergoes $S_N1$ reaction very rapidly because the resulting carbocation $(CH_3OCH_2^+)$ is highly stabilized by resonance from the lone pair on the oxygen atom.
$2$. Benzyl chloride $(C_6H_5CH_2Cl)$ also forms a resonance-stabilized carbocation,but it is less stable than the methoxymethyl carbocation.
$3$. Neopentyl chloride $((CH_3)_3CCH_2Cl)$ is a primary alkyl halide with significant steric hindrance at the $\beta$-carbon,which makes $S_N2$ reactions extremely slow.
$4$. Propyl chloride $(CH_3CH_2CH_2Cl)$ is a primary alkyl halide that undergoes $S_N2$ reaction at a moderate rate.
Therefore,methoxymethyl chloride $(1)$ is hydrolysed most easily,and neopentyl chloride $(3)$ is hydrolysed most slowly.

(B) The reaction involves the substitution of a chlorine atom in $CH_3-O-CH_2-Cl$ with a hydroxyl group.
In this substrate,the carbon atom attached to chlorine is adjacent to an oxygen atom with lone pairs.
If the $C-Cl$ bond breaks to form a carbocation,the intermediate $CH_3-O^+-CH_2$ is formed.
This carbocation is highly stabilized by resonance due to the donation of lone pair electrons from the oxygen atom $(CH_3-O^+=CH_2)$.
Due to this significant resonance stabilization,the reaction proceeds via the $S_{N}1$ mechanism.

(B) Dehydrohalogenation of alkyl halides with alcoholic $KOH$ follows the $E_2$ mechanism.
In this reaction,the rate-determining step involves the formation of an alkene.
The stability of the resulting alkene is the primary driving force for the reaction.
According to Saytzeff's rule,more substituted alkenes are more stable.
Since tertiary $(3^{\circ})$ alkyl halides form the most substituted (and thus most stable) alkenes,they react the fastest.
Therefore,the order of reactivity is $3^{\circ} > 2^{\circ} > 1^{\circ}$.

(D) In the presence of heat (boiling point conditions) or $UV$ light,chlorine reacts with the side chain of toluene via a free radical substitution mechanism.
When excess $Cl_{2}$ is used,it leads to the progressive substitution of all three hydrogen atoms on the methyl group.
The reaction proceeds as follows:
$C_6H_5CH_3$ $\xrightarrow{Cl_2, \Delta} C_6H_5CH_2Cl$ $\xrightarrow{Cl_2, \Delta} C_6H_5CHCl_2$ $\xrightarrow{Cl_2, \Delta} C_6H_5CCl_3$.
Thus,the final product is $C_6H_5CCl_3$,which is known as trichloromethylbenzene or benzotrichloride.

(B) Step $1$: Dehalogenation with $Zn/\Delta$ removes the two bromine atoms to form a double bond,resulting in $3-\text{ethyl}-4-\text{methylcyclopentene}$.
Step $2$: Electrophilic addition of $HBr$ to the alkene follows Markovnikov's rule. Protonation of the double bond forms a secondary carbocation.
Step $3$: $A$ $1,2-\text{hydride shift}$ occurs to form a more stable tertiary carbocation at the carbon bearing the methyl group.
Step $4$: The bromide ion $(Br^-)$ attacks the tertiary carbocation to form $1-\text{bromo}-1-\text{methyl}-2-\text{ethylcyclopentane}$ as the major product.

(B) The structural isomers of $C_5H_{11}Br$ are:
$1$. $1$-bromopentane: $CH_3CH_2CH_2CH_2CH_2Br$ $\rightarrow$ $1$-pentanol (achiral).
$2$. $2$-bromopentane: $CH_3CH_2CH_2CH(Br)CH_3$ $\rightarrow$ $2$-pentanol ($CH_3CH_2CH_2CH(OH)CH_3$,chiral).
$3$. $3$-bromopentane: $CH_3CH_2CH(Br)CH_2CH_3$ $\rightarrow$ $3$-pentanol (achiral).
$4$. $1$-bromo-$2$-methylbutane: $CH_3CH_2CH(CH_3)CH_2Br$ $\rightarrow$ $2$-methyl-$1$-butanol ($CH_3CH_2CH(CH_3)CH_2OH$,chiral).
$5$. $2$-bromo-$2$-methylbutane: $CH_3CH_2C(Br)(CH_3)_2$ $\rightarrow$ $2$-methyl-$2$-butanol (achiral).
$6$. $2$-bromo-$3$-methylbutane: $CH_3CH(Br)CH(CH_3)_2$ $\rightarrow$ $3$-methyl-$2$-butanol ($CH_3CH(OH)CH(CH_3)_2$,chiral).
$7$. $1$-bromo-$3$-methylbutane: $CH_3CH(CH_3)CH_2CH_2Br$ $\rightarrow$ $3$-methyl-$1$-butanol (achiral).
$8$. $1$-bromo-$2,2$-dimethylpropane: $(CH_3)_3CCH_2Br$ $\rightarrow$ $2,2$-dimethyl-$1$-propanol (achiral).
Among the products,$2$-pentanol,$2$-methyl-$1$-butanol,and $3$-methyl-$2$-butanol exhibit optical isomerism.
Thus,the number of products is $3$.

(D) The conversion of $1-\text{phenylethyl bromide}$ $(X)$ to $2-\text{phenylacetic acid}$ $(Y)$ involves the following steps:
$1$. Dehydrohalogenation: Treatment of $1-\text{phenylethyl bromide}$ with $NaOEt$ (a strong base) leads to the formation of $styrene$ $(Ph-CH=CH_{2})$ via an $E2$ elimination mechanism.
$2$. Hydroboration-Oxidation: $Styrene$ reacts with $B_{2}H_{6}$ followed by $H_{2}O_{2}/OH^{-}$ to yield $2-\text{phenylethanol}$ $(Ph-CH_{2}-CH_{2}-OH)$ via anti-Markovnikov addition of water.
$3$. Oxidation: Finally,$2-\text{phenylethanol}$ is oxidized to $2-\text{phenylacetic acid}$ $(Ph-CH_{2}-COOH)$ using a strong oxidizing agent like Jones reagent $(CrO_{3}/H_{2}SO_{4})$.
Thus,the correct sequence is $(i) NaOEt, (ii) B_{2}H_{6}/H_{2}O_{2}, (iii) \text{Jones reagent}$.

(D) The rate of an $S_{N}2$ reaction depends on the nucleophilicity of the nucleophile. Nucleophilicity is generally inversely proportional to the stability of the anion (conjugate base).
The stability order of the given anions is: $ClO_{4}^{-} > CH_{3}COO^{-} > PhO^{-} > {}^{-}OH$.
Since nucleophilicity is the reverse of the stability of the anion,the order of nucleophilicity (and thus the rate of reaction) is: ${}^{-}OH > PhO^{-} > CH_{3}COO^{-} > ClO_{4}^{-}$.

(B) The reaction of benzyl halides with $KCN$ typically proceeds via an $S_{N}1$ mechanism in polar protic solvents or $S_{N}2$ depending on conditions. However,for benzyl halides,the rate is governed by the stability of the intermediate carbocation in $S_{N}1$ or steric/electronic factors in $S_{N}2$. In both cases,electron-donating groups $(EDG)$ increase reactivity,while electron-withdrawing groups $(EWG)$ decrease it.
$a$ has $-OH$ (strong $EDG$ via resonance),$b$ has $-NH_2$ (strong $EDG$ via resonance),$c$ has $-NO_2$ (strong $EWG$ at para position),and $d$ has $-NO_2$ (strong $EWG$ at meta position).
Comparing $a$ and $b$: $-NH_2$ is a stronger $EDG$ than $-OH$ due to lower electronegativity of $N$ compared to $O$,making $b$ more reactive than $a$.
Comparing $c$ and $d$: $-NO_2$ at the para position $(c)$ exerts a stronger destabilizing effect on the carbocation than at the meta position $(d)$ due to direct resonance destabilization.
Thus,the order is $b > a > d > c$.

(C) The rate of $S_{N}1$ reaction is directly proportional to the stability of the carbocation $(C^{\oplus})$ formed.
$I$ is $1-$bromobicyclo[$2.2$.$1$]heptane and $II$ is $1-$bromobicyclo[$2.2$.$2$]octane. Both are bridgehead halides and are extremely unreactive in $S_{N}1$ reactions due to Bredt's rule,which prevents the formation of a planar carbocation at the bridgehead position.
Between $I$ and $II$,$II$ is slightly more reactive than $I$ because the bridgehead carbocation in $II$ is less strained than in $I$.
$III$ is a tert-butyl carbocation,which is more stable than the bridgehead carbocations.
$IV$ is a triphenylmethyl carbocation,which is the most stable due to extensive resonance stabilization by three phenyl rings.
Thus,the correct order of reactivity is $II < I < III < IV$.

(B) In a polar protic solvent like methanol,nucleophilicity is determined by the ability of the nucleophile to donate an electron pair.
The order of nucleophilicity is governed by basicity and polarizability.
$C_2H_5O^{-}$ is a strong base and a good nucleophile.
$C_6H_5O^{-}$ (phenoxide) is less basic than $C_2H_5O^{-}$ due to resonance stabilization of the negative charge on the oxygen atom.
$I^{-}$ is a very good nucleophile in polar protic solvents due to its high polarizability,despite being a weak base.
$F^{-}$ is a small,highly solvated ion in methanol,making it a poor nucleophile.
Thus,the correct order is $I^{-} > C_2H_5O^{-} > C_6H_5O^{-} > F^{-}$.

(B) For a molecule to show optical isomerism,it must be chiral,meaning it must possess at least one chiral center.
$A$ chiral center is a carbon atom bonded to four different groups.
In the case of a monohaloalkane with the general formula $C_nH_{2n+1}X$,we need to find the smallest alkane chain that can accommodate a chiral center.
For $n=1$ (chloromethane) and $n=2$ (chloroethane),all carbons are bonded to at least two identical hydrogen atoms.
For $n=3$ (chloropropane),$1$-chloropropane has no chiral center,and $2-$chloropropane has two identical methyl groups attached to the central carbon.
For $n=4$,$2$-chlorobutane $(CH_3-CHCl-CH_2-CH_3)$ has a chiral center at the $C-2$ position because it is bonded to four different groups: $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_3$.
Therefore,the minimum number of carbon atoms required is $4$.

(D) The reactivity of alkyl halides towards $S_N2$ reactions is primarily governed by steric hindrance.
$S_N2$ reactions proceed via a transition state where the nucleophile attacks the carbon atom from the side opposite to the leaving group.
As the steric hindrance (bulkiness) near the reactive carbon increases,the rate of $S_N2$ reaction decreases.
The order of reactivity for primary alkyl halides is: $CH_3X > R-CH_2X > R_2CH-CH_2X > R_3C-CH_2X$.
Comparing the given options:
$(A)$ $1$-Bromo-$3$-methylbutane: $CH_3-CH(CH_3)-CH_2-CH_2Br$ (Primary,but branched at $\gamma$-position).
$(B)$ $1$-Bromo-$2,2$-dimethylpropane: $(CH_3)_3C-CH_2Br$ (Primary,but highly hindered at $\beta$-position).
$(C)$ $1$-Bromo-$2$-methylbutane: $CH_3-CH_2-CH(CH_3)-CH_2Br$ (Primary,but branched at $\beta$-position).
$(D)$ $1$-Bromobutane: $CH_3-CH_2-CH_2-CH_2Br$ (Straight-chain primary alkyl halide).
Since $1$-Bromobutane has the least steric hindrance among the given options,it exhibits the highest reactivity towards $S_N2$ reaction.

(A) Trichloromethane,commonly known as chloroform $(CHCl_3)$,undergoes slow oxidation by atmospheric oxygen in the presence of light to form carbonyl chloride,also known as phosgene $(COCl_2)$,which is highly toxic.
The chemical reaction is as follows:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$

(A) chiral carbon atom is a carbon atom that is bonded to four different groups or atoms.
In $2$-chlorobutane $(CH_3-CHCl-CH_2-CH_3)$,the carbon atom at position $2$ is bonded to four different groups: a hydrogen atom $(-H)$,a chlorine atom $(-Cl)$,a methyl group $(-CH_3)$,and an ethyl group $(-CH_2CH_3)$.
Since all four groups attached to this carbon are distinct,it is a chiral center.
In the other options,the carbon atoms are either bonded to identical groups (e.g.,two methyl groups or two hydrogen atoms),making them achiral.

(B) The reaction of bromocyclohexane with magnesium in the presence of dry ether forms a Grignard reagent,which is cyclohexylmagnesium bromide $(C_6H_{11}MgBr)$.
When this Grignard reagent is treated with water $(\text{H}_2\text{O})$,it undergoes protonolysis to yield cyclohexane $(C_6H_{12})$ as the final product '$A$'.

(C) The reaction involving the coupling of an alkyl halide and an aryl halide in the presence of sodium metal and dry ether to form an alkylbenzene is known as the Wurtz-Fittig reaction.
General equation: $R-X + 2Na + X-Ar \xrightarrow{\text{dry ether}} R-Ar + 2NaX$.

(C) . Williamson synthesis involves the reaction of an alkoxide ion with an alkyl halide,which is a nucleophilic substitution $(III)$ reaction.
$B$. Friedel-Crafts reaction involves the substitution of an aromatic hydrogen by an alkyl or acyl group,which is an electrophilic aromatic substitution $(IV)$ reaction.
$C$. Bromination of vinyl benzene (an alkene) involves the addition of bromine across the double bond,which is an electrophilic addition $(I)$ reaction.
$D$. Chlorination of toluene in the presence of light involves the substitution of a hydrogen atom on the side chain via a free radical mechanism,which is a free radical substitution $(II)$ reaction.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

$(A)$ Statement $(I)$ is true because the benzyl carbocation formed as an intermediate in the $S_N1$ mechanism is resonance-stabilized by the phenyl ring, making it significantly more stable than the primary ethyl carbocation.
Statement $(II)$ is also true because resonance stabilization (involving the delocalization of electrons over the aromatic ring) provides much greater stability to the carbocation than hyperconjugation (involving the overlap of $C-H$ $\sigma$-orbitals with the empty $p$-orbital).
Since both statements are correct, the correct option is $(A)$.