Properties of Haloalkanes Questions in English

(A) The reaction proceeds in two steps:
$1$. The reaction of $4$-hydroxybenzyl alcohol with $HCl$ under heating $(\Delta)$ leads to the substitution of the alcoholic $-OH$ group with $-Cl$,forming $4$-(chloromethyl)phenol as product $A$.
$2$. The reaction of $A$ with $NaI$ in acetone (Finkelstein reaction) proceeds via an $S_N2$ mechanism,where the $-Cl$ atom is replaced by an $-I$ atom to form $4$-(iodomethyl)phenol as product $B$.

(B) The given reaction is an elimination reaction of a quaternary ammonium salt $(CH_{3}CH_{2}CH_{2}CH(N^{+}(CH_{3})_{3})CH_{3})$ using a base $(C_{2}H_{5}ONa)$.
This is an example of the Hofmann elimination reaction.
In Hofmann elimination,the less substituted alkene is formed as the major product due to steric hindrance caused by the bulky leaving group $(N^{+}(CH_{3})_{3})$.
Therefore,the terminal alkene $CH_{3}CH_{2}CH_{2}CH=CH_{2}$ is the major product $(A)$,and the internal alkene $CH_{3}CH_{2}CH=CHCH_{3}$ is the minor product $(B)$.

(A) $NaCN$ is an ionic compound,so it provides $CN^-$ ions in solution. The carbon atom is more nucleophilic than the nitrogen atom,leading to the formation of alkyl cyanide $(R-CN)$ as the major product.
$AgCN$ is a covalent compound. The nitrogen atom has a lone pair of electrons available for nucleophilic attack,while the carbon atom is involved in a covalent bond with silver. This leads to the formation of alkyl isocyanide $(R-NC)$ as the major product.
Therefore,for the reaction with $AgCN$,the product $A$ is $CH_3CH_2NC$ (ethyl isocyanide),and for the reaction with $NaCN$,the product $B$ is $CH_3CH_2CN$ (ethyl cyanide).

(C) $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate. The rate of $S_{N}1$ reaction depends on the stability of the carbocation formed.
$I$: This is a vinylic halide. The resulting vinylic carbocation is highly unstable due to the $sp$ hybridization of the positively charged carbon.
$II$: This is a tert-butyl bromide. It forms a stable $3^{\circ}$ carbocation,which is stabilized by hyperconjugation.
$III$: This is $p$-methoxybenzyl bromide. It forms a resonance-stabilized benzylic carbocation,which is further stabilized by the electron-donating $+R$ effect of the $-OCH_3$ group.
$IV$: This is a bridgehead bromide ($1$-bromobicyclo[$2.2$.$1$]heptane). Formation of a carbocation at the bridgehead position is highly unstable due to Bredt's rule,which prevents the planar geometry required for the carbocation.
Therefore,only compounds $II$ and $III$ can undergo an $S_{N}1$ reaction.

(B) The transformation involves the conversion of styrene to $(1-\text{bromoethyl})\text{benzene}$ followed by nucleophilic substitution to form $(1-\text{cyanoethyl})\text{benzene}$.
Step $1$: Styrene reacts with $HBr$ via electrophilic addition following Markownikoff's rule to yield $(1-\text{bromoethyl})\text{benzene}$.
Step $2$: The resulting alkyl bromide undergoes nucleophilic substitution ($S_N2$ or $S_N1$ depending on conditions,typically $S_N2$ with $NaCN$) to replace the bromine atom with a cyano group $(-CN)$,yielding the final product.

(B) The reaction between a tertiary alkyl halide,$(CH_{3})_{3}CCl$,and a strong base,$CH_{3}CH_{2}ONa$,in ethanol primarily undergoes an $E2$ elimination reaction.
Since $(CH_{3})_{3}CCl$ is a tertiary halide,steric hindrance makes substitution difficult.
The strong base $CH_{3}CH_{2}O^-$ abstracts a proton from the $\beta$-carbon,leading to the formation of an alkene.
The reaction is: $(CH_{3})_{3}CCl + CH_{3}CH_{2}ONa \rightarrow CH_{2}=C(CH_{3})_{2} + CH_{3}CH_{2}OH + NaCl$.
The major product is $2$-methylpropene,which corresponds to option $B$.

(C) The $S_N 1$ reaction rate depends upon the stability of the carbocation intermediate formed in the rate-determining step. Higher stability of the carbocation leads to higher reactivity.
The carbocation intermediates formed are:
$1$. $CH_3-CO-CH_2^+$ (Least stable due to the strong electron-withdrawing effect of the carbonyl group).
$2$. $CH_3-CH_2-CH_2^+$ ($1^\circ$ carbocation).
$3$. $(CH_3)_3C^+$ ($3^\circ$ carbocation,most stable due to hyperconjugation and inductive effect).
Thus,the order of stability of the carbocations is $(CH_3)_3C^+ > CH_3-CH_2-CH_2^+ > CH_3-CO-CH_2^+$.
Therefore,the order of $S_N 1$ reactivity is $3 > 2 > 1$.

(A) In the presence of $AlCl_3$,the alkyl halide $(CH_3)_2CHCH_2Cl$ undergoes ionization to form a primary carbocation $(CH_3)_2CHCH_2^+$.
This primary carbocation is unstable and undergoes a $1,2$-hydride shift to form a more stable tertiary carbocation $(CH_3)_2C^+CH_3$.
Ethylbenzene is an ortho/para directing group. Due to steric hindrance at the ortho position,the bulky tert-butyl group attacks the para position of ethylbenzene to form $1$-ethyl-$4$-tert-butylbenzene as the major product.

(B) $(R)-2-$bromobutane is a secondary $(2^{\circ})$ alkyl halide.
When treated with $aq. \, NaOH$,it undergoes a nucleophilic substitution reaction.
Since it is a secondary halide,the reaction proceeds primarily via the $S_N1$ mechanism,which involves the formation of a planar carbocation intermediate.
The nucleophile $(OH^-)$ can attack the carbocation from either side with equal probability,leading to the formation of both $(R)$ and $(S)$ enantiomers of $2-$butanol in equal amounts.
This process is known as racemisation,and the resulting product is a racemic mixture.

(A) The reaction between a haloalkane and $AgNO_3$ proceeds via the formation of a carbocation intermediate. The more stable the resulting carbocation,the more readily the reaction occurs to yield a precipitate of $AgCl$.
In the given cyclic ether structure,the $Cl^a$ atom is attached to the carbon atom adjacent to the oxygen atom.
Upon the removal of $Cl^a$ as a chloride ion,the resulting carbocation is an oxocarbenium ion,which is significantly stabilized by the resonance effect of the lone pair of electrons on the adjacent oxygen atom $(O-C^+ \leftrightarrow O^+=C)$.
This resonance stabilization makes the formation of this specific carbocation much faster and more favorable compared to the other positions $(Cl^b, Cl^c, Cl^d)$,which do not benefit from such direct resonance stabilization by the oxygen atom.
Therefore,$Cl^a$ reacts most readily with $AgNO_3$.

(C) The reaction involves an $S_{N}2$ mechanism. The nucleophile $CN^{-}$ attacks the primary alkyl chloride carbon,which is the most reactive site for $S_{N}2$ substitution in the given molecule. The $Cl^{-}$ ion acts as the leaving group. The $Br$ atom attached to the $sp^{2}$ carbon of the alkene is not easily substituted under these conditions. Therefore,the $CN$ group replaces the $Cl$ atom,resulting in product $(iii)$.

(C) The rate of solvolysis is directly proportional to the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
In option $C$,the compound is $Me-CH(Br)-CH(CH_3)-Ph$. Upon the loss of $Br^-$,it forms a secondary carbocation adjacent to a phenyl group and a methyl group.
This carbocation is stabilized by the resonance effect of the phenyl group and the inductive effect $(+I)$ of the methyl group,making it the most stable among the given options.
Therefore,the compound in option $C$ undergoes the fastest solvolysis.

(A) The compound $X$ reacts with alcoholic $AgNO_3$ to give a white precipitate,indicating the presence of a reactive halogen atom (like a benzylic bromide).
Upon oxidation,$X$ yields an acid with the formula $C_8H_6O_4$. This acid is phthalic acid $(benzene-1,2-dicarboxylic \ acid)$,which is known to form a cyclic anhydride (phthalic anhydride) upon heating.
For $X$ to produce phthalic acid upon oxidation,it must be an ortho-substituted benzene derivative with two side chains that can be oxidized to $-COOH$ groups. The structure corresponding to this is $o-xylene$ derivative,specifically $1-(bromomethyl)-2-methylbenzene$.
Thus,the compound $X$ is $o-methylbenzyl \ bromide$ (or $1-(bromomethyl)-2-methylbenzene$).

(C) The hydrolysis of alkyl chloride $(R-Cl)$ is slow because $Cl^-$ is a poor nucleophile.
When $NaI$ is added,$I^-$ acts as a strong nucleophile and attacks the alkyl chloride to form an alkyl iodide $(R-I)$ via an $S_N2$ reaction.
Since $I^-$ is a much better leaving group than $Cl^-$,the subsequent hydrolysis of $R-I$ to form the alcohol $(R-OH)$ occurs much faster.
Thus,$I^-$ acts as a catalyst by converting the alkyl chloride into a more reactive alkyl iodide.
Both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

(C) The reaction of organic halides with $AgNO_3$ proceeds via the formation of carbocations.
Only those chlorine atoms that can form stable carbocations (like benzylic or tertiary) or are easily ionizable will react with $AgNO_3$ to precipitate $AgCl$.
$1$. The chlorine atom on the benzene ring (aryl chloride) is not reactive towards $AgNO_3$ due to partial double bond character.
$2$. The vinylic chlorine atom is also not reactive.
$3$. The secondary benzylic chlorine atom forms a stable benzylic carbocation,which reacts with $AgNO_3$ to give $1$ mole of $AgCl$.
$4$. The tertiary alkyl chlorine atom forms a stable tertiary carbocation,which reacts with $AgNO_3$ to give $1$ mole of $AgCl$.
Thus,a total of $2$ moles of $AgCl$ are formed.
Therefore,$X = 2$.

(C) chiral carbon atom is one that is bonded to four different groups.
Let us analyze the structures:
$A$. $2-$Bromo$-1-$deuterobutane: $CH_3-CH_2-CH(Br)-CH_2D$. Here,$C-2$ is chiral (bonded to $H, Br, CH_3, CH_2D$). Only one chiral center.
$B$. $2-$Bromo$-2-$deuterobutane: $CH_3-CH_2-C(Br)(D)-CH_3$. Here,$C-2$ is chiral (bonded to $Br, D, CH_3, CH_2CH_3$). Only one chiral center.
$C$. $2-$Bromo$-3-$deuterobutane: $CH_3-CH(Br)-CH(D)-CH_3$. Here,$C-2$ is bonded to $H, Br, CH_3, CH(D)CH_3$ (chiral). $C-3$ is bonded to $H, D, CH_3, CH(Br)CH_3$ (chiral). This molecule has two chiral carbon atoms.
$D$. $2-$Bromo$-1-$deutero$-2-$methylpropane: $(CH_3)_2C(Br)-CH_2D$. No chiral carbon atom.
Therefore,the correct option is $C$.

(B) Analysis of the given statements:
$(A)$ Boiling point increases with an increase in the size of the alkyl group. Thus,$CH_3CH_2Cl < CH_3CH_2CH_2Cl < CH_3CH_2CH_2CH_2Cl$ is correct.
$(B)$ Density increases with the atomic mass of the halogen atom. The order of atomic mass is $Cl < Br < I$. Therefore,the correct order should be $CH_3CH_2Cl < CH_3CH_2Br < CH_3CH_2I$. Statement $(B)$ is incorrect.
$(C)$ Boiling point of isomeric alkyl halides decreases with an increase in branching. Thus,$CH_3CH_2CH_2Br > (CH_3)_2CHBr > (CH_3)_3CBr$. Statement $(C)$ is incorrect.
$(D)$ Density increases with the atomic mass of the halogen atom. Comparing the halogen atoms in the given compounds,the order of density is $CH_3CH_2CH_2Br < CH_3CH(Br)Cl < CH_3CH(I)Br$. Statement $(D)$ is incorrect.
$(E)$ Boiling point of isomeric alkyl halides decreases with an increase in branching. Thus,$n$-butyl chloride $(CH_3CH_2CH_2CH_2Cl)$ > isobutyl chloride $((CH_3)_2CHCH_2Cl)$ > tert-butyl chloride $((CH_3)_3CCl)$. Statement $(E)$ is correct.
Therefore,only statements $(A)$ and $(E)$ are correct. However,based on the provided options,the most appropriate choice is $(B)$ $A, B$ and $E$ only,assuming a potential typo in the question's provided options or the intended order in $(B)$.

(C) Dehydrohalogenation involves the removal of $H$ and $X$ from adjacent carbons to form an alkene. We analyze each option:
$A$) $1-$Bromo$-2-$methylbutane $(CH_3-CH_2-CH(CH_3)-CH_2Br)$: Only one $\beta$-hydrogen is available,yielding $2-$methylbut$-1-$ene ($1$ product).
$B$) $2-$Bromopropane $(CH_3-CH(Br)-CH_3)$: Only one type of $\beta$-hydrogen is available,yielding propene ($1$ product).
$C$) $2-$Bromopentane $(CH_3-CH_2-CH_2-CH(Br)-CH_3)$: Two types of $\beta$-hydrogens are available. Removal from $C_1$ gives pent$-1-$ene. Removal from $C_3$ gives pent$-2-$ene,which exists as $cis$ and $trans$ isomers. Total = $3$ products.
$D$) $2-$Bromo$-3,3-$dimethylpentane: Only one $\beta$-hydrogen is available at $C_3$ (if possible) or $C_1$,but given the structure,it yields only $1$ alkene product.
Thus,$2-$Bromopentane gives the maximum number of isomeric alkenes.

(C) The rate of $S_N 1$ reaction depends upon the stability of the carbocation intermediate formed after the departure of the leaving group $(Cl^-)$.
Electron-donating groups (like $-OCH_3$) increase the stability of the carbocation,while electron-withdrawing groups (like $-NO_2$ and $-Cl$) decrease its stability.
The stability order of the carbocations is:
$p-methoxybenzyl$ carbocation $(b)$ $>$ $benzyl$ carbocation $(d)$ $>$ $p-chlorobenzyl$ carbocation $(c)$ $>$ $p-nitrobenzyl$ carbocation $(a)$.
Therefore,the decreasing order of reactivity towards $S_N 1$ reaction is:
$(b) > (d) > (c) > (a)$.

(B) In option $B$,the reaction involves a primary alkyl halide $(CH_3)_3CCH_2Br$ with alcoholic $KOH$. Alcoholic $KOH$ acts as a strong base and promotes an elimination reaction ($E2$ mechanism) to form an alkene. However,the reaction shown is a substitution reaction,which is incorrect for alcoholic $KOH$. Aqueous $KOH$ would be required for substitution to form an alcohol.

(C) The given reaction is a nucleophilic substitution reaction of a tertiary alkyl halide,$(C_6H_5)_3C-Cl$,which proceeds via an $SN_1$ mechanism.
In an $SN_1$ reaction,the rate-determining step is the formation of the carbocation,which depends only on the concentration of the alkyl halide.
The rate law for this reaction is: $\text{Rate} = k[(C_6H_5)_3C-Cl]$.
Since the rate is directly proportional to the concentration of the alkyl halide,the graph of rate versus $[(C_6H_5)_3C-Cl]$ should be a straight line passing through the origin,as shown in option $C$.

(C) The reactivity of alkyl halides towards $S_{N}2$ reactions is primarily governed by steric hindrance. The order of reactivity is $Primary > Secondary > Tertiary$.
$(I)$ $CH_{3}CH_{2}CH_{2}CH_{2}Cl$ is a primary alkyl halide with minimal steric hindrance.
$(III)$ $(CH_{3})_{2}CHCH_{2}Cl$ is a primary alkyl halide but has more steric hindrance than $(I)$ due to the branching at the $\beta$-carbon.
$(II)$ $CH_{3}CH_{2}CH(Cl)CH_{3}$ is a secondary alkyl halide.
$(IV)$ $(CH_{3})_{3}CCl$ is a tertiary alkyl halide,which is the least reactive towards $S_{N}2$ due to maximum steric hindrance.
Therefore,the correct order is $I > III > II > IV$.

(A) This is the $Finkelstein$ reaction,which proceeds via an $S_N2$ mechanism.
In the $S_N2$ transition state,the negative charge is dispersed over the incoming nucleophile $(I^-)$ and the leaving group $(Br^-)$. Therefore,the transition state is less polar than the localized anion $(I^-)$,which facilitates the reaction in a polar aprotic solvent like acetone.
Acetic acid is a polar protic solvent,which solvates the nucleophile $(I^-)$ through hydrogen bonding,thereby decreasing its nucleophilicity and hindering the $S_N2$ reaction.
Acetone is a polar aprotic solvent; it does not solvate the anion effectively,keeping the nucleophile reactive.
$NaBr$ is insoluble in acetone and precipitates out,which drives the reaction forward according to $Le$ $Chatelier's$ principle,preventing $Br^-$ from acting as a competing nucleophile.
Thus,the correct statement is that the transition state is less polar than the localized anion.

(B) The rate of $SN1$ reaction depends on the stability of the carbocation formed after the departure of the leaving group.
$1$. In compound $A$,$Br_{(a)}$ forms a benzylic carbocation,which is highly stabilized by resonance.
$2$. In compound $B$,$I_{(a)}$ forms an allylic carbocation,which is also resonance stabilized but less than the benzylic one.
$3$. In compound $C$,$Br_{(a)}$ forms a tertiary carbocation.
$4$. In compound $D$,$Br_{(a)}$ forms a tertiary carbocation.
Comparing the stability,the benzylic carbocation formed by $Br_{(a)}$ in compound $A$ is the most stable. Therefore,$A-Br_{(a)}$ is the most reactive towards $SN1$ reaction.

(D) The reaction is an intramolecular Friedel-Crafts alkylation.
$1$. The Lewis acid $AlCl_3$ reacts with the alkyl chloride to form a carbocation intermediate at the secondary carbon.
$2$. The benzene ring is substituted with a $-OCH_3$ group (activating,ortho/para directing) and a $-NO_2$ group (deactivating,meta directing).
$3$. The $-OCH_3$ group is a stronger directing group than the $-NO_2$ group.
$4$. The ortho position relative to the $-OCH_3$ group is sterically hindered,so the cyclization occurs at the para position relative to the $-OCH_3$ group (which is ortho to the $-NO_2$ group).
$5$. The resulting product is a substituted tetralin derivative as shown in option $D$.

(C) In reaction $(I)$,the substrate has an electron-donating group $(-OMe)$ at the para position,which stabilizes the carbocation intermediate. This facilitates the $S_N1$ mechanism,which is a $1^{st}$ order reaction.
In reaction $(II)$,the substrate has an electron-withdrawing group $(-NO_2)$ at the para position,which destabilizes the carbocation intermediate,making the $S_N1$ pathway unfavorable. Instead,it proceeds via the $S_N2$ mechanism,which is a $2^{nd}$ order reaction.
Therefore,reaction $(I)$ is of $1^{st}$ order and reaction $(II)$ is of $2^{nd}$ order.

(A) The reaction with $AgNO_3$ involves the formation of a carbocation intermediate. Compounds that form stable carbocations react readily to produce a precipitate of $AgBr$.
$A$. $1$-Bromocyclohexene: The bromine atom is attached to an $sp^2$ hybridized carbon atom of a double bond (vinylic halide). The resulting vinylic carbocation is highly unstable due to the high electronegativity of the $sp^2$ carbon and the inability to stabilize the positive charge through resonance. Therefore,it does not react with $AgNO_3$ to form a precipitate.
$B$. Benzyl bromide: Forms a resonance-stabilized benzyl carbocation,which reacts readily.
$C$. $3$-Bromocyclopropene: Forms a cyclopropenyl cation,which is aromatic ($2\pi$ electrons) and highly stable.
$D$. $3$-Phenyl-$3$-bromoprop-$1$-ene: Forms a resonance-stabilized carbocation (allylic and benzylic stabilization).
Thus,$1$-Bromocyclohexene is the correct answer.

(A) Reaction with $C_2H_5O^-$: $C_2H_5O^-$ is a strong nucleophile. $2-$Methylpropyl bromide is a primary alkyl halide. Therefore,it undergoes an $S_N2$ reaction to form isobutyl ethyl ether $(A)$.
Reaction with $C_2H_5OH$: $C_2H_5OH$ is a weak nucleophile. The reaction proceeds via an $S_N1$ mechanism. The primary carbocation formed initially undergoes a $1,2-H$ shift to form a more stable tertiary carbocation,which then reacts with the nucleophile to form tert-butyl ethyl ether $(B)$.

(B) $1$. $CH_3-CH_2-CH_2-Br + KOH \text{ (alc)} \rightarrow CH_3-CH=CH_2$ (Alkene,$III$)
$2$. $CH_3-CH_2-CH_2-Br + KCN \text{ (alc)} \rightarrow CH_3-CH_2-CH_2-CN$ (Nitrile,$I$)
$3$. $CH_3-CH_2-CH_2-Br + AgNO_2 \rightarrow CH_3-CH_2-CH_2-NO_2 + AgBr$ (Nitroalkane,$IV$)
$4$. $CH_3-CH_2-CH_2-Br + CH_3COOAg \rightarrow CH_3COOCH_2-CH_2-CH_3 + AgBr$ (Ester,$II$)
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) The reaction sequence is as follows:
$1$. Reduction of acetaldehyde $(CH_3CHO)$ with $LiAlH_4$ followed by $H_3O^+$ gives ethanol $(CH_3CH_2OH)$ as $[A]$.
$2$. Dehydration of ethanol with $H_2SO_4$ at high temperature gives ethene $(CH_2=CH_2)$ as $[B]$.
$3$. Addition of $HBr$ to ethene gives bromoethane $(CH_3CH_2Br)$ as $[C]$.
$4$. The reaction of bromoethane with bromobenzene in the presence of $Na$ and dry ether is a Wurtz-Fittig reaction,which yields ethylbenzene as the final product $[D]$.

(D) 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 from the side opposite to the leaving group,resulting in the inversion of configuration (Walden inversion).

(D) $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity.
$(A)$ $H_2C=CH-CH_2Cl$ forms an allyl carbocation $(H_2C=CH-CH_2^+)$,which is resonance stabilized,so it shows $S_{N}1$ reaction.
$(B)$ $CH_3-CH=CHCl$ is a vinylic halide. The carbocation formed $(CH_3-CH=CH^+)$ is highly unstable due to the positive charge on an $sp$-hybridized carbon atom,so it does not show $S_{N}1$ reaction.
$(C)$ $C_6H_5CH_2Cl$ (benzyl chloride) forms a benzyl carbocation $(C_6H_5CH_2^+)$,which is highly resonance stabilized,so it shows $S_{N}1$ reaction.
$(D)$ $(CH_3)_2CHCl$ (isopropyl chloride) forms a secondary carbocation $((CH_3)_2CH^+)$,which is relatively stable due to inductive effect and hyperconjugation,so it shows $S_{N}1$ reaction.
Therefore,only $(B)$ will not show $S_{N}1$ reaction.

(D) $AgCN$ is a covalent compound. In $AgCN$,the carbon atom is not free to donate its lone pair of electrons because it is involved in a covalent bond with $Ag$. However,the nitrogen atom has a lone pair of electrons available for donation. Therefore,the nucleophilic attack occurs through the nitrogen atom,leading to the formation of alkyl isocyanide $(R-NC)$.

(D) Statement-$I$: The rate of $S_{N}2$ reaction is given by $Rate = k[Substrate][Nu^{-}]$.
$S_{N}2$ reactions are favored by high concentrations of strong nucleophiles and minimal steric hindrance in the substrate.
Statement-$II$: Solvolysis reactions,such as the reaction of an alkyl halide with a large excess of a polar protic solvent like ethanol,proceed via the $S_{N}1$ mechanism because the solvent acts as both the nucleophile and the ionizing medium.
Therefore,both statements are correct.

(A) The free radical chlorination of $2$-chlorobutane $(CH_3CHClCH_2CH_3)$ can occur at different positions:
$1$. Substitution at $C_1$: $ClCH_2CHClCH_2CH_3$ ($1$-chloro$-2-$chlorobutane). This has one chiral center at $C_2$,so it exists as $2$ enantiomers.
$2$. Substitution at $C_2$: $CH_3CCl_2CH_2CH_3$ ($2$,$2$-dichlorobutane). This is achiral (optically inactive).
$3$. Substitution at $C_3$: $CH_3CHClCHClCH_3$ ($2$,$3$-dichlorobutane). This has two chiral centers. The meso form is optically inactive,while the $(2R, 3R)$ and $(2S, 3S)$ forms are optically active ($2$ isomers).
$4$. Substitution at $C_4$: $CH_3CHClCH_2CH_2Cl$ ($1$,$3$-dichloro$-2-$chlorobutane is incorrect,it is $1,3-$dichlorobutane or $2,4-$dichlorobutane depending on numbering). The product is $CH_3CHClCH_2CH_2Cl$ ($1$,$3$-dichlorobutane). This has a chiral center at $C_2$,so it exists as $2$ enantiomers.
Total optically active isomers = $2$ (from $1-$chloro$-2-$chlorobutane) + $2$ (from $2,3-$dichlorobutane) + $2$ (from $1,3-$dichlorobutane) = $6$.

(A) $KCN$ is an ionic compound and provides cyanide ions $(CN^-)$ in solution. Since carbon is more nucleophilic,it attacks the alkyl halide to form alkyl cyanide $(R-CN)$ as the major product.
$AgCN$ is predominantly covalent in nature. In $AgCN$,the carbon atom is linked to silver,making the nitrogen atom the only site available for nucleophilic attack. Therefore,it forms alkyl isocyanide $(R-NC)$ as the major product.
Thus,Assertion $(A)$ is correct,but Reason $(R)$ is incorrect because $AgCN$ is covalent,not ionic.

(B) The reaction of toluene with $Cl_2$ in the presence of light $(hv)$ is a free radical substitution reaction.
It proceeds as follows:
$C_6H_5CH_3 + 2Cl_2 \xrightarrow{hv} C_6H_5CHCl_2 + 2HCl$
Here,$A$ is benzal chloride $(C_6H_5CHCl_2)$.
Next,the hydrolysis of benzal chloride with water at $373 \ K$ yields benzaldehyde:
$C_6H_5CHCl_2 + H_2O \xrightarrow{373 \ K} C_6H_5CHO + 2HCl$
Thus,$A$ is benzal chloride and $B$ is benzaldehyde.

(C) $1$. The reaction of $(A)$ ($4$-bromophenethyl bromide) with alcoholic $NaOH$ proceeds via an $E2$ elimination mechanism to form $(B)$ ($4$-bromostyrene).
$2$. The reaction of $(B)$ with $HBr$ in ether follows an electrophilic addition mechanism (Markovnikov's rule) to form $(C)$ ($1$-($4$-bromophenyl)$-1-$bromoethane).
$3$. Comparing $(A)$ and $(C)$:
$(A)$ is $1-$bromo$-2-$($4$-bromophenyl)ethane.
$(C)$ is $1-$bromo$-1-$($4$-bromophenyl)ethane.
These are position isomers because the bromine atom on the side chain is at different positions relative to the benzene ring.

(A) The reaction is a dehydrohalogenation reaction using alcoholic $KOH$ (an elimination reaction,$E2$).
The substrate is $1\text{-phenyl-2-bromo-3-methylbutane}$.
Elimination of $HBr$ occurs from the $\beta$-carbon atoms adjacent to the carbon bearing the $Br$ atom.
There are two possible $\beta$-hydrogens:
$1$. From $C-1$ (benzylic position),which leads to a conjugated alkene $(1\text{-phenyl-3-methylbut-1-ene})$.
$2$. From $C-3$,which leads to a less substituted alkene $(1\text{-phenyl-3-methylbut-2-ene})$.
According to Saytzeff's rule,the more substituted and conjugated alkene is the major product.
Therefore,the major product $P$ is $1\text{-phenyl-3-methylbut-1-ene}$.

(C) The reaction involves a tertiary alkyl halide ($1$-bromo-$1$-methylcyclopentane) reacting with a strong base $(\text{OH}^-)$ in an alcoholic solvent $(C_2H_5OH)$. This condition favors an $E2$ elimination reaction. The base abstracts a $\beta$-hydrogen atom,leading to the formation of the most stable alkene according to Zaitsev's rule. The most substituted alkene,$1$-methylcyclopent-$1$-ene,is the major product.

(B) The reaction proceeds via an $S_N1$ mechanism because the substrate is a secondary alkyl halide and the solvent is polar protic $(H_2O)$.
$1$. The first step is the formation of a secondary carbocation by the loss of the $Cl^-$ ion.
$2$. This secondary carbocation undergoes a $1,2-H^-$ shift to form a more stable tertiary carbocation.
$3$. The nucleophile ($OH^-$ from $NaOH$) then attacks the tertiary carbocation to form the major product,$1-propylcyclohexanol$.

(C) The rate of $S_{N}2$ reaction depends on the steric hindrance around the electrophilic carbon atom. The order of reactivity is: $Methyl \ halide > 1^{\circ} \ halide > 2^{\circ} \ halide > 3^{\circ} \ halide$.
In the given options:
$(A)$ is a $3^{\circ}$ alkyl halide.
$(B)$ is a $3^{\circ}$ alkyl halide.
$(C)$ is a $1^{\circ}$ alkyl halide (cyclobutyl-methyl bromide).
$(D)$ is a $2^{\circ}$ alkyl halide.
Since $1^{\circ}$ alkyl halides have the least steric hindrance,they undergo $S_{N}2$ reactions the fastest. Therefore,cyclobutyl-methyl bromide will undergo the fastest $S_{N}2$ reaction.

(C) $S_N2$ reactions proceed via a backside attack,leading to the inversion of configuration (Walden inversion). Since only one specific stereoisomer is formed,they are stereospecific.
$S_N1$ reactions proceed via the formation of a planar carbocation intermediate. The nucleophile can attack from either side with equal probability,leading to the formation of a racemic mixture (a mixture of both enantiomers).
Therefore,both Statement $I$ and Statement $II$ are true.

(A) $1$. The reaction of $1$-bromo-$3$-chlorocyclobutane with $Na/Et_2O$ is a Wurtz reaction. Since the $Br$ atom is a better leaving group than $Cl$,the coupling occurs at the $Br$ position,forming $1,1'$-dichloro-$3,3'$-bicyclobutane (compound $A$).
$2$. Compound $A$ reacts with $(i) Mg/Et_2O$ followed by $(ii) D_2O$. The $Mg$ forms a Grignard reagent with the $Cl$ atoms,which is then quenched by $D_2O$ to replace $Cl$ with $D$,yielding $B$ as $1,1'$-dideuterio-$3,3'$-bicyclobutane.
$3$. Compound $A$ reacts with $CoF_2$ (a fluorinating agent used in the Swarts reaction) to replace the $Cl$ atoms with $F$ atoms,yielding $C$ as $1,1'$-difluoro-$3,3'$-bicyclobutane.
$4$. Therefore,the products $B$ and $C$ are $1,1'$-dideuterio-$3,3'$-bicyclobutane and $1,1'$-difluoro-$3,3'$-bicyclobutane respectively.

(C) In an $S_N2$ reaction,the nucleophile attacks from the backside of the leaving group,leading to a trigonal bipyramidal transition state.
In the case of benzyl bromide $(C_6H_5CH_2Br)$,the transition state involves a partially bonded unhybridized $p$-orbital at the benzylic carbon.
This $p$-orbital is stabilized by conjugation with the $\pi$-electron system of the phenyl ring,which lowers the activation energy of the reaction.
Therefore,the $S_N2$ reaction of $C_6H_5CH_2Br$ occurs more readily than that of $CH_3CH_2Br$.
Both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(A) The reaction involves the participation of the phenyl ring in a Neighboring Group Participation $(NGP)$ mechanism.
$1$. The $\pi$-electrons of the phenyl ring assist in the departure of the leaving group $(Br^{-})$.
$2$. This leads to the formation of a phenonium ion intermediate.
$3$. The nucleophile $(CN^{-})$ then attacks the more electrophilic carbon of the phenonium ion.
$4$. This process results in the substitution of the $Br$ atom by the $CN$ group,leading to the product shown in option $A$.

(B) The iodoform test is given by compounds containing the $CH_3CO-$ group or compounds that can be oxidized to this group.
Treatment of $CH_3-CH_2-CCl_2-CH_3$ with aqueous $KOH$ leads to the hydrolysis of the gem-dichloride to a gem-diol $(CH_3-CH_2-C(OH)_2-CH_3)$,which is unstable and loses a water molecule to form butan$-2-$one $(CH_3-CH_2-CO-CH_3)$.
Butan$-2-$one contains the $CH_3CO-$ group,which reacts with potassium hypoiodite $(KOI)$ to form iodoform $(CHI_3)$,a yellow precipitate.
Thus,the correct compound is $CH_3-CH_2-CCl_2-CH_3$.

(C) The rate of $S_N 1$ reaction depends on the stability of the carbocation intermediate formed in the rate-determining step.
$1$. $1$-Phenylethyl bromide forms a secondary benzylic carbocation,which is highly stabilized by resonance with the benzene ring.
$2$. Bromocyclohexane forms a secondary alkyl carbocation.
$3$. (Cyclohexylmethyl) bromide forms a primary alkyl carbocation.
$4$. Bromobenzene does not undergo $S_N 1$ reaction easily as the $C-Br$ bond has partial double bond character due to resonance.
Since the secondary benzylic carbocation is the most stable among the options,$1$-phenylethyl bromide reacts the fastest.

(B) Step $1$: Reaction of $2$-methylcyclohexanol with $PBr_3$ leads to the formation of $1$-bromo-$2$-methylcyclohexane $(A)$ via an $S_N2$ mechanism.
Step $2$: Treatment of $1$-bromo-$2$-methylcyclohexane $(A)$ with alcoholic $KOH$ (a strong base) results in dehydrohalogenation via an $E2$ mechanism.
Step $3$: According to Zaitsev's rule,the more substituted alkene is the major product. The base abstracts a proton from the $C_2$ position,leading to the formation of $1$-methylcyclohexene $(B)$ as the major product.

(B) . $CH_3-CHBr-CD_3$ undergoes $E2$ elimination with alc. $KOH$ to form $CH_2=CH-CD_3$ $(A-Q)$.
$B$. $Ph-CHBr-CH_3$ shows a primary kinetic isotope effect $(k_H/k_D > 1)$ in $E2$ reactions,so it reacts faster than the deuterated analog $(B-Q)$.
$C$. $Ph-CH_2-CH_2Br$ undergoes $E1cb$ mechanism due to the acidic $\beta$-hydrogens stabilized by the phenyl ring,and it follows first-order kinetics $(C-R,S)$.
$D$. $PhCH_2CH_2Br$ and $PhCD_2CH_2Br$ reacting at the same rate suggests the rate-determining step does not involve $C$-$H$/$C$-$D$ bond breaking,characteristic of $E1$ reactions $(D-P,S)$.