Properties of Haloalkanes Questions in English

(D) The reaction is an elimination reaction (dehydrohalogenation) of $1$-chloro-$2$-methylcyclohexane in the presence of a base ($C_2H_5O^-$ from $C_2H_5OH$) and heat.
This reaction proceeds via an $E2$ mechanism.
The possible alkenes formed by removing a hydrogen atom from the adjacent carbons are:
$1$. If the hydrogen is removed from the $C_3$ position,we get $3$-methylcyclohexene.
$2$. If the hydrogen is removed from the $C_1$ position (where the methyl group is attached),we get $1$-methylcyclohexene.
$3$. If the hydrogen is removed from the $C_2$ position (the carbon with the methyl group),it is not possible to form methylenecyclohexane because the methyl group is already attached to the ring carbon,and the double bond would require five bonds on that carbon.
Therefore,methylenecyclohexane is not a product of this reaction.

(A) The given reaction is a nucleophilic substitution reaction $(S_N2)$.
In the reactant,$4$-bromobenzyl chloride,there are two halogen atoms: $Br$ attached to the benzene ring and $Cl$ attached to the side chain $(CH_2Cl)$.
$NaCN$ acts as a nucleophile $(CN^-)$. The $Cl$ atom in the $CH_2Cl$ group is part of an alkyl halide (specifically a primary benzylic halide),which is highly reactive towards nucleophilic substitution.
In contrast,the $Br$ atom is directly attached to the benzene ring (aryl halide),which is inert towards nucleophilic substitution under these conditions due to the partial double-bond character of the $C-X$ bond.
Therefore,the $CN^-$ nucleophile will selectively replace the $Cl$ atom in the $CH_2Cl$ group to form $4$-bromophenylacetonitrile $(Br-C_6H_4-CH_2CN)$.

(B) The rate of $S_N1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group.
$I$ and $II$ are $p$-methoxybenzyl halides. The $-OCH_3$ group at the para position provides strong $+M$ (mesomeric) effect,which stabilizes the benzylic carbocation significantly.
Comparing $I$ ($p$-methoxybenzyl bromide) and $II$ ($p$-methoxybenzyl chloride),the bromide ion $(Br^-)$ is a better leaving group than the chloride ion $(Cl^-)$,so $I$ reacts faster than $II$.
$IV$ is benzyl chloride,which forms a stable benzylic carbocation but lacks the additional $+M$ stabilization from the $-OCH_3$ group,so it is less reactive than $I$ and $II$.
$III$ is $p$-chloroanisole,where the $Cl$ atom is directly attached to the benzene ring. $S_N1$ reaction at this position would form an unstable aryl cation,making it the least reactive.
Thus,the order is $I > II > IV > III$.

(A) The substrate is $1$-bromo-$1,2$-dimethylcyclohexane. When treated with a bulky base like potassium $tert$-butoxide $(t-BuOK)$,the $E_2$ elimination reaction occurs.
$t-BuOK$ is a sterically hindered base,which prefers to abstract a proton from the less sterically hindered position (Hofmann product).
In this molecule,there are two types of $\beta$-hydrogens available for elimination:
$1$. Protons on the $CH_2$ group of the ring (leading to the more substituted,Zaitsev product,$1,2$-dimethylcyclohex-$1$-ene).
$2$. Protons on the exocyclic $CH_3$ group attached to the $C_1$ position or the $CH_2$ group adjacent to $C_1$ (leading to the less substituted,Hofmann product,$1$-methyl-$2$-methylenecyclohexane).
Due to the steric bulk of the $t-BuOK$ base,it preferentially abstracts the more accessible proton from the $CH_2$ group adjacent to the $C_1$ position to form the less substituted alkene,which is $1$-methyl-$2$-methylenecyclohexane.

(C) The $SN^1$ reaction mechanism proceeds via the formation of a carbocation intermediate. The rate of $SN^1$ reaction depends on the stability of the carbocation formed.
$A$. $CH_3-Br$ forms a primary methyl carbocation,which is highly unstable and does not undergo $SN^1$.
$B$. $Ph-Br$ (Bromobenzene) does not undergo $SN^1$ because the $C-Br$ bond has partial double bond character due to resonance,and the phenyl cation is extremely unstable.
$C$. $CH_2=CH-CH_2-Br$ forms an allyl carbocation $(CH_2=CH-CH_2^+)$,which is resonance-stabilized. This reaction can proceed via $SN^1$ in the presence of a polar protic solvent like $H_2O$.
$D$. $1-$bromobicyclo[$2.2$.$1$]heptane cannot undergo $SN^1$ because the bridgehead carbon cannot achieve the planar geometry required for a carbocation intermediate (Bredt's rule).
Therefore,the correct option is $C$.

(B) The reaction of $1\text{-methylcyclopentene}$ with $\text{HBr}$ in the presence of peroxide follows the anti-Markovnikov addition mechanism (Kharasch effect or peroxide effect).
In this mechanism,the bromine atom $(\text{Br}^\bullet)$ adds to the carbon atom of the double bond that has more hydrogen atoms (or less substituted carbon),while the hydrogen atom adds to the other carbon.
For $1\text{-methylcyclopentene}$,the double bond is between $C_1$ (substituted with a methyl group) and $C_2$ (which has one hydrogen atom).
According to the anti-Markovnikov rule,the $\text{Br}$ atom will attach to the $C_2$ position,resulting in $2\text{-bromo-1-methylcyclopentane}$ as the major product.

(D) The correct answer is $(d)$.
Polar aprotic solvents are solvents that do not contain a hydrogen atom attached to an electronegative atom (like $O$ or $N$).
$DMSO$ (Dimethyl sulfoxide) is a well-known polar aprotic solvent.
Crown ethers are also considered polar aprotic solvents as they can solvate cations effectively without having acidic protons.
$DMG$ (Dimethylglyoxime) is also used in specific contexts as a polar aprotic solvent.
Therefore,all the given options are examples of polar aprotic solvents.

(C) The dipole moments of the given compounds are as follows:
$1$. $CCl_4$ is a symmetrical tetrahedral molecule,so its net dipole moment is $0 \ D$.
$2$. $cis-CHCl=CHCl$ has a significant dipole moment due to the polar $C-Cl$ bonds on the same side,approximately $1.90 \ D$.
$3$. $CH_2Cl_2$ has a dipole moment of approximately $1.60 \ D$.
$4$. $CHCl_3$ has a dipole moment of approximately $1.03 \ D$.
Comparing these values,the order is $cis-CHCl=CHCl > CH_2Cl_2 > CHCl_3 > CCl_4$.

(B) molecule is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $1,3-$dibromo$-1-$chloropropane $(CH_2Br-CH_2-CH(Cl)Br)$,the carbon atom at position $1$ is bonded to a hydrogen atom $(H)$,a chlorine atom $(Cl)$,a bromine atom $(Br)$,and a $2-$bromoethyl group $(-CH_2-CH_2-Br)$.
Since all four groups attached to the $C-1$ carbon are different,it is a chiral center.
Therefore,$1,3-$dibromo$-1-$chloropropane is chiral.

(D) Grignard reagents $(RMgX)$ are highly reactive towards compounds containing active hydrogen atoms (like alcohols,water,amines) and electrophilic carbonyl groups (like ketones and esters).
$ (a) $ Acetone contains a carbonyl group and will react with $PhMgBr$.
$ (b) $ Ethyl acetate is an ester and will react with $PhMgBr$.
$ (c) $ Absolute alcohol contains an acidic hydroxyl group and will react with $PhMgBr$.
$ (d) $ Benzene is an inert hydrocarbon and does not contain any active hydrogen or reactive functional groups,making it a suitable solvent for the reaction.

(B) The reactant is phthaloyl chloride $(C_6H_4(COCl)_2)$.
Each acid chloride group $(-COCl)$ reacts with $2$ moles of Grignard reagent $(CH_3MgBr)$ to form a tertiary alcohol.
Specifically,the first mole of $CH_3MgBr$ attacks the carbonyl carbon to form a ketone intermediate,and the second mole attacks the ketone to form the tertiary alcohol after acidic workup $(H^+)$.
Since there are two $-COCl$ groups in the molecule,a total of $2 \times 2 = 4$ moles of $CH_3MgBr$ are consumed.
Therefore,the correct option is $B$.

(D) Grignard reagents $(RMgX)$ are highly reactive organometallic compounds. They require a polar aprotic solvent that can act as a Lewis base to stabilize the magnesium atom through coordination.
$A$,$B$,and $C$ are ethers,which are polar aprotic solvents capable of coordinating with the magnesium center,thus stabilizing the Grignard reagent.
$D$ (Cyclohexane) is a non-polar hydrocarbon solvent. It lacks the lone pairs of electrons necessary to coordinate with and stabilize the magnesium atom in the Grignard reagent. Therefore,it is not a suitable solvent for the Grignard reaction.

(B) The reaction involves the use of Gilman reagent,$(n-Bu)_2CuLi$,which is an organocopper reagent. Gilman reagents are known for their ability to perform nucleophilic substitution reactions with primary alkyl halides,replacing the halogen atom with an alkyl group. In this reaction,the $n-butyl$ group from the Gilman reagent attacks the benzylic carbon atom of $3,5-dimethoxybenzyl$ bromide,displacing the bromide ion. This results in the formation of $1-(3,5-dimethoxyphenyl)pentane$.

(D) Grignard reagents $(RMgX)$ are strong bases and react rapidly with compounds containing acidic hydrogen atoms (such as $-OH$,$-SH$,and $-NH_2$ groups).
In options $(a)$,$(b)$,and $(c)$,the molecules contain acidic hydrogen atoms attached to oxygen,sulfur,or nitrogen,which would immediately protonate the Grignard reagent.
In option $(d)$,the nitrogen atom is part of a tertiary amine group $(-N(CH_3)_2)$,which does not have an acidic hydrogen.
Therefore,this Grignard reagent can be prepared.

(A) $1$. The reaction of $sec$-butyl bromide with $Mg$ in $Et_2O$ forms the Grignard reagent $A$,which is $sec$-butylmagnesium bromide $(CH_3CH_2CH(CH_3)MgBr)$.
$2$. The Grignard reagent acts as a strong nucleophile and attacks the epoxide ring.
$3$. In the presence of a basic medium (Grignard reagent),the nucleophilic attack occurs at the less sterically hindered carbon atom of the epoxide.
$4$. The $sec$-butyl group attacks the less substituted carbon of the $1-$methyl$-1,2-$epoxycyclopentane ring.
$5$. Subsequent acid workup with $H_3O^+$ protonates the resulting alkoxide to form the final alcohol product $(B)$.
$6$. Based on the stereochemistry and regioselectivity,the product is the one shown in option $A$.

(C) The reaction is an $S_N2$ reaction,which proceeds with inversion of configuration at the chiral carbon atom where the leaving group $(Br^-)$ is attached. The nucleophile $(I^-)$ attacks from the side opposite to the leaving group. In the given reactant,the $Br$ atom is on the left side of the chiral carbon. Therefore,in the product,the $I$ atom will be on the right side of the same chiral carbon. Comparing the options,option $(A)$ shows the $I$ atom on the left,option $(B)$ shows the $I$ atom on the left,option $(C)$ shows the $I$ atom on the right,and option $(D)$ shows the $I$ atom on the left. Thus,the correct structure is $(C)$.

(C) The $S_N2$ mechanism requires a backside attack by the nucleophile on the carbon atom bonded to the leaving group.
In $1$-bromobicyclo$[2.2.1]$heptane (option $C$),the carbon atom attached to the bromine is at the bridgehead position.
Due to the rigid bicyclic structure,the backside of the bridgehead carbon is sterically hindered and inaccessible to the nucleophile.
Furthermore,the formation of a planar transition state (required for $S_N2$) or a carbocation (required for $S_N1$) at the bridgehead position is geometrically constrained and energetically unfavorable due to Bredt's rule.
Therefore,$S_N2$ reactions are negligible at the bridgehead position.

(A) The reaction involves the nucleophilic substitution of a bromine atom by an ammonia $(NH_3)$ molecule.
In the given molecule,there are two bromine atoms: one is a benzylic bromide $(-CH_2Br)$ and the other is an aryl bromide (attached directly to the benzene ring).
The benzylic bromide is highly reactive towards nucleophilic substitution ($S_N2$ or $S_N1$ depending on conditions) because the resulting carbocation is resonance-stabilized or the transition state is easily formed.
In contrast,the aryl bromide is very unreactive towards nucleophilic substitution due to the partial double bond character of the $C-Br$ bond caused by resonance with the benzene ring.
Therefore,the $NH_3$ molecule will selectively attack the benzylic carbon,replacing the $-Br$ group with an $-NH_2$ group to form $2$-bromobenzylamine.

(B) The reaction involves a nucleophilic substitution $(S_N2)$ mechanism using $I^-$ as the nucleophile in a polar aprotic solvent $(DMF)$.
$S_N2$ reactions are highly sensitive to steric hindrance.
In the substrate $Cl-CH_2-C(CH_3)_2-CH_2-CH_2-Cl$,there are two primary chloride sites.
The site at the end of the chain $(-CH_2-CH_2-Cl)$ is less sterically hindered compared to the site near the bulky tert-butyl group $(-CH_2-C(CH_3)_2-)$.
Therefore,the nucleophile $I^-$ preferentially attacks the less hindered primary carbon atom,resulting in the formation of $Cl-CH_2-C(CH_3)_2-CH_2-CH_2-I$ as the major product.

(B) The $S_{N}2$ reaction stands for Substitution Nucleophilic Bimolecular.
In a bimolecular reaction,the rate of the reaction depends on the concentration of both the substrate (electrophile) and the nucleophile.
Therefore,the rate law is expressed as: $Rate = k$ [electrophile] [nucleophile].

(A) The reaction is a Finkelstein reaction,which proceeds via an $S_N2$ mechanism in acetone solvent.
In the given substrate,$Br^{\ominus}$ is a better leaving group than $F^{\ominus}$.
Therefore,the $I^{\ominus}$ nucleophile will attack the carbon atom attached to the $Br$ atom,displacing it to form the corresponding iodo-derivative.
The configuration at the chiral center undergoes inversion during the $S_N2$ process.

(B) The reactivity of alkyl halides towards the $S_{N^1}$ reaction depends on the stability of the carbocation intermediate formed in the rate-determining step ($R$.$D$.$S$.).
$(1)$ $CH_3CH_2CH_2Cl$ forms a primary $(1^{\circ})$ carbocation $(CH_3CH_2CH_2^+)$,which is the least stable.
$(2)$ $CH_2=CH-CHCl-CH_3$ forms an allylic carbocation $(CH_2=CH-CH^+-CH_3)$,which is resonance-stabilized and thus the most stable.
$(3)$ $CH_3CH_2-CHCl-CH_3$ forms a secondary $(2^{\circ})$ carbocation $(CH_3CH_2-CH^+-CH_3)$,which is more stable than the primary carbocation due to hyperconjugation (more $\alpha$-hydrogens) but less stable than the resonance-stabilized allylic carbocation.
Therefore,the order of stability of carbocations is: $(2) > (3) > (1)$.
Hence,the decreasing order of $S_{N^1}$ reactivity is $2 > 3 > 1$.

(A) The reaction with $NaCN$ typically proceeds via the $S_N2$ mechanism,as $CN^-$ is a strong nucleophile.
The rate of $S_N2$ reaction is inversely proportional to the steric hindrance at the reaction center.
The order of reactivity for $S_N2$ is: primary $(1^\circ)$ > secondary $(2^\circ)$ > tertiary $(3^\circ)$.
Comparing the given compounds:
$(A)$ $CH_3CH_2CH_2CH_2Br$ is a primary alkyl halide $(1^\circ)$.
$(B)$ $CH_3CH_2CH(CH_3)Br$ is a secondary alkyl halide $(2^\circ)$.
$(C)$ $(CH_3)_2CHCH_2Br$ is a primary alkyl halide $(1^\circ)$,but it has steric hindrance at the $\beta$-carbon.
$(D)$ $CH_3CH_2C(CH_3)_2Br$ is a tertiary alkyl halide $(3^\circ)$.
Among the primary halides,$CH_3CH_2CH_2CH_2Br$ has the least steric hindrance at the reaction center,making it the fastest to undergo $S_N2$ substitution.

(C) The rate of solvolysis follows an $S_N1$ mechanism,which depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
$(1)$ $CH_2=C(CH_3)^+$ is a vinylic carbocation,which is highly unstable due to the $sp$ hybridization of the positively charged carbon.
$(2)$ $1-methylcyclohexyl$ cation is a tertiary $(3^{\circ})$ carbocation,which is highly stable due to hyperconjugation and inductive effects.
$(3)$ $CH_3-CH^+-CH_2-CH(CH_3)_2$ is a secondary $(2^{\circ})$ carbocation,which is less stable than a tertiary carbocation but more stable than a vinylic carbocation.
Therefore,the order of stability of the carbocations is $(2) > (3) > (1)$.
Thus,the rate of solvolysis is $(2) > (3) > (1)$.

(A) The reaction involves the nucleophilic substitution of the chlorine atom in $4-$bromobenzyl chloride by the cyanide ion $(CN^-)$ from sodium cyanide $(NaCN)$.
In $4-$bromobenzyl chloride,the $Cl$ atom is attached to a benzylic carbon,which is highly reactive towards nucleophilic substitution ($S_N2$ mechanism).
Conversely,the $Br$ atom is directly attached to the benzene ring,which exhibits partial double-bond character due to resonance,making it a poor leaving group and resistant to nucleophilic substitution under these conditions.
Therefore,the $CN^-$ nucleophile selectively replaces the $Cl$ atom to form $4-$bromobenzyl cyanide.

(D) The correct answer is $D$. All the given compounds do not readily undergo nucleophilic substitution reactions ($S_N1$ or $S_N2$):
$1$. $1$-Bromobicyclo$[2.2.1]$heptane: The bridgehead carbon is sterically crowded,preventing $S_N2$ attack. Furthermore,the formation of a carbocation at the bridgehead is forbidden by Bredt's rule,preventing $S_N1$ reaction.
$2$. $Ph-Br$ (Bromobenzene): The lone pair on the bromine atom is in resonance with the benzene ring,giving the $C-Br$ bond partial double bond character. This makes the bond stronger and shorter,preventing nucleophilic substitution.
$3$. $(CH_3)_3C-CH_2-Br$ (Neopentyl bromide): This is a primary alkyl halide with significant steric hindrance at the $\beta$-carbon,which blocks the backside attack required for $S_N2$. Additionally,the formation of a primary carbocation is energetically unfavorable,preventing $S_N1$.

(D) In $1-$bromotriptycene,the bromine atom is found to be virtually inert to nucleophiles.
This is because the rigid structure of the triptycene molecule prevents the formation of a planar carbocation intermediate required for an $S_N1$ reaction.
Furthermore,the backside attack by a nucleophile,which is necessary for an $S_N2$ reaction,is sterically hindered by the three benzene rings,making it impossible for the nucleophile to approach the carbon atom bonded to the bromine.
Therefore,the conversion does not take place under standard conditions.

(B) The given reaction is a $S_N1$ solvolysis reaction.
$1$. The first step is the slow ionization of the alkyl bromide to form a carbocation intermediate.
$2$. The second step is the fast nucleophilic attack by $H_2O$ to form an oxonium ion intermediate.
$3$. The third step is the fast deprotonation to form the final alcohol product.
Since there are three steps,the reaction coordinate diagram must have three transition states and two intermediates. Diagram $B$ correctly represents this $3$-step process.

(C) The reaction proceeds via Neighboring Group Participation $(NGP)$ by the phenyl ring. The $\pi$-electrons of the phenyl ring attack the carbon bearing the $OTs$ group,forming a phenonium ion intermediate. This intermediate is symmetric in terms of the two carbons of the ethyl chain,allowing the nucleophile $(RCOO^-)$ to attack either of the two equivalent carbons. Consequently,the $OCOR$ group can be attached to either the $^{14}C$ labelled carbon or the non-labelled carbon,resulting in a mixture of both products.

(A) Step $1$: $NBS$ ($N$-Bromosuccinimide) is a reagent used for allylic or benzylic bromination. In the reaction of $o$-iodotoluene with $NBS$,the benzylic hydrogen is replaced by bromine to form $o$-iodobenzyl bromide,which is product $(A)$.
Step $2$: The product $(A)$ ($o$-iodobenzyl bromide) reacts with $CH_3SNa$ (sodium methanethiolate). This is a nucleophilic substitution reaction ($S_N2$ mechanism) where the $-SCH_3$ group replaces the $-Br$ atom at the benzylic position to form $o$-iodobenzyl methyl sulfide as product $(B)$.

(A) The rate of $S_N2$ reaction is inversely proportional to the steric hindrance around the reaction center.
As the number of alkyl groups attached to the carbon bearing the leaving group increases (i.e.,from methyl to tertiary),the steric hindrance increases,which significantly decreases the rate of the $S_N2$ reaction.
Therefore,$\log(\text{rate})$ decreases as we move from $CH_3Br$ to $t-BuBr$.
The correct graph is the one showing a downward trend.

(D) The reactivity of benzylic halides toward $S_N2$ reactions is influenced by the electronic effects of substituents on the benzene ring.
An electron-withdrawing group (like $-NO_2$) at the para position stabilizes the transition state of the $S_N2$ reaction by dispersing the negative charge that develops on the carbon atom during the nucleophilic attack.
Conversely,electron-donating groups (like $-CH_3$) destabilize the transition state.
Therefore,$p$-nitrobenzyl chloride is the most reactive toward $S_N2$ reaction among the given options.

(B) The rate of an $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the leaving group departs.
In option $(c)$,the first compound is bromocyclohexane,which forms a secondary carbocation.
The second compound is $3-$bromocyclohexene,which forms an allylic carbocation upon the loss of the bromide ion.
An allylic carbocation is resonance-stabilized,making it significantly more stable than a simple secondary carbocation.
Therefore,$3$-bromocyclohexene reacts faster than bromocyclohexane.
In option $(b)$,the first compound is $CH_{3}C(Br)(CH_{3})CH_{2}CH_{3}$ (a tertiary alkyl bromide),which forms a tertiary carbocation.
The second compound is $CH_{3}CH(Br)CH(CH_{3})_{2}$ (a secondary alkyl bromide),which forms a secondary carbocation.
Since a tertiary carbocation is more stable than a secondary carbocation,the first compound reacts faster than the second.
Thus,both $(b)$ and $(c)$ are correct,but $(b)$ is the standard textbook example for tertiary vs secondary carbocation stability.

(A) The reaction of allyl alcohol with excess $HBr$ proceeds in two steps:
$1.$ Substitution of the hydroxyl group by bromine to form allyl bromide:
$CH_2=CH-CH_2-OH + HBr \rightarrow CH_2=CH-CH_2-Br + H_2O$
$2.$ Electrophilic addition of $HBr$ to the double bond of allyl bromide following Markovnikov's rule:
$CH_2=CH-CH_2-Br + HBr \rightarrow CH_3-CH(Br)-CH_2-Br$
Therefore,the major product is $1,2$-dibromopropane.

(C) $S_{N}1$ reactions involve the formation of a carbocation intermediate,which can undergo rearrangement if a more stable carbocation can be formed. $S_{N}2$ reactions occur in a single step with inversion of configuration and no rearrangement. For the products to be the same (excluding stereoisomers),the substrate must not undergo rearrangement during the $S_{N}1$ process. In $4$-methylchlorocyclohexane,the chlorine is attached to a secondary carbon. Upon ionization,it forms a secondary carbocation that does not rearrange to a more stable form,leading to the same substitution product as the $S_{N}2$ pathway.

(C) For $S_N2$ reactions,the rate depends on the strength of the nucleophile and the quality of the leaving group.
$(A)$ In pair $A$,$I$ involves $Br^-$ as a leaving group,while $II$ involves $CH_3COO^-$ as a leaving group. Since $Br^-$ is a better leaving group than $CH_3COO^-$,$I$ has a greater $S_N2$ rate.
$(B)$ In pair $B$,$III$ uses $SH^-$ (a strong nucleophile) while $IV$ uses $CH_3SH$ (a neutral,weaker nucleophile). Thus,$III$ has a greater $S_N2$ rate.
$(C)$ In pair $C$,both reactions occur in $DMSO$ (a polar aprotic solvent). In polar aprotic solvents,nucleophilicity increases as basicity increases. Since $Cl^-$ is more basic than $I^-$,$Cl^-$ is a better nucleophile in $DMSO$. Thus,$V$ has a greater $S_N2$ rate.
$(D)$ In pair $D$,the reactions occur in methanol (a polar protic solvent). In polar protic solvents,nucleophilicity increases down the group due to decreased solvation. Thus,$I^-$ is a better nucleophile than $Cl^-$. Therefore,$VIII$ has a greater $S_N2$ rate.
Conclusion: The correct choices are $I, III, V, VIII$,which corresponds to option $C$.

(C) In isomer $A$,the iodine atom is in the axial position. Due to $1,3-$diaxial interactions,the leaving group $(I^-)$ is more sterically crowded and experiences repulsion,which facilitates its departure.
In isomer $B$,the iodine atom is in the equatorial position,which is less sterically hindered and more stable,making the departure of the leaving group slower.
Therefore,the axial isomer $(A)$ undergoes $S_N2$ substitution faster due to the relief of steric strain upon the departure of the leaving group.

(C) In Reaction-$1$,the $-OTs$ group and the $-NMe_2$ group are in a cis-configuration. The nucleophilic attack by $CH_3COO^-$ occurs via an $S_N2$ mechanism,resulting in inversion of configuration at the carbon atom bearing the $-OTs$ group. Since the starting material is chiral,the product is a single enantiomer.
In Reaction-$2$,the $-OTs$ group and the $-NMe_2$ group are in a trans-configuration. The lone pair on the nitrogen atom of the $-NMe_2$ group participates in Neighboring Group Participation $(NGP)$,forming a cyclic aziridinium ion intermediate. This intermediate is then opened by the acetate nucleophile. Due to the symmetry and the nature of the $NGP$-assisted pathway,this reaction also leads to a specific stereochemical outcome,typically resulting in a single product (or a specific pair of enantiomers depending on the starting material's chirality). Given the options,the most accurate description is that a single product is obtained in both reactions due to the stereospecific nature of these pathways.

(A) The graph shows the variation of reaction rate with the degree of substitution of the alkyl halide.
$1$. For $S_{N^2}$ reactions,the rate decreases as the steric hindrance around the reaction center increases. Thus,the rate order is $CH_3-X > Me-CH_2-X > Me_2CH-X > Me_3C-X$. This corresponds to graph $B$.
$2$. For $S_{N^1}$ reactions,the rate increases as the stability of the carbocation intermediate increases. Thus,the rate order is $Me_3C-X > Me_2CH-X > Me-CH_2-X > CH_3-X$. This corresponds to graph $A$.
Therefore,$A \to S_{N^1}$ and $B \to S_{N^2}$.

(C) The reaction involves a nucleophilic substitution with $NaN_3$ (azide ion,$N_3^-$) acting as the nucleophile.
$Br^-$ is a better leaving group than $Cl^-$,so the substitution occurs at the carbon atom attached to the bromine atom.
Since $NaN_3$ is a strong nucleophile and the reaction proceeds via an $S_N2$ mechanism,the nucleophile attacks from the side opposite to the leaving group,resulting in an inversion of configuration at the chiral center.
Therefore,the $-N_3$ group replaces the $-Br$ group with inversion of configuration.

(A) Reaction $A$ represents an $S_N2$ mechanism involving a neutral nucleophile and a neutral substrate,leading to a charged transition state. Increasing solvent polarity stabilizes the charged transition state more than the reactants,thus increasing the rate.
Reaction $B$ represents an $S_N1$ mechanism where the rate-determining step involves the formation of a carbocation $(R^{+})$ and an anion $(L^{-})$. Increasing solvent polarity stabilizes the ionic transition state and the resulting ions,thereby increasing the rate of reaction $B$ as well.

(C) The reactivity of alkyl halides toward $S_{N}2$ reactions depends on the steric hindrance and the stability of the transition state.
$S_{N}2$ reactions are favored by less steric hindrance and electron-withdrawing groups that stabilize the transition state.
In $Me-O-CH_2-Cl$,the oxygen atom has lone pairs that can stabilize the transition state through resonance (or the inductive effect of the oxygen atom),making it highly reactive.
Specifically,$\alpha$-halo ethers are known to be extremely reactive in $S_{N}2$ reactions due to the stabilization of the transition state by the adjacent oxygen atom.
Therefore,$Me-O-CH_2-Cl$ is the most reactive toward $S_{N}2$ reaction.

(D) The reactivity of alkyl halides toward $S_N2$ reactions depends on steric hindrance and the nature of the leaving group.
In option $D$,the first compound is allyl chloride $(CH_2=CH-CH_2-Cl)$,which is a primary alkyl halide and highly reactive toward $S_N2$ due to resonance stabilization of the transition state.
The second compound is vinyl chloride $(CH_2=CH-Cl)$,where the $C-Cl$ bond has partial double bond character due to resonance,making it very unreactive toward nucleophilic substitution.
Therefore,the first compound is significantly more reactive than the second.

(B) The rate of an $S_{N}2$ reaction depends on the nucleophilicity of the nucleophile.
$(a)$ Charged nucleophiles $(EtO^{-})$ are much stronger than neutral nucleophiles $(EtOH)$. Thus,the first reaction is faster.
$(b)$ $EtS^{-}$ is a better nucleophile than $EtO^{-}$ because sulfur is larger and more polarizable than oxygen,leading to a more stable transition state. Therefore,the second reaction is faster.
$(c)$ The rate of $S_{N}2$ is given by $Rate = k[Substrate][Nu^{-}]$. In both cases,the product of concentrations $[Substrate] \times [Nu^{-}]$ is $1 \times 2 = 2$,so the rates are equal.
$(d)$ $Ph_{3}P$ is a better nucleophile than $Ph_{3}N$ due to the larger size of phosphorus,which allows for better charge dispersion in the transition state. Thus,the first reaction is faster.

(D) $S_{N^1}$ reaction rate depends on the following factors:
$1$. Leaving group ability: $Br^-$ is a better leaving group than $Cl^-$,so the second reaction in option $A$ is faster.
$2$. Solvent polarity: $H_2O$ has a higher dielectric constant than $CH_3OH$,which stabilizes the carbocation intermediate more effectively,making the second reaction in option $B$ faster.
$3$. Substrate concentration: The rate of $S_{N^1}$ is proportional to the concentration of the substrate. $A$ higher concentration $(2 \ M)$ in the second reaction makes it faster than the first $(1 \ M)$ in option $C$.
Therefore,all pairs satisfy the condition.

(D) The $S_N2$ reaction proceeds through a single-step concerted mechanism where the nucleophile attacks the electrophilic carbon from the side opposite to the leaving group.
In the transition state,the bond between the nucleophile and the carbon is partially formed,and the bond between the carbon and the leaving group is partially broken.
Therefore,the transition state involves both the nucleophile and the electrophile (substrate) simultaneously.
Thus,option $D$ is correct.

(A) The rate law for an $S_N2$ reaction is given by: $\text{Rate} = k [\text{Substrate}] [\text{Nucleophile}]$.
Since the rate is directly proportional to the concentration of the nucleophile,increasing the concentration of the nucleophile by a factor of $10$ will increase the reaction rate by a factor of $10$.

(C) The rate law for an $S_N2$ reaction is given by: $\text{Rate} = k[\text{Substrate}][\text{Nucleophile}]$.
Since the reaction is first-order with respect to the substrate (electrophile),the rate is directly proportional to its concentration.
If the concentration of the substrate is decreased by a factor of $3$,the reaction rate will also decrease by a factor of $3$.

(B) The rate law for an $S_N2$ reaction is given by: $Rate = k [Substrate] [Nucleophile]$.
If the concentration of the substrate is increased by a factor of $3$ and the concentration of the nucleophile is increased by a factor of $3$,the new rate becomes:
$Rate_{new} = k (3 [Substrate]) (3 [Nucleophile]) = 9 \times k [Substrate] [Nucleophile]$.
Therefore,the reaction rate increases by a factor of $9$.

(B) The given reaction involves a secondary alkyl halide reacting with a strong nucleophile,$Na^+ -SCH_2CH_3$ (sodium ethanethiolate).
This reaction proceeds via an $S_N2$ mechanism.
In an $S_N2$ reaction,the nucleophile attacks the electrophilic carbon from the side opposite to the leaving group $(-Br)$,resulting in the inversion of configuration at the chiral center (Walden inversion).
Therefore,the correct representation is the inversion of configuration.