Properties of Haloalkanes Questions in English

(A) The reaction of alkanes with halogens is a free radical substitution reaction. The reactivity of halogens follows the order: $F_2 > Cl_2 > Br_2 > I_2$. The reaction with fluorine $(F_2)$ is extremely exothermic and occurs explosively even in the dark,making it difficult to control. Therefore,$F_2$ reacts explosively with alkanes.

(B) The reaction of an alkene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule (Kharasch effect or peroxide effect).
In $CH_3-CH=CHBr$,the double bond is between $C_2$ and $C_3$.
According to the anti-Markovnikov rule,the $Br^-$ atom attaches to the carbon atom of the double bond that has more hydrogen atoms.
Here,$CH_3-CH=CHBr$ is an unsymmetrical alkene.
However,the presence of the $Br$ group on the terminal carbon influences the electron density.
Applying the anti-Markovnikov rule,$H$ adds to the carbon with fewer hydrogens and $Br$ adds to the carbon with more hydrogens.
Thus,$CH_3-CH=CHBr + HBr \xrightarrow{\text{peroxide}} CH_3-CH(Br)-CH_2Br$.

(B) The reaction of $n$-propyl bromide $(CH_3-CH_2-CH_2-Br)$ with ethanolic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,leading to the formation of a double bond between the $\alpha$ and $\beta$ carbons.
The reaction is:
$CH_3-CH_2-CH_2-Br + KOH \xrightarrow{C_2H_5OH} CH_3-CH=CH_2 + KBr + H_2O$.
Thus,the product formed is propene.

(C) The reaction of ethyl chloride $(CH_3CH_2Cl)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.
In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,leading to the elimination of $HCl$ and the formation of a double bond between the $\alpha$ and $\beta$ carbons.
The chemical equation is: $CH_3CH_2Cl + \text{alc. } KOH \rightarrow CH_2=CH_2 + KCl + H_2O$.
Thus,ethylene $(CH_2=CH_2)$ is produced.

(D) $SbCl_5$ acts as a Lewis acid.
It reacts with the alkyl chloride to abstract the chloride ion,leading to the formation of a stable carbocation intermediate.
Therefore,the reaction proceeds due to the formation of a carbocation.

(A) The dehydrobromination of $2$-bromobutane $(CH_3-CHBr-CH_2-CH_3)$ follows $Saytzeff's$ rule.
According to this rule,the more substituted alkene is the major product.
$2$-bromobutane can undergo elimination to form either $1$-butene $(CH_2=CH-CH_2-CH_3)$ or $2$-butene $(CH_3-CH=CH-CH_3)$.
$2$-butene is a disubstituted alkene,while $1$-butene is a monosubstituted alkene.
Since $2$-butene is more stable due to hyperconjugation and inductive effects,it is formed as the major product.

(A) The reaction of $1$-chlorobutane $(CH_3CH_2CH_2CH_2Cl)$ with alcoholic $KOH$ (potash) is a dehydrohalogenation reaction.
This is an elimination reaction ($E2$ mechanism) where the base $(OH^-)$ removes a hydrogen atom from the $\beta$-carbon,leading to the formation of an alkene.
$CH_3CH_2CH_2CH_2Cl + KOH (\text{alc.}) \rightarrow CH_3CH_2CH=CH_2 + KCl + H_2O$.
The product formed is $1$-butene.

(A) The reaction of $1$-chlorobutane $(CH_3CH_2CH_2CH_2Cl)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,leading to the elimination of a molecule of $HCl$.
$CH_3CH_2CH_2CH_2Cl + KOH (\text{alc.}) \rightarrow CH_3CH_2CH=CH_2 + KCl + H_2O$.
The major product formed is $1$-butene.

(A) In the presence of alcoholic $KOH$,the $KOH$ reacts with the ethanol solvent to produce ethoxide ions $(C_2H_5O^{-})$.
$KOH + C_2H_5OH \rightarrow C_2H_5O^{-} + K^{+} + H_2O$.
Since the ethoxide ion $(C_2H_5O^{-})$ is a much stronger base than the hydroxide ion $(OH^{-})$,it acts as the primary attacking reagent in the dehydrohalogenation (elimination) reaction of $CH_3CH_2Cl$ to form ethene $(CH_2=CH_2)$.

(A) Dehydrohalogenation follows the $E2$ mechanism,which is favored by the stability of the transition state and the leaving group ability.
$1$. The rate of dehydrohalogenation depends on the nature of the leaving group. Iodide $(I^-)$ is a better leaving group than bromide $(Br^-)$ because it is a weaker base.
$2$. Tertiary alkyl halides undergo dehydrohalogenation faster than secondary or primary alkyl halides due to the formation of a more stable alkene product (Saytzeff's rule).
$3$. Comparing $t$-butyl iodide and $t$-butyl bromide,$t$-butyl iodide reacts faster due to the superior leaving group ability of the iodide ion.
Therefore,$t$-butyl iodide is the fastest to undergo dehydrohalogenation.

(B) The reaction of $n$-propyl bromide $(CH_3CH_2CH_2Br)$ with ethanolic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,and the bromide ion is eliminated,resulting in the formation of an alkene.
$CH_3CH_2CH_2Br + KOH (\text{ethanolic}) \rightarrow CH_3CH=CH_2 + KBr + H_2O$.
The product formed is $CH_3CH=CH_2$,which is propene.

(A) The reaction of alkyl halides with $C_2H_5ONa/C_2H_5OH$ proceeds via an $E2$ elimination mechanism.
For $2$-halopentanes,the elimination can occur at $C_1$ or $C_3$ to form $1$-pentene or $2$-pentene,respectively.
According to Zaitsev's rule,the more substituted alkene ($2$-pentene) is the major product.
Among the halogens,iodine is a better leaving group than bromine.
Therefore,$2$-iodopentane $(CH_3CH(I)CH_2CH_2CH_3)$ undergoes faster elimination and provides a higher yield of the more stable $2$-pentene compared to $2$-bromopentane.

(C) The dehydrohalogenation of $2$-bromobutane $(CH_3-CH(Br)-CH_2-CH_3)$ follows $Saytzeff's$ rule.
According to this rule,the more substituted alkene is the major product.
$2$-butene $(CH_3-CH=CH-CH_3)$ is a disubstituted alkene,while $1$-butene $(CH_3-CH_2-CH=CH_2)$ is a monosubstituted alkene.
Therefore,$2$-butene is more stable and is formed as the major product.

(C) An alcoholic solution of $KOH$ acts as a strong base. It is used to remove a molecule of hydrogen halide $(HX)$ from an alkyl halide,a process known as dehydrohalogenation. The reaction is: $R-CH_2-CH(X)-R' + KOH (\text{alc.}) \rightarrow R-CH=CH-R' + KX + H_2O$.

(B) The reaction of $3$-phenylpropene $(C_6H_5CH_2CH=CH_2)$ with $HBr$ follows Markovnikov's rule.
In the presence of an electrophilic addition,the proton $(H^+)$ adds to the terminal carbon $(CH_2)$ to form a more stable carbocation.
The intermediate formed is a secondary benzylic carbocation $(C_6H_5CH_2CH^+CH_3)$,which is stabilized by resonance with the phenyl ring.
Subsequently,the bromide ion $(Br^-)$ attacks this carbocation to form $1$-bromo-$1$-phenylpropane $(C_6H_5CH(Br)CH_2CH_3)$.

(A) The reaction of an alkyl halide with $Na$ in dry ether is the Wurtz reaction,which produces a symmetric alkane.
For the resulting hydrocarbon to yield only one monochloro derivative,all hydrogen atoms in the alkane must be equivalent.
$t$-Butyl chloride ($CH_3)_3CCl$ reacts with $Na$/dry ether to form $2,2,3,3$-tetramethylbutane: $(CH_3)_3C-C(CH_3)_3$.
In $2,2,3,3$-tetramethylbutane,all $18$ hydrogen atoms are equivalent,so its monochlorination produces only one product.
Therefore,$A$ is $t$-butyl chloride.

(C) The reaction of $2$-bromopentane with ethanolic potassium ethoxide is a dehydrohalogenation reaction (an $E2$ elimination reaction).
According to Zaitsev's rule,the more substituted alkene is the major product.
$2$-bromopentane $(CH_3-CH(Br)-CH_2-CH_2-CH_3)$ undergoes elimination to form two possible alkenes: $1$-pentene and $2$-pentene.
$2$-pentene is more substituted than $1$-pentene and is therefore the major product.
Among the isomers of $2$-pentene,the trans-isomer is more stable than the cis-isomer due to less steric hindrance.
Thus,trans-$2$-pentene is the major product.

(B) The reaction of ethyl bromide $(CH_3CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.
In this reaction,the alcoholic $KOH$ acts as a strong base and abstracts a proton from the $\beta$-carbon,leading to the elimination of a molecule of $HBr$.
This results in the formation of an alkene.
The chemical equation is: $CH_3CH_2Br + KOH (\text{alc.}) \rightarrow CH_2=CH_2 + KBr + H_2O$.
Thus,the product formed is ethene.

(C) In $SN^2$ mechanism,the nucleophile attacks the carbon atom from the back side.
In tertiary alkyl halides,the presence of three bulky alkyl groups around the electrophilic carbon atom creates significant steric hindrance.
This steric hindrance prevents the nucleophile from approaching the carbon atom,making tertiary alkyl halides practically inert to the $SN^2$ mechanism.

(A) The rate of an $S_N2$ reaction is directly proportional to the nucleophilicity of the attacking species.
Nucleophilicity is related to the basicity of the nucleophile.
Stronger bases are generally better nucleophiles.
The basicity order of the given species is: $CH_3O^{-} > HO^{-} > PhO^{-} > AcO^{-}$.
$CH_3O^{-}$ is the strongest base due to the $+I$ effect of the $CH_3$ group.
$HO^{-}$ is a strong base.
$PhO^{-}$ is less basic than $HO^{-}$ due to resonance stabilization of the phenoxide ion.
$AcO^{-}$ is the least basic among these due to the resonance stabilization of the carboxylate ion by two oxygen atoms.
Therefore,the order of nucleophilicity and reaction rate is: $(4) CH_3O^{-} > (3) HO^{-} > (1) PhO^{-} > (2) AcO^{-}$.
Thus,the correct order is $4 > 3 > 1 > 2$.

(A) Nucleophilicity is related to the basicity of the species. Stronger bases are generally better nucleophiles.
$1$. $CH_3COO^-$: Acetate ion,resonance stabilized.
$2$. $CH_3O^-$: Methoxide ion,strong base.
$3$. $CN^-$: Cyanide ion,strong nucleophile.
$4$. $CH_3-C_6H_4-SO_3^-$: Tosylate ion,highly resonance stabilized,very weak base.
Comparing basicity: $CH_3O^- > CN^- > CH_3COO^- > CH_3-C_6H_4-SO_3^-$.
However,in terms of nucleophilicity,$CN^-$ is often a better nucleophile than $CH_3O^-$ in polar aprotic solvents,but generally,the order of nucleophilicity follows the order of basicity for similar species. The correct decreasing order is $3 > 2 > 1 > 4$.

(D) The reaction involves the ionization of the $C-Cl$ bond in the presence of the Lewis acid $SbCl_5$,which acts as a catalyst.
This leads to the formation of a planar $1$-phenylethyl carbocation intermediate.
The nucleophile $(Cl^-)$ can then attack the planar carbocation from either side with equal probability,resulting in the formation of a racemic mixture.

(A) molecule has a net dipole moment if it is polar,meaning the vector sum of its individual bond dipoles is non-zero.
$CH_2Cl_2$ (dichloromethane) is a polar molecule with a net dipole moment of $1.60 \ D$.
$CH_3Cl$ (chloromethane) is also a polar molecule with a net dipole moment of $1.87 \ D$.
For the isomers of $1-$chloroprop$-1-$ene,the cis-isomer has a non-zero dipole moment due to the alignment of the $C-Cl$ and $C-CH_3$ bond dipoles,while the trans-isomer has a significantly lower dipole moment because the bond dipoles partially cancel each other out.
Since all the given options represent molecules that possess a net dipole moment,the question likely asks for the molecule with the highest dipole moment or is a multiple-choice question where all are technically correct. However,in standard chemistry contexts,all these molecules exhibit a dipole moment.

(C) In reaction $(i)$,the reagent $C_2H_5OH$ is a weak nucleophile,which favors the $S_{N^1}$ mechanism.
In reaction $(ii)$,the reagent $C_2H_5O^-$ is a strong nucleophile,which favors the $S_{N^2}$ mechanism.
Therefore,the reaction mechanisms are $S_{N^1}$ and $S_{N^2}$ respectively.

(A) In aprotic solvents,the nucleophilicity order is determined by the basicity of the ions.
Since $F^{-}$ is the most basic among the given halide ions,it is the strongest nucleophile.
Conversely,$I^{-}$ is the least basic and therefore the weakest nucleophile in an aprotic solvent.

(B) The leaving group ability is inversely proportional to the basicity of the group. Weaker bases are better leaving groups.
The conjugate acids of the given groups are:
$I. CH_3COOH$ $(pKa \approx 4.75)$
$II. CH_3OH$ $(pKa \approx 15.5)$
$III. CH_3SO_3H$ $(pKa \approx -1.2)$
$IV. CF_3SO_3H$ $(pKa \approx -14)$
Lower $pKa$ values indicate stronger conjugate acids,which means the corresponding conjugate bases are weaker and thus better leaving groups.
The order of acidity is $IV > III > I > II$.
Therefore,the order of leaving group ability is $IV > III > I > II$.

(B) The conversion of an alkyl halide $(R-X)$ to an alcohol $(R-OH)$ is a nucleophilic substitution reaction.
In this reaction,the halide ion $(X^-)$ is replaced by a hydroxyl group $(OH^-)$.
The reaction is represented as: $R-X + OH^- \to R-OH + X^-$.
For example: $R-Cl + OH^- \to R-OH + Cl^-$.

(C) The reaction of ethanol with $PCl_5$ is:
$C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl + POCl_3 + HCl$
Here,$A$ is $C_2H_5Cl$ (ethyl chloride).
The reaction of $A$ $(C_2H_5Cl)$ with silver nitrite $(AgNO_2)$ is:
$C_2H_5Cl + AgNO_2 \rightarrow C_2H_5NO_2 + AgCl$
Here,$B$ is $C_2H_5NO_2$ (nitroethane).

(C) The reactant is $1$-bromo-$2$-methylpropane,which is a primary alkyl halide with steric hindrance at the $\beta$-carbon.
When treated with a strong base like methoxide $(CH_3O^-)$ in methanol $(CH_3OH)$,the reaction proceeds via an $E_2$ elimination mechanism rather than $S_N2$ substitution due to the bulky nature of the base and the structure of the substrate.
The elimination of $HBr$ from the substrate leads to the formation of $2$-methylpropene as the major product.
Reaction: $CH_3-CH(CH_3)-CH_2Br + CH_3O^- \rightarrow CH_3-C(CH_3)=CH_2 + CH_3OH + Br^-$

(A) The hydrolysis of trichloromethane $(CHCl_3)$ with aqueous $KOH$ proceeds as follows:
$CHCl_3 + 3KOH_{(aq)} \rightarrow CH(OH)_3 + 3KCl$
Since $CH(OH)_3$ (methanetriol) is unstable,it loses a water molecule to form formic acid $(HCOOH)$:
$CH(OH)_3 \rightarrow HCOOH + H_2O$
The formic acid then reacts with the remaining $KOH$ to form potassium formate $(HCOOK)$:
$HCOOH + KOH \rightarrow HCOOK + H_2O$
Thus,the final product is potassium formate.

(D) The oxidation of $C_6H_5CH_2Cl$ (benzyl chloride) leads to the formation of $C_6H_5COOH$ (benzoic acid).
This reaction typically involves the use of an oxidizing agent like alkaline $KMnO_4$ followed by acidification.

(C) Nucleophilicity is related to the basicity of the species. Stronger bases are generally better nucleophiles.
$1$. $CH_3O^-$ is a strong base (conjugate base of a weak acid $CH_3OH$).
$2$. $CN^-$ is a strong nucleophile due to the high electron density on carbon.
$3$. $CH_3COO^-$ is a weaker base than $CH_3O^-$ due to resonance stabilization.
$4$. $CH_3-C_6H_4-SO_3^-$ (tosylate ion) is a very weak base because its negative charge is highly delocalized over three oxygen atoms.
Comparing the basicity: $CH_3O^- > CN^- > CH_3COO^- > CH_3-C_6H_4-SO_3^-$.
However,in terms of nucleophilicity in polar protic solvents,$CN^-$ is often considered very effective. Based on standard trends of nucleophilicity (where $CH_3O^-$ is a stronger nucleophile than $CN^-$ in many contexts,but $CN^-$ is highly reactive),the order is $B > C > A > D$.

(A) . Benzyl chloride undergoes $S_N1$ mechanism most readily because it forms a resonance-stabilized benzyl carbocation $(C_6H_5CH_2^+)$ as an intermediate.
Primary alkyl halides like ethyl chloride and secondary alkyl halides like isopropyl chloride prefer $S_N2$ or $E2$ mechanisms,while chlorobenzene is highly unreactive towards nucleophilic substitution due to partial double bond character.

(B) The heterolytic bond dissociation energy of alkyl halides depends on the strength of the $C-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the $C-X$ bond length increases.
An increase in bond length leads to a decrease in the bond dissociation energy.
Therefore,the correct sequence is: $R-F > R-Cl > R-Br > R-I$.

(B) The structure of $2-$methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
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): $CH_3-C(Cl)(CH_3)-CH_2-CH_3$ (chiral,exists as $(+)$ and $(-)$ enantiomers).
$3$. At $C-3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$ (chiral,exists as $(+)$ and $(-)$ enantiomers).
$4$. At $C-4$ (terminal methyl): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ (achiral).
Thus,the chiral products are the enantiomers formed by substitution at $C-2$ and $C-3$.
Total number of chiral compounds = $2$ (from $C-2$) + $2$ (from $C-3$) = $4$.

(D) $S_N2$ reactions involve the attack of a nucleophile from the backside of 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 electrophilic carbon center.
Therefore,tertiary alkyl halides are practically inert to $S_N2$ substitution.

(B) The conversion of an alkyl halide $(R-X)$ to an alcohol $(R-OH)$ is a nucleophilic substitution reaction.
When an alkyl halide is treated with an aqueous alkali like $Aq. KOH$ or $NaOH$,the halide ion $(X^-)$ is replaced by the hydroxyl group $(OH^-)$.
Reaction: $R-X + KOH (aq) \rightarrow R-OH + KX$.
This is a nucleophilic substitution reaction ($S_N1$ or $S_N2$ depending on the substrate).
Therefore,the correct option is $B$.

(B) The given reaction is $(CH_3)_3CBr \xrightarrow{H_2O} (CH_3)_3COH$.
In this reaction,the bromine atom $(-Br)$ is replaced by a hydroxyl group $(-OH)$.
Since one functional group is substituted by another,this is a substitution reaction.

(A) The formation of a Grignard reagent $(R-Mg-X)$ involves the oxidative insertion of magnesium into the carbon-halogen bond.
The reactivity depends on the strength of the $C-X$ bond.
As the size of the halogen atom increases,the $C-X$ bond length increases and the bond dissociation energy decreases.
Therefore,the ease of reaction follows the order of the leaving group ability: $I^- > Br^- > Cl^-$.
Thus,the order of reactivity is $CH_3I > CH_3Br > CH_3Cl$.

(C) The polarity of the $C-X$ bond depends on the electronegativity difference between the carbon atom and the halogen atom $(X)$.
As we move down the group in the periodic table,the electronegativity of the halogen decreases $(F > Cl > Br > I)$.
Therefore,the electronegativity difference between $C$ and $X$ decreases in the order $C-Cl > C-Br > C-I$.
Thus,the correct order of bond polarity is $CH_3Cl > CH_3Br > CH_3I$.

(D) In an $S_{N}2$ reaction,the rate of reaction depends on the ability of the leaving group to depart.
The leaving group ability is determined by the strength of the $C-X$ bond and the stability of the halide ion $(X^-)$.
The bond strength decreases as the size of the halogen increases $(C-F > C-Cl > C-Br > C-I)$.
Consequently,the iodide ion $(I^-)$ is the best leaving group because it is the weakest base,while the fluoride ion $(F^-)$ is the poorest leaving group.
Therefore,the order of reactivity for $S_{N}2$ reactions is $RI > RBr > RCl > RF$.

(C) The strength of the $C-X$ bond depends on the bond length. As the size of the halogen atom increases from $F$ to $I$,the bond length increases,which leads to a decrease in bond strength.
The order of bond strength is: $CH_3F > CH_3Cl > CH_3Br > CH_3I$.
Therefore,the $C-F$ bond in $CH_3F$ is the strongest.

(D) In reaction $A$,a saturated alkyl halide is converted into an unsaturated alkene by the removal of $H$ and $Br$ atoms. Hence,it is an elimination reaction.
In reaction $B$,the $-Br$ group is replaced by the $-OH$ group. Hence,it is a substitution reaction.
In reaction $C$,the addition of $Br_2$ across the double bond converts an unsaturated compound into a saturated compound. Hence,it is an addition reaction.
Therefore,the correct statement is that $A$ is elimination,$B$ is substitution,and $C$ is addition reaction.

(C) In this reaction,the bromine atom $(Br)$ in the alkyl halide is replaced by the amino group $(-NH_2)$ from ammonia $(NH_3)$.
Since a nucleophile $(NH_3)$ replaces a leaving group $(Br^{-})$,it is a nucleophilic substitution reaction.
Option $(a)$ is an electrophilic addition reaction.
Options $(b)$ and $(d)$ are nucleophilic addition reactions.

(A) Nucleophilicity is the ability of a species to donate an electron pair to an electrophile.
In a polar protic solvent,nucleophilicity increases down the group because the smaller ions are more heavily solvated,which hinders their ability to act as nucleophiles.
Therefore,the order of nucleophilicity for halide ions is $Cl^{-} < Br^{-} < I^{-}$.

(D) The Friedel-Crafts reaction requires an alkyl or acyl halide that can form a carbocation or an electrophilic species in the presence of a Lewis acid like $AlCl_3$.
$1$. Aryl halides like chlorobenzene and bromobenzene,and vinyl halides like vinyl chloride $(CH_2=CH-Cl)$,are not suitable for Friedel-Crafts reactions because the carbon-halogen bond has partial double bond character due to resonance,making it very strong and difficult to break.
$2$. Alkyl halides like chloroethane $(CH_3CH_2Cl)$ and isopropyl chloride $((CH_3)_2CHCl)$ readily react with $AlCl_3$ to form carbocations,which then act as electrophiles in the Friedel-Crafts alkylation.
$3$. Both $C$ and $D$ are alkyl halides and can be used. However,in the context of typical textbook examples,both are valid. Given the options,$C$ and $D$ are both correct,but usually,such questions might have a single best answer or multiple correct options. Based on the provided image,isopropyl chloride is explicitly shown as a reactant. Therefore,$D$ is a definitive choice.

(C) The reaction $CH_3CH_2CH_2Br + NaCN \to CH_3CH_2CH_2CN + NaBr$ follows an $S_N2$ mechanism.
$S_N2$ reactions are significantly faster in polar aprotic solvents because these solvents do not solvate the nucleophile $(CN^-)$ through hydrogen bonding,thereby keeping the nucleophile highly reactive.
Among the given options,$N, N'-$dimethylformamide $(DMF)$ is a polar aprotic solvent,whereas ethanol,methanol,and water are polar protic solvents.
Therefore,the reaction will be the fastest in $DMF$.

(A) The $S_N1$ reaction proceeds via the formation of a planar carbocation intermediate.
Since the nucleophile can attack from either side of the planar carbocation,a racemic mixture is formed.
However,because the leaving group partially blocks the front side (ion-pair effect),the attack from the back side is slightly favored.
Therefore,there is inversion more than retention,leading to partial racemisation.