Properties of Haloalkanes Questions in English

(B) The conversion of alkyl chlorides or alkyl bromides to alkyl iodides is known as the Finkelstein reaction.
In this reaction,the alkyl halide is treated with sodium iodide $(NaI)$ in the presence of dry acetone.
The reaction is: $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$ (where $X = Cl, Br$).

(A) The alkaline hydrolysis of $1-$bromo$-1-$phenylethane proceeds via an $S_N1$ mechanism.
In the first,rate-determining step,the leaving group (bromide ion) departs,resulting in the formation of a carbocation intermediate.
The carbocation intermediate has a positively charged carbon atom bonded to three other groups (a methyl group,a phenyl group,and a hydrogen atom).
Since the positively charged carbon is bonded to three atoms and has no lone pairs,it is $sp^2$ hybridized,resulting in a planar geometry.
The reaction is as follows:
$CH_3-CH(Ph)-Br \rightleftharpoons CH_3-CH^+(Ph) + Br^-$
$CH_3-CH^+(Ph) + OH^- \rightarrow CH_3-CH(OH)-Ph$

(C) The conversion of $tert$-butyl bromide to isobutylene is a dehydrohalogenation reaction,which is an elimination reaction ($E1$ or $E2$).
In this reaction,a base is required to abstract a proton from the $\beta$-carbon.
Alcoholic $NH_3$ (or more commonly,alcoholic $KOH$) acts as a base to facilitate this elimination.
As shown in the mechanism,the loss of $Br^-$ leads to a carbocation,followed by the loss of $H^+$ to form the double bond in isobutylene.
Therefore,the correct reagent is $NH_3$ (alc.).

(C) The dehydrohalogenation of alkyl halides follows the $E2$ mechanism,where the stability of the transition state is determined by the stability of the alkene being formed.
According to Saytzeff's rule,more substituted alkenes are more stable.
Therefore,the ease of dehydrohalogenation follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.

(A) Tertiary butyl bromide $(CH_3)_3C-Br$ is a tertiary alkyl halide. When it reacts with an alcoholic solution of $NH_3$,it undergoes a dehydrohalogenation reaction ($\alpha, \beta-$elimination).
In this reaction,the base (ammonia) abstracts a proton from the $\beta-$carbon,leading to the elimination of $HBr$ and the formation of an alkene.
The reaction is: $(CH_3)_3C-Br \xrightarrow{alc. NH_3} CH_3-C(CH_3)=CH_2 + HBr$.
The product formed is $2-$methylpropene.

(A) The reaction of $1$-chlorobutane $(CH_3-CH_2-CH_2-CH_2-Cl)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.
In this reaction,an $H$ atom is removed from the $\beta$-carbon and the $Cl$ atom is removed from the $\alpha$-carbon,resulting in the formation of an alkene.
The reaction proceeds as follows:
$CH_3-CH_2-CH_2-CH_2-Cl + \text{alc. } KOH \xrightarrow{\Delta} CH_3-CH_2-CH=CH_2 + KCl + H_2O$
The product formed is but-$1$-ene.

(C) Chloroform ($CHCl_3$,trichloromethane) undergoes slow oxidation in the presence of air and light to form a highly poisonous gas known as phosgene $(COCl_2)$.
The chemical reaction is: $2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$.
Due to this,chloroform is stored in dark-colored,airtight bottles to prevent oxidation.

(A) Gammexane,also known as lindane,is the $\gamma$-isomer of benzene hexachloride $(BHC)$.
It is prepared by the addition of chlorine to benzene in the presence of ultraviolet light.
It is widely used as an insecticide,not a herbicide.

(C) The boiling points of haloalkanes are primarily determined by the magnitude of van der Waals forces,which increase with an increase in molecular size and mass.
For the given compounds,the molecular mass increases with the number of bromine atoms present.
$CHBr_3$ $(M \approx 253 \ g/mol)$ has the highest molecular mass,followed by $CH_2Br_2$ $(M \approx 174 \ g/mol)$,then $CH_3Br$ $(M \approx 95 \ g/mol)$,and finally $CH_3Cl$ $(M \approx 50.5 \ g/mol)$.
Therefore,the correct decreasing order of boiling point is $CHBr_3 > CH_2Br_2 > CH_3Br > CH_3Cl$.

(A) The reaction between a strong base/nucleophile like $CH_3ONa$ (sodium methoxide) and a tertiary alkyl halide like $(CH_3)_3CBr$ (tert-butyl bromide) proceeds primarily via an $E2$ elimination mechanism.
Because the substrate is sterically hindered,the nucleophilic substitution $(S_N2)$ is suppressed.
Therefore,the major product formed is the alkene,which is $2$-methylpropene (isobutylene).
Thus,the correct option is $A$.

(A) The reactions are identified as follows:
$(i)$ $R-Cl + NaI \xrightarrow{\text{dry acetone}} R-I + NaCl$ is the Finkelstein reaction $(c)$.
$(ii)$ $CH_3-Br + AgF \xrightarrow{\Delta} CH_3-F + AgBr$ is the Swarts reaction $(a)$.
$(iii)$ $R-X + Mg$ $\xrightarrow{\text{dry ether}} R-Mg-X$ $\xrightarrow{H_2O} RH + Mg(OH)X$ involves the formation of a Grignard reagent $(d)$.
Therefore,the correct match is $i$ $\rightarrow c; ii$ $\rightarrow a; iii$ $\rightarrow d$.

(D) The reactivity of compounds in an $S_N1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
The stability order of the carbocations is as follows:
$(I)$ $C_6H_5-CH_2^+$ (Benzyl carbocation,stabilized by one phenyl ring).
$(II)$ $(C_6H_5)_2CH^+$ (Diphenylmethyl carbocation,stabilized by two phenyl rings).
$(III)$ $C_6H_5-CH^+(CH_3)$ ($1$-Phenylethyl carbocation,stabilized by one phenyl ring and one methyl group).
$(IV)$ $(C_6H_5)_2C^+(CH_3)$ ($1$,$1$-Diphenylethyl carbocation,stabilized by two phenyl rings and one methyl group).
Comparing the stability: $(IV) > (II) > (III) > (I)$.
Since the rate of $S_N1$ reaction is directly proportional to the stability of the carbocation,the order of reactivity is $(IV) > (II) > (III) > (I)$.

(D) The rate of an $S_N2$ reaction depends primarily on the steric hindrance around the electrophilic carbon atom.
Less sterically hindered substrates react faster in $S_N2$ reactions.
The order of reactivity for $S_N2$ is: $\text{primary} > \text{secondary} > \text{tertiary}$.
In the given options:
$A$ is a secondary alkyl halide.
$B$ is a secondary alkyl halide (with two bulky phenyl groups).
$C$ is a tertiary alkyl halide.
$D$ $(C_6H_5CH_2Br)$ is a primary benzylic halide.
Since $D$ is a primary halide,it experiences the least steric hindrance,making it the most reactive towards $S_N2$ substitution.
Therefore,the correct option is $D$.

(D) The rate of $S_N 1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
$S_N 1$ reactivity order is: $3^{\circ} > 2^{\circ} > 1^{\circ} > CH_3-X$.
$A$: $2-$bromo-$3-$methylbutane is a $2^{\circ}$ halide.
$B$: $2-$chloro-$3-$methylbutane is a $2^{\circ}$ halide.
$C$: Chloromethane is a $1^{\circ}$ halide.
$D$: $2-$bromo-$2-$methylpropane is a $3^{\circ}$ halide.
Since $3^{\circ}$ carbocations are the most stable,$2-$bromo-$2-$methylpropane will undergo $S_N 1$ reaction the fastest.

(C) The reaction of $2,2,2-$trichloroethanol $(CCl_3CH_2OH)$ with calcium hydroxide $(Ca(OH)_2)$ is a base-catalyzed decomposition reaction.
$2CCl_3CH_2OH + Ca(OH)_2 \rightarrow 2CHCl_3 + Ca(HCOO)_2 + H_2O$.
In this reaction,$2,2,2-$trichloroethanol undergoes haloform-like cleavage to produce chloroform $(CHCl_3)$ as the main organic product.

(B) The chemical reaction for the dehydrohalogenation of ethyl chloride is:
$CH_3CH_2Cl + KOH \xrightarrow{\text{Ethanol}, \Delta} CH_2=CH_2 + KCl + H_2O$
Molar mass of $CH_3CH_2Cl = (2 \times 12) + (5 \times 1) + 35.5 = 64.5 \ g \ mol^{-1}$.
Molar mass of the main product,ethene $(CH_2=CH_2)$ $= (2 \times 12) + (4 \times 1) = 28 \ g \ mol^{-1}$.
According to the stoichiometry,$64.5 \ g$ of $CH_3CH_2Cl$ produces $28 \ g$ of ethene.
Therefore,$6.45 \ g$ of $CH_3CH_2Cl$ would theoretically produce:
$\text{Mass} = \frac{6.45 \ g}{64.5 \ g \ mol^{-1}} \times 28 \ g \ mol^{-1} = 2.8 \ g$ of ethene.
Since only $50\%$ of the reagent is used,the actual amount of product obtained is:
$2.8 \ g \times 0.5 = 1.4 \ g$.

(B) The given reaction is $CH_3CH_2Cl + NaI \xrightarrow{\text{acetone}} CH_3CH_2I + NaCl$.
This is a halogen exchange reaction where an alkyl chloride reacts with sodium iodide in the presence of acetone to form an alkyl iodide.
This specific reaction is known as the $Finkelstein$ reaction.

(C) The $S_{N}2$ reaction is a second-order reaction where the rate depends on the concentration of both the substrate and the nucleophile.
Statement $C$ is incorrect because the rate law for an $S_{N}2$ reaction is given by $\text{Rate} = k[\text{Substrate}][\text{Nucleophile}]$,meaning it is dependent on both concentrations.

(C) The Swartz reaction is a method used to prepare alkyl fluorides by heating alkyl chlorides or alkyl bromides in the presence of metallic fluorides like $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
In the given options,the reaction $CH_3Br + AgF \longrightarrow CH_3F + AgBr$ represents the Swartz reaction.
Note: The reactions in options $A$ and $B$ are Finkelstein reactions,and the reaction in option $D$ is the Wurtz reaction.

(D) The reactivity of alkyl halides towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the loss of the leaving group $(Br^-)$.
$1$. $C_{6}H_{5}CH_{2}^+$ (Primary benzylic carbocation)
$2$. $C_{6}H_{5}CH^+(CH_{3})$ (Secondary benzylic carbocation)
$3$. $C_{6}H_{5}CH^+(C_{6}H_{5})$ (Secondary benzylic carbocation with two phenyl rings)
$4$. $C_{6}H_{5}C^+(CH_{3})(C_{6}H_{5})$ (Tertiary benzylic carbocation with two phenyl rings)
The stability of carbocations follows the order: $3^\circ > 2^\circ > 1^\circ$. Furthermore,the presence of two phenyl groups provides extensive resonance stabilization to the carbocation.
Compound $D$ forms a tertiary benzylic carbocation stabilized by two phenyl rings,making it the most stable carbocation and thus the most reactive towards $S_{N}1$.

(B) The reactivity of alkyl halides in an $S_N1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
The stability order of carbocations is: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Let us analyze the carbocations formed from the given compounds:
$(I)$ $CH_3-CH_2-CH^+(CH_3)$ is a $2^{\circ}$ carbocation.
$(II)$ $(CH_3)_2CH-CH_2^+$ is a $1^{\circ}$ carbocation.
$(III)$ $(CH_3)_3C^+$ is a $3^{\circ}$ carbocation.
Since the stability order is $(II) < (I) < (III)$,the reactivity order in $S_N1$ reaction is $(II) < (I) < (III)$.

(A) The reactivity of alkyl halides towards $S_{N}2$ reaction depends on the steric hindrance around the electrophilic carbon atom. The order of reactivity is: $Primary > Secondary > Tertiary$.
Among the given options:
$A$. $1-$Bromobutane is a primary alkyl halide with the least steric hindrance.
$B$. $1-$Bromo-$2$-methylbutane is a primary alkyl halide but has more steric hindrance due to the methyl group at the $C-2$ position.
$C$. $1-$Bromo-$2,2$-dimethylpropane is a primary alkyl halide but is highly hindered due to the two methyl groups at the $C-2$ position (neopentyl halide).
$D$. $1-$Bromo-$3$-methylbutane is a primary alkyl halide with steric hindrance further away from the reaction site compared to $B$ and $C$.
Since $1-$Bromobutane has the least steric hindrance,it is the most reactive towards $S_{N}2$ reaction.

(C) The reaction of bromocyclohexane with magnesium in the presence of dry ether forms the Grignard reagent,cyclohexylmagnesium bromide $(A)$.
$C_6H_{11}Br + Mg \xrightarrow{\text{dry ether}} C_6H_{11}MgBr (A)$
Subsequent hydrolysis of the Grignard reagent with water in an acidic medium yields cyclohexane $(B)$ as the final product.
$C_6H_{11}MgBr + H_2O \xrightarrow{H^+} C_6H_{12} (B) + Mg(OH)Br$
Therefore,the product '$B$' is cyclohexane.

(A) When $1, 1-$dichloroethane $(CH_3CHCl_2)$ is heated with dilute $NaOH$,it undergoes hydrolysis to form an unstable gem-diol intermediate,$1, 1-$ethanediol $(CH_3CH(OH)_2)$.
This unstable intermediate loses a water molecule to form acetaldehyde $(CH_3CHO)$.
The reaction is as follows:
$CH_3CHCl_2 + 2NaOH \rightarrow CH_3CH(OH)_2 + 2NaCl$
$CH_3CH(OH)_2 \rightarrow CH_3CHO + H_2O$

(A) The boiling point of haloalkanes increases with an increase in the molecular mass and the number of halogen atoms due to stronger van der Waals forces of attraction.
Comparing the compounds:
$iii$. Chloromethane ($CH_{3}Cl$,molar mass $\approx 50.5 \ g/mol$)
$i$. Bromomethane ($CH_{3}Br$,molar mass $\approx 95 \ g/mol$)
$iv$. Dibromomethane ($CH_{2}Br_{2}$,molar mass $\approx 173.8 \ g/mol$)
$ii$. Bromoform ($CHBr_{3}$,molar mass $\approx 252.7 \ g/mol$)
Thus,the increasing order of boiling points is $iii < i < iv < ii$.

(A) The competing reaction for an $S_N2$ reaction is elimination,not rearrangement.
Substitution and elimination reactions often compete with each other.
Most bases act as nucleophiles and can participate in either substitution or elimination depending on the structure of the alkyl halide and the reaction conditions.
Rearrangements are typically associated with carbocation intermediates in $S_N1$ reactions,not $S_N2$ reactions.

(B) The prolonged exposure of chloroform $(CHCl_{3})$ to air and sunlight leads to its oxidation to form phosgene $(COCl_{2})$,which is a highly poisonous gas.
Phosgene causes damage to the liver and other organs upon inhalation or exposure.
The chemical reaction is:
$2CHCl_{3} + O_{2} \xrightarrow{\text{Sunlight}} 2COCl_{2} + 2HCl$

(A) The rate of $S_N 1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
The stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ}$.
In pair $(i)$,$2$-chloro$-2-$methylpropane forms a $3^{\circ}$ carbocation,which is more stable than the $2^{\circ}$ carbocation formed by $2$-chloropentane.
In pair (ii),$2$-chloropentane forms a $2^{\circ}$ carbocation,which is more stable than the $1^{\circ}$ carbocation formed by $1$-chloropentane.
Therefore,both $2$-chloro$-2-$methylpropane and $2$-chloropentane undergo faster $S_N 1$ reactions compared to their respective counterparts in the given pairs.

(C) The reaction of $1$-Bromo-$3$-chlorocyclobutane with metallic sodium in dry ether is an intramolecular Wurtz reaction. Since both halogen atoms are present on the same ring at positions $1$ and $3$,the sodium metal facilitates the removal of both halogen atoms as $NaBr$ and $NaCl$ (or $NaBr$ and $NaCl$ depending on the specific halogen reactivity,here $Br$ is more reactive),leading to the formation of a new carbon-carbon bond between the $1$ and $3$ positions. This results in the formation of bicyclo[$1.1.0$]butane.

(A) - The mechanism of nucleophilic substitution in haloalkanes,whether $S_{N}2$ or $S_{N}1$,is significantly influenced by the nature of the solvent used in the reaction.
- $S_{N}2$ reactions are favored in polar aprotic solvents (e.g.,acetone,$DMSO$,$DMF$) because they do not solvate the nucleophile strongly.
- $S_{N}1$ reactions are favored in polar protic solvents (e.g.,$H_2O$,alcohols) because they stabilize the carbocation intermediate and the leaving group through solvation.

(B) $R-X + AgCN \longrightarrow R-NC + AgX$
$R-X + KCN \longrightarrow R-CN + KX$
$KCN$ is predominantly ionic in nature,providing $CN^-$ ions in solution. The carbon atom is more nucleophilic,leading to the formation of alkyl cyanide $(R-CN)$.
$AgCN$ is predominantly covalent in nature. The nitrogen atom has a lone pair available for bonding,while the carbon atom is involved in the covalent bond with silver. This leads to the formation of alkyl isocyanide $(R-NC)$.

(A) The reactivity of halides towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed in the rate-determining step.
$S_{N}1$ reactivity order is generally $3^{\circ} > 2^{\circ} > 1^{\circ}$.
In $C_{6}H_{5}CH_{2}Cl$,the loss of $Cl^{-}$ generates a benzyl carbocation $(C_{6}H_{5}CH_{2}^{+})$,which is highly stabilized by resonance with the benzene ring.
$CH_{3}CH_{2}Cl$ and $CH_{3}CH_{2}CH_{2}CH_{2}I$ form primary carbocations,which are less stable.
$C_{6}H_{5}Cl$ does not undergo $S_{N}1$ reaction easily due to the partial double bond character of the $C-Cl$ bond and the instability of the phenyl cation.
Therefore,$C_{6}H_{5}CH_{2}Cl$ shows the highest reactivity towards $S_{N}1$ reaction.

(D) In $S_{N}2$ reactions,the nucleophile attacks the carbon atom bearing the leaving group from the backside.
In tertiary alkyl halides,the central carbon is surrounded by three bulky alkyl groups.
These bulky groups create significant $steric \text{ } hindrance$,which blocks the path of the incoming nucleophile.
Due to this,the rate of $S_{N}2$ reaction is extremely low for tertiary alkyl halides.
Therefore,the order of reactivity for $S_{N}2$ mechanism is $1^{\circ} > 2^{\circ} > 3^{\circ}$.

(A) Alkyl halides $(R-X)$ are polar in nature because the halogen atom is more electronegative than the carbon atom,creating a dipole with a partial positive charge on carbon and a partial negative charge on the halogen.
They do not form hydrogen bonds with water molecules,which makes them insoluble in water.
They typically undergo substitution or elimination reactions,not addition reactions.

(B) Dehydrohalogenation is a $\beta$-elimination reaction where an alkyl halide reacts with a strong base to form an alkene.
The rate of this reaction depends on the strength of the $C-X$ bond.
The bond dissociation energy follows the order $C-F > C-Cl > C-Br > C-I$.
Since the $C-I$ bond is the weakest,it is the easiest to break,making the alkyl iodide the most reactive.
Therefore,the order of reactivity is $R-I > R-Br > R-Cl > R-F$.

(C) Let us analyze each reaction:
$I.$ $RX + AgOH(aq) \longrightarrow ROH + AgX$. The product is an alcohol $(ROH)$,not an alkane $(RH)$. So,$I$ is incorrect.
$II.$ $RX + AgCN(alc) \longrightarrow RNC + AgX$. This is a correct reaction where alkyl isocyanide is formed.
$III.$ $RX + KCN(alc) \longrightarrow RCN + KX$. The product is alkyl cyanide $(RCN)$,not isocyanide $(RNC)$. So,$III$ is incorrect.
$IV.$ $RX + Na(ether) \longrightarrow R-R + 2NaX$. This is the Wurtz reaction,which correctly forms an alkane $(R-R)$.
Thus,pairs $II$ and $IV$ are correctly matched.

(C) $S_{N}1$ reactions proceed via the formation of a carbocation intermediate.
The rate of the reaction depends only on the concentration of the substrate,not on the concentration of the nucleophile.
$3^{\circ}$-alkyl halides are most reactive towards $S_{N}1$ due to the stability of the resulting carbocation.
$1^{\circ}$-alkyl halides generally undergo $S_{N}2$ reactions because they form unstable carbocations.
Polar solvents stabilize the transition state and the carbocation intermediate,thus favouring $S_{N}1$ reactions.
Therefore,the statement that $1^{\circ}$-alkyl halides generally react through $S_{N}1$ reaction is incorrect.

(C) The hydrolysis of $t$-butyl bromide $((CH_3)_3CBr)$ with aqueous $NaOH$ follows the $S_{N}1$ mechanism.
In the $S_{N}1$ mechanism,the rate-determining step is the formation of a carbocation intermediate,which depends only on the concentration of the substrate ($t$-butyl bromide).
Rate $= k[(CH_3)_3CBr]$.
Since the rate is independent of the concentration of the nucleophile $(OH^-)$,the rate of the reaction does not change when the concentration of alkali is doubled.
Therefore,the statement that the rate of the reaction doubles when the concentration of alkali is doubled is false.

(A) The general reaction is $R-Cl + KCN \rightarrow R-CN + KCl$.
Propanenitrile is $CH_3CH_2CN$.
For this to form,the alkyl group $R$ must be $CH_3CH_2-$.
Therefore,$R-Cl$ is $CH_3CH_2Cl$,which is chloroethane.

(D) Methyl bromide $(CH_3Br)$ is a primary alkyl halide.
In the reaction with aqueous sodium hydroxide $(NaOH)$,the nucleophile $(OH^-)$ attacks the electrophilic carbon from the backside.
Since the reaction rate depends on the concentration of both the substrate $(CH_3Br)$ and the nucleophile $(OH^-)$,it follows a second-order kinetics,which is characteristic of an $S_{N}2$ mechanism.
This process involves a single-step transition state and results in the inversion of configuration.

(A) When ethyl bromide $(C_{2}H_{5}Br)$ reacts with alcoholic $KCN$,a nucleophilic substitution reaction occurs where the cyanide ion $(CN^-)$ acts as an ambident nucleophile and attacks through the carbon atom to form an alkyl cyanide (nitrile).
The reaction is: $C_{2}H_{5}Br + KCN (\text{alc.}) \rightarrow C_{2}H_{5}CN + KBr$.
Here,$C_{2}H_{5}CN$ is propane nitrile (also known as propanenitrile).

(B) Solvolysis proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate. The rate of solvolysis depends on the stability of the carbocation formed.
$(a)$ $C_6H_5Cl$: The carbocation formed would be a phenyl cation,which is highly unstable.
$(b)$ $C_6H_5CH_2Cl$: The carbocation formed is $C_6H_5CH_2^+$,which is resonance-stabilized by the benzene ring (benzylic carbocation). This is the most stable among the options.
$(c)$ $C_6H_5CH_2CH_2Cl$: The carbocation formed is a primary alkyl carbocation,which is less stable than the benzylic carbocation.
$(d)$ $CH_2=CHCl$: The carbocation formed would be a vinyl cation,which is highly unstable due to the $sp$ hybridization of the positively charged carbon.
Therefore,$C_6H_5CH_2Cl$ readily undergoes solvolysis.

(A) The reactivity of alkyl halides in $S_N2$ reactions is primarily governed by steric hindrance. The nucleophile attacks from the side opposite to the leaving group. As the number of bulky groups attached to the electrophilic carbon increases,the steric hindrance increases,making the $S_N2$ reaction slower.
The compounds are:
$1$. $C_6H_5CH_2Br$ (Primary,least hindered)
$2$. $C_6H_5CH(CH_3)Br$ (Secondary,more hindered than primary)
$3$. $C_6H_5CH(C_6H_5)Br$ (Secondary,more hindered than $C_6H_5CH(CH_3)Br$ due to the bulky phenyl group)
$4$. $C_6H_5C(CH_3)(C_6H_5)Br$ (Tertiary,most hindered)
Thus,the order of reactivity in $S_N2$ is: $C_6H_5C(CH_3)(C_6H_5)Br < C_6H_5CH(C_6H_5)Br < C_6H_5CH(CH_3)Br < C_6H_5CH_2Br$.

(D) The bond length of $C-Cl$ depends on the hybridization of the carbon atom attached to the chlorine atom.
$1$. In $p-Nitrochlorobenzene$,$Chlorobenzene$,and $CH_2=CH-Cl$,the carbon atom attached to $Cl$ is $sp^2$ hybridized.
$2$. In $3-Chlorocyclohexene$,the carbon atom attached to $Cl$ is $sp^3$ hybridized.
$3$. The bond length increases as the $s$-character of the hybrid orbital decreases. $sp^3$ hybridized carbon has $25\%$ $s$-character,while $sp^2$ hybridized carbon has $33.3\%$ $s$-character.
$4$. Therefore,the $C-Cl$ bond in $3-Chlorocyclohexene$ is the longest due to the $sp^3$ hybridization of the carbon atom.

(D) $2-$Bromobutane is a secondary alkyl halide. In the presence of aqueous $NaOH$,the reaction proceeds primarily through the $S_N1$ mechanism.
This mechanism involves the formation of a planar carbocation intermediate.
The nucleophile $(OH^-)$ can attack the carbocation from either side with equal probability.
This leads to the formation of both enantiomers in equal amounts,resulting in a racemic mixture of $(\pm)$ butan$-2-$ol.

(D) The $Wurtz$ reaction involves the coupling of two alkyl halide molecules in the presence of metallic $Na$ and dry ether to form a symmetrical alkane.
The product $4,5-$diethyl octane has the structure $CH_{3}CH_{2}CH_{2}CH(CH_{2}CH_{3})CH(CH_{2}CH_{3})CH_{2}CH_{2}CH_{3}$.
This molecule is formed by the coupling of two $3-$bromohexane units at the $C-3$ position.
Therefore,the alkyl bromide $(X)$ is $3-$bromohexane,which is $CH_{3}CH_{2}CH_{2}CH(Br)CH_{2}CH_{3}$.

(A) The reaction proceeds in two steps as $HBr$ is in excess:
$1$. The $-OH$ group is substituted by $-Br$ to form allyl bromide: $CH_{2}=CH-CH_{2}-OH + HBr \rightarrow CH_{2}=CH-CH_{2}Br + H_{2}O$.
$2$. The second equivalent of $HBr$ adds across the double bond following Markownikoff's rule to form $1,2-\text{dibromopropane}$: $CH_{2}=CH-CH_{2}Br + HBr \rightarrow CH_{3}-CH(Br)-CH_{2}Br$.

(C) The reaction of $p$-nitroethylbenzene with $Br_2$ in the presence of $UV$ light proceeds via a free radical mechanism. The bromine radical abstracts a hydrogen atom from the benzylic position because the resulting benzylic radical is stabilized by resonance with the benzene ring. Therefore,the major product is $1$-bromo-$1$-($4$-nitrophenyl)ethane.

(B) Alcoholic $KOH$ acts as a dehydrohalogenating agent, which removes a molecule of $HBr$ from the alkyl halide to form an alkene.
When $n-$propyl bromide $(CH_{3}CH_{2}CH_{2}Br)$ reacts with alcoholic $KOH$, it undergoes dehydrohalogenation to produce propene $(CH_{3}CH=CH_{2})$.
The reaction is: $CH_{3}CH_{2}CH_{2}Br + \text{alc. } KOH \longrightarrow CH_{3}CH=CH_{2} + KBr + H_{2}O$.

(B) The reaction involves a tertiary alkyl halide,$(CH_3)_3C-Cl$,and a strong base,$C_2H_5ONa$ (sodium ethoxide).
Since the alkyl halide is sterically hindered (tertiary),the $S_N2$ pathway is unfavorable.
Instead,the base acts as a proton acceptor,leading to an $E2$ elimination reaction.
The base abstracts a proton from one of the $\beta$-carbon atoms,resulting in the formation of an alkene.
The major product is $2$-methylpropene,which is $(CH_3)_2C=CH_2$.