Properties of Haloalkanes Questions in English

(D) The given reaction is $RCl + NaI \xrightarrow{\text{Acetone}} R-I + NaCl$.
This is a halogen exchange reaction where an alkyl chloride or bromide is converted into an alkyl iodide by treatment with sodium iodide in acetone.
This specific reaction is known as the $Finkelstein$ reaction.

(B) The boiling point of alkyl halides depends on the magnitude of van der Waals forces,which increases with the increase in the size and mass of the halogen atom.
As we move from $Cl$ to $Br$ to $I$,the size and molecular mass of the halogen atom increase.
Therefore,the strength of van der Waals forces increases in the order $C_2H_5Cl < C_2H_5Br < C_2H_5I$.
Thus,the correct order of boiling points is $C_2H_5I > C_2H_5Br > C_2H_5Cl$.

(B) The density of haloalkanes increases with an increase in the number of halogen atoms and the atomic mass of the halogen atoms.
For the given chloromethanes,the order of density is $CCl_4 (1.59 \ g/mL) > CHCl_3 (1.48 \ g/mL) > CH_2Cl_2 (1.33 \ g/mL) > CH_3Cl (0.92 \ g/mL)$.
Comparing these with water ($H_2O$,density $\approx 1.00 \ g/mL$),the correct order is $CCl_4 > CHCl_3 > CH_2Cl_2 > H_2O > CH_3Cl$.

(A) The boiling point of alkyl halides depends on two main factors: molecular mass and surface area (branching).
$1$. As the molecular mass increases,the boiling point increases. Thus,$n-$butyl iodide has the highest boiling point among the given options.
$2$. For isomers,the boiling point decreases with an increase in branching because branching reduces the surface area,leading to weaker van der Waals forces.
$3$. Comparing $t-$butyl chloride and $n-$butyl chloride,$t-$butyl chloride is branched,which results in a smaller surface area compared to the straight-chain $n-$butyl chloride.
$4$. Therefore,$t-$butyl chloride has the lowest boiling point among the given options.

(C) Alkyl iodides are sensitive to light and air. On exposure to light,they undergo decomposition to liberate iodine $(I_2)$,which imparts a dark or brown color to the compound. The reaction is: $2R-I \xrightarrow{h\nu} 2R^{\bullet} + I_2$.

(B) The reaction of $n$-propyl bromide $(CH_3CH_2CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
In this reaction,a hydrogen atom is removed from the $\beta$-carbon and the bromine atom is removed from the $\alpha$-carbon,resulting in the formation of a double bond.
The reaction is: $CH_3CH_2CH_2Br + KOH (\text{alc.}) \rightarrow CH_3CH=CH_2 + KBr + H_2O$.
The product formed is propene.

(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 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.

(D) In an $S_N2$ reaction,the rate depends on the strength of the nucleophile. $A$ stronger nucleophile is typically a stronger base,provided the steric hindrance is not significant.
Comparing the given nucleophiles:
$1$. $H_2O$ is a neutral molecule and a weak nucleophile.
$2$. $Cl^-$ is a weak nucleophile in polar protic solvents due to solvation.
$3$. $HCOO^-$ (formate ion) has resonance stabilization,which reduces the availability of the lone pair,making it a weaker nucleophile compared to $OH^-$.
$4$. $OH^-$ is a strong,small,and highly reactive nucleophile.
Therefore,$OH^-$ is the best nucleophile among the given options for an $S_N2$ reaction when $E_2$ competition is ignored.

(D) The rate of $S_{N}2$ reaction depends on the steric hindrance around the carbon atom attached to the leaving group. The order of reactivity for $S_{N}2$ is: $\text{Primary} > \text{Secondary} > \text{Tertiary}$.
Additionally,the nature of the leaving group affects the rate: $I^{-} > Cl^{-}$.
Comparing the options:
$(A)$ $(CH_{3})_{3}C-CH_{2}I$ is a primary alkyl halide with steric hindrance at the $\beta$-carbon.
$(B)$ $(CH_{3})_{3}C-Cl$ is a tertiary alkyl halide,which is very slow for $S_{N}2$.
$(C)$ $(CH_{3})_{2}CH-I$ is a secondary alkyl halide.
$(D)$ $(CH_{3})_{2}CH-CH_{2}CH_{2}CH_{2}I$ is a primary alkyl halide with less steric hindrance compared to $(A)$.
Therefore,$(CH_{3})_{2}CH-CH_{2}CH_{2}CH_{2}I$ is the most reactive towards $S_{N}2$ due to being a primary alkyl halide with the best leaving group $(I^{-})$ and minimal steric hindrance at the $\alpha$-carbon.

(A) The reaction of an alkyl halide with alcoholic $KOH$ is a dehydrohalogenation reaction ($E2$ mechanism).
In $CH_3CD_2CHBr-CH_2CD_3$,the $\alpha$-carbon is the one attached to $Br$.
The $\beta$-hydrogens are available on the adjacent carbons:
$C_2$ (which has $D$ atoms) and $C_4$ (which has $H$ atoms).
However,the $H$ atoms on $C_4$ are more acidic and easier to remove than $D$ atoms on $C_2$ due to the kinetic isotope effect.
Thus,the elimination occurs at the $C_4$ position to form the alkene $CH_3CD_2CH=CH-CD_3$.

(A) An $S_{N}2$ reaction is a bimolecular nucleophilic substitution reaction that proceeds via a single-step concerted mechanism.
It is favored by primary alkyl halides due to minimal steric hindrance.
Option $A$ involves methyl bromide $(CH_3Br)$,which is a primary substrate and highly reactive towards $S_{N}2$ mechanisms.
Option $B$ is a secondary alkyl halide,which typically undergoes $S_{N}1$ or $S_{N}2$ depending on conditions,but $CH_3Br$ is the classic example of $S_{N}2$.
Option $C$ is an elimination reaction.
Option $D$ involves a tertiary alkyl halide,which undergoes $S_{N}1$ due to steric hindrance and carbocation stability.

(B) 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 $(Cl^-)$.
$1$. For $(I)$,$CH_3CH_2CH_2Cl$ forms a $1^{\circ}$ carbocation $(CH_3CH_2CH_2^+)$,which is the least stable.
$2$. For $(III)$,$CH_3CH_2CH(Cl)CH_3$ forms a $2^{\circ}$ carbocation $(CH_3CH_2CH^+CH_3)$,which is more stable than a $1^{\circ}$ carbocation.
$3$. For $(II)$,$CH_2 = CH - CH(Cl)CH_3$ forms an allylic carbocation $(CH_2 = CH - CH^+CH_3)$. This carbocation is resonance-stabilized,making it the most stable among the three.
Therefore,the order of stability of carbocations is $(II) > (III) > (I)$.
Thus,the decreasing order of reactivity towards $S_{N^1}$ is $II > III > I$.

(B) For an $E_2$ reaction to yield a single product,the substrate must have only one type of $\beta$-hydrogen available for abstraction,or the resulting alkenes must be identical due to symmetry.
In $1-$bromo$-1-$methylcyclohexane,the $\beta$-carbons are the $C_2$ and $C_6$ positions of the ring. Due to the symmetry of the molecule,abstraction of a $\beta$-hydrogen from either position leads to the same product: $1-$methylcyclohexene.
Other options like $3-$bromo$-2-$methylpentane and $2-$bromo$-3-$methylpentane have multiple non-equivalent $\beta$-hydrogens,leading to a mixture of isomeric alkenes (Saytzeff and Hofmann products).

(A) Isopropyl chloride is a secondary $(2^{\circ})$ alkyl halide.
Secondary alkyl halides can undergo hydrolysis via both $S_{N}1$ and $S_{N}2$ mechanisms depending on the reaction conditions (solvent,nucleophile concentration,etc.).
In polar protic solvents,the $S_{N}1$ pathway is favored due to the formation of a relatively stable secondary carbocation.
In polar aprotic solvents with strong nucleophiles,the $S_{N}2$ pathway is favored.

(C) Nucleophilic substitution reactions in haloalkenes depend on the stability of the transition state and the nature of the $C-X$ bond.
In $C_6H_5Cl$,$CH_2=CH-Cl$,and $CH_3CH=CH-Cl$,the chlorine atom is directly attached to an $sp^2$ hybridized carbon atom,which makes the $C-X$ bond stronger due to partial double bond character caused by resonance.
In $ClCH_2-CH=CH_2$ (allyl chloride),the chlorine is attached to an $sp^3$ hybridized carbon atom.
Upon loss of the chloride ion,the resulting allyl carbocation $(CH_2=CH-CH_2^+)$ is resonance stabilized.
Therefore,$ClCH_2-CH=CH_2$ is the most reactive towards nucleophilic substitution.

(B) The reactivity of alkyl halides towards the $E_2$ mechanism depends on the stability of the transition state and steric hindrance.
For $E_2$ reactions,the order of reactivity is generally $3^\circ > 2^\circ > 1^\circ$ alkyl halides.
However,in the given options,we must consider the structure of the $\beta$-carbon.
Option $B$ is a $2^\circ$ alkyl halide,which is more reactive than the $1^\circ$ alkyl halides provided in $A$,$C$,and $D$.
Therefore,$CH_3CH_2CH(Br)CH_3$ is the most reactive towards the $E_2$ mechanism among the choices.

(B) The reactivity of alkyl halides towards the $S_{N}1$ mechanism depends on the stability of the carbocation intermediate formed during the rate-determining step.
$S_{N}1$ reactivity order: $\text{Tertiary} > \text{Secondary} > \text{Primary}$.
Among the given options:
$A$: Secondary butyl chloride (Secondary carbocation)
$B$: Tertiary butyl chloride (Tertiary carbocation)
$C$: $n-$butyl chloride (Primary carbocation)
$D$: Allyl chloride (Allylic carbocation)
While the allyl carbocation is resonance-stabilized,the tertiary butyl carbocation is significantly more stable due to the inductive effect and hyperconjugation of nine $\alpha-$hydrogens. Therefore,tertiary butyl chloride is the most reactive towards $S_{N}1$.

(B) Dehydrohalogenation involves the elimination of a hydrogen halide $(HX)$ from an alkyl halide to form an alkene.
The rate-determining step in this reaction involves the breaking of the $C-X$ bond.
The strength of the $C-X$ bond decreases as the size of the halogen atom increases $(C-F > C-Cl > C-Br > C-I)$.
Since the $C-I$ bond is the weakest,it breaks most easily,making alkyl iodides the most reactive towards dehydrohalogenation.
Therefore,the correct order of reactivity is $R-I > R-Br > R-Cl > R-F$.

(B) The reactivity of the $C-X$ bond towards nucleophilic substitution depends on the strength of the $C-X$ bond.
In $CH_2=CH-Cl$ (Vinyl chloride) and $C_6H_5-Cl$ (Chlorobenzene),the $C-X$ bond acquires partial double bond character due to resonance,making it stronger and less reactive.
In $CH_3-CH_2-Cl$ (Chloroethane),the $C-X$ bond is a pure single bond,making it the most reactive.
Comparing Vinyl chloride and Chlorobenzene,the $C-Cl$ bond in vinyl chloride is generally considered less reactive than in chlorobenzene due to the specific hybridization and resonance effects.
Thus,the increasing order of reactivity is: $Chlorobenzene < Vinyl chloride < Chloroethane$.

(D) The rate of $S_{N^1}$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
$S_{N^1}$ reaction follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$ carbocation stability.
$(A)$ $1-$Bromopentane is a $1^{\circ}$ alkyl halide.
$(B)$ $1-$Bromo$-2-$methylbutane is a $1^{\circ}$ alkyl halide.
$(C)$ $2-$Bromopentane is a $2^{\circ}$ alkyl halide.
$(D)$ $2-$Bromo$-2-$methylbutane is a $3^{\circ}$ alkyl halide.
Since $3^{\circ}$ carbocations are the most stable due to inductive effect and hyperconjugation,$2-$Bromo$-2-$methylbutane will form the most stable carbocation and undergo $S_{N^1}$ reaction most easily.

(D) The reaction of $2-$bromopentane with sodium ethoxide $(NaOCH_2CH_3)$ in ethanol is a dehydrohalogenation reaction (an $E2$ elimination reaction).
Since $2-$bromopentane is an unsymmetrical secondary alkyl halide,it undergoes elimination to form alkenes.
The major product is the more stable alkene,which is $2-$pentene.
$2-$pentene exists as two geometric isomers: $trans-2-$pentene and $cis-2-$pentene.
$trans-2-$pentene is more stable than $cis-2-$pentene due to less steric hindrance.
Therefore,both $trans-2-$pentene and $cis-2-$pentene are formed as products.

(A) The substrate is a primary alkyl halide,$(CH_3)_2CH-CH_2Br$ (isobutyl bromide).
The reagent is $CH_3O^- / CH_3OH$,which is a strong nucleophile and a strong base.
For a primary alkyl halide,the $S_N2$ reaction is generally favored over $E2$ because there is less steric hindrance at the $\alpha$-carbon.
Therefore,the nucleophile $CH_3O^-$ attacks the $\alpha$-carbon,displacing the bromide ion to form an ether.
The major product is $(CH_3)_2CH-CH_2OCH_3$ (methyl isobutyl ether).

(B) The reaction involves a secondary alkyl fluoride $(CH_3CH_2CH(F)CH_3)$ reacting with a strong base/nucleophile $(CH_3O^-)$ in a protic solvent $(CH_3OH)$.
Since $CH_3O^-$ is a strong base,the elimination reaction $(E2)$ is favored over substitution $(S_N2)$.
According to Zaitsev's rule,the more substituted alkene is the major product.
Elimination of $H$ from $C_2$ gives $CH_3CH=CHCH_3$ (but$-2-$ene),which is more stable due to hyperconjugation.
Elimination of $H$ from $C_4$ gives $CH_3CH_2CH=CH_2$ (but$-1-$ene),which is less stable.
Therefore,the major product is $CH_3CH=CHCH_3$.

(B) The hydrogenation of the alkene gives $2-$methylbutane,which has the structure $CH_3-CH(CH_3)-CH_2-CH_3$.
Since the alkyl bromide gives only one alkene upon reaction with sodium ethoxide (dehydrohalogenation),we look for a structure that eliminates to form a single product.
$1-$Bromo$-2-$methylbutane $(CH_3-CH_2-CH(CH_3)-CH_2Br)$ would undergo elimination to form $2-$methylbut$-1-$ene.
$2-$Bromo$-2-$methylbutane would form a mixture of alkenes.
$1-$Bromo$-2,2-$dimethylpropane cannot undergo $E2$ elimination easily as it has no $\beta-$hydrogen.
Thus,the correct alkyl halide is $1-$bromo$-2-$methylbutane.

(A) The $S_N2$ reaction proceeds with inversion of configuration at the chiral center where the nucleophilic substitution occurs.
In the product shown,the $-CN$ group is on the same side as the $-H$ atom at the $C-1$ position,meaning the $-CN$ group is in the cis position relative to the $-CH_3$ group at $C-2$.
Since $S_N2$ involves inversion,the starting material must have the leaving group $(-I)$ in the trans position relative to the $-CH_3$ group.
Therefore,the starting material is trans $-1-$ iodo $-2-$ methylcyclopentane.

(B) The $S_{N}2$ reaction mechanism is highly sensitive to steric hindrance. The reactivity order for $S_{N}2$ is: $Primary > Secondary > Tertiary$.
However,allylic $(CH_2=CH-CH_2X)$ and benzylic $(PhCH_2X)$ halides are highly reactive in $S_{N}2$ reactions due to the stabilization of the transition state by resonance.
Comparing the given compounds:
$1$. $(CH_3)_2CHX$ is a secondary halide (least reactive).
$2$. $CH_3CH_2X$ is a primary halide.
$3$. $CH_2=CH-CH_2X$ (allyl halide) and $PhCH_2X$ (benzyl halide) are more reactive than primary halides due to resonance stabilization of the transition state.
Between allyl and benzyl,the benzyl group is generally more reactive due to extended conjugation.
The correct order is: $(CH_3)_2CHX < CH_3CH_2X < CH_2=CH-CH_2X < PhCH_2X$.

(D) The $S_N2$ reaction mechanism proceeds via a concerted pathway involving a backside attack by the nucleophile,which leads to the Walden inversion (inversion of configuration).
This mechanism is highly sensitive to steric hindrance.
Primary alkyl halides $(1^{\circ})$ are the most reactive towards $S_N2$ due to minimal steric hindrance.
Among the given options:
$(A) (C_2H_5)_2CH-Cl$ is a secondary $(2^{\circ})$ alkyl halide.
$(B) (CH_3)_3C-Cl$ is a tertiary $(3^{\circ})$ alkyl halide.
$(C) (CH_3)_2CH-Cl$ is a secondary $(2^{\circ})$ alkyl halide.
$(D) CH_3Cl$ is a methyl halide $(0^{\circ})$,which is the least sterically hindered and shows the most efficient inversion.
However,in the context of typical textbook examples for $S_N2$ inversion,primary alkyl halides are the standard. Since $CH_3Cl$ is the simplest,it undergoes $S_N2$ most readily.

(B) The rate of $S_{N^1}$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
More stable carbocations are formed more easily,leading to a faster $S_{N^1}$ reaction.
The order of stability of carbocations is: $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$.
Option $A$ $(C_6H_5Cl)$ is an aryl halide,which is very unreactive towards $S_{N^1}$ due to partial double bond character.
Option $B$ is a $2^{\circ}$ alkyl halide $(CH_3CH_2CH_2CH(Cl)CH_2CH_3)$,which forms a $2^{\circ}$ carbocation.
Options $C$ and $D$ are $1^{\circ}$ alkyl halides,which form less stable $1^{\circ}$ carbocations.
Therefore,the $2^{\circ}$ alkyl halide in option $B$ will undergo $S_{N^1}$ reaction most easily among the given choices.

(A) The reactivity towards nucleophilic substitution ($S_N1$ or $S_N2$) depends on the stability of the carbocation formed (for $S_N1$) or steric hindrance (for $S_N2$).
$(I)$ is a secondary alkyl chloride,which is reactive.
$(II)$ is an allylic bromide $(CH_2=CH-CH(Br)-CH_3)$. Allylic halides are highly reactive due to resonance stabilization of the carbocation.
$(III)$ is an allylic chloride $(CH_2=CH-CH(Cl)-CH_3)$. It is also reactive but less than the bromide $(II)$ because $Br^-$ is a better leaving group than $Cl^-$.
$(IV)$ is a vinylic bromide $(CH_3-CH=C(Br)-CH_3)$. Vinylic halides are extremely unreactive towards nucleophilic substitution because the $C-X$ bond has partial double bond character due to resonance.
Comparing the leaving groups,$Br^-$ is a better leaving group than $Cl^-$. Thus,the order of reactivity is $II > III > I > IV$.

(D) The solvolysis reaction in formic acid proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate.
The rate of reaction depends on the stability of the carbocation formed after the loss of the chloride ion $(Cl^-)$.
$(I)$ forms a primary allylic carbocation: $CH_2=CH-CH_2^+$.
$(II)$ forms a secondary allylic carbocation: $CH_2=CH-CH^+(CH_3)$,which is more stable than $(I)$ due to the $+I$ effect of the methyl group.
$(III)$ forms a primary allylic carbocation: $CH_2=C(CH_3)CH_2^+$,which is more stable than $(I)$ due to the $+I$ effect of the methyl group,but less stable than $(II)$ because it is primary.
$(IV)$ forms a primary allylic carbocation: $CH_3CH=CH-CH_2^+$,which is more stable than $(I)$ due to the $+I$ effect of the methyl group,but less stable than $(II)$ and $(III)$ due to the position of the methyl group.
Comparing the stability: $(II) > (III) > (IV) > (I)$.
Wait,re-evaluating: $(II)$ is secondary allylic (most stable). $(III)$ is primary allylic with a methyl group at the $2$-position (more stable than $(IV)$). $(IV)$ is primary allylic with a methyl group at the $3$-position. $(I)$ is primary allylic (least stable).
Therefore,the order is $(II) > (III) > (IV) > (I)$.

(C) $1$. The formation of a red precipitate with ammoniacal $Cu_2Cl_2$ indicates the presence of a terminal alkyne $(CH_3-C \equiv CH)$.
$2$. Reaction of $X$ with alcoholic $KOH$ (dehydrohalogenation) yields a terminal alkyne,suggesting $X$ is a geminal or vicinal dihalide.
$3$. Reaction of $X$ with aqueous $KOH$ (hydrolysis) gives an aldehyde. $A$ geminal dihalide $(R-CHX_2)$ on hydrolysis yields an aldehyde $(R-CHO)$.
$4$. For a three-carbon chain,$1,1-$dichloropropane $(CH_3-CH_2-CHCl_2)$ would yield propanal,and $2,2-$dichloropropane $(CH_3-CCl_2-CH_3)$ would yield acetone. However,the question specifies an aldehyde,and $1,1-$dichloropropane is the correct structure for $X$ to form propyne upon dehydrohalogenation $(CH_3-CH_2-CHCl_2 \xrightarrow{alc. KOH} CH_3-C \equiv CH)$.
$5$. Therefore,$X$ is $1,1-$dichloropropane.

(A) The reactivity of halides towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed.
More stable the carbocation,faster is the $S_{N}1$ reaction.
Stability order of carbocations:
$(3) \ (C_6H_5)_2C^+(CH_3)$ is a tertiary carbocation stabilized by two phenyl rings and one methyl group.
$(2) \ (C_6H_5)_2CH^+$ is a secondary carbocation stabilized by two phenyl rings.
$(1) \ (CH_3)_3C^+$ is a tertiary carbocation stabilized by three methyl groups (inductive effect).
$(4) \ (CH_3)_2CH^+$ is a secondary carbocation stabilized by two methyl groups.
$(5) \ C_2H_5^+$ is a primary carbocation.
Comparing the stability: $(3) > (2) > (1) > (4) > (5)$.
Therefore,the order of reactivity is $(3) > (2) > (1) > (4) > (5)$.

(C) The reaction involves the nucleophilic substitution of a chiral haloalkane with aqueous $NaOH$.
Since the substrate is a secondary alkyl halide,the reaction proceeds via an $S_N2$ mechanism.
In an $S_N2$ reaction,the nucleophile $(OH^-)$ attacks from the side opposite to the leaving group $(Cl^-)$,resulting in the inversion of configuration (Walden inversion).
Starting with the given Fischer projection,the $Cl$ atom is on the right.
After the $S_N2$ inversion,the $OH$ group will be on the left side,while the other groups ($H$ and $CH_3$) remain in their positions.
Therefore,the correct product is the structure with $HO$ on the left,$H$ on top,$I$ on the right,and $CH_3$ on the bottom.

(A) The total rate of reaction is the sum of the rates of $S_N2$ and $S_N1$ mechanisms.
Rate $= \text{Rate}_{S_N2} + \text{Rate}_{S_N1}$
Given: $\text{Rate}_{S_N2} = 3.20 \times 10^{-5} [2\text{-bromobutane}][OH^-]$
Given: $\text{Rate}_{S_N1} = 1.5 \times 10^{-6} [2\text{-bromobutane}]$
For $[OH^-] = 1.0\,M$:
$\text{Rate}_{S_N2} = 3.20 \times 10^{-5} [2\text{-bromobutane}] \times 1.0 = 3.20 \times 10^{-5} [2\text{-bromobutane}]$
$\text{Rate}_{S_N1} = 1.5 \times 10^{-6} [2\text{-bromobutane}]$
Total Rate $= (3.20 \times 10^{-5} + 0.15 \times 10^{-5}) [2\text{-bromobutane}] = 3.35 \times 10^{-5} [2\text{-bromobutane}]$
Percentage of $S_N2 = (\text{Rate}_{S_N2} / \text{Total Rate}) \times 100$
Percentage of $S_N2 = (3.20 \times 10^{-5} / 3.35 \times 10^{-5}) \times 100 \approx 95.52\% \approx 96\%$.

(B) The reaction is an $S_N2$ mechanism.
$S_N2$ reactions proceed with inversion of configuration at the chiral center where the leaving group is attached.
In the starting material,the $Br$ atom is on one side of the cyclohexane ring (let's assume it is 'up').
During the $S_N2$ attack by $OH^-$,the nucleophile attacks from the side opposite to the leaving group $(Br^-)$.
This results in the $OH$ group being on the opposite side of the ring relative to the original position of the $Br$ atom.
Since the $CH_3$ group is at the $4$-position,the relative stereochemistry between the $CH_3$ group and the $OH$ group will be inverted compared to the $CH_3$ and $Br$ groups.
Therefore,the product $A$ is the trans-isomer of $4$-methylcyclohexanol.

(D) The reactivity of a compound towards $S_{N^1}$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Cl^-)$.
$A$: Chlorocyclohexane forms a secondary $(2^{\circ})$ carbocation.
$B$: $3-$Chlorocyclohexene forms an allylic carbocation,which is stabilized by resonance.
$C$: $1-$Methyl$-1-$chlorocyclohexane forms a tertiary $(3^{\circ})$ carbocation.
$D$: $1-$Methyl$-3-$chlorocyclohexene forms an allylic carbocation that is also tertiary $(3^{\circ})$,making it the most stable carbocation among the given options due to both resonance stabilization and the inductive effect of the methyl group.
Therefore,the compound in option $D$ is the most reactive towards $S_{N^1}$ reaction.

(B) $1$. Compound $A$ has the formula $C_5H_9Cl$. The degree of unsaturation is $5 - (9/2) + (1/2) + 1 = 2$. Since it does not react with $Br_2 / CCl_4$,it must be a bicyclic compound or have two rings.
$2$. Reaction with a strong base gives $B$ $(C_5H_8)$,which reacts with $Br_2 / CCl_4$,indicating $B$ is an alkene. This implies $A$ is a cyclic alkyl halide that undergoes dehydrohalogenation to form a cyclic alkene.
$3$. Ozonolysis of $B$ $(C_5H_8)$ gives $C_5H_8O_2$. This indicates the ring opened to form a dicarbonyl compound.
$4$. Based on the options,$A$ is $1$-chloro-$2$-methylcyclobutane. Dehydrohalogenation gives $1$-methylcyclobutene $(B)$. Ozonolysis of $1$-methylcyclobutene gives $5$-oxopentanal $(C_5H_8O_2)$.

(D) The given molecule is $3$-iodo-$3$-methyloctane.
In an $E_2$ elimination reaction,the base abstracts a proton from the $\beta$-carbon atoms adjacent to the carbon bearing the iodine atom.
The $\beta$-carbons are:
$1$. The $CH_2$ group of the ethyl chain ($C$-$2$ position).
$2$. The $CH_2$ group of the pentyl chain ($C$-$4$ position).
$3$. The $CH_3$ group attached to the $C$-$3$ position.
Elimination from the ethyl group gives $3$-methyl-oct-$2$-ene (which can exist as $E$ and $Z$ isomers).
Elimination from the pentyl group gives $3$-methyl-oct-$3$-ene (which can exist as $E$ and $Z$ isomers).
Elimination from the methyl group gives $2$-ethyl-hept-$1$-ene.
Total products = $2$ (from ethyl) + $2$ (from pentyl) + $1$ (from methyl) = $5$ products.

(C) $1$. Reaction of $C_2H_5Br$ with alcoholic $KOH$ is a dehydrohalogenation reaction,which produces ethene $(C_2H_4)$,not ethanol $(C_2H_5OH)$.
$2$. Reaction of $C_2H_5Br$ with metallic sodium (Wurtz reaction) produces butane $(C_4H_{10})$,not ethane.
$3$. Reaction of $C_2H_5Br$ with sodium ethoxide $(C_2H_5ONa)$ is a Williamson ether synthesis,which produces diethyl ether $(C_2H_5-O-C_2H_5)$. This statement is correct.
$4$. Reaction of $C_2H_5Br$ with $AgCN$ produces ethyl isocyanide $(C_2H_5NC)$,not ethyl cyanide $(C_2H_5CN)$.

(D) The reaction of ethyl bromide $(C_2H_5Br)$ with alcoholic silver nitrite $(AgNO_2)$ is a nucleophilic substitution reaction.
$AgNO_2$ is a covalent compound,and the nitrogen atom acts as the nucleophile.
Therefore,the reaction proceeds as follows:
$C_2H_5Br + AgNO_2 \rightarrow C_2H_5NO_2 + AgBr$
The product formed is nitroethane $(C_2H_5NO_2)$.

(A) $2$-bromopentane $(CH_3CH_2CH_2CH(Br)CH_3)$ undergoes dehydrohalogenation in the presence of a strong base like potassium ethoxide $(C_2H_5OK)$ in ethanol.
According to Zaitsev's rule,the more substituted alkene is the major product.
The reaction can proceed via $E2$ mechanism to form pent-$2$-ene.
Since trans-pent-$2$-ene is more stable than cis-pent-$2$-ene due to less steric hindrance,it is the major product.

(C) The reaction of alkyl halides with $Na$ in dry ether is known as the $Wurtz$ reaction.
To obtain $isobutane$ $(CH_3-CH(CH_3)-CH_3)$,we need to combine a $methyl$ group $(CH_3-)$ and an $isopropyl$ group $(-CH(CH_3)_2)$.
Therefore,the reactants must be $methyl$ $chloride$ $(CH_3Cl)$ and $isopropyl$ $chloride$ $(CH_3-CHCl-CH_3)$.
The reaction is: $CH_3Cl + CH_3-CHCl-CH_3 + 2Na \xrightarrow{dry \ ether} CH_3-CH(CH_3)_2 + 2NaCl$.

(A) The reaction of ethyl bromide $(C_2H_5Br)$ with lead-sodium alloy $(Na_4Pb)$ is a classic method for the preparation of tetraethyl lead $(TEL)$.
The chemical equation is: $4C_2H_5Br + 4Na + Pb \rightarrow (C_2H_5)_4Pb + 4NaBr + 3Na$.
Thus,the product formed is tetraethyl lead.

(B) The reaction of $t-$butyl bromide $((CH_3)_3CBr)$ with sodium methoxide $(NaOCH_3)$ involves a tertiary alkyl halide and a strong base/nucleophile.
Since $t-$butyl bromide is a tertiary halide,it undergoes an $E2$ elimination reaction rather than an $S_N2$ substitution reaction due to steric hindrance.
Sodium methoxide acts as a base,abstracting a proton from one of the $\beta-$carbons to form $2-$methylpropene (iso-butylene) as the major product.
Therefore,the correct option is $B$.

(B) The $S_{N}2$ mechanism is a concerted,single-step process.
In this mechanism,the nucleophile attacks the carbon atom from the side opposite to the leaving group.
This leads to the formation of a pentacoordinate $S_{N}2$ transition state where the bond to the incoming nucleophile is forming and the bond to the leaving group is breaking simultaneously.
Therefore,the reaction proceeds through a transition state.

(B) The reaction of propylbenzene $(C_6H_5-CH_2-CH_2-CH_3)$ with bromine $(Br_2)$ in the presence of light $(h\nu)$ and heat is a free radical substitution reaction.
In alkylbenzenes,the benzylic hydrogen is the most reactive due to the stability of the resulting benzylic radical.
The benzylic radical is stabilized by resonance with the benzene ring.
Therefore,the bromine atom replaces the hydrogen atom at the benzylic position (the carbon atom directly attached to the benzene ring).
Thus,the product formed is $1$-phenyl-$1$-bromopropane,which is $C_6H_5-CH(Br)-CH_2-CH_3$.

(D) The reaction involves a nucleophilic substitution $(S_N2)$ reaction.
$p-$bromobenzyl chloride contains a primary alkyl halide group $(-CH_2Cl)$ attached to the benzene ring.
The $CN^-$ ion acts as a nucleophile and attacks the electrophilic carbon of the $-CH_2Cl$ group,displacing the chloride ion $(Cl^-)$.
The bromine atom attached to the benzene ring is much less reactive towards nucleophilic substitution under these conditions compared to the benzylic chloride.
Therefore,the product formed is $4-$bromobenzyl cyanide $(Br-C_6H_4-CH_2CN)$.

(D) The conversion of ethylbenzene $(C_6H_5CH_2CH_3)$ to styrene $(C_6H_5CH=CH_2)$ requires a two-step process:
$1$. Benzylic bromination using $NBS$ ($N$-Bromosuccinimide) in the presence of light or peroxide to form $C_6H_5CH(Br)CH_3$.
$2$. Dehydrohalogenation using alcoholic $KOH$ $(alc. KOH)$ to eliminate $HBr$ and form the double bond,resulting in styrene $(C_6H_5CH=CH_2)$.

(C) The $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate.
The stability of the carbocation determines the reactivity of the substrate.
In the given compounds,the carbocations formed are benzyl carbocations substituted at the $p$-position.
$(II)$ $p$-Methoxybenzyl chloride forms a $p$-methoxybenzyl carbocation,where the $-OCH_{3}$ group exerts a strong $+M$ effect,significantly stabilizing the carbocation.
$(I)$ Benzyl chloride forms a benzyl carbocation with no substituent,providing moderate stability.
$(III)$ $p$-Nitrobenzyl chloride forms a $p$-nitrobenzyl carbocation,where the $-NO_{2}$ group exerts a strong $-I$ and $-M$ effect,which destabilizes the carbocation.
Therefore,the order of stability of the carbocations (and thus reactivity) is: $p$-Methoxybenzyl carbocation > Benzyl carbocation > $p$-Nitrobenzyl carbocation.
The correct order is $II > I > III$.

(B) The starting material is $4-bromobenzyl$ chloride $(Cl-CH_2-C_6H_4-Br)$.
Step $1$: Reaction with $Mg$ in ether forms the Grignard reagent,$4-chloromethylphenylmagnesium$ bromide $(Cl-CH_2-C_6H_4-MgBr)$,which is $(X)$.
Step $2$: Reaction of the Grignard reagent with heavy water $(D_2O)$ replaces the $-MgBr$ group with a deuterium atom $(-D)$.
The final product $(Y)$ is $1-chloromethyl-4-deuteriobenzene$ $(Cl-CH_2-C_6H_4-D)$.