Properties of Haloalkanes Questions in English

(B) The reaction of $tert$-butyl bromide $((CH_3)_3CBr)$ with silver fluoride $(AgF)$ is a Swarts reaction,which is used for the synthesis of alkyl fluorides.
However,$tert$-butyl bromide is a tertiary alkyl halide. When heated with $AgF$,it undergoes an $E1$ elimination reaction rather than a nucleophilic substitution ($S_N1$ or $S_N2$) because the tertiary carbocation is highly stable and the fluoride ion acts as a base.
The major product formed is $2$-methylpropene (isobutylene) due to the elimination of $HBr$.

(C) The reaction between sodium ethoxide $(CH_3CH_2ONa)$ and isopropyl chloride $((CH_3)_2CHCl)$ is a base-promoted elimination reaction.
Since sodium ethoxide is a strong base and isopropyl chloride is a secondary alkyl halide,the $E2$ elimination mechanism predominates over the $S_N2$ substitution mechanism.
The base abstracts a $\beta$-hydrogen from the isopropyl chloride,leading to the formation of propene $(CH_3-CH=CH_2)$,ethanol $(CH_3CH_2OH)$,and sodium chloride $(NaCl)$.
Therefore,the product '$X$' is propene.

(D) The reaction of an alkyl halide with alcoholic $KCN$ is a nucleophilic substitution reaction ($S_N2$ mechanism) that produces an alkyl cyanide (nitrile).
To obtain butanenitrile $(CH_3CH_2CH_2CH_2CN)$,which contains $4$ carbon atoms in the chain plus the nitrile carbon,we need a starting alkyl halide with $4$ carbon atoms.
$n-$Propylchloride $(CH_3CH_2CH_2Cl)$ has $3$ carbons,so it would yield butanenitrile $(CH_3CH_2CH_2CN)$ which is $3$ carbons in the chain plus the nitrile carbon,totaling $4$ carbons.
Wait,butanenitrile is $CH_3CH_2CH_2CN$. This contains $4$ carbon atoms in total.
Therefore,$n-$propylchloride $(CH_3CH_2CH_2Cl)$ reacts with $KCN$ to form $CH_3CH_2CH_2CN$ (butanenitrile).
Reaction: $CH_3CH_2CH_2Cl + KCN \text{ (alc.)} ightarrow CH_3CH_2CH_2CN + KCl$.

(A) The reactivity of alkyl halides towards $S_N2$ reactions follows the order: $Primary > Secondary > Tertiary$.
This is due to the steric hindrance,which increases from primary to tertiary alkyl halides,making it difficult for the nucleophile to attack the electrophilic carbon.
Among the given options:
$(A)$ $n$-Butyl iodide is a primary $(1^{\circ})$ alkyl halide.
$(B)$ $sec$-Butyl iodide is a secondary $(2^{\circ})$ alkyl halide.
$(C)$ Isobutyl iodide is a primary $(1^{\circ})$ alkyl halide,but it has steric hindrance at the $\beta$-carbon.
$(D)$ $tert$-Butyl iodide is a tertiary $(3^{\circ})$ alkyl halide.
Since $n$-Butyl iodide is a primary alkyl halide with the least steric hindrance,it shows the highest reactivity for $S_N2$ reactions.

(B) Racemization occurs in $SN^1$ reactions when the substrate forms a stable,planar carbocation intermediate that is chiral.
Among the given options,$CH_3-CH_2-CH(Cl)-CH_3$ ($2$-chlorobutane) is a secondary alkyl halide that can form a planar carbocation $(CH_3-CH_2-CH^+-CH_3)$ upon the loss of the chloride ion.
This carbocation is prochiral and can be attacked by the nucleophile from either side,leading to a racemic mixture.
$CH_3-CH(Cl)-CH_3$ (isopropyl chloride) forms a less stable secondary carbocation and is less prone to $SN^1$ compared to $2-$chlorobutane.
$CH_3-CH_2-CH(Cl)-CH_2-CH_3$ ($3$-chloropentane) forms a carbocation that is achiral due to symmetry,so it does not result in a racemic mixture.
$(CH_3)_3C-CH_2-Cl$ is a primary alkyl halide and typically undergoes $SN^2$ reactions.

(D) When $2-$Bromobutane $(CH_3CH(Br)CH_2CH_3)$ is treated with an aqueous solution of sodium hydroxide $(NaOH)$,it undergoes a nucleophilic substitution reaction ($S_N1$ or $S_N2$ mechanism depending on conditions,but typically $S_N2$ for secondary halides in aqueous media).
The hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the bromide ion $(Br^-)$.
The reaction is: $CH_3CH(Br)CH_2CH_3 + NaOH_{(aq)} \rightarrow CH_3CH(OH)CH_2CH_3 + NaBr$.
The product formed is Butan$-2-$ol.

(D) The boiling point of haloalkanes increases with an increase in molecular mass and the number of halogen atoms present in the molecule.
This is because the magnitude of van der Waals forces increases as the size and number of halogen atoms increase.
Comparing the given compounds: $CHBr_3$ (tribromomethane) has the highest molecular mass,followed by $CH_2Br_2$ (dibromomethane),$CH_3Br$ (bromomethane),and $CH_3Cl$ (chloromethane) has the lowest molecular mass.
Therefore,the correct decreasing order is $CHBr_3 > CH_2Br_2 > CH_3Br > CH_3Cl$.

(C) The reaction of alkyl halides with ammonia is a nucleophilic substitution reaction $(S_N2)$.
In this reaction,the halide ion $(X^-)$ acts as a leaving group.
The reactivity of alkyl halides depends on the strength of the $C-X$ bond.
The bond dissociation energy decreases as the size of the halogen atom increases,making the $C-I$ bond the weakest and the $C-Cl$ bond the strongest.
Therefore,the leaving group ability follows the order: $I^- > Br^- > Cl^-$.
Thus,the order of reactivity of alkyl halides with ammonia is $R-I > R-Br > R-Cl$.

(D) The reaction of haloalkanes $(R-X)$ with silver nitrite $(AgNO_2)$ leads to the formation of nitroalkanes $(R-NO_2)$ as the major product.
This occurs because $AgNO_2$ is a covalent compound,and the nitrogen atom is more nucleophilic than the oxygen atom.
In contrast,ionic nitrites like $KNO_2$ or $NaNO_2$ primarily yield alkyl nitrites $(R-ONO)$ because the oxygen atom is more nucleophilic in these ionic species.

(D) For a given alkyl group,the boiling point of alkyl halides increases as the size and mass of the halogen atom increase.
This is due to the increase in the magnitude of van der Waals forces with an increase in the size and mass of the halogen.
Thus,the boiling point follows the order: $CH_3I > CH_3Br > CH_3Cl > CH_3F$.
Therefore,$CH_3I$ has the highest boiling point.

(A) The dehydrohalogenation of alkyl halides follows the $E2$ mechanism,where the rate depends on the stability of the resulting alkene.
According to the Saytzeff rule,more substituted alkenes are more stable.
The stability order of alkenes is: $R_2C=CR_2 > R_2C=CHR > R_2C=CH_2 > RCH=CH_2$.
Since $3^{\circ}$ alkyl halides form more substituted (stable) alkenes compared to $2^{\circ}$ and $1^{\circ}$ alkyl halides,the order of ease of dehydrohalogenation is $3^{\circ} > 2^{\circ} > 1^{\circ}$.

(A) The boiling point of haloalkanes increases with an increase in molecular mass due to the increase in the magnitude of van der Waals forces of attraction.
Comparing the given compounds:
$CH_3Cl$ $(50.5 \ g/mol)$
$CH_3Br$ $(95.0 \ g/mol)$
$CH_2Br_2$ $(173.8 \ g/mol)$
$CHBr_3$ $(252.7 \ g/mol)$
Since $CH_3Cl$ has the lowest molecular mass,it has the lowest boiling point.

(B) The bond enthalpy of the $C-X$ bond depends on the bond length. As the size of the halogen atom increases from $F$ to $I$,the $C-X$ bond length increases.
According to the inverse relationship between bond length and bond strength,as the bond length increases,the bond strength decreases.
Since $F$ is the smallest halogen,the $C-F$ bond is the shortest and strongest,requiring the highest energy to break.
Therefore,$CH_3-F$ has the highest bond enthalpy among the given alkyl halides.

(B) The reaction between an alkyl halide $(CH_3CH_2Br)$ and a silver salt of a carboxylic acid $(CH_3COOAg)$ is a nucleophilic substitution reaction.
In this reaction,the acetate ion $(CH_3COO^-)$ acts as a nucleophile and attacks the ethyl group,displacing the bromide ion $(Br^-)$.
The silver ion $(Ag^+)$ reacts with the bromide ion $(Br^-)$ to form a precipitate of silver bromide $(AgBr)$.
The reaction is: $CH_3CH_2Br + CH_3COOAg \xrightarrow{\Delta} CH_3COOCH_2CH_3 + AgBr$.
The product formed is ethyl acetate $(CH_3COOCH_2CH_3)$.

(C) The $S_{N}2$ mechanism proceeds through a transition state where the nucleophile and the leaving group are both partially bonded to the central carbon atom.
$A$ carbocation intermediate is not formed; this is a characteristic of the $S_{N}1$ mechanism.

(B) The boiling point $(B.P.)$ of haloalkanes depends on the magnitude of van der Waals forces,which increase with the size and mass of the halogen atom.
Since all the given compounds contain a single carbon atom $(CH_3X)$,the boiling point depends on the size of the halogen atom $(X)$.
The size of the halogen atoms follows the order: $F < Cl < Br < I$.
Therefore,the boiling point increases in the order: $CH_3F < CH_3Cl < CH_3Br < CH_3I$.
Thus,fluoromethane $(CH_3F)$ has the lowest boiling point.

(B) The reaction of $tert-$butyl bromide with silver fluoride $(AgF)$ is a Swarts reaction,which is used for the synthesis of alkyl fluorides.
In this reaction,the bromine atom in $tert-$butyl bromide is replaced by a fluorine atom to form $2-$fluoro$-2-$methylpropane.
The chemical equation is: $(CH_3)_3CBr + AgF \xrightarrow{\Delta} (CH_3)_3CF + AgBr$.

(D) Alkyl halides $(R-X)$ react with ionic nitrites like $KNO_2$ to form alkyl nitrites $(R-O-N=O)$ as the major product because the oxygen atom is more nucleophilic than the nitrogen atom.
Conversely,reaction with $AgNO_2$ (covalent) typically yields nitroalkanes $(R-NO_2)$.

(C) The reaction of an alkyl halide $(C_2H_5Cl)$ with alcoholic $KCN$ is a nucleophilic substitution reaction.
$KCN$ is an ionic compound that provides $CN^-$ ions in solution,which acts as an ambident nucleophile and attacks the alkyl halide to form an alkyl cyanide $(C_2H_5CN)$.
The reaction is: $C_2H_5Cl + KCN \text{ (alc.)} \xrightarrow{\Delta} C_2H_5CN + KCl$.
Therefore,the compound $Y$ is $KCN$ (alc.).

(D) In the $S_N1$ mechanism,the first step involves the ionization of the substrate to form a planar carbocation intermediate. This intermediate is $sp^2$ hybridized.

(A) The reaction of $3-$bromo$-2-$methylpentane with alcoholic $KOH$ is a dehydrohalogenation reaction ($E2$ mechanism).
According to Saytzeff's rule,the major product is the more substituted alkene.
The starting material is $CH_3-CH(CH_3)-CH(Br)-CH_2-CH_3$.
Removal of $H$ from $C-2$ gives $CH_3-C(CH_3)=CH-CH_2-CH_3$ ($2-$methylpent$-2-$ene).
Removal of $H$ from $C-4$ gives $CH_3-CH(CH_3)-CH=CH-CH_3$ ($4-$methylpent$-2-$ene).
$2-$Methylpent$-2-$ene is more substituted (trisubstituted) compared to $4-$methylpent$-2-$ene (disubstituted),hence it is the major product.

(B) The false statement about $S_N1$ (Substitution Nucleophilic Unimolecular) reaction mechanism is: $(B)$ It is a single-step mechanism.
$S_N1$ reaction mechanism is a two-step process. In the first step,the leaving group departs,generating a carbocation intermediate. In the second step,the nucleophile attacks the carbocation,leading to the formation of the substitution product.
The correct statements regarding $S_N1$ reaction mechanism are:
$(A)$ Intermediate formed during the reaction is a carbocation. This is true as the leaving group departure creates a carbocation intermediate.
$(C)$ Concentration of nucleophile does not affect the rate of the reaction. This is true because the rate-determining step is the formation of the carbocation intermediate,which does not involve the nucleophile.
$(D)$ Racemization takes place if the reaction is carried out at a chiral carbon in an optically active substance. This is true because the carbocation intermediate is planar and can be attacked by nucleophiles from either side,resulting in the formation of both enantiomers.

(B) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group.
More stable carbocations lead to faster $S_{N}1$ reactions.
The order of stability of carbocations is: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
In option $B$,the compound is $3$-chloro-$3$-methylhexane. Upon loss of the chloride ion,it forms a tertiary $(3^{\circ})$ carbocation,which is the most stable among the given options.
Therefore,it undergoes the $S_{N}1$ mechanism most rapidly.

(D) $SN^2$ mechanism is a concerted process that occurs in a single step.
It is favored in primary alkyl halides due to minimal steric hindrance.
In tertiary alkyl halides,steric hindrance is very high,which prevents the nucleophile from attacking the carbon atom.
Therefore,the statement that $SN^2$ mechanism is observed in tertiary alkyl halides is incorrect.

(C) The boiling point of haloalkanes increases with an increase in molecular mass and surface area.
Comparing the given compounds:
$A$: $CH_3-CH_2-CH_2-Cl$ $(M.W. \approx 78.5 \ g/mol)$
$B$: $CH_3-CH_2-Cl$ $(M.W. \approx 64.5 \ g/mol)$
$C$: $CH_3-CH(Cl)-CH_2-CH_3$ $(M.W. \approx 92.5 \ g/mol)$
$D$: $CH_3-Cl$ $(M.W. \approx 50.5 \ g/mol)$
Since $CH_3-CH(Cl)-CH_2-CH_3$ has the highest molecular weight,it exhibits the highest boiling point.

(A) The reaction of alkyl halides $(R-X)$ with silver nitrite $(AgNO_2)$ yields nitro alkanes $(R-NO_2)$ as the major product.
$AgNO_2$ is a covalent compound,and the nitrogen atom is the nucleophilic site,leading to the formation of a $C-N$ bond.
Conversely,potassium nitrite $(KNO_2)$ is an ionic compound,which provides the nitrite ion $(NO_2^-)$ where oxygen is the nucleophilic site,resulting in the formation of alkyl nitrites $(R-ONO)$.

(D) This reaction is known as the $Wurtz$ reaction.
It involves the coupling of two molecules of an alkyl halide in the presence of metallic $Na$ and dry ether.
The general reaction is: $2R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$.
As shown,the resulting alkane $(R-R)$ contains double the number of carbon atoms compared to the original alkyl halide $(R-X)$.

(D) The rate of $SN^2$ reaction depends on two main factors:
$1$. Steric hindrance: The rate of $SN^2$ reaction is inversely proportional to the steric hindrance around the electrophilic carbon. The order of reactivity is $CH_3-X > 1^{\circ} > 2^{\circ} > 3^{\circ}$ alkyl halides.
$2$. Leaving group ability: If the alkyl group is the same,the rate of $SN^2$ reaction is directly proportional to the leaving group ability. The order of leaving group ability is $I^- > Br^- > Cl^- > F^-$.
Comparing the given options:
- Options $A$,$B$,and $D$ are primary $(1^{\circ})$ alkyl halides.
- Option $C$ is a secondary $(2^{\circ})$ alkyl halide,which is slower than primary halides due to higher steric hindrance.
- Among the primary halides ($A$,$B$,and $D$),the leaving group ability determines the rate. Since $I^-$ is the best leaving group among $Cl^-$,$Br^-$,and $I^-$,$CH_3CH_2CH_2CH_2I$ undergoes $SN^2$ reaction most rapidly.

(C) The rate of $SN^2$ reaction depends on steric hindrance and the nature of the leaving group.
$1$. Steric hindrance: $SN^2$ reactions are fastest for primary $(1^{\circ})$ alkyl halides compared to secondary $(2^{\circ})$ or tertiary $(3^{\circ})$ alkyl halides.
$2$. Leaving group: For the same alkyl group,the rate of $SN^2$ reaction is faster with a better leaving group. Bromide $(Br^-)$ is a better leaving group than chloride $(Cl^-)$.
Comparing the options:
- Options $A$ and $B$ are secondary $(2^{\circ})$ alkyl halides.
- Options $C$ and $D$ are primary $(1^{\circ})$ alkyl halides.
Since primary alkyl halides are less sterically hindered,they react faster than secondary ones.
Between $C$ and $D$,$C$ contains a bromide leaving group,which is better than the chloride in $D$.
Therefore,$1$-bromo$-3-$methylpentane undergoes $SN^2$ reaction most rapidly.

(A) The $S_{N}1$ reaction is a unimolecular nucleophilic substitution reaction where the rate-determining step is the formation of a carbocation intermediate.
Since the nucleophile attacks only after the formation of the carbocation,the rate of the $S_{N}1$ reaction is independent of the concentration and strength of the nucleophile.
$A$ more powerful nucleophile actually favours the $S_{N}2$ mechanism,where the nucleophile attacks the substrate simultaneously with the leaving group departure.
Therefore,the statement that a more powerful nucleophile favours $S_{N}1$ is incorrect.

(B) The reaction of alkyl halides with metallic fluorides like $AgF$ is known as the Swarts reaction.
In this reaction,the halogen atom (in this case,$Br$) is replaced by a fluorine atom.
$tert$-butyl bromide is $(CH_3)_3C-Br$.
When it reacts with $AgF$,the $Br$ atom is replaced by $F$ to form $2$-Fluoro-$2$-methylpropane,which is $(CH_3)_3C-F$.

(B) The reaction of haloalkanes with $AgNO_2$ (silver nitrite) yields nitroalkanes as the major product because $AgNO_2$ is a covalent compound,and the nitrogen atom acts as the nucleophilic center.
In contrast,ionic nitrites like $NaNO_2$ or $KNO_2$ primarily yield alkyl nitrites $(R-ONO)$ because the oxygen atom is more nucleophilic in these ionic species.
Therefore,for the conversion of chloroethane $(CH_3CH_2Cl)$ to nitroethane $(CH_3CH_2NO_2)$,the reagent $A$ is silver nitrite $(AgNO_2)$.

(B) The reaction of $2-$Chlorobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction,specifically $E2$).
According to $Saytzeff's$ rule,the major product is the more substituted alkene.
$CH_3-CH_2-CHCl-CH_3 \xrightarrow{Alc. KOH} CH_3-CH=CH-CH_3$ ($But-2-ene$,major product) $+ CH_3-CH_2-CH=CH_2$ ($But-1-ene$,minor product).
Therefore,$But-2-ene$ is the major product.

(A) The $S_N1$ mechanism proceeds via the formation of a planar carbocation intermediate.
Because the carbocation is planar,the nucleophile can attack from either the front side or the back side.
This leads to both inversion and retention of configuration,resulting in a racemic mixture if the substrate is chiral.
Backside attack is a characteristic feature of $S_N2$ mechanisms,not $S_N1$.
Therefore,option $A$ is $NOT$ a feature of $S_N1$ mechanisms.

(C) When $t$-butyl bromide is treated with alcoholic ammonia,it undergoes a dehydrohalogenation reaction (elimination reaction) rather than nucleophilic substitution because $t$-butyl bromide is a tertiary alkyl halide and ammonia acts as a base.
The reaction is:
$(CH_3)_3C-Br + NH_3 \rightarrow CH_3-C(CH_3)=CH_2 + NH_4Br$
The product formed is isobutylene ($2$-methylpropene).

(C) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed.
For $(CH_{3})_{2}CHBr$ (isopropyl bromide),a $2^{\circ}$ carbocation is formed.
Relative rates for $S_{N}1$ reactions are typically defined as: $CH_{3}Br (1) < CH_{3}CH_{2}Br (1) < (CH_{3})_{2}CHBr (0.02) < (CH_{3})_{3}CBr (10^{6})$.
Therefore,the relative rate for $(CH_{3})_{2}CHBr$ is $0.02$.

(A) Step $1$: Dehydrohalogenation of $1$-iodopropane with alcoholic $KOH$ gives propene $(X)$:
$CH_{3}CH_{2}CH_{2}I + KOH(alc.) \xrightarrow{\Delta} CH_{3}CH=CH_{2} (X) + KI + H_{2}O$
Step $2$: Anti-Markovnikov addition of $HBr$ to propene in the presence of peroxide gives $1$-bromopropane $(Y)$:
$CH_{3}CH=CH_{2} + HBr \xrightarrow{\text{peroxide}} CH_{3}CH_{2}CH_{2}Br (Y)$
Step $3$: Nucleophilic substitution of $1$-bromopropane with $KCN$ gives butanenitrile $(Z)$:
$CH_{3}CH_{2}CH_{2}Br + KCN \xrightarrow{\text{alcohol}, \Delta} CH_{3}CH_{2}CH_{2}CN (Z) + KBr$
Thus,$Z$ is $CH_{3}CH_{2}CH_{2}CN$.

(B) The reaction $CH_{3}CH_{2}Br + 2[H] \xrightarrow[\text{Alcohol}]{Zn-Cu(X)} CH_{3}CH_{3} + HBr$ represents the reduction of an alkyl halide to an alkane.
In this reaction,the $Zn-Cu$ couple acts as a reducing agent in the presence of alcohol.
Ethyl bromide $(CH_{3}CH_{2}Br)$ is reduced to ethane $(CH_{3}CH_{3})$ by the $Zn-Cu$ couple,where the alcohol provides the necessary protons for the reaction.

(A) $(A)$
As the size of the halogen atom increases from $F$ to $I$, the $C-X$ bond length increases.
Since bond strength is inversely proportional to bond length, the bond strength decreases as the size of the halogen increases.
Therefore, the correct order of $C-X$ bond strength is $CH_{3}F > CH_{3}Cl > CH_{3}Br > CH_{3}I$.

(D)
The reactivity of alkyl halides towards $SN^{1}$ reactions depends on the stability of the carbocation intermediate formed.
The order of stability of carbocations is: $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$.
Therefore,the order of reactivity towards $SN^{1}$ is: $(CH_{3})_{3}CBr > CH_{3}CH(Br)CH_{3} > CH_{3}CH_{2}Br > CH_{3}Br$.
Thus,$CH_{3}Br$ (methyl bromide) is the least reactive.

(C) For isomeric haloalkanes,the boiling point decreases with an increase in branching.
$n$-butyl bromide is a straight-chain molecule with the largest surface area,leading to stronger van der Waals forces.
As branching increases,the surface area decreases,resulting in weaker intermolecular forces and lower boiling points.
Therefore,the order of boiling points is: $n$-butyl bromide > isobutyl bromide > sec-butyl bromide > tert-butyl bromide.

(A) The reactivity of alkyl halides towards nucleophilic substitution (like with ammonia) depends on the strength of the $C-X$ bond.
As the size of the halogen atom increases,the bond length increases and the bond dissociation energy decreases.
The decreasing order of $C-X$ bond strength is $C-Cl > C-Br > C-I$.
Therefore,the ease of breaking the bond follows the order $C-I > C-Br > C-Cl$.
Thus,the correct decreasing order of reactivity of alkyl halides with ammonia is $R-I > R-Br > R-Cl$.

(A) The boiling point of alkyl halides depends on the magnitude of van der Waals forces.
As the size and mass of the halogen atom increase from $F$ to $I$,the polarizability and the magnitude of van der Waals forces increase.
Therefore,the boiling point increases in the order: $RF < RCl < RBr < RI$ or $RI > RBr > RCl > RF$.

(B) The reaction useful for the exchange of halogen in alkyl chloride by iodide is the $Finkelstein$ reaction.
This reaction is used in the preparation of alkyl iodide by the reaction of alkyl chloride or alkyl bromide with $NaI$ in dry acetone.
$CH_3-CH_2-Cl + NaI \xrightarrow{\text{dry acetone}} CH_3-CH_2-I + NaCl$

(B) The reaction of alkyl halides (like ethyl bromide) with sodium iodide $(NaI)$ in the presence of dry acetone to form alkyl iodides (like ethyl iodide) is a classic example of the Finkelstein reaction.
The reaction is: $C_2H_5Br + NaI \xrightarrow{\text{dry acetone}} C_2H_5I + NaBr$.

(C) Ethanolic $KOH$ acts as a strong base and promotes the elimination of a hydrogen atom and a halogen atom from adjacent carbon atoms in a haloalkane.
This process is known as dehydrohalogenation,which results in the formation of an alkene.

(D) The reduction of chloroform $(CHCl_{3})$ with zinc dust and water $(Zn / H_{2}O)$ leads to the formation of methane $(CH_{4})$.
The chemical reaction is: $CHCl_{3} + 6[H] \xrightarrow{Zn / H_{2}O} CH_{4} + 3HCl$.
Therefore,the correct product is methane.

(C) The conversion of alkyl chlorides or alkyl bromides into 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 $X = Cl, Br$).

(B) The Finkelstein reaction is a type of halogen exchange reaction where an alkyl chloride or bromide is converted into an alkyl iodide using sodium iodide $(NaI)$ in the presence of acetone.
The reaction is: $C_2H_5Cl + NaI \xrightarrow{\text{acetone}} C_2H_5I + NaCl$.
Option $B$ represents this reaction.