Properties of Haloalkanes Questions in English

(A) $Silver$ $nitrate$ $(AgNO_3)$ reacts with halide ions ($Cl^-$,$Br^-$,$I^-$) in aqueous solution to form a precipitate of silver halide $(AgX)$.
$Ethyl$ bromide $(C_2H_5Br)$ is a covalent organic compound and does not ionize to provide free bromide ions $(Br^-)$ in water.
Therefore,it does not react with $AgNO_3$ to form a precipitate.
In contrast,$Sodium$ bromide $(NaBr)$,$Calcium$ chloride $(CaCl_2)$,and $Sodium$ chloride $(NaCl)$ are ionic compounds that dissociate in water to provide free halide ions,which react with $Ag^+$ to form precipitates of $AgBr$ or $AgCl$.

(D) The reaction for dehydrohalogenation of ethyl chloride is: $C_2H_5Cl \rightarrow C_2H_4 + HCl$
Molar mass of $C_2H_5Cl = (2 \times 12) + (5 \times 1) + 35.5 = 64.5 \ g/mol$
Molar mass of $C_2H_4 = (2 \times 12) + (4 \times 1) = 28 \ g/mol$
According to the stoichiometry,$64.5 \ g$ of $C_2H_5Cl$ produces $28 \ g$ of $C_2H_4$.
Therefore,$32.25 \ g$ of $C_2H_5Cl$ would theoretically produce: $\frac{28 \times 32.25}{64.5} = 14 \ g$ of $C_2H_4$.
Since the yield is $50\%$,the actual mass of the product obtained is: $14 \ g \times 0.50 = 7 \ g$.

(D) $SbCl_5$ is a strong Lewis acid. It reacts with the $C-Cl$ bond of $1-$chloro$-1-$phenylethane to abstract the chloride ion,forming a stable benzylic carbocation intermediate: $Ph-CH(CH_3)-Cl + SbCl_5 \rightarrow [Ph-CH^+(CH_3)] + [SbCl_6]^-$.
Since the carbocation is planar,the chloride ion can attack from either side,leading to the formation of a racemic mixture of the $(d)$ and $(l)$ forms.

(C) Alkyl halides undergo alkaline hydrolysis to form alcohols. This process occurs via a nucleophilic substitution reaction where the halogen atom $(-X)$ is replaced by a nucleophile,specifically the hydroxyl group $(-OH)$.
Primary alkyl halides typically follow the $S_{N}2$ mechanism,whereas tertiary alkyl halides follow the $S_{N}1$ mechanism.
The general reaction is: $R-X + KOH \rightarrow R-OH + KX$

(A) The given reaction involves the conversion of a tertiary alkyl halide,$(CH_3)_3CBr$,into a tertiary alcohol,$(CH_3)_3COH$,using an aqueous base.
Since the substrate is a tertiary alkyl halide,the reaction proceeds via the formation of a stable carbocation intermediate.
This mechanism is characteristic of the $SN^1$ (unimolecular nucleophilic substitution) reaction.
Therefore,the correct option is $(a)$.

(B) Dehydrohalogenation of an alkyl halide in the presence of alcoholic potash is an example of an elimination reaction.
$R-CH_2-CH_2-Cl + KOH \text{ (alc.) } \xrightarrow{\Delta} R-CH=CH_2 + KCl + H_2O$

(C) The reaction involves the replacement of a bromine atom $(-Br)$ by a hydroxyl group $(-OH)$ in a tert-butyl bromide molecule.
Since the water molecule acts as a nucleophile (providing the $OH^-$ group) to replace the leaving group $(Br^-)$,this is a nucleophilic substitution reaction.

(B) The dehydrohalogenation of $2-$bromobutane 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 $(CH_2=CH-CH_2-CH_3)$ is a monosubstituted alkene.
Therefore,$2-$butene is formed as the major product.

(D) In an $S_N2$ reaction (Substitution Nucleophilic Bimolecular):
$(1)$ The rate of reaction depends on the concentration of both the substrate and the nucleophile. Therefore,statement $(1)$ is false.
$(2)$ The nucleophile attacks the carbon atom from the side opposite to the leaving group (backside attack),which is true.
$(3)$ The reaction occurs in a single step where bond formation and bond cleavage happen simultaneously,which is true.
Thus,statements $(2)$ and $(3)$ are correct.

(B) The reaction is a dehydrohalogenation of $2$-bromobutane using alcoholic $KOH$.
When an alkyl halide can undergo dehydrohalogenation to form more than one alkene,the major product is the most highly substituted alkene (the one with the most alkyl groups attached to the double-bonded carbons).
According to Saytzeff's rule,$CH_3CH=CHCH_3$ ($2$-butene) is more substituted than $CH_2=CHCH_2CH_3$ ($1$-butene),making it the major product.
Therefore,the correct option is $(b)$.

(A) molecule has an enantiomer if it contains a chiral carbon atom (a carbon atom bonded to four different groups).
In $2$-chlorobutane $(CH_3-CH_2-CHCl-CH_3)$,the second carbon is chiral as it is attached to four different groups: $-H$,$-Cl$,$-CH_3$,and $-C_2H_5$.
Therefore,it exists as a pair of enantiomers.

(C) $S_N^1$ reaction proceeds via the formation of a carbocation intermediate. The rate of $S_N^1$ reaction is directly proportional to the stability of the carbocation formed. The stability order of carbocations is $3^\circ > 2^\circ > 1^\circ > CH_3^+$.
In the given options:
$(a)$ $CH_3-CH_2-Cl$ forms a $1^\circ$ carbocation $(CH_3-CH_2^+)$.
$(b)$ $(CH_3)_2CH-Cl$ forms a $2^\circ$ carbocation $((CH_3)_2CH^+)$.
$(c)$ $CH_3-C(CH_3)_2-Cl$ forms a $3^\circ$ carbocation $((CH_3)_3C^+)$.
$(d)$ $CH_3-CH(CH_3)-CH_2-Cl$ forms a $1^\circ$ carbocation $(CH_3-CH(CH_3)-CH_2^+)$.
Since the $3^\circ$ carbocation is the most stable,the $S_N^1$ reaction is fastest in $CH_3-C(CH_3)_2-Cl$.

(C) The reaction shows the nucleophilic substitution of a bromide ion by a hydroxide ion.
The product shows an inversion of configuration at the chiral carbon atom,which is a characteristic feature of the $S_N2$ mechanism.
In an $S_N2$ reaction (Bimolecular Nucleophilic Substitution),the nucleophile attacks from the side opposite to the leaving group,leading to Walden inversion.

(D) Dehydrohalogenation is an elimination reaction where a hydrogen atom and a halogen atom are removed from adjacent carbon atoms to form a double bond. In the presence of a base like $OH^{-}$,the base abstracts a proton $(H^{+})$ from the $\beta$-carbon,leading to the formation of a double bond and the elimination of the halide ion $(Br^{-})$. The correct mechanism shows the $OH^{-}$ attacking the $\beta$-hydrogen,the $C-H$ bond breaking to form the $C=C$ double bond,and the $C-Br$ bond breaking to release the $Br^{-}$ ion. Option $D$ correctly depicts this concerted mechanism.

(D) This reaction is a $Wurtz$ reaction,where alkyl halides react with sodium metal in the presence of dry ether to form higher alkanes.
$2C_2H_5I + 2Na \xrightarrow{\text{Dry Ether}} C_2H_5-C_2H_5 + 2NaI$
$2C_2H_5I + 2Na \xrightarrow{\text{Dry Ether}} C_4H_{10} (\text{Butane}) + 2NaI$
Therefore,the product formed is $Butane$.

(C) Sodium ethoxide $(C_2H_5ONa)$ is a strong base commonly used to perform dehydrohalogenation of alkyl halides to form alkenes.
For example: $CH_3-CH_2-CH_2-Br \xrightarrow[C_2H_5ONa]{\Delta} CH_3-CH=CH_2 + C_2H_5OH + NaBr$.

(A) Grignard reagent $(RMgX)$ is a strong base and a strong nucleophile. It reacts violently with water to form an alkane $(RH)$ and magnesium hydroxide halide $(Mg(OH)X)$.
$RMgX + H_2O \rightarrow RH + Mg(OH)X$
Therefore,it is prepared in an anhydrous ether medium to prevent decomposition by moisture.

(C) The correct answer is $(C)$.
Alcoholic $KOH$ contains alkoxide ions $(RO^-)$,which act as strong bases.
These ions abstract a proton from the $\beta$-carbon of an alkyl halide,leading to an elimination reaction known as dehydrohalogenation.
The reaction is: $CH_3-CH_2-Br + KOH_{(alc)} \xrightarrow{\Delta} CH_2=CH_2 + KBr + H_2O$.
Mechanism:
$ROH + KOH \to ROK + H_2O$
$ROK \to RO^- + K^+$
$RO^- + H-CH_2(\beta)-CH_2(\alpha)-Br \to ROH + CH_2=CH_2 + Br^-$

(C) The reaction of $C_2H_5Cl$ with hydrogen in the presence of palladium on carbon $(Pd/C)$ is a reduction reaction.
The chemical equation is: $C_2H_5Cl + H_2 \xrightarrow{Pd/C} C_2H_6 + HCl$.
In this reaction,the chlorine atom is replaced by a hydrogen atom,resulting in the formation of ethane $(C_2H_6)$.
This method is commonly used for the preparation of pure alkanes from alkyl halides.

(D) The reaction of haloalkanes with alcoholic $KOH$ is a dehydrohalogenation reaction.
$CH_3-CH_2-Cl + KOH (alc.) \rightarrow CH_2=CH_2 + KCl + H_2O$
This is an elimination reaction where an alkene is formed as the major product.

(A) When ethyl bromide $(CH_3CH_2Br)$ is treated with alcoholic $KOH$,it undergoes a dehydrohalogenation reaction.
This is an elimination reaction where a hydrogen atom from the $\beta$-carbon and the bromine atom from the $\alpha$-carbon are removed.
The reaction is: $CH_3-CH_2-Br + KOH (\text{alc.}) \to CH_2=CH_2 + KBr + H_2O$.
The final product formed is ethylene $(CH_2=CH_2)$.

(D) . $CH_3-CH(Br)-CH_2-CH_3 \xrightarrow{\text{alc. } KOH} CH_3-CH=CH-CH_3$ ($2-$butene).
The reaction involves the removal of a hydrogen atom and a bromine atom,which is known as dehydrohalogenation.

(D) The reaction of ethyl bromide $(CH_3-CH_2-Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).
In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,leading to the elimination of $HBr$ and the formation of ethene $(CH_2=CH_2)$.
The reaction is: $CH_3-CH_2-Br + KOH \text{ (alc)} \to CH_2=CH_2 + KBr + H_2O$.

(D) Both $1-$chloropropane $(CH_3-CH_2-CH_2-Cl)$ and $2-$chloropropane $(CH_3-CHCl-CH_3)$ undergo dehydrohalogenation (elimination reaction) in the presence of alcoholic $KOH$ to form propene $(CH_3-CH=CH_2)$.
$CH_3-CH_2-CH_2-Cl + KOH (\text{alc.}) \rightarrow CH_3-CH=CH_2 + KCl + H_2O$
$CH_3-CHCl-CH_3 + KOH (\text{alc.}) \rightarrow CH_3-CH=CH_2 + KCl + H_2O$
Thus,the product obtained in both cases is propene.

(A) The reaction of $1-$chlorobutane $(CH_3CH_2CH_2CH_2Cl)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.
This reaction follows the $E2$ mechanism,where the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,leading to the elimination of $HCl$.
$CH_3CH_2CH_2CH_2Cl + alc. KOH \rightarrow CH_3CH_2CH=CH_2 + KCl + H_2O$.
The major product formed is $1-$butene.

(B) The reaction of monohalides (alkyl halides) with alcoholic $KOH$ is a dehydrohalogenation reaction.
This reaction involves the elimination of a hydrogen atom from the $\beta$-carbon and a halogen atom from the $\alpha$-carbon,resulting in the formation of an alkene.
For example: $CH_3-CH_2-Cl \xrightarrow{\text{alc. } KOH} CH_2=CH_2 + KCl + H_2O$.

(B) The reaction of an alkyl halide with alcoholic $KOH$ leads to the removal of a hydrogen atom and a halogen atom from adjacent carbon atoms,resulting in the formation of an alkene. This process is known as dehydrohalogenation,which is a type of elimination reaction.
$R-CH_2-CH_2-X \xrightarrow{\text{alc. } KOH} R-CH=CH_2 + HX$

(B) The reaction of $n$-propyl bromide $(CH_3-CH_2-CH_2-Br)$ with ethanolic potassium hydroxide $(KOH)$ is a dehydrohalogenation reaction.
This is an elimination reaction ($E2$ mechanism) where the base removes a proton from the $\beta$-carbon and the bromide ion is eliminated.
$CH_3-CH_2-CH_2-Br + KOH \xrightarrow{C_2H_5OH} CH_3-CH=CH_2 + KBr + H_2O$.
The product formed is $Propene$.

(B) Neopentyl bromide $(CH_3-C(CH_3)_2-CH_2-Br)$ is a primary alkyl halide with significant steric hindrance at the $\beta-$carbon.
Upon reaction with alcoholic $KOH$,the formation of a primary carbocation is unfavorable.
Instead,it undergoes a concerted rearrangement via a $1, 2-$methyl shift to form a more stable tertiary carbocation $(CH_3-C^{+}(CH_3)-CH_2-CH_3)$.
Elimination of a proton from this intermediate according to Saytzeff's rule yields $2-$methyl$-2-$butene as the major product.

(C) The elimination of $HBr$ from $2-$bromobutane to form $2-$butene is an example of dehydrohalogenation. According to $Saytzeff's$ rule,in dehydrohalogenation reactions,the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons) is the major product. $CH_3-CH(Br)-CH_2-CH_3 \xrightarrow{alc. KOH} CH_3-CH=CH-CH_3 + HBr$.

(B) $3$-Phenylpropene $(C_6H_5CH_2CH=CH_2)$ reacts with $HBr$ via electrophilic addition.
According to Markownikoff's rule,the electrophile $(H^+)$ adds to the terminal carbon to form a more stable secondary benzylic carbocation $(C_6H_5CH^+CH_2CH_3)$.
The nucleophile $(Br^-)$ then attacks this carbocation to form the major product,$1$-bromo-$1$-phenylpropane $(C_6H_5CH(Br)CH_2CH_3)$.

(D) When toluene $(C_6H_5CH_3)$ is treated with $Cl_2$ in the presence of light ($UV$ light) and in the absence of a halogen carrier (like $FeCl_3$ or $AlCl_3$),the reaction proceeds via a free-radical mechanism.
This leads to the substitution of hydrogen atoms on the side-chain methyl group.
The reaction proceeds in stages:
$1$. Toluene reacts with $Cl_2$ to form Benzyl chloride $(C_6H_5CH_2Cl)$.
$2$. Further reaction with $Cl_2$ gives Benzal dichloride $(C_6H_5CHCl_2)$.
$3$. Final reaction with $Cl_2$ gives Benzotrichloride $(C_6H_5CCl_3)$.
Since the question asks for the product formed under these conditions,all three products are formed depending on the amount of $Cl_2$ used. Therefore,the most appropriate answer is that all of these are formed.

(A) Benzene hexachloride,commonly known as $BHC$ or $Gammexane$,is produced by the addition of chlorine to benzene in the presence of ultraviolet light.
Its chemical structure is $1, 2, 3, 4, 5, 6-$hexachlorocyclohexane,which corresponds to the formula $C_6H_6Cl_6$.

(C) The reaction of ethyl alcohol $(C_2H_5OH)$ with chlorine $(Cl_2)$ and calcium hydroxide $(Ca(OH)_2)$ is known as the haloform reaction.
First,ethyl alcohol is oxidized to acetaldehyde $(CH_3CHO)$ by chlorine.
Then,acetaldehyde is chlorinated to form chloral $(CCl_3CHO)$.
Finally,chloral reacts with $Ca(OH)_2$ to produce chloroform $(CHCl_3)$ and calcium formate.

(D) When toluene $(C_6H_5CH_3)$ reacts with chlorine $(Cl_2)$ in the presence of sunlight ($UV$ light),a free radical substitution reaction occurs at the side chain (methyl group).
This results in the formation of benzyl chloride $(C_6H_5CH_2Cl)$ and hydrogen chloride $(HCl)$ as a byproduct.
The reaction is: $C_6H_5CH_3 + Cl_2 \xrightarrow{\text{Sunlight}} C_6H_5CH_2Cl + HCl$.

(A) . Acetone forms chloroform when heated with bleaching powder.
$CaOCl_2 + H_2O \to Ca(OH)_2 + Cl_2$
$CH_3COCH_3 + 3Cl_2 \to CCl_3COCH_3 + 3HCl$
$2CCl_3COCH_3 + Ca(OH)_2 \to 2CHCl_3 + (CH_3COO)_2Ca$

(A) The reaction of ethyl alcohol $(CH_3CH_2OH)$ with bleaching powder $(CaOCl_2)$ involves the following steps:
$1$. Bleaching powder reacts with water to produce chlorine: $CaOCl_2 + H_2O \to Ca(OH)_2 + Cl_2$.
$2$. Chlorine oxidizes ethyl alcohol to acetaldehyde: $CH_3CH_2OH + Cl_2 \to CH_3CHO + 2HCl$.
$3$. Acetaldehyde is then chlorinated to form chloral $(CCl_3CHO)$: $CH_3CHO + 3Cl_2 \to CCl_3CHO + 3HCl$.
$4$. Finally,chloral reacts with calcium hydroxide to form chloroform: $2CCl_3CHO + Ca(OH)_2 \to 2CHCl_3 + (HCOO)_2Ca$.

(B) $DDT$ (dichlorodiphenyltrichloroethane) is prepared by the condensation reaction of chlorobenzene with chloral $(CCl_3-CHO)$ in the presence of concentrated sulfuric acid $(H_2SO_4)$.
The reaction is as follows:
$CCl_3CHO + 2C_6H_5Cl \xrightarrow{conc. H_2SO_4} (ClC_6H_4)_2CHCCl_3 + H_2O$
Thus,the correct option is $(b)$.

(C) The correct answer is $C$. Moist silver oxide $(Ag_2O + H_2O)$ acts as a source of $AgOH$ (silver hydroxide).
The reaction is:
$Ag_2O + H_2O \to 2AgOH$
$C_2H_5Br + AgOH \to C_2H_5OH + AgBr$

(C) The reaction represents the oxidation of chloroform $(CHCl_3)$ in the presence of light and air to form a highly poisonous gas called phosgene $(COCl_2)$.
$2CHCl_3 + O_2 \xrightarrow{\text{Light, air}} 2COCl_2 + 2HCl$
Therefore,$X$ stands for light and air.

(D) The reactivity of alkyl halides towards nucleophilic substitution depends on the stability of the carbocation intermediate (in $S_N1$) or steric hindrance (in $S_N2$).
Benzyl chloride $(C_6H_5CH_2Cl)$ is highly reactive due to the resonance stabilization of the benzyl carbocation.
Ethyl chloride $(C_2H_5Cl)$ is a primary alkyl halide and is less reactive than benzyl chloride.
Chlorobenzene $(C_6H_5Cl)$ is the least reactive due to the partial double bond character of the $C-Cl$ bond caused by resonance.
Therefore,the correct order of reactivity is $C_6H_5CH_2Cl > C_2H_5Cl > C_6H_5Cl$.
Thus,the reactivity of ethyl chloride is less than that of benzyl chloride.

(A) The reactivity of alkyl halides towards nucleophilic substitution depends on the strength of the $C-X$ bond.
Among the given options,the $C-Cl$ bond is stronger than the $C-I$ and $C-Br$ bonds due to the smaller size of the chlorine atom.
Between $n-$propyl chloride and isopropyl chloride,the $C-Cl$ bond in $n-$propyl chloride is less sterically hindered,but the reactivity is generally compared based on the bond dissociation energy and the stability of the carbocation formed.
However,comparing the halogen atoms,the $C-Cl$ bond is the strongest,making it the least reactive.
Comparing $n-$propyl chloride and isopropyl chloride,$n-$propyl chloride is generally less reactive in $S_N1$ reactions due to the formation of a less stable primary carbocation compared to the secondary carbocation formed by isopropyl chloride.
Thus,$n-$propyl chloride exhibits the minimum reactivity among the choices provided.

(A) The reactivity of alkyl halides towards nucleophilic substitution reactions depends on the stability of the carbocation intermediate (in $S_N1$) or steric hindrance (in $S_N2$).
$Methyl$ $chloride$ $(CH_3Cl)$ is a primary halide with minimal steric hindrance,making it highly reactive in $S_N2$ reactions.
$Propyl$ $chloride$ $(CH_3CH_2CH_2Cl)$ is also a primary halide but has more steric hindrance than $methyl$ $chloride$,making it slightly less reactive.
$Chlorobenzene$ is the least reactive because the $C-Cl$ bond acquires partial double bond character due to resonance,and the $sp^2$ hybridized carbon atom is more electronegative,making the bond shorter and stronger.
Therefore,the order of reactivity is: $Methyl$ $chloride$ $>$ $propyl$ $chloride$ $>$ $chlorobenzene$.

(D) The reaction of $AgNO_3$ with organic halides depends on the ease of formation of the carbocation or the reactivity of the $C-X$ bond.
However,in the context of common laboratory tests,$CHI_3$ (iodoform) reacts with $AgNO_3$ to form a yellow precipitate of $AgI$ due to the presence of weakly bonded iodine atoms.
$C_6H_5CH_2Cl$ forms a stable benzylic carbocation,but $CHI_3$ reacts very readily to give the characteristic yellow precipitate of $AgI$.

(B) $(CH_3)_3CBr + CH_3ONa \xrightarrow{\text{Elimination}} CH_2=C(CH_3)_2 + CH_3OH + NaBr$
The product is Isobutylene.
$CH_3ONa \to CH_3O^{-} + Na^{+}$
Methoxide ion $(CH_3O^{-})$ is a strong base,therefore it abstracts a proton from the $3^{\circ}$ alkyl halide and favours an elimination reaction.

(D) $CCl_4$ is a covalent compound and does not ionize to provide $Cl^-$ ions in the solution.
Since $AgNO_3$ reacts with free $Cl^-$ ions to form a white precipitate of $AgCl$,no reaction occurs between $CCl_4$ and $AgNO_3$.
Therefore,the correct option is $D$.

(B) The reaction of benzyl chloride $(C_6H_5CH_2Cl)$ with aqueous potassium cyanide $(KCN)$ is a nucleophilic substitution reaction $(S_N2)$.
The cyanide ion $(CN^-)$ acts as a nucleophile and replaces the chloride ion $(Cl^-)$ to form benzyl cyanide $(C_6H_5CH_2CN)$ and potassium chloride $(KCl)$.
The reaction is: $C_6H_5CH_2Cl + KCN(aq.) \to C_6H_5CH_2CN + KCl$.
Thus,$X$ is $C_6H_5CH_2CN$ and $Y$ is $KCl$.

(B) Ethylidene chloride $(CH_3-CHCl_2)$ reacts with aqueous $KOH$ to undergo nucleophilic substitution,replacing both chlorine atoms with hydroxyl groups to form a geminal diol $(CH_3-CH(OH)_2)$.
Geminal diols are unstable and readily lose a water molecule to form a stable carbonyl compound.
$CH_3-CHCl_2 + 2KOH(aq) \rightarrow CH_3-CH(OH)_2 + 2KCl$
$CH_3-CH(OH)_2 \rightarrow CH_3-CHO + H_2O$
Thus,the final product is acetaldehyde $(CH_3-CHO)$.

(B) The dehydrobromination of $2-bromobutane$ $(CH_3-CHBr-CH_2-CH_3)$ yields $but-2-ene$ $(CH_3-CH=CH-CH_3)$ as the major product according to $Saytzeff's$ rule.
This rule states that in an elimination reaction,the more highly substituted alkene (the one with more alkyl groups attached to the doubly bonded carbon atoms) is the preferred product.