Mix Examples-Haloalkanes and Haloarenes Questions in English

(C) The correct matches are as follows:
$A.$ Swarts reaction: Used for the preparation of alkyl fluorides,e.g.,$CH_3CH_2F$ (Ethyl fluoride).
$B.$ Sandmeyer's reaction: Used to convert diazonium salts to aryl halides or cyanides,e.g.,$C_6H_5CN$ (Cyanobenzene).
$C.$ Wurtz-Fittig reaction: Used for the synthesis of alkyl benzenes,e.g.,$C_6H_5CH_2CH_3$ (Ethyl benzene).
$D.$ Finkelstein reaction: Used for the preparation of alkyl iodides from alkyl chlorides/bromides,e.g.,$CH_3CH_2I$ (Ethyl iodide).
Therefore,the correct sequence is $A-IV, B-III, C-I, D-II$.

(A) The correct matches are:
$A$. The reaction $2C_6H_5X + 2Na \xrightarrow{\text{Dry Ether}} C_6H_5-C_6H_5 + 2NaX$ is the Fittig reaction $(III)$.
$B$. The reaction $ArN_2^{+} X^{-} \xrightarrow[HCl]{Cu} ArCl + N_2 \uparrow + CuX$ is the Gatterman reaction $(IV)$.
$C$. The reaction $C_2H_5Br + NaI \xrightarrow{\text{Acetone}} C_2H_5I + NaBr$ is the Finkelstein reaction $(II)$.
$D$. The reaction $(CH_3)_3COH \xrightarrow[ZnCl_2]{HCl} (CH_3)_3CCl + H_2O$ is the Lucas reaction $(I)$.
Therefore,the correct sequence is $A-III, B-IV, C-II, D-I$.

(C) The molar mass of $2-$bromopentane $(C_5H_{11}Br)$ is $(5 \times 12) + (11 \times 1) + 80 = 60 + 11 + 80 = 151 \ g \ mol^{-1}$.
Given mass of $2-$bromopentane = $151 \ g$,so moles of $2-$bromopentane = $1 \ mol$.
Reaction $1$: $2-$bromopentane $\xrightarrow{\text{alc. KOH}}$ pent-$2-$ene $(P)$ + $HBr$.
Since the yield of $P$ is $80 \%$,moles of $P$ formed = $0.8 \ mol$.
Reaction $2$: Pent-$2-$ene $(P)$ $\xrightarrow{Br_2}$ $2,3-$dibromopentane $(Q)$.
Since the yield of $Q$ is $100 \%$,moles of $Q$ formed = $0.8 \ mol$.
The molar mass of $Q$ $(C_5H_{10}Br_2)$ is $(5 \times 12) + (10 \times 1) + (2 \times 80) = 60 + 10 + 160 = 230 \ g \ mol^{-1}$.
Mass of $Q$ = $\text{moles} \times \text{molar mass} = 0.8 \ mol \times 230 \ g \ mol^{-1} = 184 \ g$.

(C) The reaction sequence proceeds as follows:
$1$. Free radical bromination of methylcyclohexane with $Br_2/h\nu$ occurs at the tertiary carbon to form $1$-bromo-$1$-methylcyclohexane.
$2$. Dehydrohalogenation with alcoholic $KOH$ and $\Delta$ (heat) leads to the formation of $1$-methylcyclohexene via an $E2$ mechanism.
$3$. The addition of $HBr$ in the presence of peroxide $(R-O-O-R)$ and $h\nu$ follows an anti-Markovnikov addition mechanism,resulting in the formation of $1$-bromo-$2$-methylcyclohexane as the major product.

(D) Let us analyze each reaction:
$A.$ $2$-bromomethylcyclohexane with $NaOEt$ undergoes $E2$ elimination to form an alkene.
$B.$ $2$-bromobutane with aqueous $KOH$ undergoes nucleophilic substitution ($SN^2$ or $SN^1$) to form butan-$2$-ol,not an alkene.
$C.$ $2$-bromo-$2$-methylpropane with $NaOMe$ undergoes $E2$ elimination to form $2$-methylpropene (an alkene).
$D.$ Phenol with $Na_2Cr_2O_7 / H_2SO_4$ undergoes oxidation to form $p$-benzoquinone,not an alkene.
$E.$ $2$-methylpropan-$2$-ol with $Cu$ at $573 \ K$ undergoes dehydration to form $2$-methylpropene (an alkene).
Thus,reactions $B$ and $D$ do not produce an alkene.

(A) The reactions are matched as follows:
$A$. $ROH + SOCl_2 \rightarrow RCl + SO_2 + HCl$ is the Darzen process $(P)$.
$B$. $RX + NaI \rightarrow RI + NaX$ is the Finkelstein reaction $(Q)$.
$C$. $RX + AgF \rightarrow RF + AgX$ is the Swarts reaction $(R)$.
$D$. $ArX + RX + 2Na \rightarrow Ar-R + 2NaX$ is the Wurtz-Fittig reaction $(S)$.
Therefore,the correct matching is $A$ $\rightarrow P, B$ $\rightarrow Q, C$ $\rightarrow R, D$ $\rightarrow S$.

(D) The reaction sequence is: $R-X + Mg$ $\xrightarrow{\text{Ether}} R-MgX$ $\xrightarrow{H_2O} R-H + Mg(OH)X$.
Here,the final product is Propane $(CH_3-CH_2-CH_3)$,which contains $3$ carbon atoms.
Therefore,the alkyl halide $(R-X)$ must contain $3$ carbon atoms.
Both $n-$propyl chloride $(CH_3-CH_2-CH_2-Cl)$ and isopropyl chloride $(CH_3-CHCl-CH_3)$ contain $3$ carbon atoms.
Both will form their respective Grignard reagents,which upon hydrolysis with $H_2O$ will yield Propane.

(A) The reaction of a mixture of alkyl halides with sodium metal in dry ether is known as the Wurtz reaction. When a mixture of chloroethane $(CH_3CH_2Cl)$ and $1-$chloropropane $(CH_3CH_2CH_2Cl)$ is treated with sodium in dry ether,the following products are formed due to the coupling of alkyl radicals:
$1$. Coupling of two chloroethane molecules: $CH_3CH_2-CH_2CH_3$ ($n-$butane).
$2$. Coupling of two $1-$chloropropane molecules: $CH_3CH_2CH_2-CH_2CH_2CH_3$ ($n-$hexane).
$3$. Cross-coupling of chloroethane and $1-$chloropropane: $CH_3CH_2-CH_2CH_2CH_3$ ($n-$pentane).
Propane is not formed in this reaction because it would require the loss of a carbon atom or a different mechanism not associated with the Wurtz coupling of these specific reactants.

(C) The reaction of bromobenzene $(C_6H_5Br)$ and bromoethane $(C_2H_5Br)$ with sodium in the presence of dry ether is a Wurtz-Fittig reaction.
In this reaction,a mixture of alkyl halide and aryl halide reacts with sodium to form an alkylbenzene.
Possible products are formed by the coupling of the radicals generated:
$1$. Coupling of two bromoethane molecules gives $n$-butane $(CH_3CH_2CH_2CH_3)$.
$2$. Coupling of two bromobenzene molecules gives diphenyl $(C_6H_5-C_6H_5)$.
$3$. Coupling of one bromobenzene and one bromoethane molecule gives ethylbenzene $(C_6H_5-C_2H_5)$.
Toluene $(C_6H_5CH_3)$ cannot be formed because it would require the use of bromomethane $(CH_3Br)$ instead of bromoethane $(C_2H_5Br)$.
Therefore,toluene is $NOT$ obtained.

(D) The Wurtz reaction involves the coupling of alkyl halides with sodium metal in dry ether to form higher alkanes.
When a mixture of $n$-butyl bromide $(CH_3CH_2CH_2CH_2Br)$ and ethyl bromide $(CH_3CH_2Br)$ is treated with sodium,the following products are formed:
$1$. Coupling of two ethyl bromide molecules: $CH_3CH_2-CH_2CH_3$ ($n$-butane).
$2$. Coupling of two $n$-butyl bromide molecules: $CH_3CH_2CH_2CH_2-CH_2CH_2CH_2CH_3$ ($n$-octane).
$3$. Cross-coupling of ethyl bromide and $n$-butyl bromide: $CH_3CH_2-CH_2CH_2CH_2CH_3$ ($n$-hexane).
Ethane $(CH_3CH_3)$ is not formed in this reaction mixture.

(B) The chemical structure of $DDT$ is $1,1,1-\text{trichloro}-2,2-\text{bis}(p-\text{chlorophenyl})\text{ethane}$.
Its molecular formula is $C_{14}H_9Cl_5$.
In the structure,there are two benzene rings,each containing $3$ $(\pi)$ bonds,totaling $6$ $(\pi)$ bonds.
Counting the $(\sigma)$ bonds: There are $29$ $(\sigma)$ bonds in the entire molecule,including the $C-C$,$C-H$,and $C-Cl$ bonds.
Thus,the correct answer is $29$ $(\sigma)$ and $6$ $(\pi)$ bonds.

(C) The given reaction scheme shows three different reactions of ethyl chloride $(C_2H_5Cl)$:
$1$. The conversion of $C_2H_5Cl$ to $C_2H_5F$ is a Swarts reaction,which uses a metallic fluoride like $Hg_2F_2$ or $AgF$. Thus,$X = Hg_2F_2$.
$2$. The conversion of $C_2H_5Cl$ to ethene $(CH_2=CH_2)$ is a dehydrohalogenation reaction,which occurs in the presence of alcoholic $KOH$. Thus,$Y = \text{alcoholic } KOH$.
$3$. The conversion of $C_2H_5Cl$ to butane $(C_4H_{10})$ is a Wurtz reaction,which uses $Na$ in dry ether. Thus,$Z = Na \text{ in dry ether}$.
Therefore,the correct sequence is $Hg_2F_2$,alcoholic $KOH$ and $Na$ in dry ether.

(B) The reaction of a mixture of methyl bromide $(CH_3Br)$ and bromobenzene $(C_6H_5Br)$ with sodium metal in dry ether is a combination of Wurtz and Fittig reactions,known as the Wurtz-Fittig reaction.
Possible products include:
$1$. Wurtz reaction: $2CH_3Br + 2Na \rightarrow CH_3-CH_3 + 2NaBr$ (Ethane is formed).
$2$. Wurtz-Fittig reaction: $C_6H_5Br + CH_3Br + 2Na \rightarrow C_6H_5-CH_3 + 2NaBr$ (Toluene is formed).
$3$. Fittig reaction: $2C_6H_5Br + 2Na \rightarrow C_6H_5-C_6H_5 + 2NaBr$ (Diphenyl is formed).
Propane $(C_3H_8)$ cannot be formed in this reaction as there is no source of a three-carbon chain.

(B) The reaction sequence is as follows:
$1$. Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
$2$. Benzenediazonium chloride reacts with $Cu_2Cl_2$ (Sandmeyer reaction) to form chlorobenzene $(C_6H_5Cl)$.
$3$. Chlorobenzene reacts with $Na$ in the presence of dry ether (Wurtz-Fittig reaction) to form biphenyl $(C_6H_5-C_6H_5)$.

(D) The first step is the addition of $HBr$ to propene $(C_3H_6)$ in the presence of benzoyl peroxide,which follows the anti-Markovnikov rule. This results in the formation of $1$-bromopropane $(CH_3CH_2CH_2Br)$ as $X$.
In the second step,$1$-bromopropane reacts with bromobenzene in the presence of $Na$ and dry ether. This is a Wurtz-Fittig reaction,which couples the alkyl group to the benzene ring,forming propylbenzene as $Y$.

(D) The reaction sequence is as follows:
$1.$ Propan$-1-$ol reacts with Lucas reagent $(HCl / ZnCl_2)$ to form $1-$chloropropane.
$2.$ $1-$chloropropane undergoes Finkelstein reaction with $NaI$ in dry acetone to form $1-$iodopropane.
$3.$ $1-$iodopropane undergoes Wurtz reaction in the presence of $Na$ and dry ether to form $n-$hexane $(CH_3CH_2CH_2CH_2CH_2CH_3)$.

(A) The correct matching is:
$(A)$ Benzene matches with $4$. $(4n + 2) \pi$ electrons: Benzene has $6 \pi$ electrons,which follows the Huckel rule $(4n + 2) \pi$ electrons.
$(B)$ Ethylene matches with $3$. Mustard gas: Ethylene reacts with $S_2Cl_2$ to form mustard gas.
$(C)$ Acetaldehyde matches with $2$. Silver mirror: Acetaldehyde gives a positive silver mirror test with Tollen's reagent.
$(D)$ Chloroform matches with $1$. Phosgene: Chloroform on oxidation gives phosgene,which is a poisonous gas.
Therefore,the correct sequence is $A-4, B-3, C-2, D-1$.

(B) The addition of $HI$ to propene follows Markownikoff's rule,proceeding via the formation of a $2^{\circ}$ carbocation,which is more stable than a $1^{\circ}$ carbocation. The major product is $2\text{-iodopropane}$ $(III)$.
$(B)$ The reaction of ethylbenzene $(C_6H_5CH_2CH_3)$ with $Br_2$ under heating $(\Delta)$ is a free radical substitution at the benzylic position. The $1\text{-phenylethyl}$ radical is stabilized by resonance,leading to $1\text{-bromo-}1\text{-phenylethane}$ $(IV)$.
$(C)$ The addition of $HBr$ to styrene $(C_6H_5CH=CH_2)$ follows Markownikoff's rule. The electrophilic addition of $H $ forms a stable benzylic carbocation $(C_6H_5CH^ CH_3)$,which then reacts with $Br^-$ to form $1\text{-bromo-}1\text{-phenylethane}$ $(IV)$.
Therefore,the correct match is $A-III, B-IV, C-IV$.

(A) The reactions are analyzed as follows:
$(A)$ $CH_3-CHBr-CH_2Br \xrightarrow{alc. KOH} CH_3-C \equiv CH$ or $CH_3-CH=CHBr$. This is a dehydrohalogenation reaction yielding an alkenyl bromide $(IV)$.
$(B)$ $CH_3-CH_2-CH=CH_2 \xrightarrow{HBr, \text{peroxide}} CH_3-CH_2-CH_2-CH_2Br$. This is Anti-Markovnikov addition,yielding a $1^{\circ}$-alkyl bromide $(I)$.
$(C)$ $CH_3-CH_2-CH_3 \xrightarrow{Br_2, h\nu} CH_3-CHBr-CH_3$. Free radical bromination occurs preferentially at the $2^{\circ}$ carbon to form a $2^{\circ}$-alkyl bromide $(II)$.
$(D)$ $CH_3-CH=CH_2 \xrightarrow{NBS, \Delta} BrCH_2-CH=CH_2$. $NBS$ ($N$-Bromosuccinimide) performs allylic bromination,yielding allyl bromide $(III)$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(A) In reaction $(1)$,the starting material is $3$-methylcyclohexene. Under radical bromination conditions $(Br_2, h\nu)$,the reaction proceeds via the formation of a free radical. The most stable radical is the tertiary $(3^\circ)$ radical formed at the $C-3$ position. Therefore,the major product should be $3$-bromo-$3$-methylcyclohexene,not $3$-(bromomethyl)cyclohexene. Thus,reaction $(1)$ is incorrectly represented.

(D) Statement $I$: In $CH_2=CH-Cl$ (vinyl chloride),the lone pair on chlorine atom participates in resonance with the double bond,giving the $C-Cl$ bond a partial double bond character. This makes it shorter and stronger than the $C-Cl$ bond in $CH_3-CH_2-Cl$ (ethyl chloride),where only a single bond exists. Thus,Statement $I$ is true.
Statement $II$: The given molecule is a chiral alkyl halide. Hydrolysis of such a molecule typically proceeds via an $S_N1$ mechanism,which involves the formation of a planar carbocation intermediate. Nucleophilic attack on this carbocation occurs from both sides with equal probability,leading to the formation of a racemic mixture. $A$ racemic mixture is optically inactive and cannot rotate plane polarized light $(PPL)$. Thus,Statement $II$ is false.

(A) $sec$-butyl chloride ($2$-chlorobutane) is the only compound among the given options that contains a chiral carbon atom and thus exhibits optical isomerism.
The molecular formula of $sec$-butyl chloride is $C_4H_9Cl$.
Molar mass of $C_4H_9Cl = (4 \times 12) + (9 \times 1) + 35.5 = 48 + 9 + 35.5 = 92.5 \ g/mol$.
Percentage of carbon $= \frac{\text{Mass of carbon}}{\text{Molar mass}} \times 100 = \frac{48}{92.5} \times 100 \approx 51.89 \%$.
Rounding to the nearest integer,we get $52 \%$.

A is true: alcoholic $KOH$ is a stronger base and promotes elimination. E is true: $SOCl_2$ works well for alcohols, but phenols are not effectively converted to aryl chlorides this way. C is false: aryl halides are very different (less reactive). D is false: allyl chloride is a haloalkene, not a haloalkyne.

(B) . Finkelstein reaction involves the exchange of halogen atoms,typically using $NaI$ in acetone to prepare alkyl iodides.
$B$. Swarts reaction is used to prepare alkyl fluorides by treating alkyl chlorides or bromides with metallic fluorides like $SbF_3$ or $AgF$.
$C$. Sandmeyer's reaction involves the conversion of diazonium salts into aryl halides using copper$(I)$ salts like $Cu_2Cl_2$.
$D$. Fittig reaction involves the coupling of two aryl halide molecules in the presence of $Na$ metal and dry ether to form diaryls.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) Statement $I$ is true: Boiling point increases with an increase in molecular mass due to stronger van der Waals forces. Since the molecular mass of $CH_3CH_2CH_2I$ $(170 \text{ g/mol})$ > $CH_3CH_2I$ $(156 \text{ g/mol})$ > $CH_3I$ $(142 \text{ g/mol})$,the order is correct.
Statement $II$ is true: $p$-dichlorobenzene is more symmetric than $o$-dichlorobenzene,which allows it to pack more efficiently in the crystal lattice,resulting in a higher melting point. However,boiling points are determined by intermolecular forces and molecular shape; $o$-dichlorobenzene has a higher dipole moment than $p$-dichlorobenzene,leading to stronger dipole-dipole interactions and a higher boiling point.

(A) Statement $I$: $3$-phenylpropene $(C_6H_5CH_2CH=CH_2)$ reacts with $HBr$ via electrophilic addition following Markovnikov's rule.
The intermediate carbocation formed at $C-2$ $(C_6H_5CH_2CH^+CH_3)$ is more stable due to the inductive effect and hyperconjugation compared to the primary carbocation at $C-1$.
The major product is $2$-bromo-$3$-phenylpropane $(C_6H_5CH_2CH(Br)CH_3)$,which contains a chiral center at $C-2$.
Thus,Statement $I$ is true.
Statement $II$: The Sandmeyer reaction uses $Cu_2Cl_2/HCl$ or $Cu_2(CN)_2/KCN$ to convert diazonium salts to aryl chlorides or aryl cyanides.
The Gattermann reaction is a modification of the Sandmeyer reaction that uses copper powder $(Cu)$ in the presence of the corresponding acid ($HCl$ or $HCN$).
Both reactions are standard methods for the preparation of aryl halides and cyanides.
Thus,Statement $II$ is true.

(B) $1$. The reaction of propan$-1-$ol $(CH_{3}CH_{2}CH_{2}OH)$ with phosphorus pentachloride $(PCl_{5})$ is given by:
$CH_{3}CH_{2}CH_{2}OH + PCl_{5} \rightarrow CH_{3}CH_{2}CH_{2}Cl + POCl_{3} + HCl$
Thus,$X = POCl_{3}$.
$2$. The product $CH_{3}CH_{2}CH_{2}Cl$ reacts with alcoholic $KOH$ upon heating to undergo dehydrohalogenation,forming propene $(Y = CH_{3}CH=CH_{2})$.
$3$. Propene $(CH_{3}CH=CH_{2})$ reacts with $HBr$ in the presence of peroxide $((C_{6}H_{5}CO)_{2}O_{2})$ via the anti-Markownikoff addition mechanism to yield $1-$bromopropane $(Z = CH_{3}CH_{2}CH_{2}Br)$.
Therefore,$X = POCl_{3}$ and $Z = CH_{3}CH_{2}CH_{2}Br$.