Mix Examples-Haloalkanes and Haloarenes Questions in English

(A) $(1)$ Bridgehead halides are inert because the formation of a planar carbocation $(S_{N^1})$ at the bridgehead is geometrically restricted (Bredt's rule),and $S_{N^2}$ attack is sterically hindered.
$(2)$ Both $S_{N^1}$ and $E_1$ reactions involve the rate-determining step of ionization to form a common carbocation intermediate.
$(3)$ $S_{N^2}$ reactions proceed with inversion of configuration (Walden inversion),not retention.
$(4)$ $E_2$ reactions are bimolecular and are favored by high concentrations of strong bases and solvents that do not stabilize the transition state excessively,often low polarity solvents.
Therefore,statements $(1), (2),$ and $(4)$ are correct.

(B) The reaction involves the catalytic hydrogenation of the $C=C$ double bond in $3$-phenylcyclopent-$2$-en-$1$-one using $10\% Pd/C$ under $2-3 \ atm$ pressure of $H_2$.
The benzene ring remains inert under these mild hydrogenation conditions.
The hydrogenation of the double bond creates a chiral center at the $C-3$ position of the cyclopentanone ring.
Since the catalyst can approach the planar double bond from either the top or the bottom face with equal probability,both $(R)$ and $(S)$ enantiomers are formed in equal amounts.
Therefore,the product is a racemic mixture,which is optically inactive.

(C) $1$. The reaction of $2$-bromotoluene with $Mg$ in ether forms the Grignard reagent,$2$-methylphenylmagnesium bromide.
$2$. Hydrolysis of the Grignard reagent with $H_3O^+$ yields toluene $(methylbenzene)$.
$3$. Oxidation of the methyl group in toluene using alkaline $KMnO_4$ followed by acidification with $H^+$ converts the methyl group into a carboxylic acid group,resulting in the formation of benzoic acid.

(D) The reaction with $Cl_2/hv$ follows a free radical substitution mechanism. The tertiary hydrogen on the isopropyl group is the most reactive towards free radical abstraction,leading to the formation of $2-$chloro$-2-$($4$-methylphenyl)propane as the major product $(P)$.
The reaction with $Cl_2/AlCl_3$ is an electrophilic aromatic substitution $(EAS)$. The methyl group is ortho/para directing,and the isopropyl group is also ortho/para directing. Due to the steric hindrance of the isopropyl group,the electrophile $(Cl^+)$ attacks the position ortho to the methyl group,leading to $2-$chloro$-1-$methyl$-4-$isopropylbenzene as the major product $(Q)$.

(B) $1$. Ethyl chloride $(C_2H_5Cl)$ on reduction with $Zn-Cu$ couple and alcohol gives ethane $(C_2H_6)$. This is correct.
$2$. The reaction of methyl magnesium bromide $(CH_3MgBr)$ with acetone $(CH_3COCH_3)$ gives tert-butanol ($2$-methylpropan-$2$-ol),not butanol-$2$. Thus,statement $B$ is wrong.
$3$. Alkyl halides follow the reactivity sequence $R-I > R-Br > R-Cl > R-F$ due to the decreasing bond dissociation energy of the $C-X$ bond. This is correct.
$4$. $C_2H_4Cl_2$ exists as two isomers: $1,1$-dichloroethane and $1,2$-dichloroethane. This is correct.

(D) The given reaction is a Wurtz-Fittig type reaction involving benzyl chloride $(C_6H_5CH_2Cl)$ and methyl chloride $(CH_3Cl)$ in the presence of sodium and dry ether.
This reaction proceeds via the formation of free radicals:
$1$. $C_6H_5CH_2Cl \rightarrow C_6H_5CH_2^{\bullet} + Cl^{\bullet}$
$2$. $CH_3Cl \rightarrow CH_3^{\bullet} + Cl^{\bullet}$
These radicals can combine in different ways:
- Combination of two benzyl radicals: $C_6H_5CH_2^{\bullet} + C_6H_5CH_2^{\bullet} \rightarrow C_6H_5CH_2CH_2C_6H_5$ ($1$,$2$-diphenylethane)
- Combination of two methyl radicals: $CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3-CH_3$ (ethane)
- Combination of one benzyl and one methyl radical: $C_6H_5CH_2^{\bullet} + CH_3^{\bullet} \rightarrow C_6H_5CH_2CH_3$ (ethylbenzene)
Compound $C_6H_5CH_2C_6H_5$ (diphenylmethane) is not formed in this reaction because the radical $C_6H_5CH_2^{\bullet}$ does not undergo coupling with a phenyl radical $(C_6H_5^{\bullet})$ in this specific reaction mixture.

(D) $1$. $P$ is $C_3H_7I$.
$2$. Reaction with alc. $KOH$ is a dehydrohalogenation reaction (elimination).
$3$. If $P$ is $CH_3-CH_2-CH_2-I$ ($n$-propyl iodide),$Q$ is $CH_3-CH=CH_2$ (propene).
$4$. $NBS$ ($N$-Bromosuccinimide) performs allylic bromination. $CH_3-CH=CH_2 \xrightarrow{NBS} Br-CH_2-CH=CH_2$ $(R)$.
$5$. $R$ reacts with $KCN$ to give $CH_2=CH-CH_2-CN$ (allyl cyanide).
$6$. If $P$ is $CH_3-CH(I)-CH_3$ (isopropyl iodide),$Q$ is also $CH_3-CH=CH_2$ (propene),leading to the same product.
$7$. Thus,both $A$ and $B$ can be $P$.

(D) In $1-$bromocyclohex$-1-$ene,the carbon attached to the bromine is part of a double bond,hence it is $sp^2$ hybridized.
In vinyl chloride $(CH_2=CH-Cl)$,the carbon attached to the chlorine is part of a double bond,hence it is $sp^2$ hybridized.
In bromobenzene $(C_6H_5Br)$,the carbon attached to the bromine is part of the aromatic ring,which is $sp^2$ hybridized.
Therefore,all the given compounds contain a halogen atom attached to an $sp^2$ hybridized carbon.

(D) tertiary alkyl halide is a compound where the halogen atom is attached to a carbon atom that is bonded to three other carbon atoms.
In $(1-$chloro$-1-$methyl ethyl) benzene,the structure is $C_6H_5-C(CH_3)_2-Cl$. Here,the carbon atom attached to the chlorine is bonded to two methyl groups and one phenyl group,making it a tertiary carbon.
In $2-$chloro$-2-$methyl propane,the structure is $(CH_3)_3C-Cl$. Here,the carbon atom attached to the chlorine is bonded to three methyl groups,making it a tertiary carbon.
Therefore,both $A$ and $B$ are tertiary alkyl halides.

(B) The reaction of $o$-toluic acid ($2$-methylbenzoic acid) with $Br_2/Fe$ is an electrophilic aromatic substitution reaction.
In $o$-toluic acid,the $-CH_3$ group is ortho/para-directing,and the $-COOH$ group is meta-directing.
Both groups direct the incoming electrophile $(Br^+)$ to the same position,which is the position para to the $-CH_3$ group and meta to the $-COOH$ group.
This position is the $5$-position of the ring (relative to the carboxylic acid group at position $1$ and methyl at position $2$).
Therefore,the product formed is $5$-bromo-$2$-methylbenzoic acid,which corresponds to structure $B$.

(A) The given compound is $2-$methylbromocyclohexane (specifically the trans isomer based on the wedge/dash representation).
Reaction with $alc. KOH$ is a dehydrohalogenation reaction ($E2$ mechanism).
In an $E2$ mechanism,the leaving group $(Br)$ and the $\beta$-hydrogen must be anti-periplanar.
There are two types of $\beta$-hydrogens available: one at $C-1$ (bearing the $CH_3$ group) and one at $C-3$.
Removal of the $\beta$-hydrogen from $C-1$ leads to the more substituted alkene,$1-$methylcyclohexene,which is the Saytzeff product and is more stable.
Removal of the $\beta$-hydrogen from $C-3$ leads to the less substituted alkene,$3-$methylcyclohexene,which is the Hofmann product.
According to Saytzeff's rule,the more substituted alkene is the major product.
Therefore,$1-$methylcyclohexene is the major product.

(B) The rate of $S_{N^1}$ reaction depends on the stability of the carbocation intermediate formed after the leaving group departs.
$1$. Compound $1$ is an allylic bromide. Upon loss of $Br^-$,it forms a stable allylic carbocation.
$2$. Compound $2$ is a vinylic bromide. Vinylic halides are extremely unreactive towards $S_{N^1}$ because the resulting vinylic carbocation is highly unstable due to the $sp$ hybridization of the carbon atom.
$3$. Compound $3$ is an allylic chloride. It forms an allylic carbocation,but $Cl^-$ is a poorer leaving group than $Br^-$,making it less reactive than $1$.
$4$. Compound $4$ is an allylic bromide. It forms a secondary allylic carbocation. Comparing $1$ and $4$,$1$ forms a more substituted (more stable) allylic carbocation than $4$.
Thus,the order of stability of carbocations (and reactivity) is $2 < 3 < 4 < 1$.

(A) The starting material is $1$-bromo-$3$-chlorobenzene.
$1$. Treatment with $n$-butyllithium $(n-BuLi)$ leads to a halogen-metal exchange reaction. Since bromine is more electropositive and forms a more stable organolithium intermediate than chlorine,the lithium atom selectively replaces the bromine atom,forming $3$-chlorophenyllithium.
$2$. The resulting $3$-chlorophenyllithium reacts with carbon dioxide $(CO_2)$ to form a lithium carboxylate intermediate.
$3$. Subsequent acidic workup $(H^+, H_2O)$ protonates the carboxylate to yield $3$-chlorobenzoic acid as the final product.

(B) The reactant is a secondary alkyl halide ($2^\circ$ halide) attached to a tetralin ring system.
$NaC \equiv CH$ is a strong base and a nucleophile.
Since the substrate is a secondary halide,both $S_N2$ (substitution) and $E2$ (elimination) reactions can occur.
$1$. $S_N2$ reaction: The acetylide ion acts as a nucleophile and attacks the carbon atom,resulting in the substitution of $X$ with the $-C \equiv CH$ group. This gives product $(IV)$.
$2$. $E2$ reaction: The acetylide ion acts as a base and abstracts a proton from the adjacent carbon atoms.
- Abstraction of a proton from the $C_1$ position leads to the formation of a double bond between $C_1$ and $C_2$,resulting in product $(III)$.
- Abstraction of a proton from the $C_3$ position leads to the formation of a double bond between $C_2$ and $C_3$,resulting in product $(II)$.
Product $(I)$ is not possible as it involves a rearrangement of the aromatic ring.
Therefore,the possible products are $(II)$,$(III)$,and $(IV)$.

(D) The Assertion is incorrect because alkylbenzene $CAN$ be prepared by Friedel-Crafts alkylation,although it has limitations such as polyalkylation and rearrangement.
The Reason is also incorrect because alkyl halides are generally more reactive or comparable in reactivity to acyl halides in the context of Friedel-Crafts reactions,but the primary issue with alkylation is the formation of polyalkylated products due to the activating nature of the alkyl group added to the benzene ring.
Therefore,both the Assertion and the Reason are incorrect.

(N/A) $(i)$ Cyclohexanol reacts with $SOCl_2$ to form chlorocyclohexane: $\text{Cyclohexanol} + SOCl_2 \rightarrow \text{Chlorocyclohexane} + SO_2 + HCl$.
$(ii)$ $4-\text{Ethylnitrobenzene}$ undergoes free radical bromination at the benzylic position: $4-\text{Ethylnitrobenzene} + Br_2 \xrightarrow{\text{heat or UV light}} 4-(1-\text{Bromoethyl})\text{nitrobenzene} + HBr$.
$(iii)$ $4-\text{Hydroxymethylphenol}$ reacts with $HCl$ to replace the alcoholic $-OH$ group: $4-\text{Hydroxymethylphenol} + HCl \xrightarrow{\text{heat}} 4-\text{Chloromethylphenol} + H_2O$.
$(iv)$ $1-\text{Methylcyclohexene}$ undergoes electrophilic addition with $HI$ following Markovnikov's rule: $1-\text{Methylcyclohexene} + HI \rightarrow 1-\text{Iodo}-1-\text{methylcyclohexane}$.
$(v)$ $CH_3CH_2Br$ reacts with $NaI$ in acetone (Finkelstein reaction): $CH_3CH_2Br + NaI \rightarrow CH_3CH_2I + NaBr$.
$(vi)$ Cyclohexene undergoes allylic bromination with $Br_2$ under heat or $UV$ light: $\text{Cyclohexene} + Br_2 \xrightarrow{\text{heat or UV light}} 3-\text{Bromocyclohexene} + HBr$.

(N/A) $(i)$ $CH_3CH_2OH \xrightarrow{SOCl_2, \text{Pyridine}} CH_3CH_2Cl + SO_2 + HCl$; $HC \equiv CH + NaNH_2 \xrightarrow{\text{Liq. } NH_3} HC \equiv C^{-}Na^{+}$; $CH_3CH_2Cl + HC \equiv C^{-}Na^{+} \to CH_3CH_2C \equiv CH + NaCl$
$(ii)$ $CH_3-CH_3$ $\xrightarrow{Br_2/UV\ light} CH_3-CH_2Br + HBr$ $\xrightarrow{KOH(alc), \Delta} CH_2=CH_2$ $\xrightarrow{Br_2/CCl_4} BrCH_2-CH_2Br$ $\xrightarrow{KOH(alc), \Delta} CH_2=CHBr$
$(iii)$ $CH_3-CH=CH_2 + HBr$ $\xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2Br$ $\xrightarrow{AgNO_2} AgBr + CH_3-CH_2-CH_2NO_2$
$(iv)$ Toluene $\xrightarrow{Cl_2/UV\ light} \text{Benzyl chloride}$ $\xrightarrow{NaOH(aq)} \text{Benzyl alcohol}$
$(v)$ $CH_3-CH=CH_2$ $\xrightarrow{Br_2/CCl_4} CH_3-CH(Br)-CH_2Br$ $\xrightarrow[\Delta]{2KOH(alc)} CH_3-C \equiv CH$
$(vi)$ $CH_3-CH_2-OH$ $\xrightarrow{PCl_5} CH_3-CH_2-Cl$ $\xrightarrow{AgF} CH_3-CH_2-F$
$(vii)$ $CH_3-Br$ $\xrightarrow{KCN(alc)} CH_3-CN$ $\xrightarrow{CH_3MgBr} CH_3-C(CH_3)=NMgBr$ $\xrightarrow{H_3O^{+}} CH_3-CO-CH_3$
$(viii)$ $CH_3CH_2CH=CH_2$ $\xrightarrow{HBr} CH_3CH_2CH(Br)CH_3$ $\xrightarrow{KOH(alc), \Delta} CH_3CH=CHCH_3$
$(ix)$ $2CH_3CH_2CH_2CH_2Cl + 2Na \xrightarrow{\text{dry ether}} CH_3(CH_2)_6CH_3 + 2NaCl$
$(x)$ Benzene $\xrightarrow{Br_2/Fe, \text{dark}} \text{Bromobenzene}$ $\xrightarrow{Na, \text{dry ether}} \text{Biphenyl}$

(D) The major organic products for the given reactions are as follows:

$Sr. \ No.$	Reaction and Major Organic Product
$(i)$	$CH_3-CH_2-CH_2-Cl + NaI \xrightarrow{\text{acetone, heat}} CH_3-CH_2-CH_2-I + NaCl$
$(ii)$	$CH_3-C(CH_3)_2-Br + KOH \xrightarrow{\text{ethanol, heat}} CH_3-C(CH_3)=CH_2 + KBr + H_2O$
$(iii)$	$CH_3-CH(Br)-CH_2-CH_3 + NaOH \xrightarrow{\text{water}} CH_3-CH(OH)-CH_2-CH_3 + NaBr$
$(iv)$	$CH_3-CH_2-Br + KCN \xrightarrow{\text{aq. ethanol}} CH_3-CH_2-CN + KBr$
$(v)$	$C_6H_5-ONa + C_2H_5-Cl \rightarrow C_6H_5-O-C_2H_5 + NaCl$
$(vi)$	$CH_3-CH_2-CH_2-OH + SOCl_2 \rightarrow CH_3-CH_2-CH_2-Cl + SO_2 + HCl$
$(vii)$	$CH_3-CH_2-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2-CH_2-Br$
$(viii)$	$CH_3-CH=C(CH_3)_2 + HBr \rightarrow CH_3-CH_2-C(Br)(CH_3)_2$

$(i)$ Propene $\xrightarrow{HBr/\text{Peroxide}}$ $1-$Bromopropane $\xrightarrow{Aq. KOH, \Delta}$ Propan$-1-$ol.
$(ii)$ Ethanol $\xrightarrow{PBr_3}$ Bromoethane $\xrightarrow{Mg, \text{ether}}$ Ethylmagnesium bromide $\xrightarrow{\text{Ethylene oxide}, H_3O^+}$ Butan$-1-$ol $\xrightarrow{P/I_2}$ $1-$Iodobutane $\xrightarrow{alc. KOH}$ But$-1-$ene $\xrightarrow{Br_2/CCl_4}$ $1,2-$Dibromobutane $\xrightarrow{NaNH_2}$ But$-1-$yne.
$(iii)$ $1-$Bromopropane $\xrightarrow{alc. KOH}$ Propene $\xrightarrow{HBr}$ $2-$Bromopropane.
$(vii)$ Ethanol $\xrightarrow{PBr_3}$ Bromoethane $\xrightarrow{KCN, \text{ethanol}}$ Propanenitrile.
$(ix)$ $2-$Chlorobutane $\xrightarrow{2Na, \text{dry ether}}$ $3,4-$Dimethylhexane.
$(x)$ $2-$Methylpropene $\xrightarrow{HCl}$ $2-$Chloro$-2-$methylpropane.
$(xi)$ Ethyl chloride $\xrightarrow{KCN}$ Propanenitrile $\xrightarrow{H_3O^+}$ Propanoic acid.
$(xii)$ But$-1-$ene $\xrightarrow{HBr, \text{Peroxide}}$ $1-$Bromobutane $\xrightarrow{NaI, \text{acetone}}$ $1-$Iodobutane.
$(xiii)$ $2-$Chloropropane $\xrightarrow{alc. KOH}$ Propene $\xrightarrow{HBr, \text{Peroxide}}$ $1-$Bromopropane $\xrightarrow{Aq. KOH}$ Propan$-1-$ol.
$(xiv)$ Isopropyl alcohol $\xrightarrow{[O]}$ Acetone $\xrightarrow{I_2/NaOH}$ Iodoform.
$(xvi)$ $2-$Bromopropane $\xrightarrow{alc. KOH}$ Propene $\xrightarrow{HBr, \text{Peroxide}}$ $1-$Bromopropane.
$(xvii)$ Chloroethane $\xrightarrow{2Na, \text{dry ether}}$ Butane.
$(xix)$ tert-Butyl bromide $\xrightarrow{alc. KOH}$ $2-$Methylpropene $\xrightarrow{HBr, \text{Peroxide}}$ Isobutyl bromide.

$(i)$ When $n$-butyl chloride is treated with alcoholic $KOH$,the formation of but-$1$-ene takes place. This reaction is a dehydrohalogenation reaction.
$CH_3-CH_2-CH_2-CH_2-Cl \xrightarrow[Dehydrohalogenation]{KOH(alc)/\Delta} CH_3-CH_2-CH=CH_2 + KCl + H_2O$
$(ii)$ When bromobenzene is treated with $Mg$ in the presence of dry ether,phenylmagnesium bromide is formed.
$C_6H_5Br + Mg \xrightarrow[Dry\ ether]{} C_6H_5MgBr$
$(iii)$ Chlorobenzene does not undergo hydrolysis under normal conditions. However,it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of $623 \ K$ and a pressure of $300 \ atm$ to form phenol.
$C_6H_5Cl \xrightarrow[623 \ K, 300 \ atm]{NaOH(aq)} C_6H_5OH + NaCl$
$(iv)$ When ethyl chloride is treated with aqueous $KOH$,it undergoes hydrolysis to form ethanol.
$CH_3-CH_2-Cl + KOH(aq) \rightarrow CH_3-CH_2-OH + KCl$
$(v)$ When methyl bromide is treated with sodium in the presence of dry ether,ethane is formed. This reaction is known as the Wurtz reaction.
$2CH_3-Br + 2Na \xrightarrow[Dry\ ether]{} CH_3-CH_3 + 2NaBr$
$(vi)$ When methyl chloride is treated with $KCN$,it undergoes a substitution reaction to give methyl cyanide.
$CH_3-Cl + KCN \rightarrow CH_3-CN + KCl$

(N/A) The two main differences between haloalkanes and haloarenes are as follows:
$1$. Structure: In haloalkanes,the halogen atom is bonded to an $sp^3$ hybridized carbon atom of an alkyl group. In haloarenes,the halogen atom is directly bonded to an $sp^2$ hybridized carbon atom of an aromatic ring.
$2$. Reactivity towards Nucleophilic Substitution: Haloalkanes are generally more reactive towards nucleophilic substitution reactions compared to haloarenes. Haloarenes show low reactivity due to resonance effects,which give the $C-X$ bond partial double bond character,and the instability of the phenyl cation.

(A-3, B-5, C-1, D-2, E-4) $A-3, B-5, C-1, D-2, E-4$
$A$. $S_N1$ reactions proceed through the formation of a carbocation intermediate,which leads to the formation of a racemic mixture (Racemisation).
$B$. Chlorobromocarbons are commonly used as chemicals in fire extinguishers.
$C$. Bromination of alkenes involves the addition of bromine across the double bond to form vic-dibromides.
$D$. Alkylidene halides are also known as gem-dihalides,where two halogen atoms are attached to the same carbon atom.
$E$. The elimination of $HX$ from an alkyl halide follows the Saytzeff rule,which states that the more substituted alkene is the major product.

(N/A) For reaction $(1)$:
Isopropyl bromide $(CH_3-CH(Br)-CH_3)$ undergoes dehydrohalogenation with alcoholic $KOH$ to form propene $(CH_3-CH=CH_2)$ as $X$.
Propene reacts with $HBr$ in the presence of peroxide (Kharasch effect/anti-Markovnikov addition) to form $1$-bromopropane $(CH_3-CH_2-CH_2Br)$ as $Y$.
For reaction $(2)$:
$n$-propyl alcohol $(CH_3-CH_2-CH_2OH)$ undergoes dehydration with conc. $H_2SO_4$ at $443 \ K$ to form propene $(CH_3-CH=CH_2)$ as $X$.
Propene reacts with $HBr$ (Markovnikov addition) to form $2$-bromopropane $(CH_3-CH(Br)-CH_3)$ as $Y$.

(N/A) $(1)$ $1$-Chloropropane $(CH_3CH_2CH_2Cl)$ undergoes dehydrohalogenation with alcoholic $KOH$ to form propene $(CH_3CH=CH_2)$ as $(B)$. Propene then reacts with $Br_2$ in $CCl_4$ to form $1,2$-dibromopropane $(CH_3CHBrCH_2Br)$ as $(C)$.
$(2)$ Ethane $(CH_3CH_3)$ undergoes free radical chlorination in the presence of $hv$ to form chloroethane $(CH_3CH_2Cl)$ as $(B)$. Chloroethane then undergoes the Wurtz reaction with $Na$ in dry ether to form $n$-butane $(CH_3CH_2CH_2CH_3)$ as $(C)$.

(N/A) $1$. Conversion of propene to $2$-bromopropane and $1$-bromopropane:
- For $2$-bromopropane: Propene reacts with $HBr$ in the presence of $CCl_4$ (Markovnikov addition) to form $2$-bromopropane.
$CH_3CH=CH_2 + HBr \xrightarrow{CCl_4} CH_3CH(Br)CH_3$
- For $1$-bromopropane: Propene reacts with $HBr$ in the presence of organic peroxide (Anti-Markovnikov addition) to form $1$-bromopropane.
$CH_3CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3CH_2CH_2Br$
$2$. Conversion of $1$-bromopropane to $2$-bromopropane:
- First,$1$-bromopropane undergoes dehydrohalogenation with alcoholic $KOH$ to form propene.
$CH_3CH_2CH_2Br + KOH (\text{alc.}) \rightarrow CH_3CH=CH_2 + KBr + H_2O$
- Then,propene reacts with $HBr$ (Markovnikov addition) to form $2$-bromopropane.
$CH_3CH=CH_2 + HBr \rightarrow CH_3CH(Br)CH_3$

(N/A) $(1)$ Toluene $\xrightarrow[\text{Alkali, } \Delta]{\text{KMnO}_4, [O]}$ Benzoic acid $\xrightarrow[\text{Conc. H}_2\text{SO}_4, \Delta]{\text{Conc. HNO}_3}$ $m$-Nitrobenzoic acid
$(2)$ Benzene $\xrightarrow[\text{FeCl}_3, \Delta]{\text{Cl}_2}$ Chlorobenzene $\xrightarrow[\text{H}_2\text{SO}_4, \Delta]{\text{HNO}_3}$ $m$-Chloronitrobenzene
$(3)$ $1, 2$-Dibromopropane $\xrightarrow[\Delta]{Zn, -ZnBr_2}$ Propene $\xrightarrow[\text{Peroxide}]{\text{HBr}}$ $1$-Bromopropane ($n$-Propyl bromide)

(A) The correct matching is as follows:
$(i)$ Chlorination of chlorobenzene is an electrophilic aromatic substitution reaction: $i-b$.
(ii) Addition of $HBr$ to propene follows Markovnikov's rule,which is an electrophilic addition reaction: $ii-d$.
(iii) Substitution of $I$ by $OH^-$ in $1$-phenylethanol is a nucleophilic substitution reaction $(S_N1)$: $iii-e$.
(iv) Nucleophilic aromatic substitution of $Cl$ by $OH^-$ in $p$-nitrochlorobenzene: $iv-a$.
$(v)$ Dehydrohalogenation of $2$-bromobutane to form but-$2$-ene is a Saytzeff elimination reaction: $v-c$.
Thus,the correct match is $i-b, ii-d, iii-e, iv-a, v-c$.

(A) The correct matches are as follows:
$(i)$ The reaction of an aryl halide with an alkyl halide in the presence of sodium metal is known as the Wurtz-Fittig reaction. Thus,$(i)$ matches with $(b)$.
(ii) The reaction of two moles of aryl halide with sodium metal in the presence of dry ether to form diphenyl is known as the Fittig reaction. Thus,(ii) matches with $(a)$.
(iii) The reaction of a diazonium salt with $Cu_2X_2$ to form an aryl halide is known as the Sandmeyer reaction. Thus,(iii) matches with $(d)$.
(iv) The reaction of an alkyl chloride with sodium iodide in the presence of dry acetone is known as the Finkelstein reaction. Thus,(iv) matches with $(c)$.
Therefore,the correct sequence is $(i-b, ii-a, iii-d, iv-c)$.

(A-II, B-III, C-IV, D-I) The reactions are matched as follows:
$A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4} BrCH_2-CH_2Br$ (Vicinal dihalide formation) $\rightarrow ii$
$B$. $CH_3CH=CH_2 + HI \rightarrow CH_3CHICH_3$ (Markovnikov addition) $\rightarrow iii$
$C$. $C_6H_5N_2^+X^- + KI \rightarrow C_6H_5I + N_2 + KX$ (Sandmeyer-type reaction) $\rightarrow iv$
$D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br$ (Anti-Markovnikov addition) $\rightarrow i$
Therefore,the correct matching is: $A-ii, B-iii, C-iv, D-i$.

(A-II, B-III, C-IV, D-I) The correct matches are:
$A \to ii$: Finkelstein reaction involves the exchange of halogen with $I^-$ in dry acetone.
$B \to iii$: Swarts reaction involves the synthesis of alkyl fluorides using metallic fluorides like $AgF$.
$C \to iv$: Free radical substitution occurs in alkanes under $UV$ light $(hv)$.
$D \to i$: Electrophilic addition occurs across the double bond of alkenes with halogens.
Therefore,the correct sequence is $A-ii, B-iii, C-iv, D-i$.

(A-III, B-II, C-IV, D-I) $(A \rightarrow iii)$: Electrophilic aromatic substitution of toluene with $X_2$ in the presence of $Fe$ (dark) gives ortho and para halo-substituted toluene.
$(B \rightarrow ii)$: Aniline reacts with $NaNO_2$ and $HX$ at $273-278 \ K$ to form benzene diazonium salt.
$(C \rightarrow iv)$: Sandmeyer reaction of benzene diazonium salt with $Cu_2X_2$ yields aryl halide $(C_6H_5X)$.
$(D \rightarrow i)$: Free radical bromination of ethylbenzene with $Br_2$ under light/heat occurs at the benzylic position to form $1$-bromo-$1$-phenylethane.

(A-II, B-III, C-I, D-IV) The conversion of aniline to benzene diazonium salt is known as Diazotization $(ii)$.
$(B)$ The conversion of benzene diazonium salt to haloarene using $Cu_2X_2$ is known as the Sandmeyer reaction $(iii)$.
$(C)$ The reaction of an aryl halide with an alkyl halide in the presence of sodium and dry ether is the Wurtz-Fittig reaction $(i)$.
$(D)$ The coupling of two aryl halide molecules in the presence of sodium and dry ether is the Fittig reaction $(iv)$.
Therefore,the correct matching is: $(A-ii, B-iii, C-i, D-iv)$.

(A-1, B-3, C-4, D-2) $(A)-(1), (B)-(3), (C)-(4), (D)-(2)$
$(A)$ $OH^- + CH_3Cl \to HOCH_3 + Cl^-$: This is a primary alkyl halide reacting with a strong nucleophile,following the $S_N2$ mechanism.
$(B)$ $(CH_3)_3CBr + OH^- \to (CH_3)_3COH + Br^-$: This is a tertiary alkyl halide,which undergoes substitution via the $S_N1$ mechanism due to the formation of a stable carbocation.
$(C)$ $CH_3CH_2Br \xrightarrow[ethanol]{KOH} CH_2 = CH_2$: This is an elimination reaction (dehydrohalogenation) of an alkyl halide in the presence of a base,known as $\beta$-elimination.
$(D)$ $C_6H_6 + Br_2 \xrightarrow{Fe} C_6H_5Br + HBr$: This is an electrophilic aromatic substitution reaction ($S_E2$ aromatic) where benzene reacts with bromine in the presence of a Lewis acid catalyst.

(A-II, B-III, C-IV, D-I) $(A) CH_3CH_2Br + Mg \xrightarrow{\text{dry ether}} CH_3CH_2MgBr$ (Grignard reagent formation),so $(A \rightarrow ii)$.
$(B) 2RX + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$ (Wurtz reaction),so $(B \rightarrow iii)$.
$(C) CH_3CH_2Br \xrightarrow{\text{alc. KOH}} CH_2=CH_2 + H_2O + KBr$ (Dehydrohalogenation),so $(C \rightarrow iv)$.
$(D) \text{Bromocyclohexane} + Mg \xrightarrow{\text{dry ether}} \text{Cyclohexylmagnesium bromide}$ (Grignard reagent formation),so $(D \rightarrow i)$.
Therefore,the correct matching is $(A$ $\rightarrow ii, B$ $\rightarrow iii, C$ $\rightarrow iv, D$ $\rightarrow i)$.

(C) $1$. The simplest optically active alkene is $3$-methylpent-$1$-ene.
$2$. Hydrogenation of $3$-methylpent-$1$-ene with $H_2/Ni/\Delta$ gives $3$-methylpentane.
$3$. Free radical chlorination of $3$-methylpentane $(CH_3-CH_2-CH(CH_3)-CH_2-CH_3)$ produces various monochlorinated isomers:
- Substitution at $C_1$: $Cl-CH_2-CH_2-CH(CH_3)-CH_2-CH_3$ ($1$ chiral center,$2$ enantiomers).
- Substitution at $C_2$: $CH_3-CHCl-CH(CH_3)-CH_2-CH_3$ ($2$ chiral centers,$4$ stereoisomers).
- Substitution at $C_3$: $CH_3-CH_2-CCl(CH_3)-CH_2-CH_3$ ($1$ chiral center,$2$ enantiomers).
- Substitution at $C_4$ (methyl group): $CH_3-CH_2-CH(CH_2Cl)-CH_2-CH_3$ ($1$ chiral center,$2$ enantiomers).
$4$. Total number of stereoisomers = $2 + 4 + 2 + 2 = 10$.

(A) The reaction of $p-\text{methoxyphenyl diazonium chloride}$ with ethanol $(C_2H_5OH)$ is a reduction reaction where the diazonium group is replaced by a hydrogen atom to form anisole.
The reaction is as follows:
$CH_3O-C_6H_4-N_2^+Cl^- + CH_3CH_2OH \rightarrow CH_3O-C_6H_5 + N_2 + CH_3CHO + HCl$
Here,$(A)$ is $p-\text{methoxyphenyl diazonium chloride}$,$X$ is acetaldehyde $(CH_3CHO)$,and $Y$ is hydrogen chloride $(HCl)$.

(C) . Benzenediazonium chloride reacting with $Cu_2Cl_2/HCl$ is the Sandmeyer reaction $(a \rightarrow ii)$.
$b$. Benzenediazonium chloride reacting with $Cu/HCl$ is the Gatterman reaction $(b \rightarrow iv)$.
$c$. The reaction of alkyl halides with $Na$ in ether to form alkanes is the Wurtz reaction $(c \rightarrow i)$.
$d$. The reaction of aryl halides with $Na$ in ether to form diaryls is the Fittig reaction $(d \rightarrow iii)$.
Therefore,the correct matching is $a$ $\rightarrow ii, b$ $\rightarrow iv, c$ $\rightarrow i, d$ $\rightarrow iii$.

(B) The starting material $C_8H_{10}$ is ethylbenzene.
$1$. Nitration of ethylbenzene with $HNO_3/H_2SO_4$ gives $p$-nitroethylbenzene as the major product $(A)$.
$2$. Radical bromination of $p$-nitroethylbenzene with $Br_2/\Delta$ occurs at the benzylic position to form $1-(4-nitrophenyl)ethyl$ bromide $(B)$.
$3$. Dehydrohalogenation of $B$ with alcoholic $KOH$ yields $p$-nitrostyrene $(C)$ as the major product.

(A) $1$. The starting material $Br-CH_2-CH(Br)-Ph$ undergoes dehydrohalogenation with $Alc. KOH$ followed by $NaNH_2$ to form the alkyne $X$,which is phenylacetylene $(Ph-C \equiv C-H)$.
$2$. The alkyne $X$ $(Ph-C \equiv C-H)$ undergoes acid-catalyzed hydration $(HgSO_4/dil. H_2SO_4, \Delta)$ to form an enol intermediate,which tautomerizes to acetophenone $(Ph-CO-CH_3)$.
$3$. Acetophenone is then subjected to nitration using $Conc. HNO_3/H_2SO_4$. Since the acetyl group $(-CO-CH_3)$ is a meta-directing group,the nitro group $(-NO_2)$ is introduced at the meta-position,yielding $m-nitroacetophenone$ $(Y)$.

(A) The correct matches are:
$A$. The reaction of an aryl halide with an alkyl halide in the presence of sodium is known as the Wurtz-Fitting reaction. Thus,$A-I$.
$B$. The reaction of two molecules of aryl halide with sodium to form a diaryl is known as the Fitting reaction. Thus,$B-II$.
$C$. The conversion of benzene diazonium chloride to chlorobenzene using $Cu_2Cl_2$ is the Sandmeyer reaction. Thus,$C-III$.
$D$. The reaction of an alkyl halide with $NaI$ in acetone to form an alkyl iodide is the Finkelstein reaction. Thus,$D-IV$.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(B) All the given orders are correct:
$(A)$ $S_{N}2$ reaction rate decreases with increase in steric hindrance. Primary alkyl halide (isobutyl bromide) is more reactive than tertiary alkyl halide (tert-butyl bromide).
$(B)$ $S_{N}1$ reaction rate depends on the stability of the carbocation. Benzyl carbocation is resonance stabilized,whereas the carbocation from $2-$phenylethyl bromide is less stable.
$(C)$ Electrophilic aromatic substitution $(EAS)$ rate decreases with a decrease in electron density on the ring. The $-NO_2$ group is strongly electron-withdrawing,reducing the electron density of the ring in $1-$chloro$-4-$nitrobenzene compared to chlorobenzene.
$(D)$ Nucleophilic aromatic substitution rate increases with the presence of electron-withdrawing groups at ortho and para positions. The $-NO_2$ group at the para position activates the ring towards nucleophilic substitution.

(A) The reaction of $3$-chloro-$1$-butene with $HCl$ proceeds via the formation of a carbocation intermediate.
First,the $H^+$ ion from $HCl$ adds to the double bond to form a carbocation.
This carbocation can undergo a $1,2$-hydride shift to form a more stable carbocation.
$1$. The initial carbocation reacts with $Cl^-$ to form $2,3$-dichlorobutane. This product has two chiral centers,leading to $3$ stereoisomers ($d, l$ and $meso$ forms).
$2$. The rearranged carbocation reacts with $Cl^-$ to form $2,2$-dichlorobutane,which is an achiral product ($1$ isomer).
Total number of possible isomeric products $= 3 + 1 = 4$.

(C) The reaction proceeds in three steps:
$1$. Benzenediazonium chloride reacts with $Cu_2Br_2/HBr$ (Sandmeyer reaction) to form bromobenzene.
$2$. Bromobenzene reacts with $Mg$ in the presence of dry ether to form phenylmagnesium bromide (a Grignard reagent).
$3$. Phenylmagnesium bromide undergoes hydrolysis with $H_2O$ to yield benzene as the final product.

(C) The reaction of $CH_3CH_2Br$ with $NaOH$ in the presence of $C_2H_5OH$ (alcoholic $NaOH$) undergoes dehydrohalogenation to form ethene $(CH_2=CH_2)$ as product $A$.
Product $A$ has $4$ hydrogen atoms.
The reaction of $CH_3CH_2Br$ with $NaOH$ in the presence of $H_2O$ (aqueous $NaOH$) undergoes nucleophilic substitution to form ethanol $(CH_3CH_2OH)$ as product $B$.
Product $B$ has $6$ hydrogen atoms.
The total number of hydrogen atoms in product $A$ and product $B$ is $4 + 6 = 10$.

(C) Step $1$: The starting material $CH_3-CH_2-CH_2-Br$ ($1$-bromopropane) undergoes dehydrohalogenation with concentrated alcoholic $NaOH$ at $80^{\circ} C$ to form propene $(CH_3-CH=CH_2)$.
Step $2$: Propene reacts with $HBr$ in the presence of acetic acid (or as a solvent) via electrophilic addition following Markovnikov's rule to yield $2-$bromopropane $(CH_3-CHBr-CH_3)$.

(A) $1$. The starting material $C_6H_{10}O$ is cyclohexanone. Reaction with $CH_3MgBr$ followed by $H_2O$ gives $1$-methylcyclohexanol $(Q)$.
$2$. $Q$ reacts with $conc. HCl$ to form $1$-chloro-$1$-methylcyclohexane $(S)$. This matches structure $(2)$.
$3$. $Q$ undergoes acid-catalyzed dehydration with $20\% H_3PO_4$ at $360 \ K$ to form $1$-methylcyclohexene $(R)$.
$4$. $R$ reacts with $HBr$ in the presence of benzoyl peroxide (anti-Markovnikov addition) to form $1$-bromo-$2$-methylcyclohexane $(U)$. This matches structure $(4)$.
$5$. $R$ undergoes hydrogenation $(H_2, Ni)$ to form methylcyclohexane,followed by radical bromination $(Br_2, h\nu)$ to form $1$-bromo-$1$-methylcyclohexane $(T)$.
$6$. Comparing the structures: $S$ is $(2)$,$U$ is $(4)$,and $T$ is $(1)$. The correct set is $(2, 4)$.

(A) The reaction sequence is as follows:
$1$. Nitrobenzene is reduced by $Sn + HCl$ to form aniline $(C_6H_5NH_2)$.
$2$. Aniline reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
$3$. Benzenediazonium chloride reacts with $Cu_2Cl_2$ (Sandmeyer reaction) to form chlorobenzene $(C_6H_5Cl)$.
$4$. Chlorobenzene reacts with $Na$ in dry ether (Wurtz-Fittig reaction) to form biphenyl $(C_6H_5-C_6H_5)$.
The final product $(A)$ is biphenyl $(C_{12}H_{10})$.
Molar mass of biphenyl = $(12 \times 12) + (10 \times 1) = 144 + 10 = 154 \ g \ mol^{-1}$.