Haloalkanes and Haloarenes

In presence of peroxide, addition of HBr to an alkene follows free radical mechanism (peroxide effect).

For styrene $\mathrm{Ph{-}CH{=}CH_2}$:

Major product (with peroxide)

Br• adds in such a way that the more stable radical is formed.

If Br• adds to terminal carbon ($\mathrm{CH_2}$), the radical comes on benzylic carbon:

$\mathrm{Ph{-}\dot{C}H{-}CH_2Br}$

This is benzylic radical (resonance-stabilised), hence preferred. Then it abstracts H from HBr:

$\mathrm{Ph{-}CH_2{-}CH_2Br}$

So $1$-bromo-$2$-phenylethane is the major product.

Checking each statement

A. True

The reaction proceeds via the more stable radical intermediate (benzylic radical).

B. False

Peroxide does not generate $ \mathrm{H^\bullet}$.

It generates radicals that ultimately produce $ \mathrm{Br^\bullet}$ (the chain-carrying radical), e.g. by abstraction of H from HBr.

C. True

Benzoyl peroxide can form phenyl radicals ($\mathrm{Ph^\bullet}$), which can abstract H from HBr:

$\mathrm{Ph^\bullet + HBr \rightarrow PhH + Br^\bullet}$

So benzene ($\mathrm{PhH}$) can be formed as a byproduct.

D. False

$1$-bromo-$2$-phenylethane ($\mathrm{Ph{-}CH_2{-}CH_2Br}$) is the major, not minor, product in presence of peroxide.

E. True

In absence of peroxide, HBr adds by electrophilic addition via a carbocation intermediate (here benzylic carbocation), giving Markovnikov product.

Correct option: Option C (A, C & E Only)

Vinyl, allyl, benzyl and alkyl chlorides are identified from the position of $Cl$:

Step 1: Identify each example in List-II

1. III. $CH_2 = CHCl$

   Here $Cl$ is directly attached to a double-bond carbon $\Rightarrow$ vinyl chloride.

2. I. $CH_2 = CH-CH_2Cl$

   Here $Cl$ is on the carbon next to $C=C$ (allylic position) $\Rightarrow$ allyl chloride.

3. II. $CH_3-CH(Cl)CH_3$

   This is a saturated alkyl group with $Cl$ on an $sp^3$ carbon $\Rightarrow$ alkyl chloride (2-chloropropane).

4. IV. (structure shown is $C_6H_5-CH_2Cl$)

   $Cl$ is on the carbon next to benzene ring (benzylic position) $\Rightarrow$ benzyl chloride.

Step 2: Match List-I with List-II

• A. Vinyl chloride $\rightarrow$ III

• B. Benzyl chloride $\rightarrow$ IV

• C. Alkyl chloride $\rightarrow$ II

• D. Allyl chloride $\rightarrow$ I

So the correct option is:

$\boxed{\text{Option C: A-III, B-IV, C-II, D-I}}$

In $\mathrm{Ph}-\mathrm{OMe}$, the $-\mathrm{OMe}$ group is an electron-donating group and shows $+\mathrm{M}$ (plus mesomeric) effect. It donates electron density to the benzene ring, increases the electron density on the ring, and therefore activates the ring towards electrophilic substitution such as nitration.

In $\mathrm{Ph}-\mathrm{NO}_2$, the $-\mathrm{NO}_2$ group is a strong electron-withdrawing group and shows $-\mathrm{M}$ (minus mesomeric) effect. It pulls electron density away from the benzene ring and therefore strongly deactivates the ring towards nitration.

In $\mathrm{Ph}-\mathrm{Cl}$, the $-\mathrm{Cl}$ group is overall an electron-withdrawing group (due to its $-I$ or inductive effect). Because of this electron-withdrawing nature, it decreases the reactivity of the benzene ring towards nitration, but this effect is weaker than that of the $-\mathrm{NO}_2$ group.

Therefore, the increasing order of reactivity towards nitration is: $\mathrm{Ph}-\mathrm{NO}_2 < \mathrm{Ph}-\mathrm{Cl} < \mathrm{Ph}-\mathrm{OMe}$.

In chlorocyclohexane, chlorine shows only the $-\mathrm{I}$ (electron-withdrawing inductive) effect.

So, the $\mathrm{C}-\mathrm{Cl}$ bond remains a normal single bond and is more polar (more electron density is pulled towards chlorine).

In chlorobenzene, chlorine shows two effects: $-\mathrm{I}$ and $+\mathrm{M}$ (lone pair donation by resonance into the benzene ring).

Because $-\mathrm{I}>+\mathrm{M}$, the bond is still polar, but the $+\mathrm{M}$ effect reduces the polarity compared to chlorocyclohexane.

The $+\mathrm{M}$ effect also causes resonance between chlorine lone pairs and the aromatic ring, which gives the $\mathrm{C}-\mathrm{Cl}$ bond partial double bond character.

Rate of $\mathrm{S}_{\mathrm{N}} 1 \propto$ Stability of $\mathrm{C}^{\oplus}$ formed

(I) and (II) are unstable due to Bredt's rule, (I) has more + I effect.

$ \text { (II) }<\text { (I) }<\text { (III) }<\text { (IV) } $

$ \begin{aligned} &\text { In methanol, } \mathrm{CH}_3\mathrm{Br} \text{ reacts by } S_N2 \text{ mechanism, so the rate depends mainly on the nucleophilicity of the nucleophile.} \\[4pt] &\text { In a protic solvent like methanol, smaller anions like } \mathrm{F}^{\ominus} \text{ are strongly solvated (surrounded by solvent molecules).} \\ &\text { Because of this strong solvation, } \mathrm{F}^{\ominus} \text{ becomes less available to attack the carbon, so it is the weakest nucleophile.} \\[4pt] &\text { Larger anions like } \mathrm{I}^{\ominus} \text{ are less solvated in methanol, so they can attack more easily. Hence, } \mathrm{I}^{\ominus} \text{ is the strongest nucleophile.} \\[4pt] &\text { Among } \mathrm{C}_2\mathrm{H}_5\mathrm{O}^{\ominus} \text{ and } \mathrm{PhO}^{\ominus}, \text{ phenoxide } (\mathrm{PhO}^{\ominus}) \text{ is less nucleophilic.} \\ &\text { This is because its negative charge is delocalised (resonance) into the benzene ring, reducing its tendency to attack.} \\ &\text { Ethoxide } (\mathrm{C}_2\mathrm{H}_5\mathrm{O}^{\ominus}) \text{ has a more localised negative charge, so it is a better nucleophile.} \\[6pt] &\text { Order of nucleophilicity : }\\ &\mathrm{I}^{\ominus}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{\ominus}>\mathrm{PhO}^{\ominus}>\mathrm{F}^{\ominus} \end{aligned} $

For $CH_3Br$ (a methyl halide), the reaction proceeds by $S_N2$, so the rate increases with nucleophilicity of $Nu^-$.

For oxygen nucleophiles in the same row, nucleophilicity roughly follows basicity, and any resonance stabilization of the anion reduces nucleophilicity:

• $OH^-$: strong base, charge localized $\Rightarrow$ best nucleophile

• $PhO^-$: resonance-stabilized $\Rightarrow$ less nucleophilic than $OH^-$

• $CH_3COO^-$: even more resonance delocalization over two oxygens $\Rightarrow$ weaker nucleophile

• $ClO_4^-$: highly resonance-stabilized, essentially non-nucleophilic

$ OH^- \;>\; PhO^- \;>\; CH_3COO^- \;\gg\; ClO_4^- $

Correct option: B.

Type of reaction: This reaction happens by the $S_N1$ mechanism.

What affects the rate: The speed of an $S_N1$ reaction depends on how stable the carbocation (positively charged ion) formed in the process is.

6. As the hydrocarbon chain increases, effective van der waals force also increases. This results in increase in boiling point also.

$ \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I}>\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{I}>\mathrm{CH}_3-\mathrm{I} . $

Para substituted chlorobenzene has greater packing than ortho substituted. Better packing leads to higher melting point. Ortho-substituted chlorobenzene has greater dipole - dipole interaction that para one which gives higher boiling point.

Aryl cyanide is not formed by Gattermann reaction.

A. Finkelstein reaction:

This is a halogen exchange reaction used specifically for the preparation of alkyl iodides. In this reaction, alkyl chlorides or bromides are treated with sodium iodide ($ \mathrm{NaI} $) in dry acetone.

Therefore, A matches with III.

B. Swarts reaction:

This reaction is the preferred method for the synthesis of alkyl fluorides. It involves heating an alkyl chloride or bromide in the presence of a metallic fluoride. Some of the common fluorinating agents used are $ \mathrm{AgF} $, $ \mathrm{Hg}_2\mathrm{F}_2 $, $ \mathrm{CoF}_2 $, or antimony trifluoride ($ \mathrm{SbF}_3 $).

Therefore, B matches with I.

C. Sandmeyer's reaction:

This reaction is used to prepare haloarenes (like chlorobenzene or bromobenzene) from benzene diazonium chloride. The diazonium salt is treated with a cuprous halide, such as cuprous chloride ($ \mathrm{Cu}_2 \mathrm{Cl}_2 $ or $ \mathrm{CuCl} $), to replace the diazonium group with a halogen.

Therefore, C matches with IV.

D. Fittig reaction:

This reaction involves aryl halides. When an aryl halide is treated with sodium ($ \mathrm{Na} $) in the presence of dry ether, two aryl groups join together to form a diaryl compound (like biphenyl).

Therefore, D matches with II.

Summary of Matches:

• A (Finkelstein reaction) $\rightarrow$ III ($ \mathrm{NaI} $)

• B (Swarts reaction) $\rightarrow$ I ($ \mathrm{SbF}_3 $)

• C (Sandmeyer's reaction) $\rightarrow$ IV ($ \mathrm{Cu}_2 \mathrm{Cl}_2 $)

• D (Fittig reaction) $\rightarrow$ II ($ \mathrm{Na} $, dry ether)

Comparing our findings with the given options, the correct combination is A-III, B-I, C-IV, D-II.

Option B is the correct answer.

Check each statement carefully.

For $R-X$, alcoholic KOH generally gives elimination, while aqueous KOH gives substitution.

This happens because alcohol is less polar than water, so the hydroxide ion is less solvated and behaves more like a base, favouring $\beta$-elimination.

So, $A$ is correct.

Now $B$:

In $S_N1$ reaction, reactivity depends on stability of the carbocation formed.

For benzyl chloride:

$C_6H_5-CH_2-Cl \rightarrow C_6H_5-CH_2^+$

This is a benzyl carbocation, stabilized by resonance.

For benzhydryl chloride:

$C_6H_5-CHCl-C_6H_5 \rightarrow C_6H_5-CH^+-C_6H_5$

This carbocation is stabilized by resonance with two phenyl rings, so it is more stable than benzyl carbocation.

Hence reactivity order should be

$C_6H_5-CHCl-C_6H_5 > C_6H_5-CH_2-Cl$

So statement $B$ is false.

Now $C$:

Aryl halides do not show properties similar to alkyl halides.

They are much less reactive in nucleophilic substitution because the $C-X$ bond in aryl halides has partial double bond character due to resonance.

So $C$ is false.

Now $D$:

Vinyl chloride, $CH_2=CHCl$, is a haloalkene. This part is correct.

Allyl chloride, $CH_2=CH-CH_2Cl$, is also a haloalkene, not a haloalkyne.

So $D$ is false.

Now $E$:

Alkyl chlorides can be prepared from alcohols using thionyl chloride:

$R-OH + SOCl_2 \rightarrow R-Cl + SO_2 + HCl$

But phenols do not give aryl chlorides with $SOCl_2$.

In phenol, the $C-O$ bond has partial double bond character due to resonance, so replacement of $OH$ by $Cl$ does not happen this way.

So $E$ is correct.

The correct statements are $A$ and $E$ only.

$\boxed{\text{Option C}}$

In an $S_N1$ reaction, the rate depends on the stability of the carbocation formed after the leaving group departs.

For benzyl chloride:

$ \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{CH}_2^+ + \text{Cl}^- $

The benzyl carbocation formed is highly stable because the positive charge is delocalised over the benzene ring by resonance.

For ethyl chloride:

$ \text{CH}_3\text{CH}_2\text{Cl} \rightarrow \text{CH}_3\text{CH}_2^+ + \text{Cl}^- $

The ethyl carbocation is much less stable. It gets only slight stabilization by hyperconjugation, which is weaker than resonance stabilization.

So:

• Statement (I) is true: benzyl chloride reacts faster than ethyl chloride in the $S_N1$ mechanism.

• Statement (II) is also true: ethyl carbocation is less stabilized by hyperconjugation than benzyl carbocation by resonance.

Also, Statement (II) correctly explains Statement (I).

Therefore, the correct answer is:

$ \boxed{\text{Option A}} $

Among the given options, the major product decided by stability of carbocation formed in intermediate.

Both reactions are electrophilic substitution reaction, $\mathrm{I}^{\text {st }}$ is sulphonation and $\mathrm{II}^{\text {nd }}$ is halogenation :

A nucleophilic substitution reaction is a chemical reaction in which an electron rich molecule (nucleophile) replaces a functional group in an electron deficient molecule (electrophile). Carbocation formation occurs in SN$^1$ mechanism (unimolecular nucleophilic substitution).

Carbocation is formed in the first step when the bond between carbon and the leaving group breaks. The nucleophile attack on the carbocation occurs in the second step.

The general stability order of carbocation is

Tertiory > Secondary > Primary

The stability order including allyl carbocation, benzyl carbocation, phenyl carbocation and triphenyl methyl $\left( {{{(Ph)}_3}{C^ + }} \right)$ carbocation is

Phenyl < Allyl < Benzyl < Triphenyl methyl

(1)

Carbocation formed from this halide is

Allylic carbocation stability is due to the resonance delocalization of the positive change across the adjacent double bond.

(2)

Stable due to the delocalization of positive change across all three phenyl rings.

(3)

Benzyl carbocation stability is because of the positive charge delocalization across the aromatic ring. So, it is mole stable than allylic carbocation but less stable than Ph$_3$C+ carbocation.

(4)

Delocalization is possible for the adjacent double bond. So, most stable carbocation is

The answer is option (2)

$\mathrm{\mathop {C{H_3} - C{H_2} - C{H_2} - N{O_2}}\limits_{(A)} \buildrel {Ag - N{O_2}} \over \longrightarrow C{H_3} - C{H_2} - C{H_2} - Br\buildrel {AgCN} \over \longrightarrow \mathop {C{H_3} - C{H_2} - C{H_2} - NC}\limits_{(B)}}$

Transition state (less charge density)

$\Rightarrow$ This reaction will proceed faster in less polar medium which will not increase the activation energy value.

Transition state (Higher charge density)

$\Rightarrow$ This reaction will proceed faster in more polar medium which will decrease the activation energy value.

Addition of $\mathrm{H}-\mathrm{Br}_{(\mathrm{aq})}$ to alkene follows electrophilic addition mechanism. In the rate determining step a carbocation intermediate is formed. Among P, Q, R & S compound Q will form most stable carbocation intermediate since it is resonance stabilized.

A is phenol and B is para benzoquinone.

Solvolysis or $\mathrm{S}_{\mathrm{N}} 1 \propto$ stability of carboccation Stability order

$\mathrm{RBr} \xrightarrow[\text { dry Ether }]{\mathrm{Mg}} \mathrm{RMgBr} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{R}-\mathrm{H}$

Hence, RBr Can be

Total 4 Structural isomers.

$\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_2-\mathrm{Cl}$ will undergo $\mathrm{S}_{\mathrm{N}} 1$ mechanism because $\mathrm{CH}_3-\mathrm{O}-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ is highly stable.

Atoms H and Br are added across the double bond. Finkelstein the action is 1 an SN2 (Substitution Nucleophilic Bimolecular) relation mechanism that involves the exchange of one halogen atom for another. Here, $\mathrm{Br}^{-}$is replaced by I.In this reaction, I attacks the carbon atom attached to the halogen Br. General reaction:$ \begin{aligned} & R-X+N a I \rightarrow R-I+N a X \\ & X: C l / b r\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\mathrm{NaCl}, \mathrm{NaBr}) \end{aligned} $Swart reaction is the reaction used to synthesize alkyl fluorides from alkyl chlorides or bromides. It involves heating the alkyl halide (chloride or bromide) with a metal fluoride (AgF) to replace, the halogen $(\mathrm{Cl}$ or Br$)$ with a fluorine atom.$ \text { Hero, } \mathrm{Br}^{-} \text {is replaced by } \mathrm{F}^{-} $$ \text { General reaction: } $ $ \begin{aligned} & R-X \quad + \quad \text { AgF } \quad \rightarrow R \rightarrow F \quad +\operatorname{Ag} X \\ & (X =Cl, B r)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\mathrm{AgCl}, \mathrm{AgBr}) \end{aligned} $ A) $C-X$ bond length in $P, Q$ and $R$ follows the order $Q > R > P $. Bond length order $C-F < C - B r < C-I$ Bond length increases with the size of the halogen atom. Iodine is/ larger than bromine, which is longer than fluorine. Therefore the C-I bond (Q) is longer than $C-B r$ bond $(P)$, which is longer than the C-F bond (R). The correct order is $Q>P>R$. So, the statement is not correct.B) $C-X$, bond enthalpy in $P, Q$ and $R$ flows the order $R>P>Q$ Bond enthalpy is the energy required to break a bond stronger, bonds are shorter in length and require more energy to break the bond. The C-F bond (R) is the strongest and shortest bond, followed by $C -B r$ bond $(P)$, and the $C-I$ bond $(Q)$ is the weakest and longest loons. So, C-F Fond (R) requires higher energy to bond breaking. Its bond enthalpy is higher. C - I bond (Q) is the weakest bond and its bond enthalpy is lower. The bond enthalpy of $C-B r(P)$ is intermediate between $R_1$ and $Q$.$ \text { So, the statement is correct. } $c) Relative reactivity toward $S_{N 2}$ reaction in $P, Q$ and $\ R$ follows the order $P>R>Q$ Reactivity towards. $S_{N^2}$, order is $C-I>C-B r>C-F$ $Q>P>R$$S_N 2$ reactions are favoured by less steric hindrance. The order of reactivity is $I>B r>C l>F$, because larger halogens are better leaving groups. So, the correct order is $Q>P>R$. The statement is not correct.D) PK value of the conjugate acids of the leaving groups In $P, Q$ and $R$ follows the order $\quad R>Q>P$. $ \text { Leaving group } \Rightarrow \text { P }\,\,\,\,\,\,Q \,\,\,\,\,\,\,R $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{Br}^{-} \,\,\, \mathrm{I}^{-}\quad \mathrm{F}^{-}$$ \text { Conjugate acids } \Rightarrow \mathrm{HBr}\,\,\, HI\,\,\,\,\,HF $$ \text { pka order conjugate acids } \Rightarrow \mathrm{HI}<\mathrm{HBr}<\mathrm{HF} $The acidity of the acids HBr, HI and HF, increases in the order HF < HBr < HI. So, the Orrder of pka is HI< HBr < HF orThe acidity of the acids HBr, HI and HF, increases in the order HF < HBr < HI. So, the Orrder of pka is HI< HBr < HF or$Q < P < R$ or $R > P > Q$ : So, the given statement is not correct. Correct statement is (B). Answer: Option (B)

$ \text { The complete reaction is as follows } $

Thus, the correct decreasing order will be

$ 3^{\circ}>2^{\circ}>1^{\circ} \text { or } Y>X>Z $

$ \text { The reaction sequence is as follows } $

$ \text { So, } Z \text { is propanone. } $

The complete reaction sequence is as follows.

The complete reaction mechanism is, as follows,

$\mathrm{C}_6 \mathrm{Cl}_6$ is the molecular formula of product D. Thus, its, empirical formula is CCl.

$ \text { The complete reaction sequence is } $

$\mathrm{S}_{\mathrm{N}} 1$ mechanism depends on carbocation stability. The more stable the carbocation formed after the leaving group departs, the faster is the $\mathrm{S}_{\mathrm{N}} 1$ reaction.

In

tertiary benzylic carbocation is stabilised by $+I$ effect of $\mathrm{CH}_3$ and 2 phenyl group, which is most stable among the given options.

The presence of $e^{-}$withdrawing group like $-\mathrm{NO}_2$ at meta positions decreases the reactivity of haloamines. Thus among the given options

is least reactive towards nucleophilic substitution

$ \text { The complete reaction is as follows } $

Among the given options chlorobenzene and anisole will undergo methylation with $\mathrm{CH}_3 \mathrm{Cl} /$ anhy. $\mathrm{AlCl}_3$.

Ethyl chloride is least reactive towards hydrolysis by $\mathrm{S}_{\mathrm{N}} 1$ reaction. In $\mathrm{S}_{\mathrm{N}} 1$ reaction, the rate determining step involves the formation of carbocation. The stability of carbocation directly affects the reactivity $\mathrm{CH}_3 \mathrm{CH}_2^{+}$is least stable carbocation.

The complete reaction sequence are as follows

Thus, $p$-chlorotoluene is the major product formed in II and III.

$ \text { The complete reaction sequence is as follows } $

Now number of $s p^3$ carbon are 2 while number of $s p^2$ carbons are 6 . So, their ratio is $2: 6$ or $1: 3$.

The presence of an electron withdrawing group like $-\mathrm{NO}_2$ at ortho and para positions increases the reactivity of haloarenes. Thus among the given options

is most reactive towards nucleophilic substitution with an aqueous NaOH .

The correct statement is $\mathrm{CH}_2 \mathrm{Cl}_2$ is used as propellant in aerosols.

• Option A: HCl + ZnCl₂ (Lucas Reagent)

  • Reaction: $\mathrm{ROH + HCl \xrightarrow{ZnCl_2}} \mathrm{RCl + H_2O}$

  • Byproduct: Water ($\mathrm{H_2O}$). Water needs to be separated from the alkyl chloride, which can be challenging, especially for lower molecular weight alkyl chlorides that may be soluble in water or form azeotropes. This makes obtaining pure alkyl chloride directly difficult.

• Option B: PCl₅ (Phosphorus Pentachloride)

  • Reaction: $\mathrm{ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl}$

  • Byproducts: Phosphorus oxychloride ($\mathrm{POCl_3}$) and hydrogen chloride ($\mathrm{HCl}$). $\mathrm{POCl_3}$ is a liquid, and $\mathrm{HCl}$ is a gas. The separation of these byproducts from the desired alkyl chloride may require careful distillation or other purification steps, affecting the direct purity.

• Option C: SOCl₂ (Thionyl Chloride)

  • Reaction: $\mathrm{ROH + SOCl_2 \xrightarrow{Pyridine}} \mathrm{RCl + SO_2 \uparrow + HCl \uparrow}$

  • Byproducts: Sulfur dioxide ($\mathrm{SO_2}$) and hydrogen chloride ($\mathrm{HCl}$). Both of these are gases. They readily escape from the reaction mixture as they are formed. This leaves behind a relatively pure alkyl chloride, making the purification process very simple and efficient. This is often considered the best method for preparing pure alkyl halides because the gaseous byproducts automatically separate from the desired product.

• Option D: PCl₃ (Phosphorus Trichloride)

  • Reaction: $\mathrm{3ROH + PCl_3 \rightarrow 3RCl + H_3PO_3}$

  • Byproduct: Phosphorous acid ($\mathrm{H_3PO_3}$). This is a non-volatile solid or viscous liquid. It needs to be separated from the alkyl chloride, typically by distillation or extraction, which adds purification steps.

Conclusion:

Among the given reagents, $\mathrm{SOCl_2}$ (thionyl chloride) is the preferred reagent for the preparation of pure alkyl chloride from alcohol because the byproducts ($\mathrm{SO_2}$ and $\mathrm{HCl}$) are gases that escape from the reaction mixture, leaving behind a pure alkyl chloride.

The final answer is $\boxed{\text{SOCl}_2}$.

Fittig reaction : When aryl halide react with sodium in presence of dry ether to form a biaryl compound.

Let's define each type of halide and then match it with the given examples.

Definitions:

• Vinyl Halide: A compound in which the halogen atom (X) is directly attached to an sp² hybridized carbon atom of an alkene. The general structure is R-CH=CX-R' or R₂C=CX-R'.

• Allyl Halide: A compound in which the halogen atom (X) is attached to an sp³ hybridized carbon atom that is adjacent to a carbon-carbon double bond (an allyl group). The general structure is R-CH=CH-CH₂-X.

• Benzyl Halide: A compound in which the halogen atom (X) is attached to an sp³ hybridized carbon atom that is directly attached to an aromatic ring (a benzyl group). The general structure is Ar-CH₂-X. The carbon bearing the halogen can be primary, secondary, or tertiary.

• Aryl Halide: A compound in which the halogen atom (X) is directly attached to an sp² hybridized carbon atom of an aromatic ring. The general structure is Ar-X.

Matching the Examples:

• (A) Vinyl Halide:

  • (III) 1-bromo-3-methylcyclohexene: In this molecule, the double bond can be assumed to be between C1 and C2. If the bromine is at C1, then C1 is part of the double bond (sp² hybridized). Thus, it's a vinyl halide.

  • Match: A - (III)

• (B) Allyl Halide:

  • (IV) 3-bromo-4-methylcyclohexene: If the double bond is between C1 and C2, then C3 is an sp³ hybridized carbon adjacent to the double bond. If bromine is at C3, it is attached to an allylic carbon.

  • Match: B - (IV)

• (C) Benzyl Halide:

  • (I) 1-bromo-1-phenylethane: The structure is C₆H₅-CH(Br)-CH₃. The carbon bearing the bromine is an sp³ hybridized carbon atom and is directly attached to the phenyl group (an aromatic ring). This is a secondary benzyl halide.

  • Match: C - (I)

• (D) Aryl Halide:

  • (II) 3-bromotoluene: The structure is CH₃-C₆H₄-Br (meta-bromotoluene). The bromine atom is directly attached to one of the sp2 hybridized carbon atoms of the benzene ring.

  • Match: D - (II)

Summary of Matches:

• A - III

• B - IV

• C - I

• D - II

Comparing this with the given options, Option B is the correct answer.

The final answer is $\boxed{\text{Option B}}$