Haloalkanes and Haloarenes

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 $\boxed{\text{SOCl}_2}$.

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 $\boxed{\text{Option B}}$

Aniline does not undergoes Friedel-Craft reaction due to the Lewis basicity of nitrogen atom in aniline, which react with Lewis acid catalyst $\left(\mathrm{AlCl}_3\right)$, forming complex that deactivates the benzene ring and undergoes further electrophilic substitution.

The complete reaction sequence is as follow

Now, we know that order for $\mathrm{S}_{\mathrm{N}} 1$ reaction is $3^{\circ}$ halide $>2^{\circ}$ halide $>1^{\circ}$ halide $>\mathrm{CH}_3 \mathrm{X}$ and

$ R-\mathrm{I}>R-\mathrm{Br}>R-\mathrm{Cl}>R-\mathrm{F} $

Thus, the correct order is $y>x>z$

The sequence of reagents required to convert ethylbromide to propanal is $\mathrm{CH}_3 \mathrm{COOAg}$, DIBAL-H and $\mathrm{H}_2 \mathrm{O}$.

Allylic halides are those compound 156 where the halogen group is attached to a $s p^3$ hybridised carbon atom and at $\alpha$- to the carbon-carbon double bond. 4-chlorobut-1-ene is not allylic halide.

$ \text { The product } C \text { formed will be as follows } $

In the presence of peroxide, addition of HBr to unsymmetrical alkenes takes place contrary to the Markownikoff rule. In case of 2-butene both double bonded carbons are identical. Therefore, it does not observe anti-Markownikoff addition of HBr .

Among the given reaction 1 -iodopropane can be prepared by reactions given are $B$ and $C$ only.

When alkyl chloride/bromide is heated in presence of fluoride of same heavy metal alkyl fluoride is obtained. This reaction is named as Swarts reaction.

When chlorobenzene $(X)$ is reacted with reagent NaOH $(A)$ it produce phenol.

Hence, $A=$ (i) $\mathrm{NaOH}, 623 \mathrm{~K}, 300 \mathrm{~atm}$ (ii) $\mathrm{H}^{+} B=$ (i) NaOH , 443 K (ii) $\mathrm{H}^{+}$

The correct match is A-III, B-I, C-II.

Sandmeyer reaction is a reaction used to synthesis aryl halides from aryl diazonium salt.

Finkelstien reaction is a halogen $(X)$ exchange method. It is used to prepare alkyl iodide by reaction of alkyl chlorides/bromides with NaI in dry acetone.

$ \underset{(x=\mathrm{Cl}, \mathrm{Br})}{R-X}+\mathrm{NaI} \xrightarrow{\text { Acetone }} R \mathrm{I}+\mathrm{NaX} $

Swartz reaction is used to get alkyl fluorides from alkyl chlorides.

$ R-\mathrm{Br}+\mathrm{AgF} \longrightarrow R-\mathrm{F}+\mathrm{AgBr} $

The ammonia gas released will turn red litmus to blue. Thus, the compound $X$ and $Y$ are $\mathrm{KCN} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$.

Cl has $-I$ and $+R$-effect. Due to electron donating resonance effect it is ortho/para-directing and donates the electron density to benzene ring. Thus, the incoming electrophile -Br gets attached on para-position and para disubstituted product is obtained as major product and ortho as minor product.

Alkenyl bromide reacts with Mg in dry ether to form Grignard reagent.

Grignard reagent abstracts acidic proton. So, it takes $\mathrm{D}^{+}$ form $\mathrm{D}_2 \mathrm{O}$.

Finkelstein reaction is an $\mathrm{S}_{\mathrm{N}} 2$ reaction that involves the exchange of one halogen by iodine. In this reaction, alkyl chloride/bromides are treated with sodium iodide in acetone to form alkyl iodide and $\mathrm{NaCl} / \mathrm{NaBr}$. Acetone is used because $\mathrm{NaCl} / \mathrm{NaBr}$ are less soluble in acetone hence, they can precipitate out.

$R-\mathrm{Cl}+\mathrm{NaI} \xrightarrow{\text { Acetone }} R-\mathrm{I}+\mathrm{NaCl} \downarrow$

Density is directly proportional to mass. The molar masses of the compounds are as follows:

$ \begin{aligned} \mathrm{CCl}_4 & \rightarrow 12 + 142 = 154 \ \mathrm{g/mol} \\ n-\mathrm{C}_3\mathrm{H}_7\mathrm{I} & \rightarrow (12 \times 3) + (7 \times 1) + 127 = 170 \ \mathrm{g/mol} \\ n-\mathrm{C}_3\mathrm{H}_7\mathrm{Br} & \rightarrow (12 \times 3) + (7 \times 1) + 80 = 123 \ \mathrm{g/mol} \\ n-\mathrm{C}_3\mathrm{H}_7\mathrm{Cl} & \rightarrow (12 \times 3) + (7 \times 1) + 35.5 = 78.5 \ \mathrm{g/mol} \\ \end{aligned} $

Since the molar mass of $n-\mathrm{C}_3\mathrm{H}_7\mathrm{I}$ is the highest, it will also have the highest density.

Benzyl chloride reacts with Mg in presence of dry ether to form Grignard reagent (G.R.) and on treating with $\mathrm{CO}_2$, G.R. form carboxylate ion which on hydrolysis give carboxylic acid.

$X$ on treatment with $\mathrm{LiAlH}_4$ ether followed by acidic hydrolysis give alcohol as $Y$ product.

$\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$ is used to reduce carbonyl compounds to alkane. This reaction is called Clemmenson reduction. But in presence of basic groups like $-\mathrm{OH},-\mathrm{NH}_2$ etc. side reaction also takes place. In this case substitution of alcoholic group by $\mathrm{Cl}^{\ominus}$ will take place.

Amides reacts with Br$_2$ in presence of an alkali base (KOH, NaOH) to form amine. This reaction is known as Hofmann bromamide reaction. This reaction proceeds via formation of nitrene intermediate.

First step in the given reaction is "Lucas test" in which alcohols react with $\mathrm{HCl} / \mathrm{ZnCl}_2$ to give alkyl halides.

Second step proceed via nucleophilic substitution bimolecular $(\mathrm{S}_{\mathrm{N}} 2$) mechanism. Here, $\mathrm{I}^{-}$ replaces Cl to form alkyl iodide.

Third step is "Wurtz reaction" in which two alkyl iodide reacts with Na in presence of dry ether to form alkane with even number of carbon atoms.

1-chlorobutane reacts with aq. KOH to give 1-butanol by S_N2 reaction.

However, 1-chlorobutane reacts with alc. KOH to give 1-butene by elimination reaction.

For a compound to be aromatic, it must have (4n + 12) $\pi$-electrons.

Here, two double bonds and 4$\pi$-electrons are present in the compound. Therefore, it is anti-aromatic.

Homocyclic compounds are ring structures that consist only of carbon atoms with in the ring. Bromo group is attached to ring system as substituent, so $\mathrm{C_6H_6Br_2}$ is homocyclic but not aromatic.

Hence, option (c) is correct.

With mixture of conc. HNO$_3$ and conc. H$_2$SO$_4$, anisole gives ortho and para-nitro anisole.

Methoxy group is activating and ortho-para directing group.