Haloalkanes And Haloarenes – JEE Main ChemistryPractice Questions & PYQs

The reaction of benzyl halides with KCN proceeds via an $S_N2$ mechanism, wherein the rate is primarily governed by the electrophilicity of the benzylic carbon atom. Electron-withdrawing groups increase this electrophilicity by pulling electron density away from the reactive center, thereby accelerating the reaction, while electron-donating groups decrease it and consequently retard the rate of nucleophilic attack. This fundamental electronic effect dictates the reactivity order $b > a > d > c$, aligning with a progression from the strongest electron-withdrawing substituent to the strongest electron-donating substituent.

Alcoholic KOH acts as a potent base because alcohol is a less polar solvent than water, leading to poor solvation of the base and effectively promoting elimination pathways for $R-X$ over substitution. In the context of $S_N1$ reactivity, the rate is determined by the stability of the intermediate carbocation, where the diphenylmethyl carbocation is more stable than the benzyl carbocation, thereby making the suggested order in the statement incorrect. Aryl halides function differently than alkyl halides due to $sp^2$ hybridization and resonance effects, and classifying allyl chloride as a haloalkyne is incorrect because it is a haloalkene containing a double bond. Finally, while $R-\mathrm{OH}$ reacts with $\mathrm{SOCl}_2$ to produce $R-\mathrm{Cl}$, the resonance-stabilized C-O bond in phenols prevents the formation of $Ar-\mathrm{Cl}$ from $Ar-\mathrm{OH}$ under these conditions.

In chlorobenzene, the lone pairs on the chlorine atom undergo resonance delocalization into the aromatic ring, which introduces partial double bond character to the $C-Cl$ bond and effectively reduces its dipole moment, making it less polar than the $C-Cl$ bond in chlorocyclohexane. This resonance effect acts in opposition to the inductive electron-withdrawing effect of chlorine, leading to a reduction in the magnitude of the negative charge on the chlorine atom, while the bond length is actually shorter due to the stronger covalent character and the participation of the $sp^2$ hybridized carbon orbital. These structural and electronic factors, specifically the existence of partial double bond character, the reduced polarity, and the utilization of an $sp^2$ hybridized carbon, represent the correct chemical attributes distinguishing chlorobenzene from chlorocyclohexane.

The molar mass of the gas released is determined by calculating its density at STP, where a molar volume of $22.4 \mathrm{dm}^3$ represents one mole. Since $1 \mathrm{g}$ of the gas occupies $1.4 \mathrm{dm}^3$, its molar mass is $\frac{22.4}{1.4} = 16 \mathrm{g/mol}$, confirming the substance is methane. The reaction of the Grignard reagent $\mathrm{CH}_3\mathrm{MgI}$ with water confirms the identity of the alkyl group as methyl, as described by the reaction $\mathrm{CH}_3\mathrm{MgI} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{CH}_4 + \mathrm{Mg}(\mathrm{OH})\mathrm{I}$.

Methane reacts with iodine in the presence of $\mathrm{HIO}_3$ to form iodomethane, which serves as compound (X). The subsequent treatment of iodomethane with sodium and dry ether triggers a Wurtz coupling reaction, yielding ethane as compound (Y) according to the equation $2 \mathrm{CH}_3\mathrm{I} \xrightarrow{\mathrm{Na}/\text{Dry ether}} \mathrm{CH}_3-\mathrm{CH}_3$. Since compound (Y) is ethane with a molecular formula of $\mathrm{C}_2\mathrm{H}_6$, its molar mass is calculated by summing the atomic masses of two carbon atoms and six hydrogen atoms, resulting \in a value of $30 \mathrm{g/mol}$.

In $CH_2=CH-Cl$, the chlorine atom's lone pair is in conjugation with the $\pi$ electrons of the double bond, which imparts partial double bond character to the $C-Cl$ bond, thereby making it stronger and shorter than the $sp^3$ hybridized $C-Cl$ bond in $CH_3-CH_2-Cl$. Hydrolysis of an optically active molecule often proceeds through pathways such as the formation of a planar carbocation intermediate, which results in the production of a racemic mixture consisting of equal amounts of enantiomers. Because a racemic mixture is optically inactive, it cannot rotate the plane of polarized light, rendering the second statement false.

Name reactions are characterized by specific reagents that act as diagnostic tools for identifying reaction mechanisms in organic chemistry. In this set, the Finkelstein reaction employs $\mathrm{NaI}$ to facilitate halogen exchange, whereas the Swarts reaction utilizes $\mathrm{SbF}_3$ specifically to introduce fluorine atoms into an organic framework. The Sandmeyer reaction typically uses $\mathrm{Cu}_2\mathrm{Cl}_2$ to substitute a diazonium group with a halide, and the Fittig reaction relies on the coupling of two aryl halide molecules in the presence of sodium metal and dry ether. Identifying these reagents confirms the pairing of Finkelstein with $\mathrm{NaI}$, Swarts with $\mathrm{SbF}_3$, Sandmeyer with $\mathrm{Cu}_2\mathrm{Cl}_2$, and Fittig with sodium in dry ether.

The rate of an $\mathrm{S_N1}$ reaction is primarily determined by the stability of the intermediate carbocation formed during the rate-determining step, meaning that a more stable carbocation leads to a faster reaction. When ethyl chloride ionizes, it generates an ethyl carbocation ($\mathrm{CH_3}-\mathrm{C^+H_2}$), which is stabilized by hyperconjugation with the adjacent methyl group. In contrast, benzyl chloride forms a benzyl carbocation ($\mathrm{Ph}-\mathrm{C^+H_2}$), where the positive charge is significantly delocalized across the benzene ring through resonance. Because resonance provides a much greater degree of stabilization than hyperconjugation, the benzyl carbocation is far more stable than the ethyl carbocation, confirming that benzyl chloride reacts faster and that the relative stabilization effects are correctly described.

The rate of an $S_N1$ reaction is primarily governed by the stability of the carbocation intermediate formed, which requires the cationic center to adopt a planar $sp^2$ hybridized geometry. Rigid bridgehead systems like bicyclo[2.2.1]heptane (II) and bicyclo[2.2.2]octane (I) possess high angular strain that prevents them from achieving the necessary planarity, rendering them significantly less reactive, with the more constrained bicyclo[2.2.1] system reacting slower than the bicyclo[2.2.2] system. Non-rigid tertiary carbocations (III) facilitate easier planarization, while the most reactive species (IV) typically incorporates additional stabilizing effects such as resonance, leading to the observed reactivity order.

The nucleophilicity of anions in polar protic solvents like methanol is primarily governed by the extent of solvation, where smaller, highly charged ions are stabilized by hydrogen bonding, significantly reducing their nucleophilic strength. Consequently, $I^{-}$ acts as a superior nucleophile due to minimal solvation, whereas $F^{-}$ is heavily solvated and displays the lowest reactivity. Between the oxygen-based nucleophiles, $C_{2}H_{5}O^{-}$ is more reactive than $C_{6}H_{5}O^{-}$ because the negative charge on the phenoxide ion is stabilized through delocalization into the benzene ring, thereby decreasing its electron density and nucleophilic character.

The reactivity of aromatic compounds towards nitrating mixture is dictated by the electron density of the benzene ring, which increases with electron-donating substituents ($+I$, $+M$ effects) and decreases with electron-withdrawing substituents ($-I$, $-M$ effects). Electrophilic aromatic substitution is favored by higher ring electron density, making the compound with an electron-donating group the most reactive ($c$), followed by unsubstituted benzene ($a$), and finally the compound with an electron-withdrawing group ($b$). This electronic configuration results in the increasing order of reactivity $b < a < c$.

The dehydration of alcohols via the $E2$ mechanism follows Zaitsev's rule, which dictates that the most substituted alkene is formed as the major product due to its increased thermodynamic stability resulting from hyperconjugation and steric relief. The base abstracts the most acidic $\beta$-hydrogen, leading to a concerted elimination where the transition state energy is minimized for the isomer with the double bond located between the most highly alkylated carbons. This stereoselective and regioselective pathway significantly favors the more substituted, stable alkene over less substituted alternatives, ensuring it predominates in the final product mixture.

Nucleophilicity in $S_N2$ reactions generally correlates with basicity when comparing species with the same nucleophilic atom, as stronger bases are more effective at donating electron density to an electrophile. Since the basicity order of these anions is $\mathrm{OH}^- > \mathrm{PhO}^- > \mathrm{CH_3COO}^- > \mathrm{ClO_4}^-$, which is determined by the increasing acidity of their respective conjugate acids ($\mathrm{H_2O} \approx 15.7$, $\mathrm{PhOH} \approx 10$, $\mathrm{CH_3COOH} \approx 4.8$, and $\mathrm{HClO_4} < 0$), the rate of nucleophilic attack on the electrophilic carbon follows the same sequence.

The molecular formula $\mathrm{C}_5\mathrm{H}_{11}\mathrm{Br}$ corresponds to eight distinct structural isomers, which generate eight corresponding alcohol products upon the substitution of the bromine atom with a hydroxyl group: pentan-1-ol, pentan-2-ol, pentan-3-ol, 3-methylbutan-1-ol, 3-methylbutan-2-ol, 2-methylbutan-2-ol, 2-methylbutan-1-ol, and 2,2-dimethylpropan-1-ol. Optical isomerism occurs in molecules containing at least one chiral center, which is a carbon atom bonded to four different groups.

Evaluating each alcohol product, pentan-2-ol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH(OH)CH}_3$) contains a chiral center at the C2 position, which is bonded to a hydrogen atom, a hydroxyl group, a methyl group, and a propyl group. Similarly, 3-methylbutan-2-ol ($\mathrm{CH}_3\mathrm{CH(OH)CH(CH}_3\mathrm{)}_2$) possesses a chiral center at the C2 position, as it is bonded to a hydrogen atom, a hydroxyl group, a methyl group, and an isopropyl group. Finally, 2-methylbutan-1-ol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH(CH}_3\mathrm{)CH}_2\mathrm{OH}$) features a chiral center at the C2 position, attached to a hydrogen atom, a methyl group, an ethyl group, and a hydroxymethyl group. The other five alcohol products are achiral as they lack any carbon atom bonded to four distinct groups, resulting in 3 products that can exhibit optical isomerism.

Vinyl chloride consists of a chlorine atom attached directly to an $sp^2$ hybridized carbon involved in a double bond ($CH_2=CHCl$), while benzyl chloride refers to a chlorine atom bonded to a carbon atom that is adjacent to a benzene ring. Alkyl chlorides are defined by a chlorine atom attached to a saturated $sp^3$ hybridized aliphatic carbon, such as isopropyl chloride ($CH_3-CH(Cl)CH_3$), and allyl chloride features a chlorine atom on a carbon atom adjacent to a carbon-carbon double bond ($CH_2=CH-CH_2Cl$). Applying these structural definitions to the provided lists results in the correspondence A-III, B-IV, C-II, and D-I.

The addition of $\mathrm{HBr}$ to styrene in the presence of peroxides proceeds via a free-radical chain mechanism involving a resonance-stabilized benzylic radical, which confirms statement A. Without peroxides, the reaction follows an ionic mechanism involving a resonance-stabilized benzylic carbocation intermediate, validating statement E. The thermal decomposition of the benzoyl peroxide initiator generates benzoyloxy radicals that readily release $\mathrm{CO_2}$ to produce phenyl radicals, which can abstract hydrogen from the reaction medium to form benzene as a byproduct, confirming statement C.

The addition of $HBr$ to 3-phenylpropene ($Ph-CH_2-CH=CH_2$) proceeds according to Markovnikov's rule, wherein the electrophilic proton adds to the terminal carbon to generate a secondary carbocation at the adjacent position. Upon subsequent attack by the bromide ion, this yields 2-bromo-1-phenylpropane, a molecule containing a central carbon atom bonded to a methyl group, a benzyl group, a bromine atom, and a hydrogen atom, which confirms the formation of a chiral center.

While the Sandmeyer reaction utilizes copper(I) salts to successfully synthesize both aryl chlorides and aryl cyanides, the Gattermann reaction operates with a more limited scope. Specifically, the Gattermann reaction is effective for preparing aryl chlorides but cannot be employed to synthesize aryl cyanides, rendering the second statement inaccurate.

Boiling points of haloalkanes are primarily governed by the strength of intermolecular van der Waals forces, which increase in magnitude as molecular mass and chain length grow, thus justifying the observed order $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{I} > \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I} > \mathrm{CH}_3 \mathrm{I}$. Regarding structural symmetry, more symmetric molecules exhibit higher melting points because they pack more efficiently into a crystal lattice, whereas their reduced surface area leads to weaker dispersion forces and consequently lower boiling points compared to their less symmetric counterparts. Both statements align perfectly with these fundamental principles of intermolecular forces and molecular geometry.

Stoichiometry dictates that the quantity of product formed is determined by the amount of starting material and the overall efficiency of the reaction pathway. Given that 2-bromopentane has a molar mass of 151 g per mole, the starting quantity of 151 g is equivalent to exactly 1 mole of reactant. The reaction involves an 80 percent yield of the intermediate followed by a 100 percent yield of the final product, which means the overall efficiency of the process is 80 percent. Consequently, 0.8 moles of the final product are produced. Since the molecular mass of the final product is 230 g per mole, the total mass is calculated as:

$0.8 \times 230 = 184$

The mass of product Q obtained is 184 g.

The free-radical photochemical chlorination of alkanes involves the abstraction of hydrogen atoms by chlorine radicals to generate intermediate alkyl radicals, which then react with chlorine molecules to yield chlorinated products. The di-chloro propane isomer 'x' that exhibits optical activity is $1,2$-dichloropropane, $CH_3-CHCl-CH_2Cl$, due to the presence of a chiral center at the second carbon atom.

When $1,2$-dichloropropane undergoes further photochemical chlorination, hydrogen atoms can be abstracted from three distinct carbon environments. Abstraction from the terminal methyl group ($CH_3$) leads to $1,2,3$-trichloropropane ($ClCH_2-CHCl-CH_2Cl$), while abstraction from the central methine group ($CHCl$) yields $1,2,2$-trichloropropane ($CH_3-CCl_2-CH_2Cl$), and abstraction from the terminal methylene group ($CH_2Cl$) results in $1,1,2$-trichloropropane ($CH_3-CHCl-CHCl_2$). As these represent three structurally distinct isomers formed by replacing different hydrogen atoms, a total of three tri-chloro products are obtained.

The reaction sequence proceeds via sequential dehydrohalogenation of 1,2-dibromocyclooctane to form cyclooctyne, followed by hydration to a ketone and final reduction to an alkane. Specifically, treatment with alcoholic $\mathrm{KOH}$ and subsequently $\mathrm{NaNH}_2$ effects double elimination to produce cyclooctyne, which is then hydrated in the presence of $\mathrm{Hg}^{2+}/\mathrm{H}^{+}$ to give cyclooctanone. The process concludes with the Clemmensen reduction using $\mathrm{Zn}-\mathrm{Hg}/\mathrm{H}^{+}$, which converts the carbonyl group into a methylene group, yielding cyclooctane as the major product $P$.