Haloalkanes And Haloarenes – JEE Main ChemistryPractice Questions & PYQs

The peroxide effect, also known as the Kharasch effect, initiates a free radical mechanism that directs the addition of HBr to unsymmetrical alkenes in an anti-Markovnikov fashion. The bromine radical preferentially attacks the less substituted C2 carbon of 3-methylpent-2-ene to generate the more stable tertiary free radical at the C3 position. Subsequent hydrogen atom abstraction from HBr by the radical intermediate leads to the addition of the hydrogen atom at the C3 position and the bromine atom at the C2 position, resulting in 2-bromo-3-methylpentane as the major product.

The decolorization of bromine water is a diagnostic test for the presence of carbon-carbon multiple bonds, where the reddish-brown bromine reacts via electrophilic addition to form a colorless vicinal dibromide. Treatment with the strong, bulky base $tert-BuONa$ typically induces E2 elimination in alkyl halides to produce an alkene, provided the molecule possesses $\beta$-hydrogens. Option A fails to decolorize the bromine solution because it lacks the necessary $\beta$-hydrogens to undergo elimination, preventing the formation of an alkene and leaving the molecule saturated, which renders it chemically inert toward the addition reaction with bromine.

Bulky bases, such as the tert-butoxide ion, exert substantial steric hindrance that significantly impedes their role as nucleophiles, thereby driving them to function predominantly as strong Bronsted bases. When this steric factor is combined with elevated temperatures, the reaction mechanism shifts decisively toward bimolecular elimination, where the base abstracts a beta-proton to facilitate the formation of an alkene. This pathway results in the production of $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}=\mathrm{CH}\mathrm{C}_6\mathrm{H}_5$ as the major product, as the E2 mechanism becomes the kinetically and thermodynamically favored route over any potential nucleophilic substitution.

This reaction involves the double dehydrohalogenation of 2,4-dibromohexane using a strong base such as $\mathrm{KOH}$ in an alcoholic solvent like $\mathrm{CH}_3\mathrm{OH}$. When subjected to heat, the elimination proceeds via an E2 pathway where two molecules of $\mathrm{HBr}$ are systematically removed to establish two pi bonds within the carbon chain. Because the formation of conjugated double bonds provides thermodynamic stability through the delocalization of pi electrons, the reaction selectively yields 2,4-hexadiene, represented as $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$, which is further stabilized by the presence of six alpha-hydrogens.

The free radical addition of HBr to 3-methylpent-2-ene ($CH_3-CH=C(CH_3)-CH_2-CH_3$) proceeds via a mechanism where the bromine radical ($Br^\cdot$) attacks the alkene to form the most stable intermediate carbon radical. The bromine radical attacks the C2 position, leading to the formation of a more stable tertiary radical at the C3 position, which then abstracts a hydrogen atom to produce 2-bromo-3-methylpentane. This product possesses two chiral centers at the C2 and C3 carbons, and since the molecule lacks a plane of symmetry, the total number of stereoisomers is given by $2^n = 2^2 = 4$.

The feasibility of an $\mathrm{S_N1}$ reaction is governed by the stability of the carbocation intermediate formed during the rate-determining step, as higher stability correlates to greater reactivity. Among these compounds, $\mathrm{CH_3CH_2CH_2Cl}$ (II) produces a primary carbocation, which is the least stable, while $\mathrm{CH_3CH(Cl)CH_2CH_3}$ (I) yields a more stable secondary carbocation. The halide $\mathrm{p-H_3CO-C_6H_4-CH_2Cl}$ (III) is the most reactive because it forms a benzylic carbocation stabilized by extensive resonance delocalization, which is further enhanced by the electron-donating methoxy group, leading to an increasing order of reactivity of (II) < (I) < (III).

Elimination reactions require the presence of a strong base to effectively abstract a proton, a function performed by reagents like $\mathrm{NaOH}$ and $\mathrm{NaOEt}$. Sodium iodide, however, serves as a source of iodide ions, which act as highly effective nucleophiles rather than bases. Due to this lack of basicity, $\mathrm{NaI}$ is unable to initiate the elimination pathway and instead promotes nucleophilic substitution reactions.

When cyclohexene is exposed to bromination conditions that initiate a free radical substitution reaction, the bromine atom selectively targets the allylic carbon. This preference arises because the intermediate formed after the abstraction of an allylic hydrogen is stabilized by resonance, making it energetically favorable to form the 3-bromocyclohexene product. By substituting a hydrogen atom at the allylic position, the molecule maintains its double bond while incorporating the bromine substituent.

Synthesis of alkyl fluorides presents unique challenges because the high reactivity of fluorine often leads to uncontrollable side reactions when using conventional nucleophilic substitution pathways. The Swarts reaction overcomes these limitations by employing specific metallic fluorides, such as mercury(I) fluoride, which facilitate a controlled halogen exchange as represented by the equation $2\mathrm{R-Cl}+\mathrm{Hg_2F_2}\text{xrightarrow}{\Delta}2\mathrm{R-F}+\mathrm{Hg_2Cl_2}$. This targeted use of inorganic fluorides provides the most efficient and reliable method for introducing fluorine into an alkyl chain.

The reaction of 1,1,1-trichloroethane with silver powder involves the reductive coupling of two molecules of the trihalide, where silver acts as a halogen-abstracting agent to facilitate the formation of a carbon-carbon triple bond. This process yields but-2-yne through the following chemical transformation: $2 CH_3CCl_3 + 6 Ag \rightarrow CH_3-C\equiv C-CH_3 + 6 AgCl$

The efficiency of $\mathrm{S_N2}$ reactions depends primarily on how easily a nucleophile can approach the electrophilic carbon from the backside, meaning the reaction rate is strictly governed by the amount of steric overcrowding hindering the transition state. Bulky alkyl groups surrounding this carbon create significant congestion that physically obstructs the incoming nucleophile and destabilizes the transition state. Consequently, reactivity decreases as the degree of substitution increases, resulting in the order $\mathrm{CH_3Cl} > \mathrm{CH_3CH_2Cl} > \mathrm{(CH_3)_2CHCl} > \mathrm{(CH_3)_3CCl}$.

$SbCl_5$ acts as a strong Lewis acid that coordinates with the chlorine atom of $(-)-1$-chloro-1-phenylethane, facilitating the heterolytic cleavage of the $C-Cl$ bond to generate a stable, planar benzylic carbocation. This intermediate is achiral and can be attacked by chloride ions from either face with equal probability, leading to an equal mixture of enantiomers and resulting in racemization. The ability of the Lewis acid to extract the halide ion effectively demonstrates that the mechanism proceeds through a carbocation intermediate rather than a free radical or carbanion pathway.

The formation of a white precipitate upon reaction with alcoholic $AgNO_3$ indicates that compound (A) contains a reactive, benzylic bromine atom. Oxidation of 1-(bromomethyl)-2-methylbenzene, which corresponds to the molecular formula $C_8H_9Br$, results in the formation of phthalic acid ($C_8H_6O_4$) as both alkyl substituents are oxidized to carboxylic groups. Since phthalic acid is an ortho-dicarboxylic acid, it undergoes dehydration on heating to readily yield phthalic anhydride, which perfectly aligns with the properties of compound (B).

The formation of chiral products during the monochlorination of $CH_3CH(CH_3)CH_2CH_3$ depends on whether the chlorine atom attaches to a carbon that becomes a stereocenter. When a hydrogen atom on the terminal $CH_3$ group attached to the second carbon is replaced by chlorine, it produces $ClCH_2CH(CH_3)CH_2CH_3$, which contains a chiral $C_2$ carbon atom and exists as a pair of (R) and (S) enantiomers. Similarly, monochlorination at the $CH_2$ position results in $CH_3CH(CH_3)CHClCH_3$, where the $C_3$ carbon atom becomes a stereocenter, forming another pair of (R) and (S) enantiomers. The other possible substitution products, such as $CH_3CH_2C(Cl)(CH_3)_2$ and $ClCH_2CH_2CH(CH_3)_2$, are achiral because they do not contain a stereocenter. Summing the enantiomeric pairs from the two chiral products gives a total of 4 chiral compounds.

DDT is a potent synthetic chemical that resists natural decomposition processes, meaning environmental microorganisms are unable to break it down effectively. Because it persists in ecosystems for extended periods and accumulates within the tissues of living organisms, it is classified as a non-biodegradable pollutant.

The iodoform test specifically targets compounds containing either a methyl ketone group or a methyl carbinol structural unit, characterized by a methyl group bonded directly to the carbon atom attached to an oxygen. While ethyl methyl ketone, isopropyl alcohol, and 3-methyl-2-butanone each possess these necessary configurations, isobutyl alcohol, written as $\mathrm{CH}_3 - \text{CH}(\mathrm{CH}_3) - \mathrm{CH}_2 - \mathrm{OH}$, is a primary alcohol that lacks the essential methyl group on the carbinol carbon. Since this specific structural arrangement is mandatory for the formation of iodoform, isobutyl alcohol remains the only molecule among the choices that cannot produce a positive result.

The rate of an \text{S_N 1} reaction is primarily dictated by the stability of the carbocation intermediate formed during the rate-determining step. Since the intrinsic stability of these carbocation species decreases in the order $3^{\circ} > 2^{\circ} > 1^{\circ}$, the reactivity of the associated alkyl bromides follows this exact hierarchy. Consequently, the tertiary alkyl bromide is the most reactive, followed by the secondary and then the primary, establishing that B is more reactive than C, which in turn is more reactive than A.

When a geminal dihalide like $\mathrm{CH}_3\,\mathrm{CHCl}_2$ undergoes nucleophilic substitution with aqueous $\mathrm{KOH}$, both chlorine atoms are replaced by hydroxyl groups to form an unstable gem-diol. This intermediate, $\mathrm{CH}_3\,\mathrm{CH}(\mathrm{OH})_2$, cannot persist and instead undergoes immediate dehydration to stabilize its structure:
$\mathrm{CH}_3\,\mathrm{CHCl}_2 \xrightarrow{\text{ aq. KOH}} \mathrm{CH}_3\,\mathrm{CH}(\mathrm{OH})_2 \xrightarrow{-\mathrm{H}_2\mathrm{O}} \mathrm{CH}_3\,\mathrm{CHO}$
This rapid loss of water converts the intermediate directly into acetaldehyde, $\mathrm{CH}_3\,\mathrm{CHO}$.

$\mathrm{S}_{\mathrm{N}} 2$ reactions are governed by steric hindrance, as the incoming nucleophile must approach the electrophilic carbon from the backside, a path that becomes increasingly obstructed as the size of the attached alkyl groups grows. Consequently, the reaction rate is highest for methyl halides and primary halides, where the carbon atom is least crowded, facilitating the clean stereochemical inversion characteristic of this bimolecular process. Since $\mathrm{CH}_3\mathrm{Cl}$ is the simplest structure provided and the only one that lacks bulky substituents, it exhibits the minimal steric repulsion necessary to readily undergo the $\mathrm{S}_{\mathrm{N}} 2$ mechanism.

The $\mathrm{S_N}2$ mechanism involves a backside attack by the nucleophile, creating a crowded pentavalent transition state that makes the reaction highly sensitive to spatial congestion. Because bulkier alkyl groups physically obstruct this approach, primary halides ($\mathrm{RCH}_2\mathrm{X}$) react the fastest, while tertiary halides ($\mathrm{R}_3\mathrm{CX}$) are significantly hindered and therefore the least reactive. Consequently, the reactivity decreases as the number of alkyl groups attached to the electrophilic carbon increases, following the sequence $\mathrm{RCH}_2\mathrm{X} > \mathrm{R}_2\mathrm{CHX} > \mathrm{R}_3\mathrm{CX}$.