Haloalkanes And Haloarenes – JEE Main ChemistryPractice Questions & PYQs

The rate of solvolysis, following an $S_{N 1}$ mechanism, is directly proportional to the stability of the carbocation intermediate generated during the rate-determining step. Since the energy barrier to forming this intermediate dictates the reaction speed, structures that effectively disperse positive charge through resonance, hyperconjugation, or inductive effects will undergo solvolysis more rapidly. Consequently, the reaction kinetics align perfectly with the increasing stability of the corresponding carbocations, allowing the ascending order of rates to be determined by evaluating the electronic properties of each reactant.

Aqueous potassium hydroxide facilitates nucleophilic substitution, where hydroxide ions replace the halogen atom in alkyl chlorides to produce alcohols, as represented by $R-X + OH^-_{(aq)} \rightarrow R-OH + X^-$. In contrast, alcoholic potassium hydroxide acts as a strong base and induces dehydrohalogenation by abstracting a $\beta$-hydrogen from the alkyl chloride, leading to the formation of an alkene according to the general reaction $R-CH_2-CH_2-Cl + OH^-_{(alc)} \rightarrow R-CH=CH_2 + H_2O + Cl^-$.

Stoichiometry is governed by the law of conservation of mass, which dictates that the molar amount of reactant transformed directly determines the molar amount of product produced based on the balanced chemical equation. By identifying the mole-to-mole relationship between chlorobenzene and the product, we can bridge the gap between initial mass and the final mass of the target substance.

The molar mass of chlorobenzene ($C_6H_5Cl$) is $6(12) + 5(1) + 35.5 = 112.5 \text{ g mol}^{-1}$. Converting the initial $11.25 \text{ mg}$ to grams gives $11.25 \times 10^{-3} \text{ g}$, allowing us to calculate the moles as:
$n = \frac{11.25 \times 10^{-3} \text{ g}}{112.5 \text{ g mol}^{-1}} = 10^{-4} \text{ mol}$
Since the reaction proceeds with a complete 1:1 molar conversion, $10^{-4} \text{ mol}$ of product B is formed. Multiplying this by the molar mass of product B ($93 \text{ g mol}^{-1}$) yields a mass of $10^{-4} \times 93 = 93 \times 10^{-4} \text{ g}$. Converting this value into milligrams by multiplying by $10^3$ results in $93 \times 10^{-1} \text{ mg}$.

The reactivity of $\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_2-\mathrm{Cl}$ is heavily influenced by the electronic effects of the oxygen atom. Upon the departure of the chloride ion, the resulting carbocation $\mathrm{CH}_3-\mathrm{O}-\overset{\oplus}{\mathrm{C}}\mathrm{H}_2$ gains significant stability through resonance stabilization from the adjacent lone pair, which allows the molecule to proceed via an $\mathrm{S}_{\mathrm{N}}1$ mechanism despite being a primary halide. Simultaneously, the compound is susceptible to steric and electronic hindrance, which prevents the $\mathrm{S}_{\mathrm{N}}2$ reaction from occurring easily, confirming that both statements are accurate.

The reactivity of these molecules with $\mathrm{H}-\mathrm{Br}(\mathrm{aq})$ is determined by the stability of the carbocation intermediate that forms during the reaction, as a more stable intermediate lowers the activation energy required for the process. Molecule Q reacts the fastest because the resulting carbocation is stabilized through resonance, which delocalizes the positive charge and makes the pathway to the product significantly more favorable compared to the other options.

Williamson ether synthesis is fundamentally an $S_N2$ process where an alkoxide ion acts as a nucleophile to displace a leaving group on an alkyl halide. The efficiency of this pathway depends heavily on the accessibility of the carbon atom attached to the halide, as secondary alkyl halides like sec-butyl bromide are sterically hindered enough to impede nucleophilic attack. Because of this hindrance, the alkoxide reacts as a strong base rather than a nucleophile, making an elimination reaction the dominant process over substitution. Consequently, the reaction involving sec-butyl bromide primarily produces an alkene byproduct instead of the desired ether.

The reaction proceeds via the electrophilic addition of a proton to the alkene to generate the most stable carbocation intermediate, typically following Markovnikov's rule where the electrophile adds to the less substituted carbon. This resulting carbocation, which is stabilized by inductive effects or resonance, subsequently undergoes nucleophilic attack by the bromide ion ($Br^-$) to form the major product.

The reaction sequence utilizes electrophilic aromatic substitution with $\mathrm{Br}_2 / \mathrm{Fe}$ to regioselectively introduce a bromine atom onto the aromatic ring. Treatment with $\mathrm{Cl}_2$ under heating ($\Delta$) facilitates a free radical halogenation that occurs specifically at the more stable benzylic carbon of the alkyl side chain. Finally, the addition of alcoholic $\mathrm{KOH}$ promotes an E2 dehydrohalogenation mechanism, eliminating $\mathrm{HCl}$ from the side chain to form the corresponding vinyl derivative.

The stability of a carbocation intermediate during a nucleophilic substitution is fundamentally driven by the extent to which the positive charge can be delocalized across the surrounding structure. Option B produces the triphenylmethyl carbocation, $(C_6H_5)_3C^+$, where the empty p-orbital overlaps with the $\pi$ electron systems of three distinct phenyl rings. This expansive resonance effect effectively spreads the positive charge across a large molecular surface, conferring exceptional stability that cannot be matched by the other structures.

The formation of 2-phenylhepta-2,4-diene as the major product is driven by thermodynamic control, which favors the most stable isomer through extended conjugation. By aligning the two double bonds with the phenyl ring, the molecule achieves a resonance-stabilized $\pi$-system that significantly lowers its total energy compared to less conjugated alternatives like 2-phenylhepta-2,5-diene. Consequently, the reaction equilibrium strongly favors the isomer where delocalization is maximized across the entire conjugated chain.

This reaction relies on the behavior of ambidentate nucleophiles, which possess two distinct donor atoms capable of bonding with an electrophile. In the presence of silver salts, the covalent nature of the metal-nucleophile bond directs the attack toward the nitrogen atom in both instances rather than the alternative sites. Consequently, the reaction of propyl bromide with $\mathrm{AgNO}_2$ yields the nitroalkane, $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2$, while the treatment with $\mathrm{AgCN}$ produces the isocyanide derivative, $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NC}$.

The Hofmann Bromamide degradation converts the primary amide reactant into a primary amine (A) with the loss of a carbon atom through the formation of an isocyanate intermediate. Subsequently, treatment of this aliphatic amine with nitrous acid ($HNO_2$) generates an unstable diazonium salt that rapidly undergoes nucleophilic substitution with water to yield a primary alcohol (B) and nitrogen gas.

When haloalkanes interact with silver cyanide ($AgCN$), the covalent nature of the reagent directs the nucleophilic attack through the nitrogen atom rather than the carbon. This mechanistic preference results in the formation of an isocyanide ($R-NC$) as the major product. This behavior is exclusive to aliphatic halides, because haloarenes are completely inert and do not undergo this reaction with $AgCN$.

Both reactions proceed via an $\text{S}_\text{N}2$ pathway, but the optimal solvent polarity depends on the charge state of the incoming nucleophile. For the reaction with the negatively charged $\text{OH}^-$ ion, a less polar medium is preferred because it prevents excessive solvation of the reactant, thereby keeping the nucleophile highly reactive. In contrast, the reaction with neutral $\text{R}_3\text{N}$ generates a transition state with significant charge separation, which requires a more polar solvent to stabilize the developing ions and facilitate the process.

The reaction sequence $RBr \xrightarrow{Mg, \text{dry ether}} RMgBr \xrightarrow{H_2O} RH$ converts an alkyl bromide into an alkane by effectively replacing the bromine atom with a hydrogen atom. To obtain 2-methylbutane from this process, the starting alkyl bromide must possess the same carbon skeleton, represented by the structure $CH_3-CH_2-CH(CH_3)-CH_3$. By substituting each unique hydrogen environment in this molecule with a bromine atom, we find four distinct structural isomers: 1-bromo-3-methylbutane, 1-bromo-2-methylbutane, 2-bromo-3-methylbutane, and 2-bromo-2-methylbutane. Thus, there are 4 structural isomers of $RBr$ that yield 2-methylbutane upon hydrolysis of the corresponding Grignard reagent.

The reaction sequence is governed by the principles of elimination followed by addition, which allows for the sequential conversion of a functionalized precursor into an unsaturated intermediate and subsequently into a saturated product. The initial treatment with an alcoholic base promotes dehydrohalogenation to generate an alkene as the intermediate product $[A]$, which is then subjected to catalytic hydrogenation to reduce the $\pi$-bond, resulting in the corresponding alkane as the final product $[B]$.

The formation of the major product is governed by the stability of the reactive intermediate, where the reaction pathway proceeding through the most stable transition state is kinetically favored, represented by the condition $\Delta G^\ddagger_{major} < \Delta G^\ddagger_{minor}$. In electrophilic additions involving unsymmetrical reactants, the reaction follows the principle that the nucleophile attacks the more substituted carbon atom to stabilize the developing charge through hyperconjugation and inductive effects. This energetic preference ensures the reaction overcomes the lowest possible activation energy barrier, consistently leading to the more stable isomer.

In an $S_N2$ mechanism, the reaction proceeds through a transition state where the incoming nucleophile and the departing leaving group are both partially bonded to the central carbon. This transition state features an unhybridized $p$-orbital that is partially occupied during the bond-forming and bond-breaking process. When considering $C_6H_5CH_2Br$, the adjacent phenyl ring allows this developing $p$-orbital to participate in conjugation with the extensive $\pi$-system of the benzene ring. This conjugation provides electronic stabilization to the transition state, significantly lowering the activation energy compared to $CH_3CH_2Br$, which lacks the ability to delocalize charge through resonance. Consequently, the benzyl bromide reacts more rapidly, and the stabilization effect described in the reason is the direct cause of this increased reactivity.

The Sandmeyer reaction enables the substitution of the diazonium group in arenediazonium salts with a halogen atom by utilizing cuprous halides, specifically $CuCl$ or $CuBr$, as catalysts. This chemical transformation is primarily suitable for the synthesis of chlorobenzene and bromobenzene, as iodobenzene is typically prepared through a simpler displacement using potassium iodide, and fluorobenzene necessitates the separate Balz-Schiemann reaction. Consequently, there are only 2 halobenzenes that are synthesized through this specific methodology.

In an $\mathrm{S}_{\mathrm{N}}1$ reaction, the mechanism involves the formation of a planar carbocation intermediate that allows the nucleophile to attack from either face with equal probability, thereby generating a racemic mixture. In contrast, an $\mathrm{S}_{\mathrm{N}}2$ reaction occurs via a concerted, single-step pathway where the nucleophile attacks from the side opposite the leaving group, forcing an inversion of the stereochemical configuration at the chiral center.