Haloalkanes And Haloarenes – JEE Main ChemistryPractice Questions & PYQs

Electrophilic addition of hydrogen halides to alkenes relies on the formation of carbocation intermediates, where the pathway progresses toward the most stable configuration. When $\mathrm{HI}$ reacts, the $\mathrm{H}^{+}$ ion adds to the alkene to generate a temporary carbocation, which then undergoes a 1,2-methyl shift to rearrange into a significantly more stable $3^{\circ}$ carbocation. The subsequent addition of the $I^{\ominus}$ ion to this tertiary carbon dictates the formation of 2-iodo-2,3-dimethylbutane as the major product.

Ammonolysis is a nucleophilic substitution process involving the reaction of an alkyl or benzyl halide with an ethanolic solution of ammonia. In this reaction, the lone pair on the ammonia molecule acts as a nucleophile to cleave the carbon-halogen bond, resulting in the substitution of the halogen atom by an amino ($-\mathrm{NH}_2$) group. When benzyl chloride ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{Cl}$) reacts with ammonia, the chlorine atom is displaced by the amino group to form benzylamine ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{NH}_2$), which satisfies the exact criteria for this type of transformation. Other reactions displayed, such as the formation of an amide from an acid chloride, the reduction of a nitrile to an amine, or the simple protonation of an amine, represent distinct chemical pathways that do not involve this specific mechanism of carbon-halogen bond cleavage by ammonia.

Nucleophilic substitution is inherently difficult in chlorobenzene because the carbon-chlorine bond possesses significant double-bond character due to resonance, making the aryl halide resistant to typical substitution pathways. Overcoming this substantial activation energy barrier requires extreme physical conditions to facilitate the conversion, as shown in the following reaction:

$\text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{NaCl}$

Consequently, the process must be conducted at 623 K and 300 atm to drive the formation of sodium phenoxide.

Free radical substitution reactions are governed by the stability of the intermediate radical species, where the benzylic position is particularly favorable due to resonance stabilization with the aromatic ring. When 3-ethylbenzonitrile is exposed to bromine under UV light, homolytic fission of bromine molecules generates reactive bromine radicals. These radicals abstract a hydrogen atom from the benzylic carbon of the ethyl side chain to form a stable $2^\circ$ benzylic radical intermediate. This intermediate subsequently reacts with another bromine radical to yield 1-bromo-1-(3-cyanophenyl)ethane, ensuring that the substitution occurs specifically at the site that allows for the most stable radical configuration.

Ethanol ($C_2H_5OH$) contains an oxygen atom with available lone pairs capable of acting as a nucleophilic site, and $AgCN$ similarly functions as a nucleophile via its nitrogen atom, verifying that both species can indeed generate nucleophiles. While $KCN$ is an ionic compound that facilitates substitution through the carbon center to produce alkyl nitriles, $AgCN$ is covalent in nature and promotes attack through the nitrogen atom to yield isonitriles. Consequently, the assertion that both reagents generate nitrile nucleophiles under all conditions is incorrect because their fundamentally different bonding states dictate distinct regioselectivity during substitution.

The molar mass of an unknown gaseous substance is determined by the volume it occupies at standard temperature and pressure, where one mole of any gas occupies 22400 mL. Since 448 mL of this chloro compound weighs 1.53 g, calculating the mass of a full 22400 mL sample gives a molar mass of 76.5 g. After accounting for the atomic mass of chlorine at 35.5, the remaining 41 units of mass must belong to the carbon and hydrogen portion of the molecule, which aligns with the equation $(3 \times 12) + 5 = 41$. This composition is consistent with the structure $CH_3-CH=CH-Cl$, which produces aldehydes through ozonolysis and subsequent hydrolysis. Therefore, the molecule consists of 3 carbon atoms.

The conversion occurs through the initial Markownikoff addition of ${HCl}$ across the alkene, directing the hydrogen atom to the carbon with more hydrogen atoms and placing the chlorine at the more substituted position. Subsequently, the Finkelstein reaction performs a nucleophilic substitution on the alkyl chloride, replacing the chlorine atom with an iodine atom to yield the final product.

The reaction between the Grignard reagent $\mathrm{C}_2\mathrm{H}_5\mathrm{MgBr}$ and the carbonyl group of acetophenone, represented as $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$, proceeds via nucleophilic addition to form a tertiary alkoxide, which yields the tertiary alcohol 2-phenylbutan-2-ol upon hydrolysis. Because the Lucas reagent test using $\mathrm{ZnCl}_2$ and $\mathrm{HCl}$ differentiates alcohols based on their reactivity toward substitution, the instant turbidity observed indicates the formation of a highly reactive carbocation intermediate typical of tertiary structures. This diagnostic result confirms that the hydroxyl group is replaced by a chlorine atom through an $S_N1$ mechanism, resulting in the final product $\mathrm{C}_{10}\mathrm{H}_{13}\mathrm{Cl}$, which is 2-chloro-2-phenylbutane.

Grignard reagents, denoted as $RMgX$, act as highly reactive bases because the carbon-magnesium bond possesses significant carbanionic character, making the alkyl group a potent proton acceptor. When these reagents encounter an alcohol, they immediately undergo an acid-base neutralization reaction where the alkyl group abstracts the acidic hydroxyl proton to yield an alkane and water as the primary products.

An ambident nucleophile features two distinct sites capable of donating electron pairs, which allows it to bond to electrophiles through either of the two available atoms. The cyanide ion behaves as an ambident species because its resonance structure, represented as $:\mathrm{C}\equiv\mathrm{N}:\leftrightarrow:\mathrm{C}=\ddot{\mathrm{N}}:-$, enables bond formation through either the carbon or nitrogen atom. Similarly, the nitrite ion $\mathrm{NO}_2^{-}$ provides alternative reactive centers at both the nitrogen and oxygen atoms, establishing that both the cyanide and nitrite pairs satisfy the criteria for ambident behavior.

Alcohol dehydration and elimination pathways are determined by the specific reagents, which dictate whether the reaction follows the Saytzeff or Hofmann elimination rule. When alcohols are heated with protic acids like concentrated $H_2SO_4$ or $H_3PO_4$, the reaction yields the more highly substituted, thermodynamically stable Saytzeff alkene as the primary product. Conversely, employing bulky bases like tert-butoxide creates significant steric hindrance that prevents access to more substituted sites, forcing the base to abstract a less hindered proton and resulting in the formation of the less substituted Hofmann alkene.

The reaction between isobutane and $\text{Br}_2$ at 125 °C is a classic free radical substitution process initiated by light. The reaction begins with an initiation step where the $\text{Br}_2$ molecule absorbs energy and undergoes homolytic fission to generate reactive bromine radicals. Because isobutane is present in excess, the reaction pathway is restricted to favor monosubstitution, which prevents the undesirable formation of dibrominated byproducts. Under these specific conditions, the process leads to the formation of 1-bromo-2-methylpropane as the major product.

Electrophilic aromatic substitution is highly sensitive to solvent polarity and the presence of catalysts, which dictates the extent of bromination. Using bromine water provides a polar environment that leads to the formation of tribromo products due to the highly activated nature of the substrate. In contrast, non-polar conditions provided by $\mathrm{CS}_2$ or $\mathrm{CHCl}_3$ at 273 K, or the use of a Lewis acid like $\mathrm{FeBr}_3$, moderate the reaction to produce monobromo derivatives with the para isomer as the major product.

When a haloalkane interacts with a strong base like the methoxide ion, represented as $\text{CH}_3\text{O}^-$, the reaction pathway is governed by the competition between elimination and nucleophilic substitution. In this specific chemical environment, the high electron density and compact nature of the methoxide ion allow it to act primarily as a Brønsted base, abstracting a proton to initiate an elimination reaction. This process generates an alkene featuring a double bond that gains stability through conjugation with the adjacent aromatic ring, making it the dominant major product. While the methoxide ion can also function as a nucleophile to displace the bromide group, $\text{Br}^-$, this substitution pathway leading to an ether is significantly less favorable and results only in a minor product.

The chemical transformations involving benzene diazonium salts demonstrate a clear distinction based on the reagents used for halogen substitution. Treating benzene diazonium chloride with $\mathrm{Cu}_2\mathrm{Cl}_2$ results in the Sandmeyer reaction, whereas utilizing $\mathrm{Cu}/\mathrm{HCl}$ for the same substitution identifies the Gattermann reaction. Transitioning to coupling processes in the presence of metallic sodium and dry ether, the structure of the organic halide determines the reaction nomenclature; aliphatic halides like ethyl chloride undergo the Wurtz reaction to form longer alkanes, specifically $n$-butane. Conversely, the coupling of aryl halides such as chlorobenzene with sodium in ether leads to the formation of biphenyl, a transformation defined as the Fittig reaction.

When 1-methoxynaphthalene reacts with hydroiodic acid, the reaction proceeds through an $S_N2$ mechanism. The process begins with the protonation of the ethereal oxygen atom, which enhances the leaving group ability of the oxygen-containing species. Because the oxygen is bonded directly to the aromatic naphthalene ring, resonance stabilization imparts partial double bond character to that carbon-oxygen bond, making it significantly stronger and resistant to cleavage. Consequently, the nucleophilic iodide ion selectively attacks the less sterically hindered methyl carbon, leading to the formation of 1-naphthol and methyl iodide as the main products.

The reaction between an alkyl dibromide and magnesium in the presence of $\mathrm{Et}_2\mathrm{O}$ proceeds via the oxidative insertion of the metal into the carbon-bromine bonds to generate organomagnesium species. In this specific transformation involving the given substrate, magnesium inserts into both carbon-bromine positions to yield the di-Grignard intermediate, $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{MgBr})-\mathrm{CH}_2-\mathrm{CH}(\mathrm{MgBr})-\mathrm{CH}_3$. This initial step establishes the reactive organometallic functionality across the hydrocarbon chain before any subsequent elimination processes can take place.

The hydration of propene under acidic conditions follows Markovnikov's rule to generate 2-propanol as the intermediate $A$, which possesses the specific structural requirement to undergo the haloform reaction. Upon reacting this secondary alcohol with potassium hypoiodite, generated in situ by $\mathrm{KIO}$ and dilute $\mathrm{KOH}$, the molecule undergoes oxidation and cleavage to yield a characteristic yellow precipitate of iodoform ($\mathrm{CHI}_3$) and the potassium salt of the corresponding carboxylic acid, potassium acetate ($\mathrm{CH}_3\mathrm{COOK}$). Consequently, the products $B$ and $C$ are identified as $\mathrm{CHI}_3$ and $\mathrm{CH}_3\mathrm{COOK}$.

Electrophilic halogenation of an aromatic ring is facilitated by a Lewis acid catalyst, which polarizes the halogen molecule to generate a strong electrophile. This reactive species initiates an attack on the $\pi$ electron cloud of the aromatic system, forming a carbocation intermediate known as the sigma complex. The reaction concludes as the aromatic ring regains its resonance stability through the loss of a proton, leading to the formation of the halogenated aromatic product.

The activation of this transformation relies on the homolytic fission of the chlorine-chlorine bond through the absorption of ultraviolet energy, bypassing the need for metallic catalysts like $\mathrm{AlCl}_3$. Once exposed to light, $\mathrm{Cl}_2$ dissociates into highly reactive $\mathrm{Cl}^\bullet$ radicals, which initiate a free radical substitution process by attacking the allylic position of the hydrocarbon. This mechanism is fundamentally distinct from ionic pathways, making the presence of ultraviolet radiation the critical condition for generating the necessary species to propagate the reaction.