Haloalkanes And Haloarenes – JEE Main ChemistryPractice Questions & PYQs

The rate of an $S_N2$ reaction is primarily controlled by steric hindrance, as the incoming nucleophile must perform a backside attack on the electrophilic carbon. Because bulky alkyl groups surrounding this center physically block the path of the incoming nucleophile, the reaction rate decreases significantly as the level of substitution increases, following the order $CH_3 - X > 1^{\circ} - X > 2^{\circ} - X > 3^{\circ} - X$. This inherent requirement for minimal spatial congestion ensures that the compound with the least substituted carbon atom exhibits the fastest reaction kinetics.

The reaction of 2-chlorobutane with $\mathrm{Cl}_2$ occurs via free radical substitution, where a hydrogen atom is replaced by a chlorine atom at different positions on the butane chain. The four possible constitutional isomers formed are 1,2-dichlorobutane ($CH_2Cl-CHCl-CH_2-CH_3$), 2,2-dichlorobutane ($CH_3-CCl_2-CH_2-CH_3$), 2,3-dichlorobutane ($CH_3-CHCl-CHCl-CH_3$), and 1,3-dichlorobutane ($CH_3-CHCl-CH_2-CH_2Cl$). To determine the number of optically active isomers, we must identify chiral centers in each of these products.

In 1,2-dichlorobutane, the second carbon atom is chiral, leading to two optically active enantiomers. In 1,3-dichlorobutane, the second carbon atom is similarly chiral, resulting in two additional optically active enantiomers. In 2,3-dichlorobutane, both the second and third carbon atoms are chiral; the (2R,3R) and (2S,3S) configurations represent two optically active enantiomers, while the (2R,3S) configuration is a meso compound and is optically inactive. The 2,2-dichlorobutane molecule lacks any chiral centers and is therefore optically inactive. Summing these, the total number of optically active isomers produced is $2 + 2 + 2 = 6$.

An allyl halide is characterized by the presence of a halogen atom bonded to a carbon atom that is adjacent to a carbon-carbon double bond, which is known as the allylic carbon. The structure $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{Cl}$ is a correct example of an allyl halide as it fits this definition, but the accompanying reason is false because the allylic carbon in this molecule is $\mathrm{sp}^3$ hybridized rather than $\mathrm{sp}^2$ hybridized.

The reaction sequence proceeds via an initial Friedel-Crafts alkylation of the aromatic ring to produce an alkylarene intermediate, followed by the selective oxidation of the alkyl side chain to a carboxylic acid intermediate $B$, and concludes with the dehydration of the ammonium salt to form an amide product $C$. This transformation relies on the susceptibility of the alkyl group to oxidizing agents like $KMnO_4$ and the subsequent condensation of the resulting acid with ammonia, identifying the major products as the carboxylic acid and the corresponding amide, respectively.

The nucleophilic substitution of an alkyl halide with excess ammonia produces an ammonium salt intermediate, which is subsequently converted into a free amine upon treatment with a strong base. In this specific reaction, 1,4-dichlorobutane reacts with excess ammonia to form the ammonium salt $[\mathrm{H}_3\mathrm{N}^{+}-(\mathrm{CH}_2)_4-\mathrm{NH}_3^{+}] 2\mathrm{Cl}^{-}$, which is then deprotonated by $\mathrm{NaOH}$ to liberate 1,4-butanediamine ($\mathrm{H}_2\mathrm{N}-(\mathrm{CH}_2)_4-\mathrm{NH}_2$), sodium chloride, and water.

The reaction proceeds via the Hell-Volhard-Zelinsky (HVZ) halogenation, where the $\alpha$-carbon of propanoic acid undergoes bromination in the presence of red phosphorus and bromine to form $2$-bromopropanoic acid as intermediate $A$, specifically $CH_3CH(Br)COOH$. Subsequently, the bromine atom in intermediate $A$ undergoes nucleophilic substitution when treated with excess ammonia, resulting in the formation of $2$-aminopropanoic acid as product $B$, with the formula $CH_3CH(NH_2)COOH$.

The distinct reaction pathways observed here are fundamentally determined by the bonding characteristics of the cyanide reagents, as $KCN$ acts as an ionic compound that provides free cyanide ions where the carbon atom serves as the nucleophilic center to yield alkyl cyanides. In contrast, $AgCN$ possesses significant covalent character, which restricts the nucleophilicity of the carbon atom and renders the nitrogen atom the available site for attack, ultimately resulting in the formation of alkyl isocyanides. Because the statement erroneously claims that $AgCN$ is highly ionic, the provided reason is factually incorrect despite the assertion accurately describing the reaction products.

The regioselectivity of nucleophilic substitution reactions with cyanide reagents is determined by the bond character of the metal-cyanide linkage. Ionic salts like $\mathrm{KCN}$ and $\mathrm{NaCN}$ dissociate to provide free cyanide ions where the carbon atom acts as the nucleophile, yielding alkyl cyanide as the product. Since $\mathrm{AgCN}$ is predominantly covalent, the carbon lone pair is unavailable, causing the nitrogen atom to attack the alkyl halide and produce alkyl isocyanide instead.

The reaction begins with the dehydrohalogenation of $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br}$ using alcoholic $\mathrm{KOH}$ to yield propene ($\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$) via an elimination mechanism. Subsequent electrophilic addition of $\mathrm{HBr}$ to propene follows Markovnikov's rule, yielding 2-bromopropane ($\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CH}_3$), which then undergoes nucleophilic substitution with aqueous $\mathrm{KOH}$ to produce propan-2-ol ($\mathrm{CH}_3-\mathrm{CH(OH)}-\mathrm{CH}_3$).

The dehydration of ethanol in the presence of concentrated sulfuric acid at 443 K proceeds via an $E1$ elimination mechanism, which begins with the protonation of the hydroxyl group to form an alkyloxonium ion. This intermediate loses a water molecule to generate a carbocation, which subsequently undergoes deprotonation to yield ethylene, $CH_2=CH_2$, as the final product $P$.

For secondary alkyl halides characterized by a lack of significant steric bulk, a high concentration of a strong nucleophile provides the necessary conditions to drive the reaction via the concerted $\mathrm{S_N2}$ mechanism. Conversely, conducting the reaction in a large excess of ethanol acts as a solvolysis process, where the polar protic nature of the solvent facilitates the formation of a carbocation intermediate, thereby directing the reaction through the unimolecular $\mathrm{S_N1}$ pathway.

The $\mathrm{S}_{\mathrm{N}} 1$ mechanism depends crucially on the stability of the carbocation intermediate formed after the halide leaves the molecule. While $\mathrm{H}_2\mathrm{C}=\mathrm{CH}-\mathrm{CH}_2\mathrm{Cl}$ generates a resonance-stabilized allyl carbocation, the molecule $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}$ would need to form a vinylic carbocation, $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}^+$, which is exceptionally unstable and energetically unfavorable. Because this high-energy intermediate cannot readily form under standard conditions, $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}$ is unable to participate in an $\mathrm{S}_{\mathrm{N}} 1$ reaction.

This transformation relies on an initial elimination reaction followed by an electrophilic addition to reposition the functional group. Treating 1-bromopropane with concentrated alcoholic $\mathrm{NaOH}$ at 80 °C promotes dehydrohalogenation, where the loss of hydrogen and bromine from adjacent carbons generates propene ($\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$). Subsequent reaction of this alkene with $\mathrm{HBr}$ in acetic acid follows Markovnikov's rule, wherein the hydrogen atom adds to the terminal carbon and the bromine atom adds to the more substituted secondary carbon, yielding $\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CH}_3$.

The reaction mechanism involved in this transformation proceeds through a common intermediate that effectively bypasses the initial stereochemical constraints of the starting material. Because this pathway converges regardless of the reactant's geometric configuration, both the cis and trans isomers inevitably rearrange to form the identical product containing the $\mathrm{CH}_3$ group.

Position isomerism occurs when structural isomers share an identical carbon skeleton but differ in the attachment point of a functional group or substituent along that chain. Because the structures of $A$ and $C$ vary only by the site of this attachment while maintaining the same parent framework, they strictly qualify as position isomers. This specific arrangement of atoms defines their relationship as structural isomers distinguished by the location of their functional groups.

The determination of the total hydrogen count relies on the structural analysis of products $A$ and $B$, specifically identifying all hydrogen atoms based on their chemical formulas or structural representations. Once the identity of each product is established, you systematically count the hydrogen atoms present in each molecule by considering all valence positions. Adding the individual hydrogen tallies from $A$ and $B$ together yields a final sum of 10.

Protection of the amino group via acetylation with acetic anhydride is the critical first step in this sequence, as it attenuates the high activating power of the $-NH_2$ group and minimizes ortho-substitution due to the bulky nature of the acetyl substituent. This facilitates selective para-nitration using the electrophilic nitronium ion ($NO_2^+$), ensuring that meta-directing effects and excessive poly-substitution are avoided. Subsequent acid-catalyzed hydrolysis effectively removes the acetyl protecting group to regenerate the primary amine, resulting in the desired para-substituted product.

The ratio of oxygen atoms to bromine atoms in a chemical compound is determined by the stoichiometric subscripts of the elements in the molecular formula $Q$. For the bromine oxide $\text{Br}_2\text{O}_3$, there are 3 oxygen atoms and 2 bromine atoms, which gives a numerical ratio of $\frac{3}{2} = 1.5$. This ratio is expressed in the required format as $15 \times 10^{-1}$.

In phenol, the hydroxyl group is attached directly to an $sp^2$ hybridized carbon atom of the benzene ring, where the lone pairs of electrons on the oxygen atom participate in resonance, imparting significant partial double-bond character to the $C-O$ bond. This structural configuration makes the bond much stronger and more difficult to break than the $C-O$ bond in aliphatic alcohols, thereby preventing the nucleophilic substitution reactions required to replace the hydroxyl group with a halogen. Therefore, the assertion that aryl halides cannot be prepared using this method is chemically accurate. The provided reason is false because the lack of reaction is due to this inherent bond strength and electronic stability rather than an unspecified violent reaction with halogen acids, which does not occur under standard conditions.

Nucleophilic substitution of alkyl halides with aqueous sodium hydroxide ($\mathrm{NaOH}$) proceeds via the displacement of the halide atom by a hydroxyl group ($\mathrm{-OH}$), forming an alcohol. The structure of the resulting alcohol is determined by the carbon skeleton of the starting alkyl halide, where a secondary substrate like 2-bromopropane yields the secondary alcohol 2-Propanol (B), while a primary substrate like 1-bromopropane yields the primary alcohol 1-Propanol (C). Because the reaction outcome depends strictly on the structural configuration of the alkyl group, these products represent the predictable results of nucleophilic attack on the respective substrates.