Hydrocarbons – JEE Main ChemistryPractice Questions & PYQs

The reaction of propene with $\mathrm{HOCl}$ proceeds via electrophilic addition, where the species acts as a source of the electrophilic chlorine ion, $\mathrm{Cl}^+$. This electrophile attacks the double bond of propene, bonding to the terminal carbon to ensure the development of the most stable carbocation at the central position. This process results in the formation of the secondary carbocation intermediate $\mathrm{CH_3-CH^{+}-CH_2-Cl}$, which is energetically favored due to the stabilizing effect of the adjacent methyl group.

The reaction sequence is initiated by the diazotization of the aliphatic primary amine, $CH_3CH_2NH_2$, with nitrous acid to form an unstable diazonium intermediate that immediately decomposes into ethanol ($A$), nitrogen gas, and water. Ethanol subsequently reacts with $PCl_5$ through a nucleophilic substitution mechanism to replace the hydroxyl group with a chlorine atom, yielding chloroethane as product ($B$). In the final step, the ammonolysis of chloroethane with ammonia results in the formation of ethylamine as product ($C$) through the substitution of the chlorine atom by an amino group.

Combustion reactions follow the law where, at constant temperature and pressure, the volumes of gaseous reactants and products are directly proportional to their stoichiometric coefficients in the balanced equation $C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O(l)$. Since the air provided is 20% oxygen by volume, 375 mL of air contains 75 mL of reactive $O_2$, leaving 300 mL of nitrogen that remains chemically unchanged after the reaction. Given that the final gas mixture occupies 330 mL and includes this unreacted nitrogen, the volume of $CO_2$ produced is the difference of 30 mL. Comparing the volume of the hydrocarbon to the volume of $CO_2$ produced, we arrive at the equation $15x = 30$, which gives $x = 2$. To determine the hydrogen content, we use the oxygen consumption stoichiometry $15(x + \frac{y}{4}) = 75$, which simplifies to $x + \frac{y}{4} = 5$. Substituting $x = 2$ into this expression leads to $2 + \frac{y}{4} = 5$, resulting in $y = 12$. Because a hydrocarbon with the molecular formula $C_2H_{12}$ cannot exist, the experimental data does not match any of the provided structural options.

Producing only a single mono-substituted alkyl halide is possible only when an alkane contains exclusively equivalent hydrogen atoms, as any substitution will inevitably lead to the same product. An alkane with a molecular mass of 72 u corresponds to $C_5H_{12}$, which includes structural isomers such as pentane, isopentane, and neopentane. Among these, neopentane stands out due to its highly symmetrical tetrahedral geometry, where all twelve hydrogen atoms are equivalent primary hydrogens, ensuring that halogenation results in a single, unique alkyl halide isomer.

The reduction of an internal alkyne such as 2-Hexyne to a trans-alkene is achieved through a process known as dissolving metal reduction, which utilizes alkali metals like lithium dissolved in liquid ammonia. This reaction proceeds via the anti-addition of hydrogen atoms across the carbon-carbon triple bond, as the metal acts as an electron source to generate an intermediate that places the alkyl groups on opposite sides to minimize steric repulsion. Because other common hydrogenation reagents like $\mathrm{Pt}/\mathrm{H}_2$ or $\mathrm{Pd}/\mathrm{BaSO}_4$ facilitate syn-addition rather than this specific stereochemical outcome, $\mathrm{Li}/\mathrm{NH}_3$ is the necessary reagent to produce trans-2-Hexene.

Ozonolysis functions by cleaving carbon-carbon double bonds and adding oxygen to each carbon atom involved, meaning the formation of formaldehyde ($HCHO$) indicates the original molecule possessed a terminal methylene ($=\text{CH}_2$) group. Since formaldehyde consists of a carbonyl carbon bonded to two hydrogen atoms, its presence implies that the parent alkene must have contained a carbon atom similarly bonded to two hydrogens through a double bond. This specific arrangement, where a double bond connects to a terminal $\text{CH}_2$ unit, is the defining characteristic of a vinyl group.

The dehydration of an alcohol like $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CH}(\mathrm{CH}_3)_2$ in the presence of concentrated $\mathrm{H}_2\mathrm{SO}_4$ proceeds according to the Saytzeff rule to dictate the major product. This principle establishes that the most thermodynamically stable alkene-specifically the one with the highest degree of substitution and the lowest number of hydrogen atoms on the carbons sharing the double bond-will be produced in the greatest amount. By abstracting a proton to create the most highly substituted double bond possible, the reaction demonstrates regioselectivity to prioritize the formation of the most stable structural isomer.

Reductive ozonolysis is a chemical process that cleaves a carbon-carbon double bond by adding ozone across the linkage, followed by a workup with zinc and water to produce carbonyl-containing compounds. When this process is applied to but-2-ene, $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$, the double bond breaks symmetrically, and each of the two alkene carbons forms a new double bond to an oxygen atom. Because the structure splits into two identical $-\mathrm{CH}$ fragments, the resulting product $B$ is acetaldehyde, denoted as $\mathrm{CH}_3\,\mathrm{CHO}$.

Grignard reagents behave as strong bases when they encounter molecules with acidic protons, such as the terminal hydrogen atom found in $\mathrm{CH}_3\mathrm{C}\equiv\mathrm{C}-\mathrm{H}$. Because the hydrogen on the triply-bonded carbon is sufficiently acidic, it is abstracted by the methyl group of the methylmagnesium halide, resulting in an acid-base reaction. This process effectively protonates the methyl anion to produce methane as shown in the following equation:

$\mathrm{CH}_3\mathrm{MgX} + \mathrm{CH}_3-\mathrm{C}\equiv\mathrm{C}-\mathrm{H} \longrightarrow \mathrm{CH}_3-\mathrm{C}\equiv\mathrm{CMgX} + \mathrm{CH}_4$

The methyl group attached to the benzene ring in toluene serves as an ortho- and para-directing activator, which directs the nitro group to these specific positions during the initial nitration step. When this mixture of ortho- and para-nitrotoluene undergoes reduction with $\text{Sn}/\text{HCl}$, the nitro groups are converted into amino groups, resulting in a mixture of ortho- and para-toluidines. Finally, these amines are subjected to diazotization followed by treatment with $\text{CuBr}$, which replaces the diazonium group with a bromine atom, leading to the formation of o-bromotoluene and p-bromotoluene.

The ability of a hydrocarbon to react with sodium in liquid ammonia depends on the presence of an acidic hydrogen atom bonded to an $sp$-hybridized carbon, which characterizes a terminal alkyne. This specific acidity allows the sodium metal to displace the terminal hydrogen atom, producing a sodium acetylide salt and evolving hydrogen gas. Among the provided choices, only $\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}\equiv\mathrm{CH}$ possesses this structure, enabling the reaction $\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}\equiv\mathrm{CH} \xrightarrow[\text{liquid } \mathrm{NH}_3]{\mathrm{Na}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{C}\equiv\mathrm{C}^- \mathrm{Na}^+ + \frac{1}{2}\mathrm{H}_2(\mathrm{g})$.

When an unsymmetrical reagent like $\mathrm{HBr}$ adds to an unsymmetrical alkyne such as $\mathrm{CH}_3-\mathrm{C}\equiv\mathrm{CH}$, Markovnikov's rule dictates the regioselectivity of the reaction. This principle states that the negative portion of the incoming reagent bonds to the carbon atom already attached to the fewest hydrogen atoms. As a result, the addition of two equivalents of $\mathrm{HBr}$ to $\mathrm{CH}_3-\mathrm{C}\equiv\mathrm{CH}$ occurs in two distinct stages, directing both bromine atoms to the central carbon to produce 2, 2-dibromopropane.

$\mathrm{FeCl}_3$ acts as a Lewis acid catalyst that initiates electrophilic aromatic substitution by activating the chlorine molecule to generate an electrophile. Because the methyl group attached to the toluene ring is an electron-donating activator, it directs the incoming chlorine electrophile specifically to the ortho and para positions of the aromatic ring. This mechanism leads to the preferential formation of o-chlorotoluene and p-chlorotoluene.

According to Hofmann's rule, when an elimination reaction can produce multiple alkenes, the major product is the one featuring the least alkylated double bond. This outcome is driven by the base abstracting a $\beta$-hydrogen that is less sterically hindered, making it more accessible for removal during the reaction. Consequently, the reaction kinetically favors the formation of the less substituted alkene, which corresponds to the structure shown in option B.

The molecular structure of 2,3-dichlorobutane, written as $CH_3-CHCl-CHCl-CH_3$, features two central carbon atoms that are each bonded to four distinct substituents. Because these carbons qualify as asymmetric carbon atoms, the molecule is capable of existing in configurations that rotate plane-polarized light. This specific structural characteristic is what results in the exhibition of optical isomerism.

Free radical bromination is a highly selective process governed by the stability of the intermediate alkyl radical formed during the rate-determining step. Because the ease of hydrogen atom replacement follows the stability order $3^{\circ} > 2^{\circ} > 1^{\circ}$, the reaction favors the abstraction of a hydrogen from the tertiary carbon position. In the molecule 2-methylbutane, represented as $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$, the hydrogen atom on the second carbon is tertiary, making it the most reactive site for substitution. Consequently, the bromine atom replaces this tertiary hydrogen to produce 2-bromo-2-methylbutane as the major product.

Determining the number of monochlorinated products for an alkane isomer relies on identifying the number of chemically distinct sets of hydrogen atoms present in the molecule. In the case of 2,3-dimethylbutane, represented by the structure $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CH}_3$, the molecular symmetry ensures that specific groups of hydrogen atoms are equivalent. All six hydrogen atoms on the two terminal methyl groups constitute one unique set, while the two hydrogen atoms on the central carbons form a second unique set. Because the molecule contains only these two non-equivalent types of hydrogen atoms, substituting one of them with a chlorine atom yields only two possible products: 1-chloro-2,3-dimethylbutane and 2-chloro-2,3-dimethylbutane.

The addition of $\mathrm{HBr}$ to 1,3-butadiene is governed by reaction temperature, which dictates whether the outcome is determined by kinetic or thermodynamic control. At low temperatures such as -80 °C, the process is kinetically controlled, meaning the product that forms most rapidly, 3-bromobutene, predominates due to the immediate proximity of the bromide ion to the carbocation. At 40 °C, the reaction is thermodynamically controlled, allowing the system to reach equilibrium where the most stable isomer persists. Because 1-bromo-2-butene, expressed as $\mathrm{BrCH}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$, possesses a more highly substituted internal double bond compared to the terminal double bond of the 1,2-addition product, it is the more stable thermodynamic product.

The boiling point of a hydrocarbon is determined by its molecular structure and the resulting strength of intermolecular forces, with alkenes generally having higher boiling points than alkanes due to the slight polarity introduced by $\pi$ electrons. Within the alkane family, the physical arrangement is critical because straight-chain isomers possess a larger surface area, leading to stronger van der Waals forces compared to their branched-chain counterparts. Since greater branching effectively minimizes the surface area available for these interactions, isobutane exhibits the weakest intermolecular forces among the compounds listed, resulting in the lowest boiling point.

The formation of a single monochloroalkane upon free radical chlorination is determined by the molecular symmetry of the alkane, which requires all hydrogen atoms to be chemically equivalent. In the structure of neopentane, represented as $\text{C}(\text{CH}_3)_4$, every hydrogen atom is bonded to a primary carbon and occupies an identical chemical environment. Consequently, replacing any of these twelve equivalent hydrogen atoms with a chlorine atom yields the same product, making neopentane the only molecule listed that produces a single unique monochloroalkane.