Hydrocarbons – JEE Main ChemistryPractice Questions & PYQs

Every single covalent bond is a sigma bond, and each double or triple bond contains exactly one sigma bond. In the molecule $\mathrm{H}_{3}\mathrm{C}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}\equiv\mathrm{C}-\mathrm{H}$, there are 6 carbon-hydrogen sigma bonds, consisting of 3 on the methyl group, 1 on the second carbon, 1 on the third carbon, and 1 on the terminal carbon. The carbon-carbon backbone provides 4 additional sigma bonds, including the single bond connecting the methyl group to the alkene, the sigma component of the double bond, the single bond between the alkene and the alkyne, and the sigma component of the triple bond. Adding these contributions, the total is $6 + 4 = 10$ sigma bonds.

Markownikoff's rule guides the electrophilic addition of HBr to the alkene by prioritizing the formation of the most stable carbocation intermediate. The process begins with the addition of an electrophile to create a $2^{\circ}$ carbocation, which promptly undergoes a methyl shift to reorganize into a more stable $3^{\circ}$ carbocation. Following this rearrangement, the nucleophile attacks the $3^{\circ}$ carbocation to generate the major product, which corresponds to the structure provided in D, proceeding through an $S_N1$ mechanism.

The Lindlar catalyst is a specialized reagent used to achieve the selective partial hydrogenation of alkynes, preventing the reaction from continuing all the way to an alkane. It functions by utilizing palladium-typically represented as $\mathrm{H}_2/\mathrm{Pd}$-which is supported on $\mathrm{CaCO}_3$ and deliberately poisoned with substances like quinoline to moderate its high activity. This resulting material, recognized as partially deactivated palladised charcoal, ensures the reaction proceeds specifically through syn-addition to yield an alkene.

The formation of a bright silver mirror serves as a definitive indicator for the presence of an aldehyde functional group via the Tollen's test. Ozonolysis cleaves unsaturated bonds to produce carbonyl compounds, so the target hydrocarbon must specifically yield an aldehyde as a product. While 2,3-dimethylbut-2-ene generates acetone, the isopropylidene derivative yields acetone and cyclopropanone, and but-2-yne results in butan-2,3-dione, none of which contain an aldehyde group capable of responding to this reagent. In contrast, the terminal alkyne $HC\equiv C-CH_2-CH_3$ undergoes ozonolysis to form 2-oxobutyraldehyde, which uniquely contains an aldehyde moiety that reduces silver ions into metallic silver to create the observed silver mirror.

The electrophilic addition of $Br_2$ to trans-but-2-ene follows an anti-addition mechanism, wherein bromine atoms attach to opposite faces of the carbon-carbon double bond. Due to the initial trans geometry, this stereospecific process results in a molecule possessing an internal plane of symmetry, which classifies the product as a meso compound. Because this specific addition pathway consistently generates this symmetric and optically inactive structure, the reaction yields 2 identical mesomers.

The conversion of $n$-heptane into the aromatic hydrocarbon toluene proceeds through a catalytic process known as dehydrocyclization. This transformation requires specific metallic oxides, including ${Mo}_{2}{O}_{3}$, ${V}_{2}{O}_{5}$, and ${Cr}_{2}{O}_{3}$, to function as catalysts under strictly controlled temperature and pressure conditions. Because ${Mo}_{2}{O}_{3}$ specifically acts as an effective aromatizing agent, it facilitates the structural rearrangement of the aliphatic chain into the required aromatic ring structure.

The chemical transformation begins with the application of dilute $KMnO_4$, which performs oxidative cleavage on the alkene to produce a mixture containing both a ketone and an aldehyde, together identified as Product A. Because $CrO_3$ functions as a highly selective oxidizing agent, it specifically targets the aldehyde functional group in the subsequent step, converting it into a carboxylic acid while leaving the ketone group entirely unaffected. This sequential process ensures that the final chemical structure of Product B consists of the original ketone and the newly generated acid.

Oxidative cleavage of alkenes using strong oxidizing agents like acidic $\text{KMnO}_4/\text{H}^{+}$ and ozone involves the rupture of the carbon-carbon double bond to generate carbonyl compounds. 2-Methylpropene (\text{CH}_2=\text{C}(\text{CH}_3)_2$) undergoes this process to produce acetone (\text{CH}_3\text{COCH}_3$, formula $\text{C}_3\text{H}_6\text{O}$) and carbon dioxide, identifying the ketone as the specified compound B. Other isomers such as but-2-ene would yield acetaldehyde, which has a different molecular formula and therefore does not match the required product.

Benzene undergoes electrophilic aromatic substitution when treated with bromine in the presence of an iron catalyst to produce bromobenzene. This intermediate subsequently reacts with magnesium in dry ether to form a Grignard reagent, which is then quenched by the acidic hydrogen in methanol, $\mathrm{CH_3OH}$, to regenerate benzene. This sequence highlights how an aromatic halide is converted into an organometallic species and then protonated to recover the parent hydrocarbon.

The reaction known as hydroformylation involves the treatment of an alkene with $\mathrm{H}_2$ and $\mathrm{CO}$ using a rhodium catalyst to generate an aldehyde. The process relies on the electrophilic addition of the formyl group to the alkene, which follows the Markownikoff rule to dictate the position of the new functional group. Through this chemical transformation, the alkene $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}=\mathrm{CH}_2$ is converted into the major product $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO}$.

The iodination of alkanes is inherently reversible because the hydrogen iodide ($\mathrm{HI}$) formed as a byproduct acts as a powerful reducing agent. This $\mathrm{HI}$ can react with the alkyl iodide product to revert it back to the original alkane, creating an equilibrium that prevents the reaction from going to completion. To effectively drive this process forward and make it irreversible, one must eliminate the $\mathrm{HI}$ as it is generated. The addition of a strong oxidizing agent, such as concentrated $\mathrm{HIO}_3$, serves this purpose by oxidizing the $\mathrm{HI}$ back into iodine. By continuously consuming the reducing byproduct, this reagent prevents the reverse reaction, successfully shifting the equilibrium toward the formation of the alkyl iodide.

An optically inactive product results from hydrogenation when the addition of $H_2$ creates a molecule possessing an internal plane of symmetry or a center of inversion, typically resulting in a meso compound. In compound B, the symmetrical arrangement around the double bond ensures that the saturated product retains this internal symmetry, effectively cancelling out any optical rotation caused by the newly formed stereocenters. Unlike other isomers that generate chiral or racemic products, this specific structural outcome renders the molecule achiral and thus optically inactive.

The major product is determined by the stability of the intermediate carbocation, which follows the stability order $3^\circ > 2^\circ > 1^\circ$. The reaction mechanism proceeds through the rearrangement of a less stable carbocation into a more stable tertiary carbocation, ensuring that the nucleophile attacks the most substituted carbon to yield the thermodynamically favored product.

Heat of combustion serves as an inverse indicator of stability because more stable compounds reside at lower energy states and release less heat upon oxidation. Because isomer stability follows the sequence $a > b > c$, the structural configurations that are the least stable inherently possess higher potential energy and liberate more energy during combustion. This inversion of the stability trend results in a heat of combustion order of $c > b > a$, which is equivalent to the progression $a < b < c$.

According to Zaitsev's rule, the major product in an elimination reaction is the most substituted alkene, as it is thermodynamically favored due to increased hyperconjugation and inductive stabilization from the attached alkyl groups. The structure $\mathrm{H}_3\mathrm{C}-\overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{C}}}=\overset{\mathrm{CH}_2\mathrm{CH}_3}{\overset{\vert}{\mathrm{C}}}-\mathrm{CH}_3$ represents a tetrasubstituted alkene, which possesses the lowest energy and highest stability among the potential isomeric products.

Electrophilic addition of hydrogen halides to unsymmetrical alkenes is governed by the stability of the intermediate carbocation, which dictates the regioselectivity of the reaction according to Markovnikov's rule. The formation of a tertiary carbocation intermediate, such as $(CH_3)_3C^+$, is energetically favorable due to hyperconjugation and inductive stabilization, which are significantly more effective than in secondary or primary alternatives. Consequently, the nucleophilic bromide ion rapidly attacks this highly substituted center to yield the tertiary alkyl halide as the major product.

Electrophilic aromatic substitution governs the initial nitration step, where the nitronium ion ($NO_2^+$) attacks the electron-rich benzene ring to form nitrobenzene. Subsequent reduction of the nitro group ($Ar-NO_2$) using metal-acid catalysts like $Sn/HCl$ or $Fe/HCl$ proceeds through a series of protonation and electron transfer steps to yield the primary amine [B] as the final product.

The reaction sequence proceeds via an initial electrophilic aromatic substitution where the substrate reacts with a nitrating mixture of concentrated $HNO_3$ and $H_2SO_4$ to generate the $NO_2^+$ electrophile, leading to the formation of the nitro-intermediate $A$. The subsequent transformation involves the chemical reduction of the nitro group in $A$ using a catalyst such as $Sn/HCl$ or $Fe/HCl$ to produce the final amine product $B$, demonstrating a classic two-step synthesis of an aromatic amine from a hydrocarbon precursor.

Combustion reactions rely on the principle of conservation of mass, requiring balanced chemical equations to determine the precise stoichiometric ratios of reactants and products. For propane, the combustion reaction is defined as $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ which dictates that 1 mole of propane necessitates 5 moles of $O_2$ to proceed to completion.

Similarly, the combustion of butane follows the balanced chemical equation $C_4H_{10} + 6.5O_2 \rightarrow 4CO_2 + 5H_2O$ where 1 mole of butane consumes 6.5 moles of $O_2$. Since the problem specifies using 2 moles of butane, we calculate the requirement by multiplying the stoichiometric coefficient by 2, finding that 13 moles of $O_2$ are necessary for this portion of the reaction. Aggregating these requirements, the total amount of $O_2$ needed is 5 moles plus 13 moles, which equals 18 moles.

The simplest optically active alkene is 3-methylpent-1-ene, which contains a chiral center at the third carbon. Hydrogenation of this molecule using hydrogen gas and nickel catalyst yields 3-methylpentane. When this alkane undergoes free radical halogenation, we must determine the number of distinct monohalogenated products by accounting for all non-equivalent carbon sites and the resulting stereoisomerism.

Substitution at the terminal methyl carbons (C1 and C6) occurs at equivalent positions and creates a chiral center at the C3 position, resulting in two distinct enantiomers. Replacing a hydrogen atom on the secondary methylene carbons (C2 and C5) also occurs at equivalent positions, but the substitution introduces a new chiral center at the substitution site while simultaneously making the existing C3 carbon chiral, which leads to four possible stereoisomers. Substitution at the tertiary carbon (C3) produces one achiral product, as the resulting molecule possesses two identical ethyl groups. Finally, replacing a hydrogen on the primary carbon of the methyl branch (C4) also yields only one achiral product for the same reason. Summing these individual contributions, the total number of monohalogenated products is $2 + 4 + 1 + 1 = 8$.