Hydrocarbons – JEE Main ChemistryPractice Questions & PYQs

The reaction sequence proceeds via the Rosenmund reduction, where a carboxylic acid derivative is first activated to an acyl chloride using thionyl chloride ($SOCl_2$), followed by selective catalytic hydrogenation using a poisoned palladium catalyst ($Pd-BaSO_4$). This specific catalyst prevents the complete reduction of the acyl chloride to an alcohol, allowing for the isolation of the aldehyde as the final product A. The transformation ensures high yield and selectivity for the formation of the aldehyde functional group.

The synthesis of succinic acid involves the nucleophilic substitution of $1,2-dibromoethane$ with potassium cyanide ($KCN$) to produce the intermediate succinonitrile ($NC-CH_2-CH_2-CN$). Subsequent acid-catalyzed hydrolysis of the nitrile groups ($-CN$) converts them into carboxylic acid groups ($-COOH$), yielding butanedioic acid, which is commonly known as succinic acid ($HOOC-CH_2-CH_2-COOH$).

The regioselectivity of alkene hydration depends on the specific reaction pathway, as acid-catalyzed hydration follows Markovnikov's rule while hydroboration-oxidation yields the anti-Markovnikov product. During the acid-catalyzed hydration of propene, the electrophilic addition of a proton creates a stable secondary carbocation intermediate, which is subsequently attacked by water to form propan-2-ol, identified as product A, $\mathrm{CH}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3$. In contrast, the hydroboration-oxidation sequence proceeds by adding boron to the less substituted terminal carbon, followed by an oxidative workup that replaces the boron atom with a hydroxyl group to produce the primary alcohol propan-1-ol, designated as product B, $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2-\mathrm{OH}$.

Reduction of an internal alkyne with sodium in liquid ammonia produces a trans-alkene via a radical anion intermediate. Subsequent treatment of this trans-alkene with cold dilute alkaline $\mathrm{KMnO}_4$ results in the syn-addition of two hydroxyl groups to the double bond.

The transformation proceeds as follows:
$\mathrm{CH}_3\mathrm{C}\equiv\mathrm{C}-\mathrm{CH}_3 \xrightarrow{\text{Na/liq. NH}_3} \text{trans-CH}_3\mathrm{CH}=\text{CHCH}_3$
$\text{trans-CH}_3\mathrm{CH}=\text{CHCH}_3 \xrightarrow{\text{dil. KMnO}_4, 273\,\mathrm{K}} \mathrm{CH}_3\mathrm{CH(OH)CH(OH)CH}_3$

The final product $P$ is 2,3-butanediol, which has the chemical formula $\mathrm{C}_4\mathrm{H}_{10}\mathrm{O}_2$. Therefore, the molecule contains 2 oxygen atoms.

The total count of $\pi$ electrons is determined by summing the electrons involved in multiple bonds, where each double bond contributes 2 $\pi$ electrons and each triple bond contributes 4 $\pi$ electrons, or by applying the Hückel rule $(4n+2)$ for aromatic systems. Product A is an aromatic structure containing 6 $\pi$ electrons, while Product B contains a single $\pi$ bond accounting for 2 $\pi$ electrons. Adding these values together yields a sum of $6 + 2 = 8$ $\pi$ electrons.

The stereochemical outcome of nucleophilic substitution reactions is dictated by their underlying mechanisms. In an $S_N2$ reaction, the nucleophile undergoes a backside attack on the chiral center, forcing an inversion of configuration that produces only one specific stereoisomer, which is the hallmark of a stereospecific reaction.

Conversely, an $S_N1$ reaction proceeds through a planar carbocation intermediate, which allows the nucleophile to attack from either the front or the back face with roughly equal probability. This non-selective approach yields both enantiomers in nearly equal proportions, resulting in the formation of a racemic mixture.

A secondary ($2^{\circ}$) carbon atom is defined as a carbon atom that is directly bonded to exactly two other carbon atoms. In the structure $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_3$, the terminal methyl groups are primary carbons as they attach to only one carbon, while the two branching methine groups are tertiary carbons because they attach to three. The central methylene group is the only carbon in the chain attached to exactly two other carbon atoms, confirming that there is exactly one $2^{\circ}$ carbon atom present.

The nitration of benzene relies on the formation of the nitronium ion, $\mathrm{NO}_2^+$, which serves as the active electrophile produced by the reaction between concentrated $\mathrm{H}_2\mathrm{SO}_4$ and $\mathrm{HNO}_3$, making the first statement accurate. In contrast, electrophilic substitution requires the generation of electron-deficient species, a role typically fulfilled by Lewis acids that accept electron pairs to facilitate the reaction. Since Lewis bases act as electron-pair donors, they are chemically unsuitable for promoting the creation of the electrophiles required for this process, rendering the second statement incorrect.

Bond strength in organic molecules is fundamentally dictated by bond order and the hybridization state of the involved carbon atoms. Because ethyne ($HC \equiv CH$) contains a carbon-carbon triple bond with $sp$ hybridization, it features a higher bond order and greater s-character compared to the carbon-carbon double bond found in ethene ($H_2C = CH_2$), which utilizes $sp^2$ hybridization. Consequently, the bond in ethyne is significantly shorter and stronger, directly contradicting the claim that the carbon-carbon linkage in ethyne is weaker than that in ethene.

Counting $\pi$-bonds in organic molecules relies on identifying every multiple bond, as each double bond contributes exactly one $\pi$-bond and each triple bond contributes two. Visualizing the structure of product $B$ after the reaction reveals an arrangement of functional groups that necessitates specific saturation levels throughout the carbon skeleton. By identifying all double and triple bonds within the resulting compound, the summation of these values confirms that there are exactly 5 $\pi$-bonds present.

The oxidation of toluene to benzaldehyde is achieved using either $\mathrm{CrO}_3$ or $\mathrm{CrO}_2\mathrm{Cl}_2$ through distinct reaction mechanisms. Toluene reacts with $\mathrm{CrO}_3$ in the presence of acetic anhydride to form benzal diacetate, which is subsequently hydrolyzed to yield benzaldehyde. Alternatively, the Etard reaction utilizes chromyl chloride, $\mathrm{CrO}_2\mathrm{Cl}_2$, to oxidize toluene directly into a chromium complex that yields benzaldehyde upon hydrolysis.

The reaction between Xenon hexafluoride and silica proceeds according to the balanced equation $2XeF_6 + SiO_2 \rightarrow 2XeOF_4 + SiF_4$. This chemical transformation yields Xenon oxytetrafluoride ($XeOF_4$) as the primary oxygen-containing product. Based on the molecular formula $XeOF_4$, there is exactly one oxygen atom present per molecule.

Product A is formed via an electrophilic addition mechanism that follows the principle of carbocation stability, where the reaction pathway proceeds through the most stable intermediate to minimize the activation energy. The electrophile attacks the alkene to form a more stable carbocation intermediate-typically secondary or tertiary-rather than a less stable primary one, as the increased hyperconjugation and inductive effect of the adjacent alkyl groups stabilize the positive charge. This regioselective control ensures that the nucleophile binds to the more substituted carbon, yielding the final structure shown in option B.

Dissolving metal reduction of internal alkynes using sodium in liquid ammonia ($Na/liq. NH_3$) proceeds via a vinyl radical anion intermediate, where the bulky alkyl substituents adopt an anti-configuration to minimize steric repulsion, yielding the trans-isomer. In the given chemical transformations, this stereoselective anti-addition accounts for the formation of trans-2-butene from the corresponding internal alkyne, distinguishing it from the cis-selective syn-addition observed during catalytic hydrogenation with poisoned catalysts. Consequently, these established stereochemical preferences in alkyne reduction confirm the identity of 2-pentyne as the specific starting material and trans-2-butene as the corresponding product.

The reaction proceeds via the electrophilic addition of hydrogen bromide to an alkene, where the formation of the major product is governed by the stability of the intermediate carbocation. Upon the protonation of the double bond, a secondary carbocation is initially formed, which quickly undergoes a 1,2-hydride shift to generate a significantly more stable tertiary carbocation. The final step involves the nucleophilic attack of the bromide ion ($Br^-$) onto this tertiary carbocation, resulting in the formation of the tertiary alkyl bromide as the major product.

The first step involving $O_3$ followed by $Zn/H_2O$ represents the ozonolysis of an alkene, which cleaves the carbon-carbon double bond to yield specific carbonyl compounds such as aldehydes or ketones. The second reagent sequence $NaOH/I_2$ is the standard condition for the iodoform reaction, which selectively oxidizes compounds containing a terminal methyl ketone moiety ($CH_3CO-$) or a group that forms one upon oxidation, producing a characteristic yellow precipitate of iodoform ($CHI_3$).

Free radical bromination of ethane proceeds by replacing hydrogen atoms with bromine atoms to generate all possible structural isomers for each substitution stage. The number of structural isomers for the halogenated products of ethane is 1 isomer for $\mathrm{C_2H_5Br}$, 2 isomers for $\mathrm{C_2H_4Br_2}$ (1,1- and 1,2-dibromoethane), 2 isomers for $\mathrm{C_2H_3Br_3}$ (1,1,1- and 1,1,2-tribromoethane), 2 isomers for $\mathrm{C_2H_2Br_4}$ (1,1,2,2- and 1,1,1,2-tetrabromoethane), 1 isomer for $\mathrm{C_2HBr_5}$ (1,1,1,2,2-pentabromoethane), and 1 isomer for $\mathrm{C_2Br_6}$ (hexabromoethane). Summing these individual counts yields $1 + 2 + 2 + 2 + 1 + 1 = 9$ distinct bromo derivatives.

When 3-chloro-1-butene reacts with $\mathrm{HCl}$, the process occurs through a carbocation intermediate. This mechanism facilitates the formation of 2,3-dichlorobutane, which features two chiral centers and thus exists as three distinct stereoisomers: a meso form and a pair of enantiomers. Accounting for an additional isomeric product generated through a different pathway involving the carbocation, the total number of possible isomeric products is one plus three, resulting in four.

The hydrocarbon $\mathrm{C}_6\mathrm{H}_8$ has a degree of unsaturation of three, which, given the consumption of two moles of $\mathrm{H}_2$ during catalytic hydrogenation, indicates the structure contains one ring and two double bonds. Ozonolysis of cyclohexa-1,4-diene cleaves the double bonds at the 1 and 4 positions of the cyclohexane ring, resulting in the formation of two identical moles of propane-1,3-dial, which is described as methane dicarbaldehyde ($\mathrm{OHC-CH}_2\mathrm{-CHO}$). This symmetry and the resulting cleavage product specifically identify the hydrocarbon as cyclohexa-1,4-diene.

The number of isomeric alkenes formed during dehydrohalogenation is determined by the availability of non-equivalent $\beta$-hydrogen atoms and the potential for creating cis-trans stereoisomers. For 1-bromo-2-methylbutane, the reaction yields only one alkene, just as 2-bromopropane produces a single alkene due to its equivalent $\beta$-hydrogens. Similarly, 2-bromo-3,3-dimethylpentane is constrained by its structural configuration to yield only one isomeric product. In contrast, 2-bromopentane ($\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}_3$) allows for elimination at different sites, resulting in the formation of pent-1-ene and both the cis and trans forms of pent-2-ene, which totals three distinct isomeric alkenes.