Hydrocarbons – JEE Main ChemistryPractice Questions & PYQs

The molecular formula of a saturated hydrocarbon follows the general expression $C_nH_{2n+2}$, where the molar mass is determined by the equation $12n + 1(2n+2) = 72$. Solving this gives $14n = 70$, resulting in $n=5$, which indicates the compound is a pentane isomer with the molecular formula $C_5H_{12}$. Among the possible isomers, 2-methylbutane has three primary carbon atoms because it features two methyl groups at the ends of the butane chain and one methyl group as a branch on the second carbon. In contrast, n-pentane contains only two terminal primary carbons, and 2,2-dimethylpropane contains four methyl groups attached to a central carbon, each representing a primary carbon.

The bromination of a cycloalkene involves the electrophilic addition of one mole of bromine to the carbon-carbon double bond. The general formula for a cycloalkene is $C_n H_{2n-2}$, which reacts to form a product $Y$ with the formula $C_n H_{2n-2} Br_2$. Given that the atomic ratio of carbon to bromine in the product is 3 to 1, the equation $\frac{n}{2} = \frac{3}{1}$ gives $n = 6$, resulting in the molecular formula $C_6 H_{10} Br_2$. The molar mass of the product is determined by the equation $6 \times 12 + 10 \times 1 + 2 \times 80 = 242$. The percentage of bromine in the product is calculated by the equation $\frac{160}{242} \times 100 = 66.11$. Rounding to the nearest integer, the percentage of bromine in the product is 66%.

Ozonolysis involves the oxidative cleavage of carbon-carbon double bonds, transforming each unsaturated carbon atom into a carbonyl group ($C=O$). To identify the original alkene, this process is reversed by removing the carbonyl oxygen atoms and establishing a double bond between the carbon atoms that were previously part of the carbonyl groups. By strategically joining the carbonyl centers of the provided fragments $HCHO$, $OHC-CO-CHO$, and $CH_3-CO-CO-CO-CH_3$, the molecular framework is reassembled to form the specific cyclic alkene structure identified in the correct option.

The mass percentage of an element in a compound is calculated using the formula $\frac{\text{Total mass of the element \in one mole}}{\text{Molar mass of the compound}} \times 100$. For the identified compound $(Q)$ with the molecular formula $C_{11}H_{12}O$, the molar mass is determined as $(11 \times 12) + (12 \times 1) + 16 = 160 \text{ g/mol}$. Applying the formula, the percentage of oxygen is $\frac{16}{160} \times 100 = 10\%$.

The synthesis proceeds via the nucleophilic addition of the reagent to the carbonyl center, creating a tetrahedral intermediate that, upon acidic workup, yields the stable alcohol $[A]$. This alcohol subsequently reacts through a substitution mechanism, where the hydroxyl group is protonated to facilitate its departure as water, allowing the nucleophile to attack and establish the final product $[B]$ with the most thermodynamically stable configuration.

The described transformations represent standard chemical behaviors of linear alkanes like heptane. In the presence of anhydrous $AlCl_3$ and HCl, linear alkanes undergo isomerization to produce branched isomers such as 2,4-dimethyl pentane, whereas catalytic aromatization and cyclization of the straight-chain heptane backbone lead to the formation of toluene and methyl cyclohexane, respectively.

The stability of alkenes increases with the number of alkyl groups attached to the $sp^2$ hybridized carbons, as these substituents stabilize the double bond through both hyperconjugation and the inductive effect. This hierarchy dictates that stability follows the decreasing order of tetra-substituted > tri-substituted > disubstituted > monosubstituted, with trans-isomers generally being more stable than cis-isomers due to reduced steric repulsion.

Compounds with the molecular formula $\mathrm{C}_{5}\mathrm{H}_{10}$ that do not decolorize $\mathrm{KMnO}_{4}$ are saturated cycloalkanes, specifically cyclopentane, methylcyclobutane, ethylcyclopropane, 1,1-dimethylcyclopropane, and 1,2-dimethylcyclopropane. Determining the number of monochloro structural isomers involves identifying all distinct hydrogen environments available for substitution in each of these five parent structures. Cyclopentane provides 1 unique hydrogen environment, methylcyclobutane has 4, ethylcyclopropane has 4, 1,1-dimethylcyclopropane has 2, and 1,2-dimethylcyclopropane has 3. Summing these counts, $1 + 4 + 4 + 2 + 3$, yields a total of 14 monochloro structural isomers.

The photochemical monochlorination of n-butane generates two constitutional isomers, 1-chlorobutane and 2-chlorobutane, with the latter possessing a chiral center and serving as the optically active compound "P". When 2-chlorobutane, represented as $CH_3-CH_2-CHCl-CH_3$, undergoes further chlorination, a chlorine atom can replace a hydrogen at any of the four chemically distinct carbon positions. Substitution on the terminal methyl group attached to the chiral center yields 1,2-dichlorobutane, whereas replacing a hydrogen on the adjacent methylene group produces 2,3-dichlorobutane. Furthermore, replacing the lone hydrogen on the chiral carbon results in 2,2-dichlorobutane, and substituting a hydrogen on the remaining terminal methyl group leads to 1,3-dichlorobutane. These four constitutional isomers account for all possible dichloro products when stereoisomers are ignored, confirming that there are exactly 4 such compounds.

The complete combustion of an alkane follows the stoichiometric relationship:
$C_nH_{2n+2} + \left(\frac{3n+1}{2}\right)O_2 \rightarrow nCO_2 + (n+1)H_2O$
Given that 8 moles of oxygen are required, the relation $\frac{3n+1}{2} = 8$ leads to $n = 5$, identifying the alkane as $C_5H_{12}$. Among the pentane isomers, only neopentane, also known as 2,2-dimethylpropane, yields a single monochlorinated product because all of its hydrogen atoms are chemically equivalent. This structure consists of a central quaternary carbon atom bonded to four methyl groups, each of which is a primary carbon, resulting in a total of 4 primary carbon atoms.

Carbocation rearrangements occur primarily when a reaction proceeds via a reactive intermediate that allows for rapid 1,2-hydride or alkyl shifts to attain a more stable electronic configuration. When a reaction proceeds without rearrangement, it is because the mechanism avoids the formation of these transient, structurally mobile intermediates, often occurring through concerted pathways where bond breaking and formation happen simultaneously. Consequently, the connectivity of the carbon skeleton is preserved throughout the process, ensuring the final product reflects the initial reactant's structure without structural migration.

The complete combustion of any alkane follows a general stoichiometric relationship defined by the balanced equation:
$C_n H_{2n+2} + \left(\frac{3n+1}{2}\right) O_2 \rightarrow n\mathrm{CO}_2 + (n+1) \mathrm{H}_2\mathrm{O}$
Given that the reaction requires 8 moles of oxygen for one mole of the alkane, setting the coefficient of oxygen to 8 produces the expression $\frac{3n+1}{2} = 8$. Solving this yields $3n + 1 = 16$, which confirms $n = 5$ and identifies the hydrocarbon as $C_5 H_{12}$. Adding the 5 carbon atoms and 12 hydrogen atoms provides the final sum of 17.

Nitrobenzene does not undergo Friedel-Crafts acylation because the strongly electron-withdrawing $-\mathrm{NO}_2$ group deactivates the benzene ring and coordinates with the $\mathrm{AlCl_3}$ catalyst, making Statement I incorrect. The $-\mathrm{NO}_2$ group is a strongly deactivating substituent due to its combined inductive (-I) and resonance (-M) effects, which deplete electron density preferentially from the ortho and para positions, thereby directing incoming electrophiles to the meta position and confirming Statement II is correct.

Friedel-Crafts alkylation of benzene with $CH_3CH_2CH_2Cl$ involves the formation of a primary carbocation that undergoes a 1,2-hydride shift to produce a more stable secondary carbocation, causing isopropylbenzene to be the major product rather than $n$-propylbenzene. Because the newly introduced alkyl group is electron-donating, it increases the electron density on the aromatic ring and activates it toward further electrophilic substitution, ensuring that multiple substitution products are inevitable under standard conditions. Furthermore, electron-donating substituents enhance the nucleophilicity of the benzene ring, thereby facilitating rather than inhibiting the Friedel-Crafts reaction, which renders the statement regarding the lack of alkylbenzene production incorrect.

The stability of an alkene is directly proportional to the number of $\alpha$-hydrogens present, as these atoms provide hyperconjugative stabilization to the double bond. By counting these substituents for each structure, 2,3-dimethylbut-2-ene contains 12 $\alpha$-hydrogens, 2-methylbut-2-ene contains 7, cis-but-2-ene contains 6, and prop-1-ene contains 3. This establishes the stability order $C > A > B > D$.

The treatment of the gem-dibromo compound $[P]$, $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CHBr}_2$, with excess sodamide (\text{NaNH}_2$) results \in double dehydrohalogenation to produce the terminal alkyne$\text{C}_6\text{H}_5\text{CH}_2\text{C}\equiv\text{CH}$as compound$[Q]$. Subsequent acid-catalyzed hydration of$[Q]$\in the presence of$\text{HgSO}_4$and$\text{H}_2\text{SO}_4$yields phenylacetone,$\text{C}_6\text{H}_5\text{CH}_2\text{COCH}_3$, which is compound$[R]$. Since this product is a methyl ketone, it undergoes the iodoform reaction to give a positive result while failing to react with Tollen's reagent, consistent with the properties described for$[R]$.

The formation of 2-methylhex-3-yne proceeds through the nucleophilic $S_N2$ alkylation of a terminal alkyne, specifically 3-methylbut-1-yne, which is deprotonated by a strong base like sodium amide ($NaNH_2$) to generate a reactive acetylide anion. This nucleophilic species subsequently attacks an ethyl halide, such as ethyl bromide ($CH_3CH_2Br$), resulting in the coupling of the alkyl group to the terminal alkyne carbon to yield 2-methylhex-3-yne ($CH_3-CH(CH_3)-C \equiv C-CH_2-CH_3$). This synthetic pathway maintains the triple bond at the 3-position while extending the carbon chain to the configuration specified in the product.

The catalytic hydrogenation of but-2-yne with $\mathrm{Pd}/\mathrm{C}$ yields cis-but-2-ene (X), whereas reduction with $\mathrm{Na}/\text{liq. } \mathrm{NH}_3$ produces trans-but-2-ene (Y). Cis-but-2-ene possesses a net dipole moment due to the arrangement of its substituents, making the statement regarding its dipole moment being zero incorrect. Additionally, ozonolysis of both isomers results in the oxidative cleavage of the double bond to generate acetaldehyde as the sole product, making the claim that they form different products also incorrect.

The hydrocarbon $P$ ($C_8H_{14}$) undergoes ozonolysis to cleave the internal carbon-carbon double bond, yielding a dicarbonyl compound $Q$ (2,7-octanedione). This dicarbonyl molecule subsequently undergoes an intramolecular aldol condensation in the presence of an alkali to form a cyclic $\alpha, \beta$-unsaturated ketone $R$, which possesses the methyl ketone moiety required for further reactions.

The treatment of the cyclic ketone $R$ with $I_2/NaOH$ initiates the haloform reaction, characterized by the cleavage of the methyl ketone group to produce a yellow precipitate of iodoform ($CHI_3$) and the corresponding carboxylic acid salt. Acidification of the solution converts this salt into the final carboxylic acid product $S$. Given the molecular formula and the necessity for the resulting enone to contain a methyl ketone function for a positive iodoform test, the structure $P$ is identified as 1,2-dimethylcyclohexene.

The combustion reaction of a hydrocarbon $C_xH_y$ follows the general equation $C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O(l)$, where the volumes of reacting gases are proportional to their stoichiometric coefficients. Treatment with KOH absorbs the $CO_2$ produced, so the volume of $CO_2$ formed is the difference between the residual gas volume and the unreacted oxygen volume, which is $160$ mL ($224 - 64$). Given that 80 mL of hydrocarbon results in 160 mL of $CO_2$, the stoichiometric coefficient $x$ is determined by $80x = 160$, yielding $x = 2$.

The total oxygen consumed during the reaction is the initial oxygen volume minus the volume remaining after KOH treatment, calculated as $200$ mL ($264 - 64$). Substituting $x = 2$ into the oxygen consumption relation $80(x + \frac{y}{4}) = 200$ yields $80(2 + \frac{y}{4}) = 200$, which simplifies to $2 + \frac{y}{4} = 2.5$. Solving for $y$ gives $\frac{y}{4} = 0.5$, so $y = 2$, identifying the hydrocarbon as $C_2H_2$.