Hydrocarbons – JEE Main ChemistryPractice Questions & PYQs

Decarboxylation of sodium ethanoate and the hydrolysis of $\mathrm{CH}_3\mathrm{MgBr}$ are valid methods to produce methane, but Kolbe's electrolysis of sodium acetate specifically forms ethane ($\mathrm{CH}_3-\mathrm{CH}_3$) through radical coupling, which invalidates the claim made in Statement-I. Regarding Statement-II, methane cannot be synthesized via the Wurtz reaction because that process requires coupling two alkyl groups to form an alkane with at least two carbons, and unsaturated hydrocarbons similarly contain at least two carbon atoms, making the synthesis of a single-carbon methane molecule impossible from those substrates.

Every single covalent bond constitutes a $\sigma$ bond, while any additional bond within a multiple bond represents a $\pi$ bond formed by lateral orbital overlap. To analyze hex-1-en-4-yne, consider the structural formula $CH_2=CH-CH_2-C \equiv C-CH_3$. The total number of $\sigma$ bonds is determined by counting all single connections, which includes 5 $\sigma$ bonds within the carbon backbone and 8 $\sigma$ bonds between the carbon and hydrogen atoms, summing to 13 $\sigma$ bonds. The $\pi$ bonds are exclusively located within the multiple bonds: the double bond contributes 1 $\pi$ bond, and the triple bond adds 2 $\pi$ bonds, yielding a total of 3 $\pi$ bonds.

Neopentane features a highly symmetric, tetrahedral arrangement where all twelve hydrogen atoms are chemically equivalent, ensuring that any substitution reaction results in only one unique derivative. Furthermore, its nearly spherical shape allows the molecules to pack much more efficiently within a solid crystal lattice compared to the elongated chains of n-pentane, which facilitates stronger intermolecular attractions in the solid state and leads to a higher melting point.

The regioselectivity of free radical bromination is governed by the relative stability of the intermediate alkyl radicals, where the reaction strongly favors the abstraction of hydrogen from the most substituted carbon atom. In sec-butylcyclohexane, the tertiary hydrogen atom at the junction of the cyclohexane ring and the sec-butyl side chain is significantly more reactive than secondary or primary hydrogens, leading to the formation of the most stable tertiary radical. This preference ensures that the bromine atom preferentially substitutes at this tertiary carbon, resulting in the structure shown in the correct option.

The compound $\mathrm{C}_6\mathrm{H}_6$ requires four moles of $\mathrm{H}_2$ for complete hydrogenation, which indicates a total of four sites of unsaturation within its structure. Since each site of unsaturation represents a $\pi$ bond containing two electrons, the calculation is based on the moles of hydrogen consumed. Thus, the total count is $4 \times 2 = 8$ $\pi$ electrons.

Ozonolysis involves the cleavage of carbon-carbon double bonds to form carbonyl products; specifically, cis-2-butene ($CH_3CH=CHCH_3$) breaks down into two molecules of ethanal ($CH_3CHO$), confirming that Statement I is true. The ozonolysis of 3,6-dimethyloct-4-ene ($CH_3CH_2CH(CH_3)CH=CHCH(CH_3)CH_2CH_3$) cleaves the double bond at the C4-C5 position to yield two molecules of 2-methylbutanal ($CH_3CH_2CH(CH_3)CHO$). Since 2-methylbutanal contains a carbon atom bonded to four distinct groups-a methyl group, an ethyl group, a hydrogen atom, and a formyl group-this product possesses a chiral center, which contradicts the assertion in Statement II.

The prediction of the major product in a multi-step reaction sequence is governed by the relative stability of the reactive intermediates, such as carbocations, which dictates the regioselectivity of the addition or substitution process. According to Markovnikov's rule, the electrophile adds to the less substituted carbon of a double bond to form the more stable tertiary or secondary carbocation, which is then attacked by the nucleophile to yield the primary product. This energetic preference ensures that the reaction proceeds through the pathway with the lowest activation energy, consistently leading to the formation of the most stable constitutional isomer.

Terminal alkynes such as propyne contain an acidic hydrogen atom bonded to an sp-hybridized carbon, allowing them to act as weak acids. When propyne reacts with excess sodium metal, the sodium replaces this acidic proton to form a sodium acetylide salt, releasing hydrogen gas according to the equation:

$\mathrm{CH}_3-\mathrm{C}\equiv\mathrm{CH} + \mathrm{Na} \rightarrow \mathrm{CH}_3-\mathrm{C}\equiv\mathrm{C}^{-}\,\mathrm{Na}^{+} + \frac{1}{2}\mathrm{H}_2$

Since one mole of propyne contains exactly one acidic hydrogen atom, the stoichiometry dictates that one mole of the alkyne will generate half a mole of $\mathrm{H}_2$ gas, confirming that the first statement is correct.

The reaction between propyne and sodium amide involves the amide ion abstracting the acidic proton to produce ammonia gas, with one mole of propyne yielding one mole of ammonia. To evaluate the second statement, consider that the molar mass of propyne is 40 g/mol, meaning that 4 g of propyne corresponds to 0.1 moles. Consequently, the reaction will produce 0.1 moles of ammonia gas. Given that one mole of any ideal gas occupies 22.4 L at STP, the produced ammonia will occupy a volume of $0.1 \times 22.4 = 2.24$ L, which is equivalent to 2240 mL. As this value differs from the 224 mL mentioned in the statement, the second statement is incorrect.

The degree of unsaturation for $\mathrm{C}_9\mathrm{H}_{12}$ is calculated as $9 + 1 - \frac{12}{2} = 4$, which corresponds to the four degrees of unsaturation inherent in a stable benzene ring. A negative Baeyer's test further confirms this aromatic nature, as the molecule lacks unsaturated side chains susceptible to oxidation by potassium permanganate. For an isomer to possess four different non-aliphatic substitution sites, the alkyl substituents must be arranged on the benzene ring such that no planes of symmetry exist, rendering the four remaining ring hydrogens chemically distinct. This specific asymmetric substitution pattern is found in exactly two isomers, 1-ethyl-2-methylbenzene and 1-ethyl-3-methylbenzene.

Effective alkene synthesis via elimination requires the abstraction of a $\beta$-hydrogen atom by a base, facilitating the formation of a $\pi$-bond through well-defined $E_1$ or $E_2$ mechanisms. Reactions B and D are excluded from this process because they either utilize substrates lacking the necessary $\beta$-hydrogens or employ reagents that preferentially drive nucleophilic substitution rather than the required elimination pathway. These inherent chemical limitations prevent the formation of the alkene functional group in these specific instances.

Secondary hydrogens are those directly attached to a carbon atom that is bonded to exactly two other carbon atoms, which means identifying them requires locating the methylene (\text{-CH}_2\text{-}$) groups within the molecular chain. When assessing 2,2,3,3-tetramethylpentane, the structural analysis reveals only one methylene group located at the fourth position of the pentane backbone, which accounts for exactly two secondary hydrogens. Other compounds in the list, such as 2,2,4,4-tetramethylhexane, 4-ethyl-3,4-dimethyloctane, and 2,2,4,5-tetramethylheptane, contain more complex arrangements that introduce additional secondary carbon sites, resulting in higher counts of four, ten, and four secondary hydrogens, respectively.

Alkynes undergo selective catalytic hydrogenation to form alkenes, which can then be oxidatively cleaved via ozonolysis to yield specific carbonyl compounds. When propyne reacts with $\text{H}_2$ over a $\text{Pd/C}$ catalyst, the triple bond is reduced to a double bond to generate propene (\text{CH}_3-\text{CH}=\text{CH}_2$) as product$[A]$. Subsequent treatment of this alkene with ozone followed by a reductive workup using zinc and water breaks the carbon-carbon double bond, resulting \in the formation of acetaldehyde (\text{CH}_3\text{CHO}$) as product $[B]$ and formaldehyde (\text{HCHO}$) as product$[C]$.

The Fittig reaction is defined by the coupling of two aryl halide molecules in the presence of metallic sodium to synthesize biaryl compounds. The Gatterman reaction facilitates the replacement of a diazonium group on an aromatic ring with a chlorine atom by treating an aryl diazonium salt, such as $\mathrm{ArN}_2^{+} \mathrm{X}^{-}$, with $\mathrm{HCl}$ and copper powder. The Finkelstein reaction involves a halogen exchange where an alkyl bromide is converted to an alkyl iodide using sodium iodide in dry acetone. Finally, the Lucas reaction serves as a standard diagnostic test for classifying alcohols, where a tertiary alcohol reacts with concentrated $\mathrm{HCl}$ and anhydrous $\mathrm{ZnCl}_2$ to rapidly produce an alkyl chloride.

The Swarts reaction is a specialized halogen exchange method primarily used to introduce fluorine into an organic molecule, such as converting an alkyl halide into ethyl fluoride. Sandmeyer's reaction involves the substitution of a diazonium group on an aromatic ring, serving as the standard route for synthesizing cyanobenzene from diazonium salts. The Wurtz-Fittig reaction couples an aryl halide with an alkyl halide using metallic sodium in dry ether to form an alkylated benzene derivative, such as the synthesis of ethyl benzene from $Ph - Br$ and $C_2H_5Br$. Finally, the Finkelstein reaction serves as a classic halogen exchange process, utilizing sodium iodide in acetone to replace a chloride or bromide with iodine, thereby producing ethyl iodide from precursors like $C_2H_5Cl$.

The formation of product $B$ relies on the nucleophilic or electrophilic attack of the reagent on intermediate $A$, which is governed by the electronic and steric environment of the existing functional groups. The reaction proceeds through a transition state where the regioselectivity is determined by the stability of the intermediate species, often following specific rules such as Markovnikov addition or standard substitution patterns depending on the reagent's nature. Consequently, the final structure of $B$ represents the most energetically favorable configuration resulting from the sequence of transformations applied to the starting material.

Hex-1,3-dien-5-yne can be represented by the structural formula $CH_2=CH-CH=CH-C \equiv CH$, where each double bond contains one $\pi$ bond and the triple bond contains two $\pi$ bonds, totaling 4 $\pi$ bonds. The carbon skeleton is held together by five C-C single bonds, and there are six additional C-H bonds, resulting in 11 $\sigma$ bonds. The sum of these values is $11 + 4 = 15$.

Reductive ozonolysis, which typically employs $\mathrm{Zn}/\mathrm{H}_2\mathrm{O}$ workup, cleaves the carbon-carbon double bond within a cyclic alkene to produce a dicarbonyl structure, effectively transforming the original olefinic carbons into aldehyde or ketone moieties. Reversing this process involves joining the terminal aldehyde carbon and the ketone carbon of 3-methyl-6-oxoheptanal with a $\text{C=C}$ bond to close the ring. This chemical pathway demonstrates how the positioning of the methyl substituent at the third carbon naturally corresponds to the geometry of the precursor.

The acid-catalyzed hydration of propyne in the presence of $\text{Hg}^{2+}$ leads to the formation of propanone through the tautomerization of an enol intermediate. This ketone then reacts with $\text{HCN}$ via nucleophilic addition to the carbonyl group to yield 2-hydroxy-2-methylpropanenitrile, $\text{CH}_3\text{C(OH)(CN)}\text{CH}_3$.

The reaction sequence begins with the acid-catalyzed hydration of a terminal alkyne, which proceeds according to Markovnikov's rule to form a ketone. Specifically, the reaction of $CH_3CH_2C \equiv CH$ with water in the presence of $Hg^{2+}$ and $H_2SO_4$ converts the terminal alkyne into butan-2-one as product A. Subsequently, sodium borohydride acts as a selective reducing agent that converts this ketone into a secondary alcohol, resulting in the formation of butan-2-ol as product B.

The protection of the highly reactive amino group in aniline via acetylation with acetic anhydride is essential to reduce the electron-donating power of the nitrogen atom, thereby minimizing over-reaction and directing subsequent electrophilic aromatic substitution to the para position. Upon reacting aniline with acetic anhydride in the presence of pyridine, the nucleophilic nitrogen atom attacks the carbonyl carbon to yield acetanilide as product A, which contains an $-NHCOCH_3$ group that sterically hinders the ortho positions. Consequently, when acetanilide undergoes nitration with a mixture of concentrated $HNO_3$ and $H_2SO_4$, the incoming nitro electrophile ($NO_2^+$) is directed to the para position, resulting in p-nitroacetanilide as product B.