Hydrocarbons – NEET UG ChemistryPractice Questions & PYQs

Heating sodium ethanoate with sodium hydroxide in the presence of calcium oxide triggers a soda lime decarboxylation reaction, which produces methane as the organic byproduct. Since the chemical formula for methane is $\mathrm{CH_4}$, its molar mass is 16 grams per mole. To determine the weight of two moles, multiply the molar mass by the quantity of moles, yielding the calculation $2 \times 16 = 32$ grams.

Huckel's rule dictates that for a species to be aromatic, it must be planar, exhibit complete delocalization of $\pi$ electrons, and contain $(4n+2)\pi$ electrons, where $n$ is an integer. Applying these conditions requires checking if each ring system supports a closed loop of conjugation and possesses a total $\pi$ electron count that fits the $(4n+2)$ series. Based on these stringent requirements, four of the presented species satisfy all criteria and are thus classified as aromatic.

Reductive ozonolysis involves the cleavage of an alkene's carbon-carbon double bond, resulting in the formation of two separate carbonyl-containing molecules. To reconstruct the starting material, one essentially reverses this process by removing the oxygen atoms from formaldehyde ($HCHO$) and 2-methylpropanal ($(CH_3)_2CHCHO$) and linking the two carbonyl carbons with a double bond. Combining the $(CH_3)_2CH-CH$ fragment with the $=CH_2$ group produces the structure $(CH_3)_2CH-CH=CH_2$, which corresponds to the alkene 3-methylbut-1-ene.

Boiling points of alkanes are primarily determined by molar mass, as larger molecules possess stronger van der Waals forces that require more heat to overcome. Furthermore, molecular structure influences boiling point, where straight-chain isomers generally exhibit higher boiling points than branched ones because their geometry allows for a larger surface area available for intermolecular contact.

Ranking the given alkanes follows directly from these principles, starting with heptane ($C_7H_{16}$) having the highest boiling point due to its superior molar mass, followed by hexane ($C_6H_{14}$) and 2-methylbutane ($C_5H_{12}$). Between the four-carbon isomers, the straight-chain butane ($C_4H_{10}$) results in a higher boiling point than the branched 2-methylpropane ($C_4H_{10}$), thereby establishing the final decreasing order of heptane, hexane, 2-methylbutane, butane, and 2-methylpropane.

The synthesis of alkenes fundamentally relies on elimination reactions, where reagents act to remove substituents and create a double bond between carbon atoms. While alcoholic potassium hydroxide facilitates dehydrohalogenation to produce an alkene, the use of aqueous potassium hydroxide promotes a nucleophilic substitution reaction instead. As illustrated by the reaction $\mathrm{R}-\mathrm{Cl} \xrightarrow{\text{aq. KOH}} \mathrm{R}-\mathrm{OH}$, the polar aqueous environment encourages the replacement of the halide with a hydroxyl group, yielding an alcohol rather than the intended unsaturated alkene.

For a compound to be considered aromatic, it must be cyclic, planar, and fully conjugated, while also containing a total number of $\pi$ electrons defined by the formula $(4n+2)$, where $n$ is an integer. Any structure that fails to meet these rigid requirements, such as by violating planarity or falling outside the specified electron count, is strictly non-aromatic. The compound in option D does not satisfy these necessary criteria, which distinguishes it from the remaining options that maintain aromatic character.

IUPAC nomenclature dictates that the parent chain must be the longest carbon chain containing the double bond, with numbering starting from the end that assigns the lowest possible locants to the double bond and the substituents. For 2,6-dimethyl-dec-4-ene, the structure requires a ten-carbon backbone with a double bond situated between the fourth and fifth carbons and methyl groups ($-\text{CH}_3$) positioned at the second and sixth carbons to satisfy the specified naming requirements.

Ozonolysis involves the oxidative cleavage of a carbon-carbon double bond ($C=C$), resulting in the conversion of each double-bonded carbon into a carbonyl group. The production of methanal ($HCHO$) as a cleavage product indicates that the reactant alkene must possess a terminal methylene group ($=CH_2$), as only this specific structural fragment yields formaldehyde upon reaction with ozone.

Electrophilic substitution proceeds through the attack of an electrophile ($E^+$) on an electron-rich aromatic system, creating a positively charged carbocation intermediate known as a sigma complex or Wheland intermediate. The aromaticity of the ring is subsequently restored via the elimination of a proton, effectively replacing a hydrogen atom with the incoming electrophilic group. This pathway is characteristic of reactions like the nitration or halogenation of benzene, where the electron-deficient species targets the high electron density of the conjugated $\pi$ system.

The cyclic trimerization of propyne ($\mathrm{CH}_3\mathrm{C}\equiv\mathrm{CH}$) over a red-hot iron tube at $873\,\text{K}$ yields $1,3,5\text{-trimethylbenzene}$ (mesitylene), which has the molecular formula $\mathrm{C}_9\mathrm{H}_{12}$. The structure consists of a central benzene ring where three methyl groups are attached to the ring carbons at positions 1, 3, and 5. To determine the number of sigma ($\sigma$) bonds, count the 6 C-C bonds within the cyclic backbone of the benzene ring, the 3 C-C bonds connecting the methyl groups to the ring, the 3 C-H bonds attached to the ring carbons at positions 2, 4, and 6, and the 9 C-H bonds within the three methyl groups. Summing these individual components yields $6 + 3 + 3 + 9 = 21$ sigma bonds.

The number of mono-chloro products obtained from the free radical chlorination of an alkane is determined by the number of chemically non-equivalent sets of hydrogen atoms available for substitution. Neopentane, represented as $\mathrm{C(CH_3)_4}$, contains a central quaternary carbon atom bonded to four symmetrical methyl groups, making all twelve hydrogen atoms chemically equivalent. Because every hydrogen position is identical, the substitution of any one of them by a chlorine atom results in the formation of only one distinct product, 1-chloro-2,2-dimethylpropane.

Reductive ozonolysis of an alkene involves cleavage of the carbon-carbon double bond to form carbonyl compounds; reversing this process by removing the oxygen atoms and joining the carbonyl carbons of propanone and ethanal with a double bond yields 2-methylbut-2-ene, represented as $\mathrm{CH_3-C(CH_3)=CH-CH_3}$. The addition of $\mathrm{HCl}$ to this unsymmetrical alkene proceeds according to Markovnikov's rule, where the electrophilic hydrogen adds to the less substituted carbon to generate the more stable tertiary carbocation intermediate, $\mathrm{CH_3-C^+(CH_3)-CH_2-CH_3}$. Subsequent nucleophilic attack by the chloride ion on this tertiary carbocation results in the major product, 2-chloro-2-methylbutane, which has the structure $\mathrm{H_3C-CH_2-C(Cl)(CH_3)-CH_3}$.

The combination of $\mathrm{H}_2$ with palladium poisoned by quinoline, known as Lindlar's catalyst, facilitates the stereoselective partial hydrogenation of alkynes to *cis*-alkenes. This catalyst system allows for the *syn*-addition of hydrogen atoms across the carbon-carbon triple bond while preventing further reduction of the alkene to an alkane.

Since the hydrocarbon undergoes a substitution reaction with bromine rather than an addition reaction, it must be an alkane. When methane ($CH_4$) reacts with bromine, it yields methyl bromide ($CH_3Br$), which subsequently undergoes the Wurtz reaction to form ethane ($CH_3-CH_3$). As ethane consists of only two carbon atoms, it meets the requirement of being a gaseous hydrocarbon containing fewer than four carbon atoms.

The acidity of hydrocarbons correlates directly with the s-character of the carbon atom holding the negative charge, where higher s-character leads to greater electronegativity and increased stabilization of the conjugate base. This principle makes $CH \equiv CH$ the strongest acid among the group, while $CH_3-C \equiv CH$ is slightly less acidic due to the electron-donating inductive effect of the methyl group. Ethene ($CH_2=CH_2$) ranks next because its $sp^2$ hybridized carbon is less electronegative than those in the alkynes, and ethane ($CH_3-CH_3$) is the weakest acid because its $sp^3$ carbon provides the least stability for the negative charge.

When $C_2H_6$ rotates around its carbon-carbon single bond, it transitions through various spatial arrangements known as conformers. This rotational movement merely alters the dihedral angle between hydrogen atoms rather than disrupting the underlying covalent structure, which ensures that bond lengths and bond angles remain identical across all conformations. Consequently, the fundamental geometric parameters of the molecule are preserved regardless of the rotation.

The acid-catalyzed hydration of propyne proceeds via the Markovnikov addition of water to the alkyne, facilitated by $\mathrm{Hg}^{2+}$ ions, to form an unstable enol intermediate with the structure $\mathrm{H}_3\mathrm{C}-\mathrm{C}(\mathrm{OH})=\mathrm{CH}_2$. This enol intermediate rapidly undergoes keto-enol tautomerization, a process where the double bond and a hydrogen atom rearrange to convert the vinyl alcohol into the significantly more stable ketone, propanone (acetone), with the structure $\mathrm{H}_3\mathrm{C}-\mathrm{C}(=\mathrm{O})-\mathrm{CH}_3$.

When bromine exists in the gaseous phase at high temperatures, it initiates reactions through a free-radical substitution mechanism rather than the standard electrophilic addition observed at room temperature. The rate of this process depends directly on the stability of the free radical intermediate formed after a hydrogen atom is abstracted. Propene ($\mathrm{C}_3\mathrm{H}_6$) reacts most readily because its allylic hydrogen abstraction generates an allyl radical ($\dot{\mathrm{C}}\mathrm{H}_2-\mathrm{CH}=\mathrm{CH}_2$), which is highly stabilized by resonance delocalization. This stabilization is superior to the inductive stability found in the alkyl radicals produced from saturated hydrocarbons like butane, making the path for propene the most energetically favorable.

Friedel-Crafts alkylation relies on the generation of a reactive carbocation intermediate from an alkyl halide when treated with a Lewis acid, such as $AlCl_3$, which then acts as an electrophile to substitute onto the aromatic ring. Aryl halides like chlorobenzene and bromobenzene, as well as vinyl halides like chloroethene, are chemically unsuitable for this process because the carbon-halogen bond possesses partial double-bond character due to resonance, which inhibits the formation of the necessary carbocation. In contrast, isopropyl chloride functions as a suitable alkylating agent because it effectively generates a reactive carbocation, $(CH_3)_2CH^+$, allowing the substitution reaction to proceed.

The reactivity of pyrrole is governed by the resonance effect, where the lone pair on the nitrogen atom is delocalized into the ring to contribute to its aromatic character. By drawing the resonance contributors, one can observe that the resulting negative charge is primarily stabilized at the carbon atoms immediately adjacent to the nitrogen, known as the 2 and 5 positions. Because these specific carbons possess the highest electron density, they serve as the most favorable sites for electrophilic attack.