Hydrocarbons – JEE Main ChemistryPractice Questions & PYQs

The conversion of an unsaturated alkene into a saturated alkane is accomplished through catalytic hydrogenation, which adds hydrogen across the carbon-carbon double bond. When butene-1 is treated with hydrogen gas in the presence of a palladium catalyst, the reaction $\mathrm{C}_4\mathrm{H}_8 + \mathrm{H}_2 \xrightarrow{\mathrm{Pd}} \mathrm{C}_4\mathrm{H}_{10}$ occurs to produce butane. This specific combination of hydrogen and a metal catalyst is essential to break the double bond and fully saturate the hydrocarbon chain.

The terminal hydrogen atoms in acetylene ($\mathrm{CH} \equiv \mathrm{CH}$) possess significant acidic character because the $sp$-hybridized carbon atoms strongly attract their bonding electrons, which allows the molecule to participate in acid-base reactions with stronger bases or displace metals. Because of this property, acetylene reacts readily with metallic sodium ($\mathrm{Na}$) or ammoniacal $\mathrm{AgNO}_3$ to form metal acetylides, and it undergoes electrophilic addition with $\mathrm{HCl}$ to produce vinyl chloride. In contrast, sodium hydroxide ($\mathrm{NaOH}$) is not a strong enough base to deprotonate the terminal hydrogens of acetylene, resulting in no observable reaction.

The reaction proceeds via the electrophilic addition of hypochlorous acid to the acetylene triple bond to form an unstable chlorovinyl alcohol intermediate, which rapidly tautomerizes into chloroacetaldehyde. Since the resulting aldehyde is highly reactive toward further electrophilic attack in the presence of excess hypochlorous acid, the process continues until the chlorine substitution yields dichloroacetaldehyde, $\mathrm{Cl_2CHCHO}$, as the final product.

The reaction proceeds via the formation of a nucleophilic acetylide intermediate followed by its alkylation through an $\text{S}_\text{N}2$ mechanism. Terminal alkynes, such as propyne, react with metallic sodium to generate the sodium acetylide species $\text{CH}_3 - \text{C} \equiv \text{C}^- \text{Na}^+$, which is identified as compound $A$. This acetylide ion subsequently acts as a strong nucleophile and undergoes an $\text{S}_\text{N}2$ substitution reaction with the alkyl bromide $B$ to displace the bromide ion, resulting in the formation of the substituted alkyne product.

Elimination reactions generally favor the formation of the more substituted and thermodynamically stable alkene, known as the Saytzeff product, particularly when using small bases or acidic conditions that allow for rearrangement to the most stable intermediate. The use of a sterically bulky base such as potassium tert-butoxide, $(\mathrm{CH}_3)_3\mathrm{CO}^{\ominus}\mathrm{K}^{\oplus}$, disrupts this preference because steric hindrance prevents the base from effectively abstracting a proton from the more substituted carbon. Consequently, the base removes a proton from the more accessible, less hindered position, leading to the formation of the less substituted Hofmann product instead of the Saytzeff product.

Hydroboration-oxidation of styrene, $\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2$, proceeds through an anti-Markovnikov addition to produce 2-phenylethanol, $\mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$. The subsequent treatment with $\mathrm{HBr}$ replaces the hydroxyl group with a bromine atom to form 1-bromo-2-phenylethane, which reacts with magnesium in ether to generate the corresponding Grignard reagent, $\mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{MgBr}$. Finally, the nucleophilic attack of the Grignard reagent on formaldehyde followed by acidic hydrolysis extends the carbon chain by one carbon atom, yielding the primary alcohol 3-phenylpropan-1-ol, $\mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$.

The hydrogenation of but-2-yne using $\mathrm{Pd} / \mathrm{C}$ proceeds via syn-addition to produce cis-but-2-ene, denoted as $A$, whereas reduction using sodium in liquid ammonia proceeds via anti-addition to yield trans-but-2-ene, denoted as $B$. Regarding physical properties, the cis isomer $A$ possesses a net dipole moment while the trans isomer $B$ has a dipole moment of zero due to its symmetry, making the claim that $A$ is more polar because its dipole moment is zero logically contradictory. Furthermore, electrophilic addition reactions like bromination generally occur more readily with the cis isomer $A$, which means the assertion that $B$ adds bromine more easily than $A$ is incorrect.

The reagent system $\mathrm{Zn}(\mathrm{Hg})/\mathrm{HCl}$ is typically employed to perform a Clemmensen reduction, which converts an aldehyde or ketone functional group into a methylene or methyl group. In this particular reaction, the aldehyde group in $\mathrm{C}_6\mathrm{H}_5-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{CH}_3)-\mathrm{CH}_2\mathrm{CHO}$ is reduced to a methyl group, while the acidic conditions of the reaction mixture simultaneously facilitate the acid-catalyzed dehydration of the benzylic alcohol. This elimination is thermodynamically favored because it results in the formation of an alkene conjugated with the phenyl ring, ultimately yielding $\mathrm{Ph}-\mathrm{CH}=\mathrm{C}(\mathrm{CH}_3)-\mathrm{CH}_2-\mathrm{CH}_3$ as the stable major product.