Hydrocarbons practice · Mixed difficulty

Option A (Ethylene) is correct because the reaction of ethylene with sulfuric acid results in the formation of ethyl hydrogen sulfate.

Option D (acetylene) is correct because calcium carbide ($CaC_2$) reacts with water to produce acetylene ($C_2H_2$) and calcium hydroxide ($Ca(OH)_2$).

Option D ($60^\circ$) is correct because the dihedral angle between the two methyl groups in the gauche conformation of n-butane is 60 degrees.

Option D (LPG) is correct because LPG (Liquefied Petroleum Gas) generally produces less pollution compared to petrol, CNG, and kerosene due to its cleaner combustion.

Option A (infinite) is correct because alkanes can rotate around their carbon-carbon single bonds, leading to an infinite number of conformations.

Option B (${\mathrm{C}_2}{\mathrm{H}_2}$) is correct because acetylene (C2H2) is commonly used in oxy-acetylene welding due to its high flame temperature.

The correct answer is B (IV $>$ II $>$ I $>$ III) because the reactivity of alkenes with HBr increases with the stability of the carbocation formed, and is decreased by electron-withdrawing groups.

Option B (Raises its octane number) is correct because tetraethyl lead (TEL) is an antiknock agent that increases the octane number of gasoline, preventing premature detonation.

Option A (Statement-I & II are true) is correct because acetylenic hydrogens are acidic and can be replaced by active metals.

Option D (Solution of $B{r_2}$ in $CC{l_4}$) is correct because bromine in carbon tetrachloride decolorizes in the presence of alkenes due to addition across the double bond.

Option C ($\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}$) is correct because terminal alkynes have an acidic hydrogen due to the sp hybridization of the carbon atom.

Option C (n-Heptane) is correct because Wurtz reaction is best suited for symmetrical alkanes. n-Heptane would require using two different alkyl halides, leading to a mixture of products and a poor yield.

The correct answer is B (7.84 L) because it is calculated using stoichiometry from the balanced combustion equation of ethane.

Option B (Aliphatic hydrocarbons with small amounts of aromatic compounds) is correct because petroleum is primarily composed of aliphatic hydrocarbons with smaller amounts of aromatic hydrocarbons.

Option B ($\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_3$) is correct because the reaction sequence involves deprotonation of the terminal alkyne followed by alkylation with methyl iodide.

Option D is correct because ozonolysis can be used to identify the position of double or triple bonds, distinguish between branched and unbranched alkenes, and differentiate between isomers like 1-butene and 2-butene.

The correct answer is C because in the presence of peroxide, HBr adds to the alkene via anti-Markovnikov's rule.

Option C (${R_3}\mathrm{C}-\mathrm{OH}$) is correct because alkaline $KMnO_4$ oxidizes the C-H bond of a tertiary carbon to a hydroxyl group.

Option C (Producer gas) is correct because producer gas is primarily composed of carbon monoxide and nitrogen, not methane.

The correct answer is B ($\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{CHC}{\mathrm{H}_3}$) because symmetrical alkenes containing two carbon atoms on either side of the double bond yield only acetic acid upon vigorous oxidation.

Option A ($44^\circ C$) is correct because the flash point in India is fixed at $44^\circ C$.

Option D ($\ge$ 6 carbon) is correct because Baeyer's strain theory is not valid for cycloalkanes with larger rings due to the puckering of the rings.

Option A (99 kcal) is correct because the sigma bond energy of C-H bond in ${\mathrm{C}_2}{\mathrm{H}_6}$ is approximately 99 kcal.

The correct answer is A (Both A and R are correct, R is the correct explanation of A) because benzene is aromatic due to the presence of 6$\pi$ electrons, which satisfies Huckel's rule (4n+2).

Option C (CH₄) is correct because methane (CH₄) is the primary constituent of marsh gas, which is produced by anaerobic decomposition of organic matter in marshes.

Option A (Tetraethyl lead) is correct because tetraethyl lead (TEL) was historically used as an anti-knock compound in gasoline.

Option C (Tollen's reagent) is correct because propyne, being a terminal alkyne, reacts with Tollen's reagent to form a precipitate, while propene does not.

The correct answer is $\mathrm{C}{\mathrm{H}_2}\mathrm{ClC}{\mathrm{H}_3}$ because the reaction of $\mathrm{C}{\mathrm{H}_2} = \mathrm{C}{\mathrm{H}_2}$ with $\mathrm{HCl}$ results in an addition reaction forming $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{Cl}$.

Option B is correct because Statement II is true (neo-pentane has one quaternary carbon), but Statement I is false (bromination of neo-pentane does not form optical isomers).

Option B (${\mathrm{C}_6}{\mathrm{H}_6}$) is the correct answer because the reaction sequence involves oxidation of toluene to benzoic acid, followed by conversion to sodium benzoate, and finally decarboxylation to benzene.