Hydrocarbons practice · Easy

Option B ($Pd/BaS{O_4}$ or $CaC{O_3}$ in quinoline or xylene) is correct because it accurately describes Lindlar's catalyst, which is used for the partial reduction of alkynes to cis-alkenes.

Option A (Markownikoff’s rule) is correct because the statement accurately describes Markownikoff's rule, which governs the addition of protic acids to alkenes.

Option B (Gilman reagent) is correct because $R_2CuLi$ is the general formula for a Gilman reagent, which is used in Corey-House synthesis.

Option B (1.08) is correct because the C-H bond length in benzene is approximately 1.08 $\mathop \mathrm{A}\limits^\mathrm{o}$.

Option C (Dark red colour) is correct because benzene does not undergo nitration at room temperature, but the solution turns dark red due to the absorption of $\mathrm{N}{\mathrm{O}_2}$ released by $\mathrm{HN}{\mathrm{O}_3}$.

Option A (n-Hexane) is correct because straight-chain alkanes have higher boiling points than branched alkanes due to increased surface area and stronger van der Waals forces.

Option B is correct because while the reaction of methane with chlorine does involve a free radical mechanism, the reaction *does* occur, but only in the presence of light (hv). Therefore, the assertion that it doesn't react in the dark is correct, but the reason is not the explanation.

Option C (non-aromatic) is correct because cyclobutadiene has 4$\pi$ electrons, making it anti-aromatic, which is a type of non-aromatic compound.

Option D (${\mathrm{C}_2}{\mathrm{H}_2}$) is correct because acetylene (${\mathrm{C}_2}{\mathrm{H}_2}$) has acidic hydrogens due to the sp hybridization of the carbon atoms, allowing it to react with metals by displacing the H-atom.

Option B (Acetylene molecule has three ‘pi’ ($\pi$) bonds and three sigma bonds) is false because acetylene has two pi bonds and three sigma bonds.

Option A (Cracking) is correct because thermal decomposition of alkanes in the absence of air is specifically called cracking, also known as pyrolysis.

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 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.

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 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 (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 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.

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 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.

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.

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 (${\mathrm{C}_6}{\mathrm{H}_6}$) 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.

Option D (176 g) is correct because it is calculated based on the stoichiometry of the combustion reaction of ethane.

Option A is correct because passing the hydrocarbon through ammoniacal AgNO3 will identify terminal alkynes (containing a triple bond at the end of the chain).