Hydrocarbons practice · Mixed difficulty

The correct answer is $\mathrm{C}{\mathrm{H}_2}\mathrm{OH}-\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ because ozonolysis of ethyne followed by reduction yields ethylene glycol.

Option D (Both (2) and (3)) is correct because both petroleum and coal are significant sources of aromatic compounds.

Option B (Elimination reaction) is correct because alkyl halides undergo dehydrohalogenation (elimination of HX) to form alkenes.

Option B (Hexadecane) is correct because hexadecane (cetane) is the reference fuel for determining the cetane number of diesel fuel.

Option A ($n$ -alkanes) is correct because straight-chain alkanes (n-alkanes) have lower octane numbers and are more prone to causing knocking in engines compared to branched alkanes or aromatic compounds.

Option D ($n$ -heptane) is correct because $n$-heptane is assigned an octane number of 0, representing very poor antiknock properties.

Option B is correct because $NaNH_2$ deprotonates the terminal alkyne, and then $CH_3CH_2I$ alkylates the resulting acetylide anion.

Option D is correct because propyne is a terminal alkyne and reacts with ammoniacal silver nitrate to form a precipitate, while propene does not react. This difference allows for their distinction.

Option C (124) is correct because 2,3,3-trimethylbutane has a research octane number (RON) greater than 100, and is approximately 124.

Option B (cis-4-octene) is correct because Lindlar's catalyst ($Pd/CaCO_3$ quinoline) is used for the partial hydrogenation of alkynes to cis-alkenes.

Option B (propene) is correct because ozonolysis of propene yields formaldehyde (HCHO) and acetaldehyde (CH3CHO).

Option D (Coal gas) is correct because coal gas is a gaseous fuel produced during the destructive distillation of coal.

Option D (resonance stabilized system is to be preserved) is correct because the stability gained through resonance in benzene makes it less prone to addition reactions.

Option A (Markownikoff’s rule) is correct because Markownikoff's rule states that in the addition of a protic acid HX to an unsymmetrical alkene, the acid hydrogen (H) becomes attached to the carbon with more hydrogens, and the halide (X) group becomes attached to the carbon with fewer hydrogens.

Option A (Statement-I & II are true) is correct because both vicinal and geminal dihalides can undergo dehydrohalogenation to form alkynes.

Option C is correct because 2-methyl-3-hexyne, when reacted with alkaline KMnO4 followed by acidification, gives isobutyric acid and propanoic acid, matching the given products.

Option A ($I > II > III$) is correct because boiling point decreases with increased branching due to reduced surface area and weaker van der Waals forces.

Option C (2, 3-dimethylbutane) is correct because 2,3-dimethylbutane yields three monochloro derivatives after chlorination.

Option D ($B{r_2}$ water) is correct because ethene, being an unsaturated compound, will decolorize bromine water, while ethane will not react.