Hydrocarbons practice · Mixed difficulty

Option D (Hexane) is correct because hexane is a nonpolar molecule and does not possess a dipole moment, hence it is not deflected by a non-uniform electrostatic field.

Option B (cycloalkanes) is correct because cycloalkanes are hydrocarbons in which the carbon atoms form a closed ring or chain.

Option A is correct because benzene is indeed stable due to resonance and does not decolorize alkaline KMnO4 under normal conditions.

Option B (combustion) is correct because polynuclear hydrocarbons are formed during the incomplete combustion of organic materials.

Option D (Both (2) and (3)) is correct because both petroleum and coal are major sources of aromatic compounds, obtained through refining and processing.

Option C ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}+\mathrm{CO}_2$) is correct because cold alkaline KMnO4 oxidizes 1-butyne to propanoic acid and carbon dioxide.

Option A (Hydrocarbons) is correct because petrol or gasoline is primarily a mixture of various hydrocarbons, mainly alkanes, with carbon chain lengths typically ranging from C6 to C11.

Option C ($E_{\text {act}}=0$) is correct because chain-terminating steps involve the combination of radicals, which requires little to no activation energy.

Option D (All are correct) is correct because ethylene is used for ripening of food, preparing ethylene oxide, and preparing ethylene chloride.

The correct answer is A (2,3-dimethyl-2-butene) because the major product formed is the most stable alkene according to Zaitsev's rule, which involves the elimination of water from the alcohol.

Option C ($p\pi - d\pi$ bonding) is correct because Ag+ ion increases the solubility of alkenes by forming $p\pi - d\pi$ bonds with the alkene's pi system.

Option C (1) is correct because butyne-1 has only one hydrogen atom directly attached to the sp hybridized carbon.

Option C is correct because but-2-ene undergoes reductive ozonolysis to produce two moles of ethanal (molar mass 44).

Option C is correct. Acetylene reacts with sodamide to produce H2 gas, but acetylene is a stronger acid than ammonia, not weaker.

The correct answer is Reduction of $\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$ with HI and red P because it will produce propane, not ethane.

Option D (Ethane has an infinite number of conformation of which the staggered conformation is possessed by majority of the molecules at room temperature) is correct because the staggered conformation is more stable and thus more populated at room temperature.

The correct answer is C (Cracking) because cracking is the process of breaking down large hydrocarbon molecules into smaller, more volatile ones.

Option B (Ethanol) is correct because the reaction of ethylene with dilute sulfuric acid in the presence of mercuric sulfate leads to the addition of water, forming ethanol.

Option A (Friedel-Crafts reaction) is correct because the reaction of benzene with methyl chloride in the presence of anhydrous aluminum chloride is a Friedel-Crafts alkylation.

Option B is correct because the staggered conformation of butane is the most stable due to minimized torsional strain and steric hindrance.

Option D (Wax) is correct because Wax is not obtained by heating wood or coal in the absence of air.

Option C (Alkylene dihalide) is correct because when two halogen atoms are attached to two adjacent carbon atoms, the dihaloalkane is called an alkylene dihalide (or vicinal dihalide).

Option B (75) is correct because the octane number is defined as the percentage of iso-octane in the fuel mixture.

Option A ($\mathrm{C}{\mathrm{l}_2}$) is correct because chlorine is the most reactive halogen in the halogenation of alkanes due to its lower bond dissociation energy compared to bromine and iodine.

Option C (inorganic graphite) is correct because diborane reacts with ammonia at high temperatures to form boron nitride, which has a graphite-like structure.

Option C (3, 5) is correct because the methyl group is ortho- and para-directing, making meta-substitution (3,5) less favorable.

Option B (2-methylpentane) is correct because it follows IUPAC naming rules, prioritizing the lowest possible number for the substituent.

Option A (100) is correct because n-hexadecane (cetane) is the reference fuel for cetane number determination and is assigned a cetane number of 100.

Option B ($R'\mathrm{H}$) is correct because the Grignard reagent ($R'\mathrm{Mg}X$) abstracts the acidic proton from the terminal alkyne ($R\mathrm{C} \equiv \mathrm{CH}$), forming an alkynyl magnesium halide and an alkane ($R'\mathrm{H}$).

Option D (128) is correct because the lowest possible alkane with ethyl groups as substituents is tetraethylmethane, which has a molecular mass of 128.