Hydrocarbons practice · Mixed difficulty

Option A ($NaOH$) will not react with acetylene because it is a strong base and acetylene is a weak acid. Acetylene requires a stronger base than NaOH to deprotonate it.

Option D (Zero) is correct because BHC (Benzene hexachloride or gammexane) is a saturated cyclic compound with no double bonds.

Option A is correct because the addition of bromine to ethene results in the formation of 1,2-dibromoethane through a vicinal dibromination reaction.

Option B ($\left( {4n + 2} \right)\pi$-electrons) is correct because Huckel's rule states that aromatic compounds must possess $\left( {4n + 2} \right)\pi$-electrons.

Option B ($11.43\mathrm{ }\;g$) is correct because it is calculated based on the stoichiometry of the reaction between pentene and bromine.

Option C (${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CH} - \mathrm{C} \equiv \mathrm{CH}$) is correct because it is a terminal alkyne that reacts with ammoniacal AgNO3 to give a white precipitate and yields the given acid upon oxidation.

Option C ($\mathrm{C}{\mathrm{l}_2}\mathrm{CH}\mathrm{.CHO}$) is correct because acetylene reacts with hypochlorous acid to form dichloroethanal.

Option D is correct because PF5 exhibits sp3d hybridization, not sp3d2 hybridization.

Option C (${C_2}{H_5}OH$) is correct because the sequence of reactions involves hydration of calcium carbide to acetylene, followed by hydration of acetylene to acetaldehyde, and finally, reduction of acetaldehyde to ethanol.

Option C is correct because n-hexane isomerizes to both 2-methylpentane and 3-methylpentane in the presence of anhydrous aluminium chloride and hydrogen chloride gas.

Option B (Ethane) is correct because decarboxylation removes a COONa group, resulting in a hydrocarbon with one less carbon atom.

The correct answer is D (If both Assertion & Reason are false) because Acetylene gives glyoxal on oxidative ozonolysis, not reductive, and benzene gives glyoxal on oxidative ozonolysis.

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 (${CHCl}={CHAsCl}_2$) is correct because the formula for Lewisite, a poisonous gas, is ${CHCl}={CHAsCl}_2$.

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 B ($C{H_3} - CH = CH - C{H_3}$) is correct because acid-catalyzed dehydration of butan-1-ol favors the formation of the more substituted alkene (2-butene) according to Zaitsev's rule.

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 B (2-butene) is correct because sec-butyl chloride undergoes dehydrohalogenation in the presence of alcoholic KOH to form the most stable alkene, which is 2-butene (Zaitsev's rule).

Option A (Both S-I $\&$ S-II are correct) is correct because n-Hexane does form benzene upon aromatization, and during aromatization of alkenes, the carbon percentage increases.

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 is correct because both statements are correct. Isomerization of alkanes does occur in the presence of $AlCl_3$ and $HCl$, and the boiling point of branched alkanes is generally lower than that of their straight-chain isomers.

Option B ($C{H_3} - C{H_2} - C{H_2}Br$) is correct because in the presence of benzoyl peroxide, HBr adds to propene via anti-Markovnikov's rule, resulting in the bromine atom attaching to the less substituted carbon.

The correct answer is C2H6 because when it burns in excess oxygen, the volume of CO2 evolved is double that of the hydrocarbon taken, as shown in the balanced equation.

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