Hydrocarbons practice · Medium

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

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

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

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 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.

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

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.

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

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 ($\mathrm{C}{\mathrm{H}_4}$) is correct because the reaction of CS2 and H2S over heated Cu produces methane and sulfur.

Option D ($C{H_3} - C{H_2} - C{H_2} - \begin{array}{cccccccccccccccccccc} {Br} \\ | \\ C \\ | \\ {C{H_3}} \end{array} - C{H_3}$) is correct because it follows Markovnikov's rule, where the bromine adds to the more substituted carbon.

Option D is correct because passing an electric discharge through graphite in the presence of hydrogen can lead to the formation of methane, ethane, and acetylene.

Option A (100) is correct because n-hexadecane (cetane) is assigned a cetane number of 100 by definition.

Option C (1-butyne) is correct because terminal alkynes react with ammoniacal silver nitrate due to the acidity of the terminal hydrogen.

Option A (2 – chloropropane) is correct because the reaction follows Markovnikov's rule in the presence of peroxide.

Option D (${\mathrm{O}_3},\mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}$) is correct because ozonolysis followed by reductive workup produces different carbonyl compounds for 2-methylpropene and butane-1.