Hydrocarbons practice · Easy

Option A (2-butene) is correct because anti-Markownikoff's rule applies only to asymmetric alkenes, and 2-butene is a symmetric alkene.

Option B (3) is correct because there are three possible isomers of xylene: ortho-xylene, meta-xylene, and para-xylene.

Option A (Two $s{p^2}$ and one $\;sp$-hybrid carbon) is correct because the propadiene molecule has two terminal carbons with $s{p^2}$ hybridization and one central carbon with $\;sp$ hybridization.

Option B (Deactivates the benzene ring more via inductive effect than activating it via resonance effect) is correct because the inductive effect of chlorine is stronger than its resonance effect, leading to overall deactivation.

The correct answer is B because both the assertion and the reason are correct, but the reason is not the correct explanation for the assertion.

Option D (Kolbe's synthesis) is correct because the electrolysis of aqueous potassium acetate results in the formation of ethane via Kolbe's electrolysis.

Option A (CaC2) is correct because calcium carbide (CaC2) reacts with water to produce ethyne (acetylene).

Option B (Benzene) is correct because paraffin (alkanes) and benzene are both non-polar substances, and \"like dissolves like.\"

Option C (${\mathrm{H}_2}$) is correct because sodium formate, when heated with soda lime, undergoes decarboxylation to produce hydrogen gas.

Option D (Both (b) and (c)) is correct because petroleum is formed from the chemical changes in both vegetable and animal matter over millions of years.

Option C ($C{l^ + }$) is correct because the electrophile in the chlorination of benzene is the chloronium ion ($C{l^ + }$).

Option A (Pitch) is correct because pitch is a viscous, black, thermoplastic material derived from coal tar or petroleum and is used as a binder in road construction.

Option B ($\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}-\mathrm{CH} = \mathrm{C}{\mathrm{H}_2} + \mathrm{HBr}$) is correct because Kharasch effect (anti-Markovnikov addition) operates with HBr in the presence of peroxides.

Option B (1.08) is correct because the typical C-H bond length in benzene is approximately 1.08 Å.

Option A (2 – methyl – 2 - butene) is correct because ozonolysis of 2-methyl-2-butene yields acetaldehyde and acetone.

Option B (Alkanes) is correct because kerosene is a mixture of alkanes, primarily those with carbon chain lengths between C11 and C16.

Option A (n-hexane) is correct because Kolbe's electrolysis of sodium butanoate results in the formation of n-hexane.

Option D (Planar form) is correct because cyclohexatriene, also known as benzene, has a planar structure due to resonance and sp2 hybridization.

Option C (Elimination reaction) is correct because alkyl halides are converted to alkenes through dehydrohalogenation, which is an elimination reaction.

Option D (Coal gas) is correct because coal gas is one of the products obtained when coal undergoes destructive distillation.

Option D (All of the above are correct) is correct because all the given reactions involve the removal of a molecule from the substrate, which is the defining characteristic of an elimination reaction.

Option C is correct because butanoyl chloride with AlCl3 followed by Zn, HCl will result in butylbenzene, not 2-phenylbutane, due to acylation followed by reduction.

Option B (Acetylene) is correct because acetylene has the most acidic hydrogen due to the sp hybridization of the carbon atoms, resulting in a higher s-character and greater electronegativity.

Option C (${C_6}{H_6}C{l_6}$) is correct because Gammaxane, also known as Lindane, is the gamma isomer of benzene hexachloride, with the formula ${C_6}{H_6}C{l_6}$.

Option B ($\mathrm{P}{\mathrm{H}_3}$) is correct because the garlic smell in impure acetylene is due to the presence of phosphine ($\mathrm{P}{\mathrm{H}_3}$) as an impurity.

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

Option C (Thioalcohols) is correct because thioalcohols (mercaptans) have a strong, unpleasant odor and are added to LPG to detect leaks.

The correct answer is D (2,2,3-trimethyl pentane) because it is the only option that contains primary, secondary, tertiary, and quaternary carbon atoms.

Option C ($CH_4$) is correct because the Wurtz reaction involves the coupling of two alkyl halides to form a higher alkane, and methane cannot be formed by this method.

Option C (Glucose) is correct because acetylene is not directly used in the production of glucose.