Hydrocarbons practice · Easy

Option C (Butanoyl chloride + $\mathrm{AlCl}_3$ then $\mathrm{Zn}, \mathrm{HCl}$) is correct because it undergoes Friedel-Crafts acylation followed by reduction, resulting in a straight chain alkyl group instead of 2-phenylbutane.

Option D (Heat/photochemical substitution) is correct because alkanes are saturated hydrocarbons and undergo substitution reactions under heat or photochemical conditions.

Statement I is correct but statement II is incorrect. 2-methylbutane can be oxidized to 2-methylbutan-2-ol with $KMnO_4$, but n-alkanes are not easily oxidized to alcohols with $KMnO_4$.

Option C (both (1) and (2) are correct) is correct because the addition of HBr to 1-butyne in the presence of peroxide follows anti-Markovnikov's rule via a free radical mechanism.

Option A (n-hexane) is correct because Kolbe's electrolysis of butanoic acid's sodium salt results in the formation of a symmetrical alkane with double the carbon atoms.

Option D (All of these) is correct because ethylene is used in the production of anti-freeze, as a solvent, and as a fumigant.

Option C ($C{H_4}$) is correct because Kolbe's electrolytic method involves the electrolysis of a salt of a carboxylic acid, leading to the formation of an alkane with an even number of carbon atoms or alkenes/alkynes. Methane cannot be formed.

Option B (Gilman reagent) is correct because ${R_2}CuLi$ is the general formula for a Gilman reagent, used in the Corey-House synthesis.

Option A (Passed through ammoniacal $\mathrm{AgNO_3}$) is correct because terminal alkynes react with ammoniacal silver nitrate to form a precipitate, indicating the presence of a triple bond at the end of the chain.

Option B (a secondary alkyl bromide) is correct because the reaction proceeds via a free radical mechanism, and the bromine adds to the less substituted carbon, resulting in a secondary alkyl bromide.

Option B ($CH_3CH_2C \equiv CH$) is correct because terminal alkynes react with Tollens' reagent ($Ag(NH_3)_2^+ OH^-$) to form a silver acetylide precipitate.

Option B is correct because both the assertion and the reason are true, but the reason is not the correct explanation of the assertion. The volume of gases evolved can be calculated, and the mechanism does involve free radicals.

Option A ($\mathrm{Ca}{\mathrm{C}_2}$) is correct because calcium carbide reacts with water to produce ethyne (acetylene).

Option D (1 – bromo – 2 – methyl propane) is correct because in the presence of peroxide, HBr adds to alkenes via an anti-Markovnikov mechanism, where the bromine atom attaches to the less substituted carbon.

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

Option C (n-propyl bromide) is correct because the Wurtz reaction of n-propyl bromide with sodium metal will yield n-hexane.

Option B (neo pentane) is correct because neo-pentane has all its hydrogens equivalent, leading to only one possible monochloro derivative.

Option B (${\mathrm{C}_2}{\mathrm{H}_2}$) is correct because Narcylene is the commercial name for acetylene, which was formerly used as an anesthetic.

Option B (${\mathrm{C}_2}{\mathrm{H}_2}$) is correct because acetylene is used for the artificial ripening of fruits.

Option B ($n$-propane) is correct because the Wurtz reaction of ethyl iodide and n-propyl iodide will produce n-butane, n-pentane, and n-hexane, but not n-propane.

Option D ($n$-propyl bromide) is correct because the reaction of HBr with propene in the presence of peroxide follows anti-Markovnikov's rule, leading to the addition of bromine to the less substituted carbon.

Option C ($\mathrm{Zn},alcohol/\Delta$) is correct because zinc in alcohol is used to dehalogenate vicinal dihalides to form alkenes.

Option B ($\mathrm{C}{\mathrm{H}_2}\mathrm{ClC}{\mathrm{H}_3}$) is correct because the reaction of ethene with HCl follows Markovnikov's rule, where hydrogen adds to the carbon with more hydrogens.

Option B (Elimination reaction) is correct because alkyl halides undergo dehydrohalogenation (elimination of HX) to form 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 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 ($I > II > III$) 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.