Hydrocarbons practice · Mixed difficulty

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 B ($\mathrm{HgO }\;\mathrm{or}\;\mathrm{ HI}{\mathrm{O}_3}$) is correct because it acts as a scavenger for HI, preventing the reverse reaction and promoting iodination of alkanes.

Option A (2 equivalents of $NaN{H_2}$, excess of ${C_2}{H_5}I$) is correct because it involves deprotonation of the terminal alkyne followed by alkylation with ethyl iodide to form Hex-3-yne.

Option A is correct because when acetylene is passed through a red hot iron tube, it polymerizes to form benzene. Benzene can be obtained from phenol by distillation with zinc dust.

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

Option D ($n$-propyl bromide) is correct because the reaction of HBr with propene in the presence of peroxides follows anti-Markovnikov's rule.

Option A ($\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu}$) is correct because acetylene reacts with $\mathrm{Cu}_2 \mathrm{Cl}_2$ in $\mathrm{NH}_4 \mathrm{Cl}$ to form copper acetylide.

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

Option D ($\mathrm{H}_3 \dot{\mathrm{C}}+\dot{\mathrm{Cl}} \longrightarrow \mathrm{CH}_3 \mathrm{Cl}$) is correct because termination steps involve the combination of two free radicals to form a stable molecule, thus ending the chain reaction.

Option D is correct because alkynes are generally less reactive than alkenes towards electrophilic addition due to the sp hybridization of the carbon atoms in the triple bond, which makes them more electronegative and less likely to donate electrons. Also, the carbocation formed in alkene is more stable than the vinyl carbocation formed in alkyne.

Option B is correct because the vigorous oxidation of 2-butene ($\mathrm{CH_3CH=CHCH_3}$) with KMnO4 cleaves the double bond, resulting in two molecules of acetic acid ($\mathrm{CH_3COOH}$).

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

Option D (n - pentane) is correct because, among the given options, n-pentane is the straight-chain alkane with the highest molecular weight, leading to the strongest van der Waals forces and thus the highest boiling point.

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 B is correct because the ozonolysis of an internal alkyne followed by hydrolysis yields carboxylic acids. The triple bond is cleaved, and each carbon of the triple bond becomes a carboxylic acid carbon.

Option D (Meso-2,3 Dibromobutane) is correct because the syn addition of hydrogen followed by anti addition of bromine to but-2-yne leads to the formation of the meso compound.

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 B ($Pd/BaS{O_4}$ or $CaC{O_3}$ in quinoline or xylene) is correct because Lindlar's catalyst is palladium supported on barium sulfate or calcium carbonate, poisoned with quinoline or xylene.

Option A (1,4-dibromo-2-butene) is correct because the reaction of 1,3-butadiene with $Br_2$ at higher temperatures favors 1,4-addition due to thermodynamic control.

Option A is correct because a hydrocarbon with the molecular formula C5H10 can be a monosubstituted, disubstituted, or trisubstituted alkene. This depends on the arrangement of the carbon atoms and the position of the double bond.

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

Option C (Acetone) is correct because propyne undergoes hydration in the presence of $H_2SO_4$ and $HgSO_4$ to form acetone via enol intermediate.

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 (${Al}_4 {C}_3, {Al}({OH})_3$) is correct because aluminum carbide ($Al_4C_3$) reacts with water to produce methane ($CH_4$) and aluminum hydroxide (${Al}({OH})_3$).