Hydrocarbons practice · Mixed difficulty

Option C ($\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2$) is correct because the terminal carbons of allene will be oxidized to CO2 upon reductive ozonolysis.

Option B (but – 2 - ene) is correct because but-2-ene is a symmetrical alkene, and the addition reaction follows Markownikoff's rule regardless of the presence of peroxides.

The correct answer is ${C_2}{H_5}OH$ because calcium carbide reacts with water to form ethyne, which is then converted to ethanol.

Option B (${\mathrm{C}_2}{\mathrm{H}_6}$) is correct because the combustion of ethane produces twice the volume of CO2 as the volume of ethane taken.

Option A ($iii > ii > i$) is correct because phenol is more reactive than toluene, which is more reactive than benzene towards electrophilic reagents.

Option D is correct because a solution of bromine in carbon tetrachloride ($\mathrm{Br_2}$ in $\mathrm{CCl_4}$) is a common reagent used to test for unsaturation in alkenes. The bromine adds across the double bond, causing the reddish-brown color of the bromine solution to disappear.

Option B (${C_2}{H_4},{H_2}$ and $C{H_4}$) is correct because cracking of ethane produces a mixture of ethene, hydrogen, and methane.

Option A (Friedel-Crafts reaction) is correct because the reaction of benzene with methyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.

Option C (${\mathrm{C}_3}{\mathrm{H}_8}$) is correct because the complete combustion of propane produces 3 moles of CO2, matching the problem statement.

Option A (decreased surface area for intermolecular attraction) is correct because branching in alkanes reduces the surface area available for van der Waals interactions, leading to lower boiling points.

Option B (Ethane) is correct because ethane has only one type of hydrogen atom, leading to only one possible monochloro product upon chlorination.

Option D (neo-pentane) is correct because all hydrogen atoms in neo-pentane are equivalent, leading to only one possible monochloro derivative.

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

Option C ($2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}$) is correct because hot $\mathrm{KMnO}_4 / \mathrm{NaIO}_4$ oxidizes 1,3-butadiene to glyoxal and then further oxidizes glyoxal to oxalic acid and formaldehyde to carbon dioxide.

Option B (combustion) is correct because incomplete combustion of organic materials leads to the formation of polynuclear hydrocarbons.

Option B ($-\mathrm{NO}_2,-\mathrm{CHO},-\mathrm{SO}_3 \mathrm{H},-\mathrm{COR}$) is correct because all these groups are electron-withdrawing and meta-directing due to resonance effects.

Option B (trans-2-butene $>$ cis-2-butene $>1$-butene) is correct because trans-2-butene is more stable than cis-2-butene due to reduced steric hindrance, and both are more stable than 1-butene due to greater alkyl substitution on the double bond.

Option B ($R - C{H_2} - C{H_2} - OH$) is correct because hydroboration-oxidation of alkenes results in anti-Markovnikov addition of water, yielding an alcohol.

Option A (Light oil) is correct because benzene is primarily obtained from the light oil fraction during the fractional distillation of coal tar.

Option A ($\mathrm{C}{\mathrm{H}_3}\mathrm{I}$) is correct because the reaction of methyl iodide with sodium metal in the Wurtz reaction produces both methane (via reduction) and ethane.

The correct answer is $\mathrm{NaN}{\mathrm{H}_2}$ because it can deprotonate 1-butyne but not 2-butyne.

Option D (Iso-hexane) is correct because Iso-hexane yields four mono chloro derivatives.

Option D (acetylene) is correct because calcium carbide ($\mathrm{CaC}_2$) reacts with water to produce acetylene ($\mathrm{C}_2\mathrm{H}_2$).

Option D (A mixture of 2 – methyl pentane & 3 – methyl pentane) is correct because isomerization of n-hexane in the presence of anhydrous $AlCl_3$ and $HCl$ yields a mixture of branched isomers.

Option A (${\mathrm{C}_2}{\mathrm{H}_4}$) is correct because the reaction of ethyl chloride with alcoholic KOH leads to dehydrohalogenation, resulting in the formation of ethene.

Option A ($55.5 \mathrm{~L}$) is correct because it is the calculated volume of ethane required considering the yields of bromination and Wurtz reaction.

Option D (1,3-butadiene) is correct because conjugated dienes like 1,3-butadiene are more stable due to resonance, thus having the smallest heat of hydrogenation per mole.

Option D ($- CN$) is correct because the cyano group (-CN) is a meta-directing group due to its electron-withdrawing nature.

Option C (sodium fumarate) is correct because Kolbe's electrolysis of sodium fumarate yields an alkyne (acetylene).

Option D (1,4-dibromobutane) is correct because the reaction of 1,4-dibromobutane with sodium metal forms a cyclic alkane (cyclobutane) via intramolecular Wurtz reaction.