Hydrocarbons practice · Easy

Option A (Detection of double bonds) is correct because Baeyer's reagent (cold, dilute alkaline KMnO4) oxidizes alkenes, causing the purple color to disappear.

The correct answer is C ($\mathrm{CO} + 3{\mathrm{H}_2}$) because synthetic gas, also known as syngas, is a mixture primarily composed of carbon monoxide (CO) and hydrogen ($H_2$). The ratio can vary, but the option closest to the typical composition is CO + 3H2.

Option D (Saturated hydrocarbon) is correct because paraffin wax consists of alkanes, which are saturated hydrocarbons.

Option B (${\mathrm{C}_n}{\mathrm{H}_{2n}}$) is correct because it represents the general formula for cycloalkanes, which have a ring structure and thus two fewer hydrogen atoms than alkanes.

Option C (6) is correct because in benzene, each π-electron is delocalized over all six carbon nuclei.

Option B (50) is correct because the energy of a $\pi$-bond is approximately 50 kcal.

The correct answer is C (Bromine water) because bromine water is used to test for unsaturation. The disappearance of the bromine color indicates the presence of a double or triple bond.

Option B (acetylene) is correct because calcium carbide reacts with water to produce acetylene ($C_2H_2$) and calcium hydroxide.

Option A (Activating groups) is correct because $o,p-$directing groups are generally activating groups.

Option B (90) is correct because the cetane number represents the percentage of n-hexadecane in the mixture, which is 100 - 10 = 90.

Option C (Alkyne>alkene>alkane) is correct because alkynes have stronger intermolecular forces due to their linear structure and greater polarizability compared to alkenes and alkanes.

Option D (Heat/photochemical substitution) is correct because alkanes undergo substitution reactions in the presence of heat or light due to their non-polar nature.

Option B (4) is correct because the formula for a saturated hydrocarbon (alkane) is ${\mathrm{C}_n}{\mathrm{H}_{2n + 2}}$, and solving $2n + 2 = 10$ gives $n = 4$.

Option B (0) is correct because Sc3+ has no unpaired electrons, resulting in a zero magnetic moment.

Option B (Baeyer test) is correct because the disappearance of the purple color of KMnO4 in its reaction with an alkene is known as the Baeyer test, which is used to test for unsaturation.

Option A ($\mathrm{Pd} - \mathrm{CaC}{\mathrm{O}_3}$ deactivated by lead acetate) is the correct definition of Lindlar's catalyst, used for the partial hydrogenation of alkynes to alkenes.

Option D (All of these) is correct because oxides of Al, Cr, and Mo can be used as catalysts in cracking.

Option B ($\mathrm{C}_2 \mathrm{H}_6$) is correct because the product obtained at the anode during Kolbe's electrolysis of sodium ethanoate is ethane.

The correct answer is C (2, 3-dimethylbutane) because it forms three monochloro derivatives due to its symmetry.

The correct answer is D. The major product is determined by Saytzeff's rule, which favors the more substituted alkene.

Option B (75) is correct because the octane number is defined as the percentage of iso-octane in the fuel mixture.

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

Option C (Ethane) is correct because it is a saturated compound with no pi bonds and therefore does not react with bromine solution.

Option B (chloronium ion) is correct because the chlorination of benzene involves the electrophilic attack of the chloronium ion ($Cl^+$) on the benzene ring.

Option D is correct because ${\mathrm{C}_6}{\mathrm{H}_{12}}$ can be hexene-3 or 2,3-dimethylbut-2-ene (symmetrical alkenes), and addition of HBr can give different products depending on the presence or absence of peroxide.

Option D (Anti-termite) is correct because chlorodane is a chlorohydrocarbon specifically used as an anti-termite agent.

Option A ($NaOH$) will not react with acetylene because it is a strong base and acetylene is a weak acid. Acetylene requires a stronger base than NaOH to deprotonate it.

Option D (Zero) is correct because BHC (Benzene hexachloride or gammexane) is a saturated cyclic compound with no double bonds.

Option B ($\left( {4n + 2} \right)\pi$-electrons) is correct because Huckel's rule states that aromatic compounds must possess $\left( {4n + 2} \right)\pi$-electrons.

Option B (Ethane) is correct because decarboxylation removes a COONa group, resulting in a hydrocarbon with one less carbon atom.