HydrocarbonsNEET MCQs with solutions

Not quite — the answer is D.

Peroxide effect enables free-radical addition of HBr, giving anti-Markovnikov product. Br• adds to the terminal carbon of propene, generating a more stable secondary carbon radical at C2. Subsequent H• addition to C2 gives 1-bromopropane (n-propyl bromide). Option A is the Markovnikov product; Option B is a vinyl bromide formed by HBr elimination rather than addition; Option C is allyl bromide, a different compound.

Not quite — the answer is D.

Ozonolysis cleaves the C=C bond and converts each carbon into a carbonyl group. In 2-methyl-2-butene, one double-bond carbon bears two methyl groups and gives propanone (acetone); the other bears one methyl and one H and gives ethanal. 3-Methyl-1-butene gives methanal and 3-methylbutanal; cyclopentene gives pentanedial; 2-methyl-1-butene gives methanal and butanone.

Not quite — the answer is B.

Modern chemistry represents benzene as a resonance hybrid — not a single Kekulé structure with alternating single and double bonds. All C–C bonds are equivalent (139 pm), intermediate between single (154 pm) and double (134 pm). Dewar benzene is a real but high-energy isomer, not the ground-state structure.

Not quite — the answer is B.

Aromaticity requires four simultaneous conditions: cyclic, planar, fully conjugated (continuous p-orbital overlap), and (4n+2) π electrons (Hückel's rule, n = 0,1,2...). Option C describes anti-aromatic compounds; Option D omits conjugation and electron count.

Not quite — the answer is A.

Staggered conformation places H atoms at maximum 60° dihedral angle, minimizing torsional strain and giving lowest potential energy. Eclipsed has maximum H-H repulsion. "Anti-eclipsed" is not a standard conformation — staggered anti is the actual low-energy form.

Not quite — the answer is A.

Isobutane has a 3-carbon longest chain with a methyl group at C2, giving 2-methylpropane. "1-methylpropane" violates the lowest locant rule; "2-methylbutane" is isopentane, not isobutane.

Not quite — the answer is B.

Alkene carbons form one C=C (one σ + one π) and two other σ bonds — three groups total, requiring sp² hybridisation. sp³ is for alkanes; sp is for alkynes.

Not quite — the answer is B.

sp² hybridisation produces three equivalent orbitals at 120° in a plane, giving trigonal planar geometry. Tetrahedral (109.5°) is sp³; linear (180°) is sp.

Not quite — the answer is A.

Soda-lime decarboxylation uses a mixture of NaOH and CaO (soda lime) heated with the sodium salt. Sodium propionate (CH3CH2COONa) loses its carboxyl carbon as carbonate and gives ethane (CH3CH3). NaNH2 is a base for alkyne deprotonation; H2/Ni is for hydrogenation.

Not quite — the answer is C.

The Wurtz reaction gives best yields with symmetrical coupling of identical alkyl halides. n-Heptane (C7) requires coupling two different chain lengths, producing a mixture from the crossed Wurtz reaction. n-Butane, n-hexane, and 2,3-dimethylbutane can each be made from a single identical halide symmetrically.

Not quite — the answer is D.

Only a terminal alkyne contains an acidic hydrogen on an sp-hybridised carbon. But-1-yne is a terminal alkyne and forms an insoluble silver acetylide precipitate with ammoniacal AgNO3. But-2-yne is an internal alkyne with no terminal hydrogen. But-2-ene and propene are alkenes and do not form acetylides.

Not quite — the answer is C.

But-2-ene (CH3CH=CHCH3) is symmetrical about its double bond. Reductive ozonolysis produces two identical ethanal molecules so only one carbonyl compound is formed. Propene gives ethanal and methanal. 2-Methylpropene gives propanone and methanal. 2-Methylbut-2-ene gives propanone and ethanal.

Not quite — the answer is B.

Alkenes contain π bonds that react readily with NOₓ and free radicals under UV light to form smog components. Alkanes lack π bonds and are far less reactive. Methane (an alkane) is a trap — reactive in smog but far less so than alkenes.

Not quite — the answer is C.

Photochemical smog forms when unburnt hydrocarbons react with NOₓ in the presence of UV sunlight, producing ozone, PAN, and other oxidants. SO₂ + water vapour produces classical (sulphurous) smog, not photochemical smog.

Not quite — the answer is A.

Alkanes contain only C–C and C–H sigma bonds with no pi bonds, making them fully saturated. Alkenes and alkynes contain pi bonds (unsaturated). Aromatic hydrocarbons have delocalized pi electrons and are classified separately from saturated compounds.

Not quite — the answer is B.

Alkenes are unsaturated hydrocarbons defined by the presence of at least one C=C double bond. The key trap is cycloalkanes: they share the general formula CₙH₂ₙ with alkenes but contain only single bonds and are saturated.

Not quite — the answer is A.

Carbon atoms in alkynes form two sigma bonds and two pi bonds using sp hybrid orbitals with 50% s-character. sp² gives trigonal planar geometry (alkenes); sp³ gives tetrahedral (alkanes).

Not quite — the answer is B.

sp hybridisation produces two bond pairs oriented 180° apart, giving a strictly linear geometry. Trigonal planar (120°) belongs to sp² hybridised systems such as alkenes.

Not quite — the answer is C.

Alkynes are defined by at least one C≡C triple bond (one sigma + two pi bonds). Cycloalkenes are a trap — they are unsaturated but contain double, not triple, bonds. Cycloalkanes are fully saturated.

Not quite — the answer is D.

Aromaticity requires a cyclic, planar system with fully delocalized pi electrons satisfying Hückel's rule (4n+2). Option C is the classic trap — cyclic alternating double bonds alone (as in hypothetical cyclobutadiene) do not confer aromaticity without satisfying the electron count rule.

Not quite — the answer is A.

Aliphatic hydrocarbons are open-chain (acyclic) compounds. Propane is acyclic. Benzene and toluene are aromatic. Cyclohexane is the key trap — it is cyclic but alicyclic (not aromatic), yet it is still NOT open-chain, so it is not aliphatic.

Not quite — the answer is B.

Cyclohexane is a cyclic saturated hydrocarbon (alicyclic) with no pi electrons. Benzene is aromatic with 6 delocalized pi electrons satisfying Hückel's rule (n=1). The terms alicyclic and aromatic are NOT interchangeable for cyclic compounds.

Not quite — the answer is C.

Cyclohexane has a closed ring of 6 carbons, making it cyclic (alicyclic). All other options — propane, butane, pentane — are straight-chain alkanes with open-chain (acyclic) structures.

Not quite — the answer is B.

DoU = (2×4 + 2 − 6)/2 = (8+2−6)/2 = 4/2 = 2. A DoU of 2 indicates either two double bonds, one triple bond, or one double bond plus one ring. Option A (DoU=1) is the trap for students who miscalculate by not applying the formula correctly.