NEET Hydrocarbons PYQs & Practice Questions

10 free NEET Chemistry questions on Hydrocarbons, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Geometrical Isomerism · Preparation of Benzene · Boiling Points of Isomeric Alkanes · Reactivity of Halogens with Alkanes · Reduction of Alkyl Halides · Reactivity of Hydrogen Halides with Alkenes · C-H Bond Dissociation Energy / Radical Stability · Reaction with Bromine Water · Conjugated Dienes · Alkyne Reduction Stereochemistry (Lindlar vs Birch/Dissolving Metal) · IUPAC Nomenclature · Electrophilic Addition Reactivity

1. Q1NEET 2026Conjugated Dienes

   Among the following, the compound having conjugated double bonds is:

   • A. hepta-1,4-diene
   • B. hepta-1,5-diene
   • C. hepta-1,6-diene
   • D. hepta-1,3-diene
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   Answer: C — hepta-1,6-diene

   Conjugated double bonds are two double bonds separated by exactly one single bond. In hepta-1,3-diene, the double bonds are at C1-C2 and C3-C4, separated by the single bond C2-C3, making them conjugated. In the other options, the double bonds are separated by more than one single bond (isolated, not conjugated).

2. Q2NEET 2026Alkyne Reduction Stereochemistry (Lindlar vs Birch/Dissolving Metal)

   Consider the following reaction sequences starting from Ph-C≡C-Me (1-phenylpropyne): Path 1: Ph-C≡C-Me --[Na/liq. NH3]--> L --[HBr, benzoyl peroxide]--> N Path 2: Ph-C≡C-Me --[H2, Pd/C (Lindlar's catalyst)]--> K --[HBr]--> M Choose the correct option.

   • A. K and L are enantiomers
   • B. M and N are geometrical isomers
   • C. M and N are stereoisomers
   • D. K and L are geometrical isomers
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   Answer: C — M and N are stereoisomers

   Na/liq.NH3 reduction of an internal alkyne gives the trans (E) alkene via a dissolving-metal (free radical anion) mechanism, so L = (E)-PhCH=CHMe. Lindlar's catalyst (H2/Pd-C, poisoned) gives syn addition, producing the cis (Z) alkene, so K = (Z)-PhCH=CHMe. K and L are thus the cis and trans isomers of the same compound, i.e., geometrical isomers of each other.

3. Q3NEET 2025Geometrical Isomerism

   Which one of the following compounds can exist as cis-trans isomers?

   • A. Pent-1-ene
   • B. 2-Methylhex-2-ene
   • C. 1,1-Dimethylcyclopropane
   • D. 1,2-Dimethylcyclohexane
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   Answer: C — 1,1-Dimethylcyclopropane

   For cis-trans isomerism in a cyclic compound, each ring carbon bearing substituents must have two different groups attached. In 1,2-dimethylcyclohexane, each substituted carbon bears a CH3, an H, and two different ring paths, allowing cis (same side) and trans (opposite side) arrangements. Pent-1-ene (terminal alkene) and 1,1-dimethylcyclopropane (geminal substitution) cannot show cis-trans isomerism; 2-methylhex-2-ene has two identical groups on one carbon.

4. Q4NEET 2025Preparation of Benzene

   Which one of the following reactions does NOT give benzene as the product? (1) sodium benzoate + sodalime, Δ; (2) n-hexane with Mo2O3 catalyst at 773K, 10-20 atm; (3) acetylene (HC≡CH) through red hot iron tube at 873K; (4) benzene diazonium chloride with H2O, warm

   • A. sodium benzoate + sodalime, Δ
   • B. n-hexane with Mo2O3 catalyst (aromatization)
   • C. acetylene through red hot iron tube (cyclic trimerization)
   • D. benzene diazonium chloride + H2O, warm
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   Answer: C — acetylene through red hot iron tube (cyclic trimerization)

   Reactions (1), (2), (3) all produce benzene: decarboxylation of sodium benzoate, catalytic aromatization/dehydrogenation of n-hexane, and cyclic trimerization of acetylene. Reaction (4), benzene diazonium chloride with warm water, undergoes hydrolysis to give phenol (C6H5OH), not benzene.

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5. Q5NEET 2025C-H Bond Dissociation Energy / Radical Stability

   Among the given compounds I-III, the correct order of bond dissociation energy of the C-H bond marked with * is: I. Phenyl C-H (aromatic), II. Phenylacetylene terminal C-H (Ph-C≡C-H), III. Cyclopropyl C-H.

   • A. II > I > III
   • B. I > II > III
   • C. III > II > I
   • D. II > III > I
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   Answer: —

   Bond dissociation energy relates inversely to the stability of the radical formed. sp-hybridized vinyl/alkynyl radical (II, from Ph-C≡C-H) is the least stable, so its BDE is highest. The aryl radical (I) is also quite unstable (high BDE) due to sp2 character, more stable than the alkynyl one's radical being even less stable makes BDE(II) highest. Cyclopropyl radical (III) is comparatively more stabilized (sp3, with some s-character effects), giving lowest BDE. Order: II > I > III.

6. Q6NEET 2025Reaction with Bromine Water

   Which one of the following compounds does not decolourize bromine water? (1) Cyclohexane, (2) Phenol, (3) Styrene (C6H5-CH=CH2), (4) Aniline (C6H5-NH2)

   • A. Cyclohexane
   • B. Phenol
   • C. Styrene
   • D. Aniline
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   Answer: —

   Cyclohexane is a saturated cyclic hydrocarbon with no π bond, so it does not react with (decolourize) bromine water. Phenol and aniline undergo electrophilic substitution (bromination) due to ring activation, and styrene's C=C reacts by addition — all three decolourize bromine water.

7. Q7NEET 2022Aromaticity

   Which compound amongst the following is not an aromatic compound? (1) Cycloheptatrienyl cation (tropylium cation, planar, 6 pi electrons) (2) Cyclopropenyl cation (planar, 2 pi electrons) (3) Cycloheptatriene with an exocyclic sp3 carbon bearing a positive charge substituent (non-planar at that carbon) (4) Cyclopentadienyl anion (planar, 6 pi electrons)

   • A. Cycloheptatrienyl cation (tropylium cation)
   • B. Cyclopropenyl cation
   • C. Compound with sp3 hybridized carbon bearing substituent, making ring non-planar
   • D. Cyclopentadienyl anion
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   Answer: B — Cyclopropenyl cation

   The compound in option 3 has an sp3 hybridized carbon in the ring (not fully conjugated/planar), so it is cyclic but not planar, and therefore cannot be aromatic despite being cyclic. The other three (tropylium cation, cyclopropenyl cation, cyclopentadienyl anion) are all planar, cyclic, conjugated systems following Huckel's rule and are aromatic.

8. Q8NEET 2022Ozonolysis - Identifying Alkene

   Compound X on reaction with O3 followed by Zn/H2O gives formaldehyde and 2-methyl propanal as products. The compound X is:

   • A. 3-Methylbut-1-ene
   • B. 2-Methylbut-1-ene
   • C. 2-Methylbut-2-ene
   • D. Pent-2-ene
   Show answer

   Answer: —

   Ozonolysis cleaves the double bond, with each carbon becoming a carbonyl. Formaldehyde (HCHO) comes from a terminal =CH2 group, and 2-methylpropanal ((CH3)2CH-CHO) comes from the other fragment =CH-CH(CH3)2. Combining these fragments reconstructs the alkene 3-methylbut-1-ene: CH2=CH-CH(CH3)-CH3... actually (CH3)2CH-CH=CH2, which is 3-methylbut-1-ene.

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9. Q9NEET 2021IUPAC Nomenclature

   The correct structure of 2,6-Dimethyl-dec-4-ene is: (four structural options with differing double bond and methyl group positions on a decane chain)

   • A. Structure with double bond between C3-C4 and methyl branches misplaced
   • B. Structure with double bond at C4-C5, methyl groups at C2 and C6 of the decane chain (correctly matching 2,6-dimethyldec-4-ene)
   • C. Structure with double bond at C7-C8 position
   • D. Structure with double bond at C4-C5 but methyl groups misplaced at different carbons
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   Answer: A — Structure with double bond between C3-C4 and methyl branches misplaced

   Numbering the 10-carbon chain to give lowest locants: methyl groups at C2 and C6, double bond between C4-C5, matching the name 2,6-Dimethyl-dec-4-ene

10. Q10NEET 2020Ozonolysis of alkenes

    An alkene on ozonolysis gives methanal as one of the products. Its structure is:

    • A. Cyclohexenyl-CH=CH-CH3 structure
    • B. Cyclohexenyl-CH2-CH2-CH3 structure (cyclohexane ring with CH2CH2CH3 and a =CH2 group)
    • C. Cyclohexyl-CH2-CH=CH2 structure
    • D. Cyclohexenyl-CH2CH2CH3 structure
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    Answer: A — Cyclohexenyl-CH=CH-CH3 structure

    Ozonolysis cleaves the C=C double bond, and methanal (HCHO) is produced from a terminal =CH2 group. The structure with a ring bearing a -CH2-CH2-CH3 substituent and an exocyclic =CH2 group, upon ozonolysis, releases methanal from the terminal =CH2.