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Hydrocarbons JENPAS MCQ & Practice Questions

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Hydrocarbons Revision Guide for JENPAS Chemistry

Hydrocarbons is part of JENPAS Chemistry preparation. A strong revision plan should combine concept review with exam-style questions so that definitions, relationships, calculations and exceptions are recalled accurately.

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Important Topics

  • Core definitions and principles
  • Important equations or relationships
  • Common applications and examples
  • Frequently confused concepts
  • Exam-style multiple-choice questions
  • Error review and targeted revision

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What to practise in Hydrocarbons

Concept Revision

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Top 20 Hydrocarbons MCQs with Answers

These public questions are selected from the exact Hydrocarbons chapter, screened for topic relevance and deduplicated so repeated versions of the same MCQ are not shown publicly. Use them for quick revision, then sign in to attempt the complete test experience with more questions, results and performance review.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

1. The general formula of acyclic alkanes is:

  • A. CnH2n-2
  • B. CnH2n-6
  • C. CnH2n+2
  • D. CnH2n
Answer: C
Explanation: Open-chain alkanes follow the formula CnH2n+2.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

2. Methane has which geometry?

  • A. Square planar
  • B. Tetrahedral
  • C. Trigonal planar
  • D. Linear
Answer: B
Explanation: Carbon in methane is sp3 hybridised with tetrahedral geometry.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

3. The C-C bond in ethane is a:

  • A. Sigma bond
  • B. Pi bond
  • C. Coordinate bond
  • D. Hydrogen bond
Answer: A
Explanation: A carbon-carbon single bond is a sigma bond.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

4. Wurtz reaction is used mainly to prepare:

  • A. Alkenes
  • B. Alkynes
  • C. Alcohols
  • D. Higher alkanes
Answer: D
Explanation: Alkyl halides couple with sodium in dry ether to give higher alkanes.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

5. Decarboxylation of sodium carboxylate with soda lime gives:

  • A. An alcohol
  • B. An aldehyde
  • C. An alkane with one carbon less
  • D. An alkene with same carbon number
Answer: C
Explanation: Soda-lime decarboxylation removes CO2 and shortens the chain by one carbon.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

6. Methane reacts with chlorine in sunlight by:

  • A. Elimination
  • B. Free-radical substitution
  • C. Electrophilic addition
  • D. Nucleophilic substitution
Answer: B
Explanation: Photochemical chlorination of methane follows a radical chain mechanism.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

7. The first step in free-radical chlorination is:

  • A. Initiation
  • B. Propagation
  • C. Termination
  • D. Elimination
Answer: A
Explanation: Initiation generates chlorine radicals by homolysis of Cl2.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

8. Complete combustion of methane gives:

  • A. CO and H2
  • B. C and H2
  • C. CH3OH
  • D. CO2 and H2O
Answer: D
Explanation: Complete combustion oxidises carbon to CO2 and hydrogen to water.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

9. Cracking of alkanes generally produces:

  • A. Only ketones
  • B. Only aromatic compounds
  • C. Smaller alkanes and alkenes
  • D. Only alcohols
Answer: C
Explanation: Cracking breaks large hydrocarbons into smaller hydrocarbons.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

10. The general formula of acyclic alkenes is:

  • A. CnH2n-6
  • B. CnH2n
  • C. CnH2n+2
  • D. CnH2n-2
Answer: B
Explanation: Acyclic monoalkenes contain one degree of unsaturation and follow CnH2n.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

11. The carbon atoms of the C=C bond in ethene are:

  • A. sp2 hybridised
  • B. sp3 hybridised
  • C. sp hybridised
  • D. dsp2 hybridised
Answer: A
Explanation: Each alkene carbon is trigonal planar and sp2 hybridised.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

12. The C=C bond contains:

  • A. Two sigma bonds
  • B. Two pi bonds
  • C. One sigma bond only
  • D. One sigma and one pi bond
Answer: D
Explanation: A double bond consists of one sigma and one pi bond.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

13. Ethene decolourises bromine solution due to:

  • A. Condensation reaction
  • B. Hydrolysis
  • C. Addition reaction
  • D. Substitution reaction
Answer: C
Explanation: Bromine adds across the double bond.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

14. Hydrogenation of ethene gives:

  • A. Ethanal
  • B. Ethane
  • C. Ethyne
  • D. Ethanol
Answer: B
Explanation: H2 adds across the double bond to form ethane.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

15. A common catalyst for hydrogenation of alkenes is:

  • A. Ni
  • B. NaCl
  • C. H2SO4 only
  • D. CaCO3
Answer: A
Explanation: Nickel catalyses hydrogen addition to C=C bonds.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

16. Hydration of ethene gives:

  • A. Ethanal
  • B. Ethanoic acid
  • C. Ethane
  • D. Ethanol
Answer: D
Explanation: Acid-catalysed addition of water to ethene forms ethanol.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

17. Markovnikov's rule applies to addition of:

  • A. Br2 to benzene
  • B. O2 to methane
  • C. HX to an unsymmetrical alkene
  • D. H2 to ethene only
Answer: C
Explanation: It predicts the regioselectivity of polar addition to unsymmetrical alkenes.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

18. According to Markovnikov's rule, H generally adds to the carbon already bearing:

  • A. No hydrogen atoms
  • B. More hydrogen atoms
  • C. Fewer hydrogen atoms
  • D. A halogen
Answer: B
Explanation: The usual ionic mechanism proceeds through the more stable carbocation.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

19. Peroxide effect is classically observed with:

  • A. HBr
  • B. HCl
  • C. HI
  • D. HF
Answer: A
Explanation: Only HBr gives the standard anti-Markovnikov radical addition under peroxide conditions.

JENPAS Chemistry · Hydrocarbons · Very Very Hard

20. Ozonolysis of an alkene is useful for determining:

  • A. Molar mass only
  • B. Boiling point
  • C. Acidity
  • D. Position of the double bond
Answer: D
Explanation: Carbonyl fragments reveal the location and substitution of C=C.

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Hydrocarbons Practice Tests

Hydrocarbons Frequently Asked Questions

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