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Class XI · Chemistry

Hydrocarbons: NEET MCQ practice

349 practice questions on Hydrocarbons, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

349

Practice questions

145

Easy

182

Medium

22

Hard

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What Hydrocarbons covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Classification, isomerism, IUPAC nomenclature, general methods of preparation, properties and reactions
  • Alkanes- conformations (Sawhorse and Newman projections of ethane), mechanism of halogenation
  • Alkenes- geometrical isomerism, mechanism of electrophilic addition, ozonolysis and polymerization
  • Alkynes- acidic character, addition reactions, polymerization
  • Aromatic hydrocarbons- nomenclature, benzene structure and aromaticity, electrophilic substitution
  • Friedel-Crafts alkylation and acylation, directive influence of functional group in mono-substituted benzene

Sample questions on Hydrocarbons

Try each one, then open the answer and explanation.

Question 1 · Practice question

Racemic mixture is obtained due to the halogenation of

  1. A.n-pentane
  2. B.isopentane
  3. C.neopentane
  4. D.Both (a) and (b)
Show answer

Answer: D. Both (a) and (b)

Halogenation of alkanes via free-radical substitution can create new chiral centers if the carbon atom after substitution has four different substituents. In n‑pentane, substitution at the 2‑ or 3‑position yields a chiral carbon, giving a racemic mixture. In isopentane (2‑methylbutane), substitution at the CH₂ group (not the tertiary carbon) also produces a chiral center, leading to a racemic mixture. Neopentane has only equivalent primary carbons; no chiral center can form, so it does not give a racemic mixture.

Question 2 · Practice question

Addition of bromine on propene in the presence of brine yields a mixture of

  1. A.CH₃CHCICH₂Br and CH₃CHBrCH₂CI
  2. B.CH₃CHCICH₂CI and CH₃CHBrCH₂Br
  3. C.CH₃CHCICH₂Br and CH₃CHBrCH₂Br
  4. D.CH₃CHCICH₂Cl and CH₃CHBrCH₂Cl
Show answer

Answer: C. CH₃CHCICH₂Br and CH₃CHBrCH₂Br

In the presence of brine (NaCl solution), bromine adds to propene via a bromonium ion intermediate, which can be opened by either Br⁻ or Cl⁻ ions. This yields both 1,2‑dibromopropane (from Br⁻ attack) and 1‑bromo‑2‑chloropropane (from Cl⁻ attack). Option (c) includes both these products, while other options either omit the dibromo product or wrongly suggest dichloro products.

Question 3 · Practice question

One or More than One Options Correct Type Direction (Q. Nos. 8-12) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct. Q. Which of the following reagents can bring about free radical chlorination of propane?

  1. A.SOCI₂
  2. B.SO₂CI₂
  3. C.none
  4. D.PCI₃
Show answer

Answer: B. SO₂CI₂

SO2Cl2 decomposes to chlorine radicals on heating or photolysis, which can abstract a hydrogen from propane and proceed via free‑radical chain substitution. The other reagents (SOCl2 and PCl3) give ionic substitution, not free‑radical chlorination, so only option (b) is correct.

Question 4 · Practice question

One Integer Value Correct Type Direction (Q, Nos. 16-19) This section contains 4 questions. When worked out will result in an integer from 0 to 9 (both inclusive). Q. If 2,4-dimethyl pentane is subjected to free radical chlorination reaction, how many different monochlorinated products would be formed?

  1. A.4
  2. B.344
  3. C.5423
  4. D.1427
Show answer

Answer: A. 4

The structure of 2,4-dimethylpentane has four distinct types of hydrogen atoms (terminal methyl, branch methyl, tertiary H on C2/C4, and secondary H on C3), giving four possible monochlorinated products. Common mistakes include missing the branch methyl hydrogens as a separate type or miscounting symmetry, but careful environment analysis confirms exactly four unique positions.

Question 5 · Practice question

On free radical chlorination reaction of butane, how many different, optically active, dichloroalkanes would be formed?

  1. A.6
  2. B.346
  3. C.5623
  4. D.1627
Show answer

Answer: A. 6

In free radical chlorination of butane, both n-butane and isobutane yield dichloro substitution products. Only n-butane gives optically active (chiral) dichloroalkanes: 1,2-dichlorobutane and 1,3-dichlorobutane each have one chiral centre (giving one pair of enantiomers), while 2,3-dichlorobutane has two chiral centres but only the (R,R) and (S,S) forms are optically active, adding another enantiomeric pair. Counting each enantiomer separately gives 2 + 2 + 2 = 6 optically active stereoisomers. Isobutane yields no chiral dichloro derivatives due to symmetry.

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