NEETAceToday's NEET Daily

Alkenes: Markovnikov/peroxide effect, ozonolysis and polymerization: NEET MCQs with answers

50 questions on this topic of Hydrocarbons (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Hydrocarbons freeAll Hydrocarbons questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. Which of the following alkenes yields only a single ketone upon reductive ozonolysis with O₃ followed by Zn/H₂O?

  1. 2,3-Dimethylbut-2-ene
  2. Hex-3-ene
  3. 2-Methylbut-2-ene
  4. 3-Methylpent-2-ene
Show answer and explanation

Answer: (A) 2,3-Dimethylbut-2-ene

Reductive ozonolysis of symmetrical tetrasubstituted alkene 2,3-dimethylbut-2-ene ((CH₃)₂C=C(CH₃)₂) cleaves the double bond to yield two moles of acetone (propan-2-one) as the exclusive product. Hex-3-ene gives propanal (an aldehyde), while 2-methylbut-2-ene gives a mixture of acetone and ethanal.

medium

2. One mole of an unknown alkene 'X' consumes one mole of ozone. Subsequent treatment with Zn and water gives one mole of acetone and one mole of ethanal. What is the product formed when 'X' is treated with HBr in the presence of an organic peroxide?

  1. 2-Bromo-2-methylbutane
  2. 1-Bromo-2-methylbutane
  3. 2-Bromo-3-methylbutane
  4. 1-Bromo-3-methylbutane
Show answer and explanation

Answer: (C) 2-Bromo-3-methylbutane

The ozonolysis products acetone ((CH₃)₂C=O) and ethanal (CH₃CHO) indicate the original alkene 'X' is 2-methylbut-2-ene, (CH₃)₂C=CH-CH₃. Addition of HBr in the presence of peroxide involves attack of Br• on the less substituted double-bonded carbon (=CH-) to generate the more stable tertiary radical (CH₃)₂C•-CH(Br)-CH₃, leading to 2-bromo-3-methylbutane.

hard

3. Which of the following sets of products is formed when 1-methylcyclopentene undergoes reductive ozonolysis with O₃ followed by treatment with Zn and H₂O?

  1. Hexane-1,5-dione
  2. Heptane-2,6-dione
  3. Heptanedioic acid
  4. 6-Oxoheptanal
Show answer and explanation

Answer: (D) 6-Oxoheptanal

1-Methylcyclopentene contains a double bond between C1 (bearing -CH₃) and C2. Reductive cleavage of this double bond oxidizes C1 to a methyl ketone (-COCH₃) and C2 to an aldehyde (-CHO). Counting the 5 ring carbons plus the methyl group gives 7 carbons in total: CH₃-CO-(CH₂)₃-CHO, which is 6-oxoheptanal.

medium

4. Given below are two statements: Statement I: Addition of HCl to unsymmetrical alkenes does not show the peroxide effect. Statement II: The addition of Cl• radical to the alkene double bond is an endothermic step. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect.

Statement I is correct because only HBr exhibits the peroxide effect. Statement II is incorrect because the attack of Cl• radical on the alkene double bond is actually exothermic; rather, the subsequent step (abstraction of H from HCl by the carbon radical) is strongly endothermic due to the high H-Cl bond energy (430.5 kJ/mol).

easy

5. Which of the following statements is INCORRECT regarding Low Density Polythene (LDP)?

  1. It is prepared by the free radical addition polymerization of ethene under 1000-2000 atm pressure.
  2. It has a highly branched structure, which prevents molecules from packing closely.
  3. It is chemically inert, tough, flexible, and a poor conductor of electricity.
  4. It is prepared using coordination catalysts and possesses high tensile strength and density.
Show answer and explanation

Answer: (D) It is prepared using coordination catalysts and possesses high tensile strength and density.

The statement describing coordination catalysts and high tensile strength applies to High Density Polythene (HDP), not Low Density Polythene (LDP). LDP is synthesized via free-radical polymerization at high pressure (1000-2000 atm) with trace O₂ or peroxide initiators and has a branched structure.

hard

6. An alkene with molecular formula C₅H₁₀ does not exhibit geometrical isomerism. Upon reductive ozonolysis (O₃, Zn/H₂O), it produces two distinct organic compounds. When the same alkene reacts with HBr in the presence of peroxide, the major product obtained is:

  1. 1-Bromopentane
  2. 2-Bromopentane
  3. 2-Bromo-2-methylbutane
  4. 1-Bromo-2-methylbutane
Show answer and explanation

Answer: (D) 1-Bromo-2-methylbutane

C₅H₁₀ has several isomers. Pent-1-ene gives two distinct carbonyl products (methanal and butanal) and lacks geometrical isomerism, but its peroxide-catalyzed HBr addition gives 1-bromopentane. 2-Methylbut-1-ene also has no cis-trans isomerism and yields methanal and butanone upon ozonolysis; reaction of 2-methylbut-1-ene with HBr/peroxide yields 1-bromo-2-methylbutane (attack of Br• at C1 gives the stable tertiary radical at C2). 2-Methylbut-2-ene exhibits no stereoisomers but 1-bromo-2-methylbutane is derived from 2-methylbut-1-ene. Among options, 2-methylbut-1-ene gives 1-bromo-2-methylbutane.

easy

7. During the free radical addition of HBr to an unsymmetrical alkene in the presence of peroxides, why does the bromine free radical (Br•) attack the double bond first rather than a hydrogen atom (H•)?

  1. The addition of H• is endothermic while addition of Br• is exothermic
  2. The peroxy radical preferentially abstracts Br from HBr to generate an H• radical
  3. The Br• radical is less reactive and thus forms a weaker bond with carbon
  4. The H-Br bond is cleaved homolytically by the peroxy radical to generate Br•, not H•
Show answer and explanation

Answer: (D) The H-Br bond is cleaved homolytically by the peroxy radical to generate Br•, not H•

In the initiation step, the radical generated from the peroxide abstracts hydrogen from HBr (forming C₆H₅COOH/ROH and Br•) because the H-Br bond is cleaved to release the more stable and accessible Br• radical. Thus, Br• is the attacking radical in the chain propagation step.

easy

8. High Density Polythene (HDP) is obtained by polymerization of ethene in a hydrocarbon solvent at 333-343 K under 6-7 atm pressure in the presence of:

  1. Concentrated H₂SO₄ at room temperature
  2. Benzoyl peroxide at 1000-2000 atm
  3. Dioxygen trace at 473 K
  4. Triethylaluminium and titanium tetrachloride
Show answer and explanation

Answer: (D) Triethylaluminium and titanium tetrachloride

High Density Polythene (HDP) is produced under mild conditions (333-343 K and 6-7 atm pressure) using a coordination catalyst known as the Ziegler-Natta catalyst, which consists of triethylaluminium ((C₂H₅)₃Al) and titanium tetrachloride (TiCl₄).

medium

9. Consider the addition of HBr to 1-methylcyclohexene. What are the major products obtained in the absence of peroxides (Reaction I) and in the presence of benzoyl peroxide (Reaction II), respectively?

  1. Reaction I: 1-bromo-1-methylcyclohexane; Reaction II: 1-bromo-2-methylcyclohexane
  2. Reaction I: 1-bromo-2-methylcyclohexane; Reaction II: 1-bromo-1-methylcyclohexane
  3. Reaction I: 1-(bromomethyl)cyclohexane; Reaction II: 1-bromo-1-methylcyclohexane
  4. Reaction I: 1-bromo-1-methylcyclohexane; Reaction II: 1-(bromomethyl)cyclohexane
Show answer and explanation

Answer: (A) Reaction I: 1-bromo-1-methylcyclohexane; Reaction II: 1-bromo-2-methylcyclohexane

In the absence of peroxides, addition proceeds via Markovnikov's rule through the more stable tertiary carbocation to form 1-bromo-1-methylcyclohexane. In the presence of benzoyl peroxide, addition proceeds via a radical mechanism where the Br• radical attacks the less substituted carbon, yielding a stable tertiary radical that abstracts H to form 1-bromo-2-methylcyclohexane.

easy

10. Polythene is formed by:

  1. Ozonolysis of ethene
  2. Addition polymerisation of ethene at high temperature and pressure
  3. Condensation of ethanol
  4. Polymerisation of ethyne
Show answer and explanation

Answer: (B) Addition polymerisation of ethene at high temperature and pressure

Large numbers of ethene molecules join under high temperature and pressure with a catalyst to form the addition polymer polythene.

NCERT Class 11

medium

11. Given below are two statements: Statement I: Addition of HBr to propene in the presence of peroxides proceeds via a free-radical intermediate rather than a carbocation intermediate. Statement II: Alkoxy radicals generated from peroxides react with propene first to form a carbon-centred free radical. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
Show answer and explanation

Answer: (C) Statement I is true but Statement II is false.

Statement I is true because the Kharasch effect proceeds via a free-radical mechanism. Statement II is false because the alkoxy radical generated from peroxide homolysis reacts with HBr to generate a bromine free radical (Br•), and it is this Br• radical that adds to propene, not the alkoxy radical directly.

medium

12. Consider the addition polymerization of vinyl chloride. How many moles of vinyl chloride monomer are required to produce 1.25 kg of poly(vinyl chloride) (PVC)? (Molar mass of vinyl chloride = 62.5 g mol⁻¹)

  1. 10 mol
  2. 20 mol
  3. 25 mol
  4. 40 mol
Show answer and explanation

Answer: (B) 20 mol

Addition polymerization occurs without loss of any small molecules, so the total mass of the monomer equals the mass of the polymer formed. Number of moles = mass / molar mass = 1250 g / 62.5 g mol⁻¹ = 20 mol.

Keep going with Hydrocarbons. All 379 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Hydrocarbons