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Friday, 9 October

NEET UG 2019 · Chemistry · Question 74

NEET 2019 Chemistry, question 74

Question 74 of the NEET 2019 paper (code P1) is a Chemistry question from Hydrocarbons. Its idea, Reverse Ozonolysis (Parent Alkene Identification), was asked twice, in 2019 and 2022, across NEET 2017–2026. The NTA final answer key gives option 3.

Updated 2026-10-08

Q74 · Chemistry

An alkene "A" on reaction with O3O_3 and Zn−H2OZn-H_2O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is :

The figure, described in words

Options (1)-(4) are drawn structural formulas of chloroalkane isomers (branched condensed structures with CH3/Cl above/below the chain).

  1. 1Cl−CH2−CH2−CH(CH3)2Cl-CH_2-CH_2-CH(CH_3)_2
  2. 2H3C−CH2−CH(CH2Cl)−CH3H_3C-CH_2-CH(CH_2Cl)-CH_3
  3. 3H3C−CH2−C(Cl)(CH3)−CH3H_3C-CH_2-C(Cl)(CH_3)-CH_3Official answer
  4. 4H3C−CH(Cl)−CH(CH3)−CH3H_3C-CH(Cl)-CH(CH_3)-CH_3

Official answer: option 3 · NTA final answer key

Same idea

Reverse Ozonolysis (Parent Alkene Identification), in other papers

NEET asked this idea twice, in 2019 and 2022. The other question below.

  • 2022 Q91Compound X on reaction with O3\mathrm{O_3} followed by Zn/H2O\mathrm{H_2O} gives formaldehyde and 2-methyl propanal as products. The…One-line MCQ · Medium

Reverse Ozonolysis (Parent Alkene Identification): every question and how it was framed

The chapter

Hydrocarbons in ten years of NEET

25 questions from 2017–2026, asked in 10 of 10 papers, about 2.4 a paper. Hydrocarbons (Alkenes) alone had 12.

Chapter analysisStudy this chapterPractise this chapter

Hydrocarbons: questions per paper

0363’171’182’192’204’212’222’231’245’253’26