JEE Main and Advanced Chemistry MCQs - Organic Chemistry Basics: IUPAC, Isomerism and Reaction Mechanisms (Set 2)

JEE Main and Advanced Chemistry MCQs - Organic Chemistry Basics: IUPAC, Isomerism and Reaction Mechanisms (Set 2) - CodingHumans

General Organic Chemistry is the foundation for every organic chapter in JEE. If you understand electronic effects and intermediates, reaction questions become logical instead of something to memorise. This set covers IUPAC naming, isomerism, inductive and resonance effects, hyperconjugation, carbocation stability, acidity and substitution and addition mechanisms.

How to practise: read each question and tap the option you think is correct. You will see right away whether you got it, along with a short explanation, and your score is tracked at the bottom of the screen. This set has 15 questions.

JEE Chemistry series: Atomic Structure and Bonding | Organic Chemistry Basics | Competitive Exams Hub

Topics covered in this set

  • IUPAC nomenclature
  • Structural, geometrical and optical isomerism
  • Inductive, resonance and hyperconjugation effects
  • Carbocation stability and acidity trends
  • SN1 vs SN2 mechanisms
  • Markovnikov and anti-Markovnikov addition

JEE Chemistry MCQs with answers and explanations

Q1. What is the IUPAC name of CH₃–CH(CH₃)–CH₂–CH₃?

a) 2-Ethylpropane

b) 2-Methylbutane

c) 3-Methylbutane

d) Pentane

Q2. How many structural isomers does C₅H₁₂ have?

a) 5

b) 3

c) 2

d) 4

Q3. Which compound shows geometrical (cis–trans) isomerism?

a) But-2-ene

b) But-1-ene

c) Propene

d) 2-Methylpropene

Q4. Which compound is optically active?

a) Propan-2-ol

b) Ethanol

c) Butan-2-ol

d) 2-Methylpropan-2-ol

Q5. What is the correct order of carbocation stability?

a) 3° > 2° > 1° > methyl

b) 1° > 2° > 3° > methyl

c) 2° > 3° > 1° > methyl

d) Methyl > 1° > 2° > 3°

Q6. Which group shows the strongest −I (electron-withdrawing inductive) effect?

a) –CH₃

b) –OCH₃

c) –NO₂

d) –C₂H₅

Q7. Why is phenol more acidic than ethanol?

a) The phenoxide ion is stabilised by resonance, delocalising the negative charge into the ring

b) Phenol is a gas

c) Ethanol is aromatic

d) Phenol has more hydrogen atoms

Q8. Which is the strongest acid?

a) ClCH₂COOH

b) Cl₃CCOOH

c) CH₃COOH

d) Cl₂CHCOOH

Q9. Which statement about the SN2 mechanism is correct?

a) It gives a racemic mixture

b) It is a single step with inversion of configuration, and its rate depends on both substrate and nucleophile

c) It proceeds through a carbocation

d) It is fastest for tertiary halides

Q10. Which conditions favour an SN1 reaction?

a) Primary substrate, polar aprotic solvent

b) Tertiary substrate, polar protic solvent

c) Methyl substrate, strong nucleophile

d) Gas phase

Q11. What is the major product when HBr adds to propene in the absence of peroxide?

a) 1-Bromopropane

b) 1,2-Dibromopropane

c) 2-Bromopropane

d) Propane

Q12. In the presence of peroxide, HBr adds to propene to give mainly:

a) 1-Bromopropane

b) 2-Bromopropane

c) No reaction

d) Propan-2-ol

Q13. Which alkene is the most stable?

a) Propene

b) 2,3-Dimethylbut-2-ene

c) 2-Methylpropene

d) Ethene

Q14. What is the hybridisation of each carbon atom in benzene?

a) sp

b) sp³d

c) sp³

d) sp²

Q15. Which functional group does the suffix "-al" indicate in IUPAC naming?

a) Alcohol

b) Carboxylic acid

c) Ketone

d) Aldehyde

Quick revision notes

  • C₄H₁₀ has 2 isomers; C₅H₁₂ has 3.
  • Carbocation stability: 3° > 2° > 1° > methyl (+I and hyperconjugation).
  • More −I groups near COOH means a stronger acid.
  • SN2: one step, inversion, best for 1°; SN1: carbocation, racemisation, best for 3°.
  • HBr + alkene: Markovnikov; with peroxide: anti-Markovnikov.

Frequently asked questions

Are these JEE previous year questions?

No, they are original practice questions written for the JEE General Organic Chemistry syllabus.

How should I study GOC?

Learn the electronic effects first, then use them to explain every stability and acidity order instead of memorising lists.

Does the peroxide effect work with HCl?

No. It is seen with HBr only, because the radical steps are not favourable for HCl or HI.

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