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JEE Main Organic Chemistry: practice questions with solutions

14 questions on Alcohols, Phenols & Acids, Aldehydes & Ketones, Amines, Aromatic Compounds, Biomolecules, General Organic Chemistry, Haloalkanes, Hydrocarbons, Isomerism. Try each one first, then open the solution.

Tip: solve on paper before opening a solution. Once you open it, that question is counted as practised and won't appear in your HeyGyan tests.
General Organic ChemistryMedium

Q1. Which carbocation is the most stable?

  1. (CH₃)₃C⁺
  2. C₆H₅CH₂⁺
  3. (C₆H₅)₃C⁺
  4. CH₂=CH–CH₂⁺
Show answer and solution

Answer: (C) (C₆H₅)₃C⁺

SolutionThe triphenylmethyl cation spreads its positive charge by resonance over three benzene rings, more delocalisation than any other option.
Common trapPicking tert-butyl, which is stabilised only by hyperconjugation.
HydrocarbonsEasy

Q2. Propene reacts with HBr in the presence of benzoyl peroxide. What is the main product?

  1. 2-Bromopropane
  2. 1-Bromopropane
  3. 1,2-Dibromopropane
  4. Propan-2-ol
Show answer and solution

Answer: (B) 1-Bromopropane

SolutionPeroxide causes a free-radical (anti-Markovnikov) addition, called the Kharasch effect. Br adds to the terminal carbon, giving 1-bromopropane.
Common trap2-Bromopropane is the product without peroxide (Markovnikov addition).
IsomerismMediumNumerical

Q3. How many structurally isomeric alcohols have the molecular formula C₄H₁₀O? (Exclude ethers and stereoisomers.)

Numerical value question: type the answer, no options.

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Answer: 4

SolutionButan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol: 4 alcohols. (With the 3 ethers, C₄H₁₀O has 7 structural isomers in all.)
Common trapAnswering 7 by including ethers.
Aldehydes & KetonesEasy

Q4. Which compound gives a positive iodoform test?

  1. Methanol
  2. Pentan-3-one
  3. Pentan-2-one
  4. Benzaldehyde
Show answer and solution

Answer: (C) Pentan-2-one

SolutionThe iodoform test needs a CH₃–CO– group (or CH₃–CH(OH)–). Pentan-2-one has CH₃–CO–; pentan-3-one has ethyl groups on both sides.
Common trapAssuming every ketone gives the test.
Aldehydes & KetonesEasy

Q5. Which compound cannot undergo aldol condensation?

  1. Ethanal
  2. Propanal
  3. Propanone
  4. Benzaldehyde
Show answer and solution

Answer: (D) Benzaldehyde

SolutionAldol condensation needs an α-hydrogen. Benzaldehyde has none (the CHO is attached directly to the ring), so it undergoes the Cannizzaro reaction instead.
Common trapCounting the aldehyde hydrogen as an α-hydrogen.
Alcohols, Phenols & AcidsMedium

Q6. What is the correct order of acidic strength?

  1. Acetic acid > p-nitrophenol > phenol > ethanol
  2. p-Nitrophenol > acetic acid > phenol > ethanol
  3. Phenol > acetic acid > p-nitrophenol > ethanol
  4. Acetic acid > phenol > p-nitrophenol > ethanol
Show answer and solution

Answer: (A) Acetic acid > p-nitrophenol > phenol > ethanol

SolutionApproximate pKa values: acetic acid 4.8, p-nitrophenol 7.2, phenol 10, ethanol 16. A lower pKa means a stronger acid.
Common trapAssuming the nitro group makes p-nitrophenol stronger than a carboxylic acid.
AminesMedium

Q7. Benzamide is heated with Br₂ and aqueous NaOH. What is the product?

  1. Benzylamine
  2. Aniline
  3. Benzoic acid
  4. Bromobenzene
Show answer and solution

Answer: (B) Aniline

SolutionThis is the Hofmann bromamide degradation. The amide loses its carbonyl carbon as CO₃²⁻, giving an amine with one carbon fewer: C₆H₅CONH₂ → C₆H₅NH₂.
Common trapBenzylamine keeps the extra carbon. The Hofmann reaction removes it.
BiomoleculesMedium

Q8. Glucose is heated with HI for a long time. What is the product?

  1. n-Hexane
  2. Gluconic acid
  3. Sorbitol
  4. Glucaric acid
Show answer and solution

Answer: (A) n-Hexane

SolutionProlonged heating with HI reduces all the –OH and –CHO groups, giving n-hexane. This proves glucose has a straight chain of six carbons. (Glucaric acid forms with nitric acid.)
Common trapGluconic acid forms with bromine water (oxidation), not HI.
HaloalkanesEasy

Q9. 2-Bromobutane is heated with alcoholic KOH. What is the major product?

  1. But-1-ene
  2. But-2-ene
  3. Butan-2-ol
  4. Butane
Show answer and solution

Answer: (B) But-2-ene

SolutionAlcoholic KOH causes elimination. By Saytzeff's rule, the more substituted (more stable) alkene, but-2-ene, is the major product.
Common trapButan-2-ol forms with aqueous KOH (substitution), not alcoholic KOH.
HaloalkanesMedium

Q10. What is the correct order of reactivity towards SN2 reaction?

  1. CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr
  2. (CH₃)₃CBr > (CH₃)₂CHBr > CH₃CH₂Br > CH₃Br
  3. CH₃CH₂Br > CH₃Br > (CH₃)₂CHBr > (CH₃)₃CBr
  4. All react at the same rate
Show answer and solution

Answer: (A) CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr

SolutionIn SN2, the nucleophile attacks from the back. Bulky groups block this approach, so the least crowded methyl bromide reacts fastest and the tertiary bromide slowest.
Common trapThe reverse order is correct for SN1, which goes through a carbocation.
Aldehydes & KetonesEasy

Q11. What are the products when formaldehyde (HCHO) undergoes the Cannizzaro reaction with concentrated NaOH?

  1. Methanol and sodium formate
  2. Ethanol and sodium acetate
  3. Methanol only
  4. Formic acid only
Show answer and solution

Answer: (A) Methanol and sodium formate

SolutionOne HCHO molecule is reduced to methanol and another is oxidised to formate (as HCOONa in base). This is self oxidation-reduction.
Common trapExpecting a single product. Cannizzaro always gives an alcohol and a carboxylate together.
General Organic ChemistryEasyNumerical

Q12. What is the degree of unsaturation (double bond equivalent) of benzene, C₆H₆?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 4

SolutionDBE = C − H/2 + 1 = 6 − 3 + 1 = 4. That is one ring plus three double bonds.
Common trapAnswering 3 by counting only the double bonds and forgetting the ring.
Aromatic CompoundsEasy

Q13. Which substituent on a benzene ring directs an incoming electrophile to the meta position?

  1. –CH₃
  2. –OH
  3. –NO₂
  4. –Cl
Show answer and solution

Answer: (C) –NO₂

Solution–NO₂ withdraws electrons by resonance, making the ortho and para positions especially electron-poor. Attack therefore goes to meta.
Common trapPicking –Cl because it deactivates the ring. Halogens deactivate but still direct ortho/para.
Alcohols, Phenols & AcidsMedium

Q14. Which reagent oxidises a primary alcohol to an aldehyde without oxidising it further to a carboxylic acid?

  1. Acidified KMnO₄
  2. Acidified K₂Cr₂O₇ (excess)
  3. PCC
  4. Concentrated HNO₃
Show answer and solution

Answer: (C) PCC

SolutionPyridinium chlorochromate (PCC) is a mild oxidant used in a non-aqueous solvent, so the reaction stops at the aldehyde.
Common trapStrong aqueous oxidants like KMnO₄ carry the oxidation all the way to the carboxylic acid.

Frequently asked questions

Which Organic Chemistry chapters are covered here?

This page covers Alcohols, Phenols & Acids, Aldehydes & Ketones, Amines, Aromatic Compounds, Biomolecules, General Organic Chemistry, Haloalkanes, Hydrocarbons, Isomerism. Questions follow the JEE Main pattern, with multiple-choice and numerical value types.

Are these previous year JEE questions?

No. These are original JEE Main-level practice questions written on the latest pattern, with solutions and the common trap for each.