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MDCAT Chemistry MCQs 2026 — Answers & Explanations

Chemistry is 45 of the 180 MDCAT marks — a quarter of the paper, and the section where preparation shows most clearly. Unlike Biology, where wide reading eventually pays off, Chemistry rewards a smaller set of well-understood ideas applied repeatedly.

The PMDC 2026 curriculum spans 20 units across the three traditional branches. Physical chemistry — atomic structure, chemical bonding, thermodynamics, equilibrium, reaction kinetics, electrochemistry — is where most calculation questions live, and where most marks are lost to arithmetic rather than ignorance. Organic chemistry — alcohols and phenols, aldehydes and ketones, alkyl halides, carboxylic acids, hydrocarbons — is largely about recognising functional groups and predicting what a named reagent does to them. Inorganic chemistry covers periodicity, s- and p-block elements and transition elements, and is the most straightforwardly memorisable of the three.

A recurring mistake is treating organic chemistry as a memorisation exercise. There are far too many named reactions to hold individually, but the number of underlying mechanisms is small. Students who learn why a nucleophile attacks where it does can answer questions they have never seen; students who memorise reaction lists cannot.

The questions below are drawn from across all three branches, with answers and worked explanations. Where a question involves a calculation, the explanation shows the reasoning rather than only the result.

40 free Chemistry MCQs with answers

Attempt calculation questions with paper rather than mentally — most Chemistry marks are lost to slips, not to gaps in knowledge, and you cannot train against a slip you never made visible.

Alcohols and Phenols

Q1. Propan-2-ol is classified as which type of alcohol?

  1. A Primary
  2. B Phenol
  3. C Tertiary
  4. D Secondary

Answer: D — The carbon bearing the hydroxyl group is attached to two other carbon atoms; which defines a secondary alcohol.

Classification and Nomenclature

Q2. Phenol is more acidic than ethanol because:

  1. A Phenol has an OH group
  2. B Phenol has higher molar mass
  3. C The phenoxide ion (C₆H₅O⁻) is stabilized by resonance with the benzene ring, while ethoxide is not
  4. D Ethanol has more hydrogen bonds

Answer: C — The phenoxide anion (C₆H₅O⁻) is resonance-stabilized — negative charge delocalized into the ring. Ethoxide (C₂H₅O⁻) cannot be resonance-stabilized → phenol is ~10⁶ times more acidic.

Difference between Alcohol and Phenol

Q3. The IUPAC name of CH₃OH is:

  1. A Propanol
  2. B Ethanol
  3. C Butanol
  4. D Methanol

Answer: D — CH₃OH is a one-carbon alcohol. IUPAC name: methanol (common name: methyl alcohol).

Nomenclature, Structure, Reactivity of Alcohols

Q4. Phenol is more reactive than benzene toward electrophilic substitution because:

  1. A The −OH group is electron-withdrawing
  2. B The −OH group activates the ring by donating electron density through resonance
  3. C Phenol is an alkane
  4. D Phenol has no pi electrons

Answer: B — The −OH group donates electrons to the ring via resonance (+M effect), increasing electron density and activating the ring.

Nomenclature, Structure, Reactivity of Phenols

Q5. Ethanol is produced industrially from ethene by:

  1. A Catalytic hydration with steam
  2. B Reduction with hydrogen
  3. C Oxidation with dichromate
  4. D Fermentation

Answer: A — Ethene and steam pass over a phosphoric acid catalyst at high temperature and pressure; water adds across the double bond.

Preparation

Q6. Phenol is more acidic than ethanol because the phenoxide ion is:

  1. A Stabilised by delocalisation into the benzene ring
  2. B Not formed at all
  3. C Positively charged
  4. D Less stable

Answer: A — The negative charge is spread over the ring by resonance; this stabilisation makes loss of the proton much more favourable than from an alkoxide.

Properties of Phenol

Q7. Sodium ethoxide dissolved in water gives a solution that is:

  1. A Alkaline because the ethoxide ion is hydrolysed
  2. B Unable to conduct electricity
  3. C Exactly neutral
  4. D Strongly acidic

Answer: A — The ethoxide ion is a stronger base than hydroxide; so it removes a proton from water and generates hydroxide ions; making the solution alkaline.

Reactions of Alcohols

Q8. Benzene requires a halogen carrier such as iron(III) bromide for bromination, whereas phenol needs none, because in phenol the:

  1. A Ring is deactivated
  2. B Ring is activated enough to react with molecular bromine directly
  3. C Hydroxyl group is removed first
  4. D Bromine becomes more electronegative

Answer: B — Electron donation from the hydroxyl oxygen raises the ring's electron density so much that it attacks bromine without needing a catalyst to polarise it.

Reactions of Phenol

Aldehydes and Ketones

Q9. The IUPAC name of CH₃CHO is:

  1. A Butanal
  2. B Ethanal
  3. C Propanal
  4. D Methanal

Answer: B — CH₃CHO is a 2-carbon aldehyde. IUPAC name: ethanal (common name: acetaldehyde).

Nomenclature and Structure

Q10. The characteristic reaction of aldehydes and ketones is:

  1. A Electrophilic substitution
  2. B Nucleophilic addition
  3. C Electrophilic addition
  4. D Free radical substitution

Answer: B — The carbonyl group (C=O) is polar; nucleophiles attack the electrophilic carbon, making nucleophilic addition the characteristic reaction.

Nucleophilic Addition Reactions

Q11. Tollen's test is positive for:

  1. A Both aldehydes and ketones
  2. B Ketones only
  3. C Aldehydes only
  4. D Carboxylic acids

Answer: C — Tollen's reagent (ammoniacal AgNO₃) is reduced by aldehydes to give a silver mirror. Ketones do not give this test.

Oxidation Reactions

Q12. Aldehydes can be prepared by mild oxidation of:

  1. A Primary alcohols using PCC
  2. B Tertiary alcohols
  3. C Secondary alcohols
  4. D Carboxylic acids

Answer: A — PCC (pyridinium chlorochromate) is a mild oxidizing agent that oxidizes primary alcohols to aldehydes without over-oxidation.

Preparation

Q13. Which is more reactive toward nucleophilic addition: formaldehyde (HCHO) or acetone ((CH3)2CO)?

  1. A Acetone, due to inductive effect of methyl groups
  2. B Formaldehyde, due to less steric hindrance and no electron-donating groups
  3. C Both are equally reactive
  4. D Acetone, due to resonance stabilization

Answer: B — Formaldehyde is more reactive than acetone in nucleophilic addition because: (1) it has minimal steric hindrance (only H atoms), and (2) it lacks electron-donating methyl groups that would decrease electrophilicity of the carbonyl carbon.

Reactivity and Comparison

Q14. Reduction of an aldehyde with NaBH4 gives a:

  1. A Secondary alcohol
  2. B Carboxylic acid
  3. C Ketone
  4. D Primary alcohol

Answer: D — Reduction of an aldehyde (R-CHO) adds two hydrogen atoms across C=O to give a primary alcohol (R-CH2OH). NaBH4 (sodium borohydride) is a mild, selective reducing agent for carbonyl groups.

Reduction to Alcohols

Alkyl Halides

Q15. 2-bromopropane is classified as which type of alkyl halide?

  1. A Primary
  2. B Aromatic
  3. C Tertiary
  4. D Secondary

Answer: D — The carbon bearing the bromine is attached to two other carbon atoms; a halide on such a carbon is secondary.

Classification and Nomenclature

Q16. In E2 elimination, the reaction is:

  1. A Bimolecular
  2. B Zero order
  3. C Unimolecular
  4. D Termolecular

Answer: A — E2 is a one-step bimolecular process: Rate = k[substrate][base]. A strong base removes a β-hydrogen simultaneously with leaving group departure.

Elimination Reactions

Q17. The IUPAC name of CH₃CH₂Cl is:

  1. A Chloromethane
  2. B Methyl chloride
  3. C Ethyl chloride
  4. D Chloroethane

Answer: D — IUPAC: the parent chain is ethane, and Cl is at position 1. The name is chloroethane.

Nomenclature, Structure, Reactivity

Q18. In a nucleophilic substitution reaction of an alkyl halide; the halogen leaves as:

  1. A A halogen atom
  2. B A halogen molecule
  3. C A hydrogen halide
  4. D A halide ion

Answer: D — The nucleophile displaces the halogen; which departs with the bonding pair of electrons as a stable halide ion; the leaving group.

Nucleophilic Substitution

Q19. SN2 reactions are favored by:

  1. A Polar protic solvents
  2. B Weak nucleophiles
  3. C Tertiary substrates
  4. D Primary substrates and strong nucleophiles

Answer: D — SN2 is favored by primary (less steric hindrance) substrates, strong nucleophiles, and polar aprotic solvents.

Nucleophilic Substitution Reactions

Q20. Which carbon-halogen bond is the weakest and therefore most easily broken?

  1. A Carbon-bromine
  2. B Carbon-fluorine
  3. C Carbon-iodine
  4. D Carbon-chlorine

Answer: C — Bond strength falls down the group as the orbitals overlap less effectively; the carbon-iodine bond is the weakest so iodides are the most reactive in substitution.

Reactivity

Atomic Structure

Q21. The proton was discovered by:

  1. A Chadwick
  2. B J.J. Thomson
  3. C Rutherford
  4. D Goldstein

Answer: D — Goldstein discovered canal rays (protons) in 1886 using a perforated cathode in a discharge tube.

Discovery of Proton

Q22. According to Aufbau principle, electrons fill orbitals in order of:

  1. A Decreasing size
  2. B Random order
  3. C Decreasing energy
  4. D Increasing energy

Answer: D — The Aufbau principle states that electrons fill orbitals starting from the lowest energy level.

Electronic Configuration

Q23. According to Planck's quantum theory, energy is emitted or absorbed in discrete packets called:

  1. A Electrons
  2. B Protons
  3. C Photons
  4. D Neutrons

Answer: C — Planck proposed that energy is emitted/absorbed in quanta (photons), not continuously.

Planck's Quantum Theory

Q24. The principal quantum number (n) determines the:

  1. A Orientation in space
  2. B Shape of the orbital
  3. C Energy level and size of the orbital
  4. D Spin of the electron

Answer: C — The principal quantum number n determines the energy level and size of the orbital.

Quantum Numbers

Q25. The shape of an s orbital is:

  1. A Dumbbell
  2. B Double dumbbell
  3. C Cloverleaf
  4. D Spherical

Answer: D — All s orbitals are spherically symmetrical in shape.

Shapes of Orbitals

Q26. The Balmer series of hydrogen spectrum lies in the:

  1. A Ultraviolet region
  2. B X-ray region
  3. C Infrared region
  4. D Visible region

Answer: D — The Balmer series corresponds to transitions to n=2 and lies in the visible region.

Spectrum of Hydrogen

Carboxylic Acids

Q27. Which reagent distinguishes a carboxylic acid from a phenol in the laboratory?

  1. A Aqueous sodium hydrogen carbonate
  2. B Neutral ferric chloride solution
  3. C Bromine water
  4. D Ammoniacal silver nitrate

Answer: A — Only the carboxylic acid is strong enough to displace carbon dioxide from hydrogen carbonate; a phenol gives no effervescence although it does dissolve in sodium hydroxide.

Acidity

Q28. Acyl halides (acid chlorides) are prepared from carboxylic acids using:

  1. A NaCl
  2. B Cl₂ gas
  3. C SOCl₂ (thionyl chloride)
  4. D HCl

Answer: C — R−COOH + SOCl₂ → R−COCl + SO₂ + HCl. Thionyl chloride is preferred because gaseous byproducts are easily removed.

Conversion to Derivatives (acyl halides, anhydrides, esters)

Q29. The linkage present in an ester is formed between a carboxylic acid and:

  1. A An alcohol
  2. B A ketone
  3. C An amine
  4. D A halide

Answer: A — The alkoxy group of the alcohol replaces the hydroxyl of the acid; giving the characteristic ester linkage often responsible for fruity odours.

Derivatives

Q30. The IUPAC name of CH₃COOH is:

  1. A Methanoic acid
  2. B Ethanoic acid
  3. C Butanoic acid
  4. D Propanoic acid

Answer: B — CH₃COOH is a 2-carbon carboxylic acid. IUPAC: ethanoic acid (common name: acetic acid).

Nomenclature, Structure, Preparation

Q31. A primary alcohol on vigorous oxidation with acidified potassium dichromate finally gives:

  1. A An ester
  2. B A carboxylic acid
  3. C An aldehyde only
  4. D A ketone

Answer: B — Oxidation proceeds through the aldehyde; but under vigorous conditions with excess oxidising agent the reaction continues to the carboxylic acid.

Preparation

Q32. Which pair of reactants would give the ester ethyl ethanoate?

  1. A Methanol and propanoic acid
  2. B Ethanol and ethanoic acid
  3. C Ethanal and ethanoic acid
  4. D Ethanol and methanoic acid

Answer: B — The alkyl part of an ester name comes from the alcohol and the acyl part from the acid; so ethyl ethanoate needs ethanol together with ethanoic acid.

Reactions

Q33. Carboxylic acids are more acidic than alcohols because:

  1. A The carboxylate anion is stabilized by resonance
  2. B The O−H bond is shorter
  3. C They are less soluble in water
  4. D They have a higher molecular weight

Answer: A — The carboxylate ion (RCOO⁻) is resonance stabilized with the negative charge delocalized over two oxygen atoms.

Reactivity

Q34. The functional group characteristic of a carboxylic acid is:

  1. A A carboxyl group
  2. B An ether linkage
  3. C A carbonyl group only
  4. D A hydroxyl group only

Answer: A — The carboxyl group combines a carbonyl and a hydroxyl on the same carbon; the interaction between them accounts for the acidity of these compounds.

Structure and Nomenclature

Chemical Bonding

Q35. Bond energy is defined as the energy required to:

  1. A Convert a solid to gas
  2. B Break a bond between two atoms in gaseous state
  3. C Ionize an atom
  4. D Form a bond between two atoms

Answer: B — Bond energy (bond dissociation energy) is the energy needed to break one mole of a particular bond in gaseous molecules.

Bond Energy

Q36. Which molecule has zero dipole moment?

  1. A CO₂
  2. B H₂O
  3. C NH₃
  4. D HCl

Answer: A — CO₂ is linear with two equal C=O dipoles pointing in opposite directions, so they cancel giving zero net dipole moment.

Dipole Moment

Q37. The hybridization of carbon in methane (CH₄) is:

  1. A sp
  2. B sp³
  3. C sp²
  4. D sp³d

Answer: B — Carbon in CH₄ forms 4 equivalent bonds using sp³ hybridization, resulting in tetrahedral geometry.

Hybridization

Q38. A sigma (σ) bond is formed by:

  1. A No overlap of orbitals
  2. B Head-on overlap of orbitals
  3. C Lateral overlap of orbitals
  4. D d-orbital overlap only

Answer: B — Sigma bonds result from head-on (axial) overlap of orbitals along the internuclear axis.

Sigma and Pi Bond

Q39. According to VSEPR theory, the shape of CH₄ is:

  1. A Tetrahedral
  2. B Linear
  3. C Square planar
  4. D Trigonal pyramidal

Answer: A — CH₄ has 4 bond pairs and no lone pairs around carbon, giving a tetrahedral geometry with 109.5° angles.

VSEPR Theory

Chemical Equilibrium

Q40. A buffer solution resists changes in:

  1. A Temperature
  2. B pH
  3. C Concentration
  4. D Volume

Answer: B — A buffer solution resists changes in pH upon addition of small amounts of acid or base.

Buffer Solution

Where to go next

These questions are a sample. The full Chemistry question bank is available in the topic-wise Chemistry practice tool, which lets you filter by chapter and tracks which questions you have already answered. If a question here exposed a gap, the MDCAT notes section covers the underlying theory chapter by chapter, and a full-length timed mock test is the best way to find out whether you can hold that accuracy under exam conditions.

Frequently asked questions

How many Chemistry questions are on the MDCAT 2026?

Chemistry accounts for 45 of the 180 MCQs — 25% of the paper. It is the second-largest section after Biology.

Is organic or physical chemistry more important for the MDCAT?

Both appear substantially. Physical chemistry produces more calculation questions and therefore more avoidable errors; organic chemistry produces more questions overall but is more predictable once the mechanisms are understood.

Are calculators allowed in the MDCAT?

No. Calculators are not permitted, so MDCAT chemistry calculations are designed to resolve with mental arithmetic or short working. If a calculation is becoming long, you have usually taken a wrong route.

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