Organic Mechanisms MCQs – 40 Practice Questions with Answers and Explanations

Master the mechanisms that A-Level, IB and AP examiners love to test. These 40 original multiple-choice questions cover bond breaking and reactive species, free-radical substitution, electrophilic addition, nucleophilic substitution and elimination, electrophilic substitution on benzene, and carbonyl chemistry – every question comes with the correct answer and a clear explanation.

Work through each section, try the question before opening the answer, and use the explanations to fix the reasoning, not just the result. All questions are free to attempt as many times as you like.

Quick formula recap: keep the key formulas beside you while you practise – open a formula sheet for your board:

Foundations: Bond Breaking & Reactive Species

Q1. A covalent bond breaks by homolytic fission. What is formed?

  • (A) Two ions, one positive and one negative
  • (B) Two free radicals, each keeping one of the shared electrons
  • (C) One free radical and one ion
  • (D) Two neutral molecules
Show Answer

Answer: B – In homolytic fission the shared pair splits evenly, one electron going to each atom, so two radicals form. Heterolytic fission gives ions instead.

Q2. The H−Cl bond breaks heterolytically. Which pair of products is formed?

  • (A) H• and Cl•
  • (B) H− and Cl+
  • (C) H+ and Cl−
  • (D) H2 and Cl2
Show Answer

Answer: C – Chlorine is more electronegative, so it takes both bonding electrons, giving H+ and Cl−. This uneven split is heterolytic fission.

Q3. Which species acts as a nucleophile?

  • (A) NO2+
  • (B) H+
  • (C) AlCl3
  • (D) OH−
Show Answer

Answer: D – A nucleophile donates a pair of electrons to form a new bond. OH− has a lone pair and a negative charge; the others are electron-poor electrophiles.

Q4. Which species is an electrophile?

  • (A) NO2+
  • (B) CN−
  • (C) NH3
  • (D) H2O
Show Answer

Answer: A – An electrophile accepts a pair of electrons. The positively charged NO2+ is strongly electron-seeking; CN−, NH3 and H2O all donate lone pairs, so they behave as nucleophiles.

Q5. In the reaction of bromoethane with potassium cyanide, the CN− ion attacks the carbon attached to bromine. CN− is acting as…

  • (A) an electrophile
  • (B) a nucleophile
  • (C) a free radical
  • (D) a catalyst
Show Answer

Answer: B – CN− donates its lone pair into a new C−C bond, which is nucleophilic behaviour. This is the first step of nucleophilic substitution.

Q6. In a reaction mechanism, what does a curly (curved) arrow show?

  • (A) The movement of a single electron
  • (B) The movement of an atom from one molecule to another
  • (C) The movement of a pair of electrons
  • (D) The direction of heat flow
Show Answer

Answer: C – A double-headed curly arrow shows a pair of electrons moving, from a bond or lone pair to a new position. Single-electron movement (in radical mechanisms) is shown with a single-headed “fishhook” arrow.

Q7. Which carbocation is the most stable?

  • (A) (CH3)3C+ (tertiary)
  • (B) CH3CH2+ (primary)
  • (C) CH3+
  • (D) (CH3)2CH+ (secondary)
Show Answer

Answer: A – Stability increases primary < secondary < tertiary. The tertiary carbocation, with three alkyl groups around the positive carbon, is the most stable.

Q8. Why is a tertiary carbocation more stable than a primary carbocation?

  • (A) It has more hydrogen atoms to lose
  • (B) Its positive charge is spread over a larger molecule by chance
  • (C) It is formed faster because the C−Br bond is weaker
  • (D) Three alkyl groups push electron density towards the positive carbon (positive inductive effect), reducing its charge density
Show Answer

Answer: D – Alkyl groups donate electron density through the sigma bonds (the +I effect and hyperconjugation), stabilising the positive charge. Three alkyl groups stabilise more than one.

Q9. Ammonia, :NH3, reacts with halogenoalkanes. It behaves as a nucleophile because…

  • (A) it is a small molecule
  • (B) the nitrogen lone pair is donated to form a new bond
  • (C) it contains hydrogen atoms that are lost as H+
  • (D) it is a gas at room temperature
Show Answer

Answer: B – Nucleophiles are electron-pair donors. Ammonia’s lone pair on nitrogen attacks electron-deficient carbon, forming a bond – the first step towards amines.

Q10. A free radical is best described as…

  • (A) an ion with a full outer shell of electrons
  • (B) a species formed only in the presence of UV light
  • (C) a species with an unpaired electron, making it highly reactive
  • (D) a molecule with a permanent dipole
Show Answer

Answer: C – Radicals such as Cl• and CH3• have an unpaired electron and react rapidly to pair it up, which is why they drive chain reactions.

Free-Radical Substitution & Electrophilic Addition

Q11. In the chlorination of methane, what happens in the initiation step?

  • (A) Cl2 → 2Cl•, using UV light to break the Cl−Cl bond homolytically
  • (B) Cl• + CH4 → HCl + CH3•
  • (C) CH3• + Cl2 → CH3Cl + Cl•
  • (D) CH3• + Cl• → CH3Cl
Show Answer

Answer: A – Initiation creates the first radicals: UV light supplies the energy for homolytic fission of Cl2. The other equations are propagation (B, C) and termination (D) steps.

Q12. Which equation shows a propagation step in the chlorination of methane?

  • (A) Cl2 → 2Cl•
  • (B) Cl• + CH4 → HCl + CH3•
  • (C) CH3• + CH3• → C2H6
  • (D) CH3Cl + Cl2 → CH2Cl2 + HCl
Show Answer

Answer: B – In propagation, a radical reacts to give a product plus a new radical, so the chain continues. Termination (C) removes radicals without replacing them.

Q13. Which equation is a termination step?

  • (A) Cl• + CH4 → HCl + CH3•
  • (B) CH3• + Cl2 → CH3Cl + Cl•
  • (C) Cl• + Cl• → Cl2
  • (D) Cl2 → 2Cl•
Show Answer

Answer: C – Termination happens when two radicals collide and combine into a stable molecule, removing radicals from the chain. No new radical is produced.

Q14. When methane reacts with excess chlorine in UV light, a mixture of CH3Cl, CH2Cl2, CHCl3 and CCl4 forms. Why?

  • (A) Chlorine is a very strong oxidising agent
  • (B) The products react with each other to exchange chlorine atoms
  • (C) Methane contains four different types of hydrogen atom
  • (D) The partially chlorinated products still contain hydrogen atoms, so free-radical substitution continues on them too
Show Answer

Answer: D – Radicals keep attacking any molecule that still has hydrogen atoms, so substitution proceeds step by step through all four products. This poor selectivity is a major limitation of free-radical substitution.

Q15. What happens first in the electrophilic addition of bromine to ethene?

  • (A) A bromide ion attacks the double bond
  • (B) The pi electrons of the C=C bond are attracted to the electrophile, a bond forms to one carbon, and a carbocation forms on the other
  • (C) The C=C bond breaks homolytically to give two radicals
  • (D) Both bromine atoms add at the same instant
Show Answer

Answer: B – The electron-rich pi bond attracts the electrophile (the δ+ end of a polarised Br−Br molecule). The pi pair forms a bond to it, leaving a carbocation, which Br− then attacks.

Q16. Propene reacts with HBr to give mainly 2-bromopropane. This follows Markovnikov’s rule because…

  • (A) the H+ electrophile adds to the carbon that gives the more stable secondary carbocation
  • (B) bromine atoms are larger than hydrogen atoms
  • (C) the double bond in propene is stronger in the middle
  • (D) 2-bromopropane has a higher boiling point
Show Answer

Answer: A – H+ adds to the CH2 carbon so that the positive charge ends up on the secondary carbon, which is more stable than the primary alternative. Br− then attacks that carbon, giving 2-bromopropane.

Q17. When propene reacts with HBr in the presence of peroxides, the major product is 1-bromopropane instead. Why?

  • (A) Peroxides act as a catalyst that speeds up the ionic mechanism
  • (B) Peroxides reverse Markovnikov’s rule by changing the carbocation stability order
  • (C) Peroxides start a free-radical mechanism in which Br• adds first, forming the more stable secondary radical
  • (D) Peroxides oxidise propene to a different alkene
Show Answer

Answer: C – This is the peroxide (Kharasch) effect: a radical chain replaces the ionic mechanism. Br• adds to the double bond so as to form the more stable secondary radical, so bromine ends up on the terminal carbon – anti-Markovnikov product.

Q18. In the addition of Br2 to an alkene, the attacking electrophile is best described as…

  • (A) a free Br• radical
  • (B) a Br− ion
  • (C) an unpolarised Br2 molecule acting as a nucleophile
  • (D) the δ+ bromine of a Br−Br bond polarised by the approaching pi bond (effectively Br+)
Show Answer

Answer: D – Br2 is non-polar on its own, but the electron-rich double bond induces a dipole; the nearer bromine becomes δ+ and is accepted as the electrophile. It is not a radical process – no UV light is needed.

Q19. Hydrogen bromide adds to but-2-ene to give only one bromoalkane product. Why is there no Markovnikov choice here?

  • (A) But-2-ene is symmetrical, so H+ adding to either double-bond carbon gives the same secondary carbocation
  • (B) But-2-ene does not react with HBr
  • (C) The reaction follows a radical mechanism
  • (D) Both carbons of the double bond are primary
Show Answer

Answer: A – CH3CH=CHCH3 is symmetrical: whichever carbon receives H+, the carbocation formed is identical, so a single product, 2-bromobutane, results.

Q20. In free-radical substitution, the role of UV light is to…

  • (A) heat the reaction mixture so molecules collide harder
  • (B) ionise the chlorine molecules
  • (C) provide the energy to break the Cl−Cl bond homolytically, producing the first radicals
  • (D) act as a catalyst for the propagation steps
Show Answer

Answer: C – The Cl−Cl bond is weak enough for UV photons to split it evenly (photolysis). Without this initiation, the chain reaction cannot start; UV is not a catalyst because the propagation steps then sustain the chain.


Nucleophilic Substitution & Elimination

Q21. Bromoethane is warmed with aqueous sodium hydroxide. The main organic product is…

  • (A) ethene
  • (B) ethanol
  • (C) ethane
  • (D) ethanoic acid
Show Answer

Answer: B – OH− is a nucleophile and replaces bromine on the carbon, giving ethanol. Aqueous conditions favour substitution; ethene would need concentrated ethanolic KOH (elimination).

Q22. Bromoethane is heated with potassium cyanide in ethanol. The product is…

  • (A) ethylamine
  • (B) ethanoic acid
  • (C) propanenitrile, CH3CH2CN – the chain is lengthened by one carbon
  • (D) ethane
Show Answer

Answer: C – CN− substitutes the bromine and adds its carbon to the chain. Nitriles are useful because they can be hydrolysed to carboxylic acids or reduced to amines.

Q23. Bromoethane reacts with excess concentrated ammonia under pressure. The first nitrogen-containing product is…

  • (A) ethylamine, CH3CH2NH2
  • (B) a nitrile
  • (C) an amide
  • (D) nitroethane
Show Answer

Answer: A – Ammonia’s lone pair attacks the carbon, and after loss of H+ the product is ethylamine. Because ethylamine is also a nucleophile, further substitution can continue to diethylamine, triethylamine and a quaternary ammonium salt unless ammonia is in large excess.

Q24. A tertiary halogenoalkane such as 2-bromo-2-methylpropane hydrolyses rapidly by an SN1 mechanism. Why does this substrate favour SN1?

  • (A) The C−Br bond in tertiary compounds is polarised the opposite way
  • (B) Three alkyl groups block nucleophilic attack completely
  • (C) Tertiary halogenoalkanes are always liquids
  • (D) Loss of Br− gives a stable tertiary carbocation, so the slow step is unimolecular ionisation
Show Answer

Answer: D – In SN1 the rate-determining step is formation of the carbocation. A tertiary carbocation is stabilised by three alkyl groups, so this pathway is fast; the bulky groups also hinder the backside attack needed for SN2.

Q25. Primary halogenoalkanes usually react by the SN2 mechanism because…

  • (A) they form very stable carbocations
  • (B) there is little steric hindrance to backside attack, and a primary carbocation would be too unstable for SN1
  • (C) they are insoluble in water
  • (D) their C−X bonds cannot break heterolytically
Show Answer

Answer: B – SN2 is a single-step mechanism in which the nucleophile attacks from the side opposite the halogen. An open primary carbon allows this; the SN1 alternative is disfavoured because primary carbocations are unstable.

Q26. Bromoethane is heated under reflux with concentrated potassium hydroxide in ethanol. The main product is…

  • (A) ethene
  • (B) ethanol
  • (C) ethanoic acid
  • (D) diethyl ether
Show Answer

Answer: A – Hot, concentrated ethanolic KOH acts as a base and removes H+ from the carbon next to the C−Br bond while Br− leaves: elimination to ethene. Solvent and concentration decide substitution (aqueous, dilute) versus elimination (ethanolic, concentrated).

Q27. Which halogenoalkane hydrolyses fastest with aqueous sodium hydroxide?

  • (A) 1-chlorobutane
  • (B) 1-bromobutane
  • (C) 1-iodobutane
  • (D) 1-fluorobutane
Show Answer

Answer: C – The rate order is RI > RBr > RCl > RF. The C−I bond is the weakest, so it breaks most easily in the rate-determining step.

Q28. The reactivity order C−I > C−Br > C−Cl in hydrolysis is explained by…

  • (A) the increasing electronegativity of the halogen
  • (B) the increasing size of the molecules
  • (C) the solubility of the silver halides formed
  • (D) carbon-halogen bond enthalpies: C−I is the weakest bond and C−Cl (and especially C−F) the strongest
Show Answer

Answer: D – Bond strength falls down the group as the halogen atom gets larger and overlap with carbon worsens. Weaker bonds break faster, so bond enthalpy – not electronegativity – controls the rate. This is also shown by how quickly AgX precipitates appear in the silver nitrate/ethanol test.

Q29. Which statement about the SN2 mechanism is correct?

  • (A) It has two steps, with a carbocation intermediate
  • (B) It is a single step: the nucleophile attacks from the side opposite the leaving group as the C−X bond breaks
  • (C) The rate depends only on the nucleophile concentration
  • (D) It is fastest for tertiary substrates
Show Answer

Answer: B – SN2 is bimolecular and concerted: bond making and bond breaking happen together through a transition state, with backside attack causing inversion of configuration. The rate depends on both substrate and nucleophile.

Q30. Fluorocarbons and chlorofluorocarbons (CFCs) are very unreactive towards nucleophiles mainly because…

  • (A) the C−F bond is extremely strong, so it does not break under normal conditions
  • (B) fluorine has no lone pairs
  • (C) fluorocarbons cannot dissolve in any solvent
  • (D) fluorine is the least electronegative halogen
Show Answer

Answer: A – The C−F bond is the strongest carbon-halogen bond, so heterolytic fission is very difficult and substitution is extremely slow. Fluorine is actually the most electronegative halogen – reactivity here is governed by bond strength.

Electrophilic Substitution & Carbonyl Chemistry

Q31. In the nitration of benzene, the electrophile that attacks the ring is…

  • (A) HNO3
  • (B) NO2
  • (C) H2SO4
  • (D) NO2+, the nitronium ion
Show Answer

Answer: D – The active electrophile is the linear nitronium ion, NO2+, generated in the acid mixture. Neutral HNO3 is not electrophilic enough to attack benzene.

Q32. Concentrated sulfuric acid is mixed with concentrated nitric acid for nitration. In generating the electrophile, the sulfuric acid acts as…

  • (A) an acid: it protonates HNO3, which then loses water to form NO2+
  • (B) a base: it removes a proton from HNO3
  • (C) an oxidising agent only
  • (D) a solvent with no chemical role
Show Answer

Answer: A – Sulfuric acid is the stronger acid here: HNO3 behaves as a base, is protonated, and loses H2O to give the nitronium ion: HNO3 + H2SO4 → NO2+ + HSO4− + H2O.

Q33. Iron(III) bromide, FeBr3, is needed to brominate benzene with Br2. Its role is to…

  • (A) oxidise benzene to make it more reactive
  • (B) act as a halogen carrier: it accepts a lone pair from Br2, polarising the Br−Br bond and generating a stronger electrophile (effectively Br+)
  • (C) remove the HBr formed in the reaction
  • (D) provide Br− ions for the second step
Show Answer

Answer: B – Benzene’s pi system is less reactive than an alkene’s, so Br2 alone is not electrophilic enough. FeBr3 (a Lewis acid) polarises Br−Br, and the resulting electrophile attacks the ring; FeBr3 is regenerated when H+ is lost.

Q34. Benzene undergoes substitution rather than addition reactions with electrophiles because…

  • (A) benzene has no pi electrons
  • (B) addition reactions are always too slow at room temperature
  • (C) addition would destroy the delocalised pi system and its stabilisation; substitution replaces H by the electrophile and restores the aromatic ring
  • (D) the hydrogen atoms of benzene cannot be removed
Show Answer

Answer: C – Benzene’s delocalisation gives it extra stability (delocalisation energy). In substitution the ring is temporarily disrupted in the intermediate but the aromatic system is restored when H+ leaves; addition products would lose that stability permanently.

Q35. In a Friedel-Crafts alkylation of benzene using chloroethane and an AlCl3 catalyst, which statement is correct?

  • (A) The electrophile is a Cl• radical
  • (B) The product is chlorobenzene
  • (C) AlCl3 is used up in the reaction
  • (D) AlCl3 generates a carbocation electrophile (CH3CH2+) from the halogenoalkane, and the product is ethylbenzene
Show Answer

Answer: D – Friedel-Crafts alkylation attaches an alkyl group to the ring. AlCl3 accepts Cl− from chloroethane, producing the ethyl carbocation electrophile; the catalyst is regenerated later in the mechanism.

Q36. Ethanal reacts with HCN only slowly, but the reaction becomes practical when a little alkali or KCN is added. Why?

  • (A) The nucleophile is CN−; HCN is a weak acid and is barely ionised, so added alkali/KCN raises the CN− concentration
  • (B) The alkali protonates the carbonyl oxygen
  • (C) Alkali converts ethanal into an alkene
  • (D) HCN needs to be oxidised before it can react
Show Answer

Answer: A – In nucleophilic addition to a carbonyl, CN− does the attacking. Pure HCN contains very little CN−, so a trace of base (or KCN) is used to generate enough of the nucleophile.

Q37. The product of the addition of HCN to ethanal is…

  • (A) propanoic acid
  • (B) propanamide
  • (C) ethanenitrile
  • (D) 2-hydroxypropanenitrile, CH3CH(OH)CN – a hydroxynitrile
Show Answer

Answer: D – CN− adds to the carbonyl carbon and the oxygen is protonated, so -CN and -OH end up on the same carbon. The product is a hydroxynitrile (cyanohydrin) with one more carbon than the starting carbonyl compound.

Q38. In nucleophilic addition to a carbonyl compound, the first step is…

  • (A) the carbonyl oxygen attacks the nucleophile
  • (B) the nucleophile donates its lone pair to the δ+ carbonyl carbon, and the C=O pi electrons move onto the oxygen, forming an alkoxide intermediate
  • (C) the C=O bond breaks homolytically
  • (D) a proton is transferred to the carbonyl carbon
Show Answer

Answer: B – The carbonyl carbon is electrophilic because oxygen pulls the pi electrons towards itself. The nucleophile forms a bond to carbon while the pi pair moves onto oxygen; the alkoxide is then protonated (for example by HCN or water) to give the addition product.

Q39. Ethanoyl chloride reacts vigorously with water to give ethanoic acid and HCl. The mechanism is classified as…

  • (A) electrophilic addition
  • (B) free-radical substitution
  • (C) nucleophilic addition-elimination: water adds to the carbonyl, then Cl− is eliminated as the C=O reforms
  • (D) electrophilic substitution
Show Answer

Answer: C – Acyl chlorides react by nucleophilic addition-elimination: the nucleophile adds to the carbonyl carbon, then the tetrahedral intermediate collapses, expelling Cl− and reforming C=O. The same pattern with alcohols gives esters and with amines/ammonia gives amides; acyl chlorides are the most reactive carboxylic acid derivatives.

Q40. Which reagent will reduce a carboxylic acid to a primary alcohol?

  • (A) NaBH4
  • (B) Acidified K2Cr2O7
  • (C) Tollens’ reagent
  • (D) LiAlH4 in dry ether
Show Answer

Answer: D – LiAlH4 is a powerful reducing agent that reduces carboxylic acids (and esters) to primary alcohols, as well as aldehydes and ketones. NaBH4 is milder: it reduces aldehydes and ketones only, and does not reduce carboxylic acids or esters. K2Cr2O7 and Tollens’ are oxidising agents.

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