IB Chemistry Quiz: Apply Electron Pair Sharing Reactions
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Apply Electron Pair Sharing ReactionsQuestion 1 of 20

The reaction of an unknown alkyne with two equivalents of hydrogen chloride gas produces 2,2-dichlorobutane as the sole product. What is the identity of the alkyne?

But-1-yne
But-2-yne
Propyne
Ethyne
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Electron Pair Sharing Reactions

Practice Apply Electron Pair Sharing Reactions in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Electron Pair Sharing Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Question 1

The reaction of an unknown alkyne with two equivalents of hydrogen chloride gas produces 2,2-dichlorobutane as the sole product. What is the identity of the alkyne?

  1. But-1-yne (correct answer)
  2. But-2-yne
  3. Propyne
  4. Ethyne
Explanation: The reaction is a double electrophilic addition. For but-1-yne (CH₃CH₂C≡CH), the first addition of HCl follows Markovnikov's rule, adding H to C1 and Cl to C2 to form 2-chlorobut-1-ene. The second addition of HCl to this alkene also follows Markovnikov's rule: the H adds to C1 (which has more H's), forming a carbocation at C2, and the Cl⁻ then attacks C2. This results exclusively in 2,2-dichlorobutane. But-2-yne would produce 2,3-dichlorobutane, and the other alkynes would produce products with different carbon chain lengths.

Question 2

The rates of nucleophilic substitution for a series of 1-halobutanes were measured under identical SN2 conditions. Which compound would be expected to react the fastest?

  1. 1-fluorobutane
  2. 1-chlorobutane
  3. 1-bromobutane
  4. 1-iodobutane (correct answer)
Explanation: In an SN2 reaction, the rate is influenced by the nature of the leaving group. A better leaving group is one that is a more stable anion and corresponds to a weaker C-X bond. The C-I bond is the weakest among the carbon-halogen bonds, and the iodide ion (I⁻) is the most stable halide ion due to its large size, which disperses the negative charge effectively. Therefore, 1-iodobutane will have the fastest rate of substitution.

Question 3

Which reaction is best classified as an electrophilic addition?

  1. CH₄ + Cl₂ → CH₃Cl + HCl (in the presence of UV light)
  2. CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr
  3. C₆H₆ + Br₂ → C₆H₅Br + HBr (in the presence of a catalyst)
  4. CH₂=CH₂ + H₂O → CH₃CH₂OH (in the presence of an acid catalyst) (correct answer)
Explanation: An electrophilic addition reaction involves an electrophile attacking an electron-rich C=C or C≡C bond, causing the pi bond to break and the atoms of a small molecule to be added across it. In the acid-catalyzed hydration of ethene (D), the H⁺ electrophile attacks the pi bond, followed by the addition of water, resulting in an alcohol. Option A is free-radical substitution. Option B is nucleophilic substitution. Option C is electrophilic substitution.

Question 4

The rate of an SN1 reaction is found to be dependent on the concentration of the halogenoalkane but independent of the concentration of the nucleophile. Which mechanistic step provides the best explanation for this kinetic observation?

  1. The heterolytic fission of the carbon-halogen bond to form a carbocation, which is the slow rate-determining step. (correct answer)
  2. The attack of the nucleophile on the carbocation, which is the fast step and does not involve the halogenoalkane.
  3. The formation of a five-coordinate transition state involving both the halogenoalkane and the nucleophile.
  4. The rearrangement of the initially formed carbocation to a more stable carbocation before nucleophilic attack.
Explanation: The rate of a reaction is determined by its slowest step, the rate-determining step (RDS). In an SN1 mechanism, the RDS is the unimolecular, heterolytic cleavage of the C-X bond to form a carbocation intermediate. Since only the halogenoalkane molecule is involved in this step, the reaction rate depends solely on its concentration. The subsequent attack by the nucleophile is fast and does not affect the overall rate.

Question 5

When excess aqueous ammonia is added to a solution containing copper(II) ions, the deep blue complex ion [Cu(NH₃)₄(H₂O)₂]²⁺ is formed. Which statement correctly describes the roles of the species in this reaction?

  1. Cu²⁺ acts as a nucleophile and NH₃ acts as an electrophile.
  2. Cu²⁺ acts as a Lewis base and NH₃ acts as a Lewis acid.
  3. Cu²⁺ acts as a Lewis acid and NH₃ acts as a Lewis base. (correct answer)
  4. Cu²⁺ acts as a Brønsted-Lowry acid and NH₃ acts as a Brønsted-Lowry base.
Explanation: In the formation of a complex ion, ligands form coordinate (dative covalent) bonds with a central metal ion. The ammonia (NH₃) molecules are ligands, using the lone pair of electrons on the nitrogen atom to form a bond with the copper(II) ion. Species that donate an electron pair are Lewis bases (or nucleophiles). The copper(II) ion accepts these electron pairs, so it is a Lewis acid (or an electrophile).

Question 6

Ethene undergoes rapid electrophilic addition with bromine water, while benzene only undergoes electrophilic substitution with bromine in the presence of a catalyst. What is the best explanation for this difference in reactivity?

  1. The pi electrons in ethene are localized, making them more available for reaction than the delocalized pi electrons in benzene. (correct answer)
  2. Benzene is a non-polar molecule, which prevents the induction of a dipole in the approaching bromine molecule.
  3. The C-C bonds in benzene are shorter and stronger than the C=C bond in ethene, making them harder to break.
  4. The activation energy for addition to benzene is higher because it would require breaking C-H bonds.
Explanation: The key difference lies in the nature of the pi electron systems. In ethene, the two pi electrons are localized in a region of high electron density between two carbon atoms, making them readily available to attack an electrophile. In benzene, the six pi electrons are delocalized over the entire ring, a configuration that confers significant aromatic stability. An addition reaction would destroy this stable aromatic system, so a substitution reaction, which preserves it, is favoured. This requires a much stronger electrophile, hence the need for a catalyst.

Question 7

Benzene is converted to nitrobenzene using a mixture of concentrated nitric acid and concentrated sulfuric acid. What is the role of the sulfuric acid and what is the electrophilic species that attacks the benzene ring?

  1. Role: Catalyst; Electrophile: NO₂⁻
  2. Role: Dehydrating agent; Electrophile: HNO₃
  3. Role: Brønsted-Lowry acid; Electrophile: NO₂⁺ (correct answer)
  4. Role: Oxidizing agent; Electrophile: HSO₄⁻
Explanation: In the nitration of benzene, sulfuric acid is a stronger acid than nitric acid. It acts as a Brønsted-Lowry acid, protonating the nitric acid molecule. The protonated nitric acid then loses a molecule of water to generate the nitronium ion, NO₂⁺. This highly reactive nitronium ion is the electrophile that attacks the electron-rich benzene ring in the subsequent step. The sulfuric acid is regenerated, so it also acts as a catalyst.

Question 8

Propanone reacts with hydrogen cyanide, in the presence of a catalytic amount of KCN, to form a cyanohydrin. In the first mechanistic step, the cyanide ion attacks the carbonyl carbon. Which statement correctly describes the roles of the reactants in this step?

  1. Propanone acts as a Lewis base and the carbonyl carbon is a nucleophilic centre.
  2. The carbonyl carbon is an electrophilic centre and the cyanide ion is a nucleophile. (correct answer)
  3. Propanone acts as a Brønsted-Lowry acid and the cyanide ion is a Brønsted-Lowry base.
  4. The carbonyl oxygen is an electrophilic centre and the cyanide ion is a Lewis acid.
Explanation: This reaction is a nucleophilic addition. The carbonyl group (C=O) is polar, with the carbon atom being electron-deficient (δ+) and the oxygen atom being electron-rich (δ-). The electron-deficient carbonyl carbon is therefore an electrophilic centre. The cyanide ion (CN⁻) has a lone pair of electrons and a negative charge, making it an excellent electron-pair donor, which is the definition of a nucleophile (and a Lewis base). The nucleophilic CN⁻ attacks the electrophilic carbon.

Question 9

The electrophilic addition of HBr to 3-methylbut-1-ene produces 2-bromo-2-methylbutane as the major product, rather than the expected 2-bromo-3-methylbutane. What is the best explanation for the formation of this product?

  1. The reaction proceeds via a primary carbocation which is the most stable and readily formed intermediate.
  2. A secondary carbocation is formed initially and then rearranges via a 1,2-hydride shift to form a more stable tertiary carbocation. (correct answer)
  3. The reaction follows an anti-Markovnikov addition pathway due to the branched structure of the alkene substrate.
  4. Steric hindrance from the methyl group on the third carbon prevents the bromide ion from attacking that position.
Explanation: According to Markovnikov's rule, the initial addition of H⁺ to the double bond forms a secondary carbocation at C2. However, this secondary carbocation can rearrange to a more stable tertiary carbocation. A hydrogen atom with its pair of electrons (a hydride ion) shifts from the adjacent C3 to C2. This 1,2-hydride shift transforms the secondary carbocation into a more stable tertiary carbocation at C3. The bromide nucleophile then attacks this tertiary carbocation, leading to the formation of 2-bromo-2-methylbutane as the major product.

Question 10

During the SN2 reaction between OH⁻ and CH₃Cl, a curly arrow is drawn from the lone pair on the oxygen of OH⁻ to the carbon atom of CH₃Cl, and a second curly arrow is drawn from the C-Cl bond to the Cl atom. What do these two arrows represent happening in a single step?

  1. The heterolytic fission of the C-Cl bond followed by the formation of a C-O bond in a subsequent step.
  2. The formation of a C-O bond and the homolytic fission of the C-Cl bond to form radical products.
  3. The formation of a C-O bond and the heterolytic fission of the C-Cl bond occurring concertedly. (correct answer)
  4. The formation of an O-H bond and the breaking of a C-H bond on the methyl group.
Explanation: The SN2 mechanism is a concerted process, meaning that bond-making and bond-breaking occur simultaneously in a single transition state. The curly arrow from the OH⁻ lone pair to the carbon shows the formation of the new C-O bond. The curly arrow from the C-Cl bond to the chlorine atom shows the simultaneous heterolytic cleavage of the C-Cl bond, with the electron pair moving onto the chlorine to form a chloride ion leaving group.

Question 11

The reaction of 2-chloro-2-methylpropane with aqueous sodium hydroxide is compared to the reaction of 1-chlorobutane under the same conditions. Which statement correctly predicts the dominant mechanism and a key feature for these reactions?

  1. Both react via an SN2 mechanism, but 1-chlorobutane is faster due to less steric hindrance around the reaction centre.
  2. 2-chloro-2-methylpropane reacts via SN1 forming a racemic mixture, while 1-chlorobutane reacts via SN2 with inversion of configuration. (correct answer)
  3. 1-chlorobutane reacts via an SN1 mechanism due to primary carbocation stability, while 2-chloro-2-methylpropane reacts via SN2.
  4. Both react via an SN1 mechanism, but 2-chloro-2-methylpropane is faster due to the stability of its tertiary carbocation intermediate.
Explanation: 2-chloro-2-methylpropane is a tertiary halogenoalkane, which favours the SN1 mechanism due to the formation of a stable tertiary carbocation intermediate. This planar intermediate can be attacked from either side, leading to a racemic mixture of products if the starting material were chiral. 1-chlorobutane is a primary halogenoalkane, which favours the SN2 mechanism due to minimal steric hindrance, allowing for backside attack by the nucleophile. The SN2 mechanism proceeds via a single concerted step that results in the inversion of stereochemical configuration.

Question 12

In the formation of [Co(NH3)6]3+[Co(NH_3)_6]^{3+} from Co3+Co^{3+} and NH3NH_3, six coordinate bonds are formed. If this complex is then treated with excess CN−CN^- ions under appropriate conditions, ligand exchange occurs to form [Co(CN)6]3−[Co(CN)_6]^{3-}. What is the primary thermodynamic factor driving this ligand substitution reaction?

  1. Cyanide ions are smaller than ammonia molecules, reducing steric repulsion and allowing closer approach to the cobalt center for stronger coordinate bonding
  2. Cyanide is a stronger σ-donor and π-acceptor than ammonia, forming more stable coordinate bonds through enhanced orbital overlap and back-bonding stabilization (correct answer)
  3. The negative charge on cyanide ions creates favorable electrostatic interactions with the positively charged cobalt center compared to neutral ammonia molecules
  4. Cyanide ions have higher electronegativity than nitrogen in ammonia, resulting in more polar Co-ligand bonds and greater ionic stabilization of the complex
Explanation: CN−CN^- is both a stronger σ-donor and π-acceptor ligand than NH3NH_3. The π-acceptor ability of cyanide allows for back-bonding (metal d-orbitals donating to π* orbitals of CN), providing additional stabilization beyond simple σ-donation. Option A is incorrect because size is not the primary factor. Option C oversimplifies by focusing only on electrostatic interactions. Option D incorrectly describes the bonding as primarily ionic rather than coordinate covalent.

Question 13

The complex ion [Fe(CN)6]4−[Fe(CN)_6]^{4-} exhibits different magnetic properties compared to [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+} despite both containing Fe2+Fe^{2+}. Based on electron-pair sharing principles and ligand field effects, what best explains the difference in magnetic behavior between these two complexes?

  1. Cyanide ligands participate in π-bonding while water ligands only participate in σ-bonding, resulting in different hybridization states and magnetic properties for the iron center
  2. Water molecules are neutral ligands while cyanide ions are charged, creating different electrostatic environments that affect the magnetic moments of unpaired d-electrons
  3. The stronger coordinate bonds in [Fe(CN)6]4−[Fe(CN)_6]^{4-} reduce the effective nuclear charge on iron, while weaker bonds in [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+} maintain higher nuclear charge
  4. Cyanide ions cause greater d-orbital splitting than water molecules, leading to low-spin configuration in [Fe(CN)6]4−[Fe(CN)_6]^{4-} and high-spin configuration in [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+} (correct answer)
Explanation: When you encounter questions about magnetic properties of transition metal complexes, focus on how different ligands affect d-orbital energy splitting and electron pairing. Both complexes contain Fe2+Fe^{2+} with a d6d^6 electron configuration, but they exhibit dramatically different magnetic behaviors due to ligand field strength. Cyanide is a strong field ligand that creates large energy gaps between d-orbitals, while water is a weak field ligand producing smaller gaps. In [Fe(CN)6]4−[Fe(CN)_6]^{4-}, the strong field forces electrons to pair up in lower energy orbitals before occupying higher ones (low-spin, diamagnetic). In [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}, the weak field allows electrons to occupy higher orbitals singly before pairing occurs (high-spin, paramagnetic). Option A incorrectly emphasizes π-bonding versus σ-bonding and hybridization. While cyanide does participate in π-backbonding, the key factor is field strength, not bonding type. Option B focuses on ligand charge rather than field strength—many neutral ligands are strong field (like CO) while some charged ligands are weak field. Option C misrepresents how coordinate bonding affects nuclear charge; the magnetic behavior stems from orbital splitting, not changes in effective nuclear charge on iron. Remember: strong field ligands (CN⁻, CO) create low-spin complexes with paired electrons, while weak field ligands (H₂O, halides) create high-spin complexes with maximum unpaired electrons. The ligand field strength, not charge or bonding type, determines magnetic properties.

Question 14

Consider the following equilibrium involving electron-pair sharing: PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) ⇌ PCl_3(g) + Cl_2(g). When this system is at equilibrium at 250°C, analysis shows that phosphorus maintains different coordination geometries in the reactant versus product. What best explains the molecular geometry changes and their relationship to the electron-pair sharing mechanism?

  1. PCl5PCl_5 has trigonal bipyramidal geometry due to five bonding pairs, while PCl3PCl_3 has trigonal pyramidal geometry due to three bonding pairs and one lone pair on phosphorus (correct answer)
  2. PCl5PCl_5 has square pyramidal geometry due to expanded octet formation, while PCl3PCl_3 has trigonal planar geometry due to three equivalent P-Cl bonds
  3. PCl5PCl_5 has trigonal bipyramidal geometry due to sp3dsp^3d hybridization, while PCl3PCl_3 has tetrahedral geometry due to sp3sp^3 hybridization of phosphorus
  4. PCl5PCl_5 has octahedral geometry due to coordination with five chlorine atoms, while PCl3PCl_3 has trigonal pyramidal geometry due to three bonding pairs and one lone pair
Explanation: PCl5PCl_5 has trigonal bipyramidal geometry (five bonding pairs, no lone pairs). PCl3PCl_3 has trigonal pyramidal geometry because phosphorus has three bonding pairs and one lone pair. Option B is wrong about PCl5PCl_5 geometry (not square pyramidal) and PCl3PCl_3 geometry (not trigonal planar due to lone pair). Option C incorrectly states PCl3PCl_3 is tetrahedral (ignores lone pair effect). Option D incorrectly states PCl5PCl_5 is octahedral.

Question 15

The compound SF4SF_4 can act as both a Lewis acid and Lewis base depending on reaction conditions. When SF4SF_4 reacts with SbF5SbF_5, it forms [SF3]+[SbF6]−[SF_3]^+[SbF_6]^-. When SF4SF_4 reacts with CsFCsF, it forms Cs+[SF5]−Cs^+[SF_5]^-. What best explains SF4SF_4's dual acid-base behavior in terms of electron-pair sharing mechanisms?

  1. SF4SF_4 can expand its coordination sphere beyond eight electrons due to sulfur's position in period 3, accommodating additional coordinate bonds in either direction
  2. The electronegativity difference between sulfur and fluorine creates partial charges that can interact favorably with both electron-rich and electron-deficient species
  3. SF4SF_4 has both vacant d-orbitals for accepting electron pairs and lone pairs on sulfur for donating electron pairs, allowing it to function as either Lewis acid or base (correct answer)
  4. The see-saw geometry of SF4SF_4 creates both electron-deficient and electron-rich regions in the molecule, enabling coordination with different types of Lewis acids and bases
Explanation: When you encounter questions about amphoteric behavior in Lewis acid-base chemistry, focus on the specific electron-pair mechanisms that enable a compound to both donate and accept electron pairs. SF4SF_4 demonstrates classic amphoteric behavior because sulfur possesses both the structural features needed for Lewis acid and base activity. As a Lewis base with CsFCsF, the lone pair on sulfur donates to form the [SF5]−[SF_5]^- anion. As a Lewis acid with SbF5SbF_5, sulfur accepts electron density, leading to fluoride transfer and formation of [SF3]+[SF_3]^+. This dual capability stems from sulfur having both lone pairs available for donation and vacant d-orbitals that can accept electron pairs. Option A incorrectly emphasizes coordination sphere expansion rather than the fundamental electron-pair donation/acceptance mechanisms. While sulfur can exceed the octet, this doesn't directly explain the amphoteric behavior observed in these specific reactions. Option B focuses on electronegativity differences creating partial charges, but this describes general polarity rather than the specific Lewis acid-base mechanisms. Partial charges alone don't explain why SF4SF_4 can both donate and accept electron pairs. Option D mentions molecular geometry creating different regions, but the see-saw shape doesn't inherently create distinct electron-rich and electron-poor zones that would enable amphoteric behavior. The geometry is a consequence of electron arrangement, not the cause of Lewis acid-base activity. Remember: amphoteric Lewis behavior requires both electron-donating capability (lone pairs) and electron-accepting capability (vacant orbitals). Look for both features when identifying amphoteric compounds.

Question 16

When SO3SO_3 reacts with H2OH_2O to form H2SO4H_2SO_4, the mechanism involves initial coordinate bond formation followed by proton transfer. If SO3SO_3 instead reacts with NH3NH_3, a different product forms. Comparing these two reactions, what best explains the different outcomes in terms of electron-pair sharing mechanisms?

  1. Water acts as both a Lewis base and Brønsted base, allowing for subsequent proton transfer, while ammonia can only act as a Lewis base, forming a stable coordinate complex (correct answer)
  2. Sulfur trioxide acts as a Lewis acid with water but as a Lewis base with ammonia due to different electronegativity relationships between the reactants
  3. Water's higher electronegativity promotes immediate bond formation and rearrangement to H2SO4H_2SO_4, while ammonia's lower electronegativity results in simple coordinate bond formation
  4. The oxygen in water can accommodate proton transfer after coordinate bond formation due to its ability to form hydrogen bonds, while nitrogen in ammonia lacks this capability
Explanation: With water, SO3SO_3 first forms a coordinate bond (H₂O acting as Lewis base), then rapid proton rearrangement occurs to form H2SO4H_2SO_4 because water can act as both Lewis and Brønsted base. With NH3NH_3, coordinate bond formation occurs but subsequent proton transfer is not favorable, leading to a different product. Option B incorrectly states SO3SO_3 changes its acid/base role. Option C incorrectly attributes the difference to electronegativity rather than proton transfer capability. Option D incorrectly focuses on hydrogen bonding rather than acid-base behavior.

Question 17

In the mechanism for the reaction between ethene and bromine, a curly arrow is drawn from the C=C double bond to one of the bromine atoms in a Br₂ molecule. What does this curly arrow represent?

  1. The homolytic fission of the C=C pi bond to form two radical centres on the carbon atoms.
  2. The movement of a pair of electrons from the pi bond to form a new covalent bond with a bromine atom. (correct answer)
  3. The movement of a single electron from the sigma bond of ethene to the bromine molecule.
  4. The heterolytic fission of the Br-Br bond, with the electron pair moving to the δ+ bromine atom.
Explanation: A curly arrow in a mechanism represents the movement of a pair of electrons. When the arrow originates from a bond (like the C=C pi bond), it signifies that those bonding electrons are being used to form a new bond. In this case, the electron-rich pi bond of ethene acts as a nucleophile, attacking the electrophilic bromine atom. The arrow shows the formation of a new C-Br bond using the two electrons from the pi bond.

Question 18

Which species would be the most effective nucleophile in a polar, protic solvent such as water?

  1. F⁻
  2. Cl⁻
  3. Br⁻
  4. I⁻ (correct answer)
Explanation: In polar, protic solvents, the nucleophile is surrounded by solvent molecules (solvated). Smaller ions with high charge density, like F⁻, are more strongly solvated through hydrogen bonding, which hinders their ability to attack an electrophilic centre. Larger ions, like I⁻, have a more diffuse charge and are less tightly solvated. This, combined with its high polarizability, makes I⁻ the strongest nucleophile among the halides in a protic solvent.

Question 19

Consider the reaction of (CH₃)₃C-I with sodium ethoxide (NaOCH₂CH₃) in ethanol. Which option correctly identifies the major organic product and the dominant mechanism?

  1. (CH₃)₃C-OCH₂CH₃ via an SN2 mechanism.
  2. (CH₃)₂C=CH₂ via an E2 mechanism. (correct answer)
  3. (CH₃)₃C-OCH₂CH₃ via an SN1 mechanism.
  4. (CH₃)₂C=CH₂ via an E1 mechanism.
Explanation: The substrate, (CH₃)₃C-I, is a tertiary haloalkane, which is sterically hindered and disfavours the SN2 mechanism. The reagent, sodium ethoxide, is a strong, hindered base. When a tertiary haloalkane reacts with a strong base, the major pathway is elimination, specifically the bimolecular E2 mechanism, which does not require the formation of a carbocation. This reaction removes a proton from a beta-carbon and the iodide leaving group simultaneously to form an alkene, 2-methylpropene, (CH₃)₂C=CH₂.

Question 20

Which species is the most effective electrophile?

  1. Br₂
  2. HBr
  3. NO₂⁺ (correct answer)
  4. H₂O
Explanation: An electrophile is a species that is electron-deficient and accepts an electron pair. While Br₂ and HBr can act as electrophiles, they are neutral molecules with polar or polarizable bonds. The nitronium ion, NO₂⁺, is a cation with a full positive charge, making it extremely electron-deficient and therefore a very strong electrophile. H₂O is a nucleophile due to its lone pairs on oxygen.