IB Chemistry Quiz: Understand Electron Pair Sharing Reactions
20 questions · exam conditions
0:00
Understand Electron Pair Sharing ReactionsQuestion 1 of 20

Propene reacts with hydrogen bromide, HBr. Which species acts as the electrophile in the initial step of this reaction mechanism?

The propene molecule.
A bromide ion, Br⁻.
The hydrogen atom in the H–Br molecule.
The bromine atom in the H–Br molecule.
← Back to quizzes

IB Chemistry Quiz

IB Chemistry Quiz: Understand Electron Pair Sharing Reactions

Practice Understand 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.

What this quiz covers

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

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Propene reacts with hydrogen bromide, HBr. Which species acts as the electrophile in the initial step of this reaction mechanism?

  1. The propene molecule.
  2. A bromide ion, Br⁻.
  3. The hydrogen atom in the H–Br molecule. (correct answer)
  4. The bromine atom in the H–Br molecule.
Explanation: In the electrophilic addition reaction between propene and HBr, the electron-rich π-bond of the propene molecule acts as a nucleophile. The H-Br bond is polar (Hδ+—Brδ-), making the hydrogen atom electron-deficient. The propene's π-bond attacks this electron-deficient hydrogen atom, which accepts the pair of electrons. Therefore, the hydrogen atom of the H-Br molecule is the electrophilic centre and acts as the electrophile. Propene (A) is the nucleophile. The bromide ion (B) is a nucleophile formed after the initial step. The bromine atom (D) is electron-rich (δ-) and is not the electrophile.

Question 2

An unknown reagent reacts with but-2-ene to form 2,3-dibromobutane. What is the electrophile in this reaction?

  1. A bromine atom, Br•
  2. A bromide ion, Br⁻
  3. A bromonium ion, Br⁺
  4. A bromine molecule, Br₂ (correct answer)
Explanation: The reaction is the electrophilic addition of bromine to but-2-ene. The electrophile is the species that is first attacked by the electron-rich π-bond of the alkene. As the alkene approaches a non-polar Br₂ molecule, its π-electron cloud induces a dipole in the Br-Br bond, making one bromine atom slightly positive (δ+) and thus electrophilic. The alkene attacks this δ+ bromine atom. Therefore, the Br₂ molecule itself is the reagent that acts as the source of the electrophile. A bromine atom (A) is a radical. A bromide ion (B) is a nucleophile. A bromonium ion (D) is a cyclic intermediate formed in the AHL mechanism, but the initial attacking species is the Br₂ molecule.

Question 3

A curly arrow in a reaction mechanism represents the movement of a pair of electrons. Which statement best describes the bond-breaking process shown by a curly arrow starting from a C–X bond and ending on the atom X?

  1. Homolytic fission, forming a carbon radical and an X radical.
  2. Heterolytic fission, where the carbon atom receives both electrons from the bond, forming a carbanion and X⁺.
  3. Heterolytic fission, where the atom X receives both electrons from the bond, forming a carbocation and X⁻. (correct answer)
  4. Homolytic fission, where the atom X receives both electrons from the bond, forming a carbocation and X⁻.
Explanation: A curly arrow shows the movement of an electron pair. An arrow originating from a bond and pointing to an atom (X) signifies that both electrons from the covalent bond move to that atom. This process is called heterolytic fission. Since X gains the electron pair, it becomes a negative ion (X⁻), and the carbon atom, having lost its share of the electrons, becomes a positive ion (a carbocation, C⁺). Therefore, statement C is the correct description. Statement A describes homolytic fission. Statement B describes heterolytic fission but with the electron pair moving to the carbon atom. Statement D incorrectly mixes the terms homolytic fission and heterolytic fission.

Question 4

Which molecule will undergo an addition reaction with bromine water but will NOT undergo a substitution reaction with aqueous sodium hydroxide?

  1. Cyclohexane
  2. Cyclohexene (correct answer)
  3. Chlorocyclohexane
  4. Benzene
Explanation: An addition reaction with bromine water is characteristic of a compound containing a C=C double bond (an alkene). A substitution reaction with aqueous sodium hydroxide is characteristic of a halogenoalkane. Cyclohexene contains a C=C double bond and will undergo addition with Br₂, but it is not a halogenoalkane, so it will not react with NaOH(aq). Cyclohexane (A) is a saturated alkane and reacts with neither. Chlorocyclohexane (C) is a halogenoalkane that undergoes substitution but not addition. Benzene (D) does not undergo addition with bromine water under normal conditions; it undergoes electrophilic substitution (AHL content).

Question 5

In the reaction between BF3\text{BF}_3 and NH3\text{NH}_3, the Lewis acid-base adduct F3B-NH3\text{F}_3\text{B-NH}_3 is formed. Which statement best explains why this electron-pair sharing reaction occurs spontaneously?

  1. The electronegativity difference between B and N drives the formation of an ionic bond
  2. The empty p orbital on boron accepts the lone pair from nitrogen, completing boron's octet (correct answer)
  3. The hydrogen atoms on ammonia form hydrogen bonds with the fluorine atoms on boron trifluoride
  4. The formal charges on both molecules become zero after the coordinate bond formation
Explanation: B is correct because BF₃ has an incomplete octet (only 6 electrons around B), creating an empty p orbital that can accept the lone pair from NH₃, forming a coordinate covalent bond. A is wrong because this forms a covalent coordinate bond, not an ionic bond. C is wrong because hydrogen bonding would be much weaker and wouldn't result in adduct formation. D is wrong because the formal charges don't become zero - boron gains a negative formal charge and nitrogen gains a positive formal charge.

Question 6

Consider the reaction: (CH3)3C++H2O→(CH3)3COH+\text{(CH}_3\text{)}_3\text{C}^+ + \text{H}_2\text{O} \rightarrow \text{(CH}_3\text{)}_3\text{COH}^+. Based on the electron-pair sharing mechanism, why does this reaction occur rapidly despite the positive charge on the carbocation?

  1. The positive charge is delocalized through hyperconjugation, making the carbon center more nucleophilic
  2. The methyl groups donate electron density through inductive effects, neutralizing the positive charge before reaction
  3. The tertiary carbocation undergoes rearrangement to a more stable secondary form before reacting with water
  4. Water molecules are attracted to the positive charge, and oxygen's lone pairs readily coordinate to the electron-deficient carbon (correct answer)
Explanation: When analyzing carbocation reactions, focus on the fundamental driving forces: electrostatic attraction and electron movement from electron-rich to electron-poor sites. The tertiary carbocation (CH3)3C+\text{(CH}_3\text{)}_3\text{C}^+ has an empty p-orbital, making it highly electrophilic (electron-seeking). Water molecules possess two lone pairs of electrons on oxygen, making them nucleophilic (electron-donating). The reaction occurs rapidly because opposite charges attract—the positively charged carbon strongly attracts water's electron-rich oxygen, which then donates a lone pair to form a coordinate covalent bond. This creates the protonated alcohol product. Option A incorrectly suggests hyperconjugation makes carbon more nucleophilic. While hyperconjugation does stabilize the carbocation by delocalizing the positive charge, it doesn't change carbon's fundamental electron-deficient nature—it remains electrophilic, not nucleophilic. Option B misunderstands inductive effects. The methyl groups do provide some electron density through hyperconjugation, but they don't "neutralize" the positive charge before reaction. The carbocation retains its positive charge throughout the mechanism. Option C proposes rearrangement to a secondary carbocation, which is backwards. Tertiary carbocations are more stable than secondary ones, so this rearrangement would be energetically unfavorable and wouldn't explain the rapid reaction rate. The correct answer is D—water's attraction to the positive charge and subsequent lone pair donation drives this nucleophilic attack. Study tip: In carbocation reactions, always identify the electrophilic (electron-poor) and nucleophilic (electron-rich) partners first. The mechanism follows electron flow from nucleophile to electrophile.

Question 7

When SO3\text{SO}_3 reacts with H2O\text{H}_2\text{O} to form H2SO4\text{H}_2\text{SO}_4, the initial step involves electron-pair sharing. Which factor most significantly influences the rate of this reaction?

  1. The high electronegativity of sulfur creates a strong dipole that attracts water molecules electrostatically
  2. The presence of π bonding in SO₃ makes the sulfur center highly electron-rich and nucleophilic
  3. The electron deficiency at sulfur due to formal charge distribution makes it susceptible to nucleophilic attack (correct answer)
  4. The ability of water to form multiple hydrogen bonds with the oxygen atoms in sulfur trioxide
Explanation: C is correct because in SO₃, sulfur has a formal charge and electron deficiency that makes it electrophilic and susceptible to nucleophilic attack by water's lone pairs. A is wrong because while sulfur is electronegative, the reaction mechanism depends on electrophilicity, not just electronegativity. B is incorrect because the π bonding actually makes sulfur electron-deficient, not electron-rich. D is wrong because hydrogen bonding is not the primary mechanism - the reaction involves nucleophilic attack at the electrophilic sulfur center.

Question 8

Consider the nucleophilic substitution reaction: CH3CH2Br+OH−→CH3CH2OH+Br−\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^-. If this reaction proceeds via an SN2 mechanism, what is the most likely sequence of electron pair movements?

  1. The C-Br bond breaks first, forming a carbocation, then OH⁻ donates its electron pair to carbon
  2. OH⁻ donates its electron pair to carbon while simultaneously the C-Br bonding pair moves to bromine (correct answer)
  3. The OH⁻ forms hydrogen bonds with adjacent hydrogen atoms before attacking the carbon center
  4. The electron pair from OH⁻ forms a coordinate bond with carbon while the C-Br bond remains intact
Explanation: B is correct because SN2 mechanisms involve concerted electron pair movements - the nucleophile (OH⁻) attacks carbon while the leaving group (Br⁻) departs simultaneously. A describes an SN1 mechanism with carbocation formation. C describes hydrogen bonding which is not the primary mechanism. D is incorrect because the C-Br bond must break for substitution to occur, and a stable pentavalent carbon intermediate is not formed.

Question 9

In organometallic chemistry, when CO\text{CO} coordinates to a transition metal center, both σ-donation and π-back-donation occur. Which statement correctly describes this electron-pair sharing interaction?

  1. The carbon atom donates its lone pair to the metal while the metal donates d electrons to CO's π* orbitals (correct answer)
  2. The oxygen atom forms the primary coordinate bond while carbon participates in secondary π interactions
  3. Both carbon and oxygen donate electron pairs simultaneously to create a bidentate chelating interaction
  4. The triple bond in CO breaks to allow independent coordination of both carbon and oxygen atoms
Explanation: A is correct because CO coordination involves σ-donation from carbon's lone pair to the metal and π-back-donation from filled metal d orbitals to CO's π* antibonding orbitals. B is wrong because carbon, not oxygen, forms the coordinate bond. C is incorrect because CO is a monodentate ligand that coordinates through carbon only. D is wrong because the C≡O triple bond remains intact during coordination.

Question 10

Consider the conversion of ethene to 1,2-dichloroethane using Cl₂(g). Which terms best describe this reaction?

  1. Electrophilic substitution
  2. Nucleophilic substitution
  3. Electrophilic addition (correct answer)
  4. Nucleophilic addition
Explanation: The reaction involves breaking the π-bond of the ethene molecule (an alkene) and adding two chlorine atoms across the double bond. This is an addition reaction. Because the reaction is initiated by the attack of the electron-rich π-bond of ethene on an electrophile (the polarized Cl₂ molecule), it is classified as electrophilic addition. Substitution reactions would involve replacing an atom or group, which does not happen here. Nucleophilic addition is typical for carbonyl compounds, not alkenes.

Question 11

What is the major organic product when 1-chloropropane is heated under reflux with aqueous sodium hydroxide?

  1. Propene
  2. Propan-1-ol (correct answer)
  3. Propan-2-ol
  4. Propane
Explanation: This reaction is the nucleophilic substitution of a primary halogenoalkane. The hydroxide ion (OH⁻) from sodium hydroxide acts as a nucleophile and replaces the chlorine atom. Since the chlorine is on the first carbon of 1-chloropropane, the resulting hydroxyl group will also be on the first carbon, forming propan-1-ol. Propene (A) is the product of elimination, which is favoured by hot, ethanolic hydroxide, not aqueous. Propan-2-ol (C) would form from 2-chloropropane. Propane (D) is a reduction product and is not formed in this reaction.

Question 12

In the mechanism for the reaction between hydroxide ions and 1-bromobutane, which description of curly arrows is correct for the main step?

  1. An arrow from a lone pair on the oxygen of OH⁻ to the C of the C–Br bond, and an arrow from the C–Br bond to the Br atom. (correct answer)
  2. An arrow from the C–Br bond to the C atom, and an arrow from the C atom to a lone pair on the oxygen of OH⁻.
  3. An arrow from a lone pair on the oxygen of OH⁻ to the Br atom, and an arrow from the C–Br bond to the C atom.
  4. An arrow from the H atom of the OH⁻ ion to the C of the C–Br bond, and an arrow from the C–Br bond to the Br atom.
Explanation: This is a nucleophilic substitution reaction. The nucleophile is the hydroxide ion, OH⁻. The source of the electron pair for the attack is a lone pair on the oxygen atom. This pair attacks the electron-deficient (δ+) carbon atom bonded to the bromine. This is shown by a curly arrow from the O lone pair to the C atom. Simultaneously, the C-Br bond breaks, with the more electronegative bromine atom taking both electrons. This is shown by a curly arrow from the C-Br bond to the Br atom. Option A correctly describes this electron movement. All other options show incorrect electron flow.

Question 13

During the addition reaction between propene and hydrogen chloride (HCl), which bond undergoes heterolytic fission?

  1. A C–H bond in the methyl group of propene.
  2. The C=C double bond in the propene molecule.
  3. A C–C single bond in the propene molecule.
  4. The H–Cl bond in the hydrogen chloride molecule. (correct answer)
Explanation: The reaction is initiated by the nucleophilic attack of the propene's π-bond on the electrophilic hydrogen atom of HCl. As the new C-H bond forms, the existing H-Cl bond must break. Since chlorine is much more electronegative than hydrogen, it takes both electrons from the shared pair, undergoing heterolytic fission to form a chloride ion, Cl⁻. The C-H and C-C bonds (A, D) in propene remain intact. The π-bond of the C=C bond (B) breaks, but it does so to form a new bond, it does not undergo fission in the same sense as the H-Cl bond.

Question 14

Which statement correctly compares the reaction of ethane with chlorine and the reaction of ethene with chlorine?

  1. Both are addition reactions occurring in the presence of UV light.
  2. Ethane reacts via substitution and ethene reacts via addition; both reactions occur readily at room temperature without a catalyst.
  3. Ethane undergoes free-radical substitution in UV light, while ethene undergoes electrophilic addition at room temperature. (correct answer)
  4. Ethene undergoes nucleophilic addition, while ethane undergoes free-radical substitution.
Explanation: Ethane is a saturated alkane and reacts with chlorine via a free-radical substitution mechanism, which requires initiation by UV light. Ethene is an unsaturated alkene and reacts with chlorine via an electrophilic addition mechanism, which occurs readily at room temperature and pressure without the need for UV light. Statement C accurately describes both reactions and their respective conditions. A is incorrect as ethane undergoes substitution. B is incorrect as the ethane reaction requires UV light. D is incorrect as ethene undergoes electrophilic, not nucleophilic, addition.

Question 15

In the reaction between bromoethane and cyanide ions, CN⁻, which statement correctly describes the roles of the species involved?

  1. Bromoethane acts as a nucleophile because the carbon atom bonded to bromine is electron-deficient.
  2. The cyanide ion acts as a nucleophile because it donates a lone pair of electrons to an electron-deficient carbon atom. (correct answer)
  3. Bromoethane acts as an electrophile because the bromine atom accepts a pair of electrons from the cyanide ion.
  4. The cyanide ion acts as an electrophile because it is attracted to the δ+ charge on the carbon atom in bromoethane.
Explanation: The C-Br bond in bromoethane is polar, with the carbon atom being electron-deficient (δ+) and thus electrophilic. The cyanide ion, CN⁻, has a lone pair of electrons and a negative charge, making it an electron-pair donor, which is the definition of a nucleophile. It attacks the electrophilic carbon atom. Therefore, statement B is correct. Statement A incorrectly identifies bromoethane as the nucleophile. Statement C incorrectly suggests the bromine atom accepts the electrons in the reaction; it leaves as the bromide ion. Statement D correctly identifies the attraction but incorrectly labels the cyanide ion as an electrophile; electrophiles are electron-pair acceptors.

Question 16

An alkene X with the molecular formula C₄H₈ undergoes electrophilic addition with HBr to form a single organic product Y. Y is a secondary halogenoalkane. What is the identity of alkene X?

  1. But-1-ene
  2. But-2-ene (correct answer)
  3. Methylpropene
  4. Cyclobutane
Explanation: The reaction forms a single product, which implies that the starting alkene must be symmetrical with respect to the double bond. Of the C₄H₈ alkene isomers, but-1-ene and methylpropene are unsymmetrical, while but-2-ene is symmetrical. Reaction of symmetrical but-2-ene (CH₃CH=CHCH₃) with HBr can only form one product, 2-bromobutane, as addition to either carbon of the double bond results in the same molecule. 2-bromobutane is a secondary halogenoalkane. But-1-ene (A) would form two products (major 2-bromobutane and minor 1-bromobutane). Methylpropene (C) would form a tertiary and a primary bromoalkane. Cyclobutane (D) is not an alkene.

Question 17

When 2-chloro-2-methylpropane is warmed with a concentrated solution of ammonia in ethanol, what is the principal organic product?

  1. 2-amino-2-methylpropane (correct answer)
  2. 2-methylprop-1-ene
  3. 2-methylpropan-2-ol
  4. N,N-dimethylpropan-2-amine
Explanation: This reaction involves a tertiary halogenoalkane and the nucleophile ammonia (NH₃). The ammonia molecule uses its lone pair of electrons to attack the electrophilic carbon atom, displacing the chloride ion in a nucleophilic substitution reaction. The amino group (-NH₂) replaces the chlorine atom. The product is 2-amino-2-methylpropane. 2-methylprop-1-ene (B) would be the product of an elimination reaction, which can compete but substitution is significant. 2-methylpropan-2-ol (C) would form if water or hydroxide were the nucleophile. N,N-dimethylpropan-2-amine (D) would require reaction with dimethylamine, not ammonia.

Question 18

Which pair of species contains one nucleophile and one electrophile?

  1. OH⁻ and CH₃⁺ (correct answer)
  2. H₂O and NH₃
  3. C₂H₄ and C₂H₆
  4. Br⁻ and I⁻
Explanation: A nucleophile is an electron-pair donor, often having a lone pair and/or a negative charge. An electrophile is an electron-pair acceptor, often being positively charged or having an electron-deficient centre. In pair A, OH⁻ is a nucleophile (negative charge, lone pairs), and CH₃⁺ is a carbocation, which is electron-deficient and a strong electrophile. In pair B, both H₂O and NH₃ are nucleophiles due to lone pairs on O and N, respectively. In pair C, C₂H₄ (ethene) is a nucleophile due to its π-bond, while C₂H₆ (ethane) is generally unreactive and not classified as either. In pair D, both Br⁻ and I⁻ are anions with lone pairs, making them both nucleophiles.

Question 19

Which reagent is required to convert bromoethane into ethanenitrile (CH₃CN)?

  1. Hydrogen cyanide, HCN
  2. Potassium cyanide, KCN, dissolved in ethanol (correct answer)
  3. Ammonia, NH₃, dissolved in ethanol
  4. Sodium hydroxide, NaOH, dissolved in water
Explanation: This conversion is a nucleophilic substitution reaction where the bromide ion is replaced by the cyanide ion (CN⁻), which acts as the nucleophile. Potassium cyanide (KCN) is an ionic salt that provides a high concentration of CN⁻ ions. It is typically dissolved in ethanol, which also serves as a solvent for the organic bromoethane. Hydrogen cyanide (A) is a covalent molecule and a weak acid, making it a poor source of the nucleophilic CN⁻ ions. Ammonia (C) would produce ethylamine. Sodium hydroxide (D) would produce ethanol.

Question 20

Which species would be the most effective nucleophile in a substitution reaction with 1-iodopropane?

  1. H₂O
  2. NH₃
  3. CH₃COOH
  4. OH⁻ (correct answer)
Explanation: Nucleophilic strength is a measure of how readily a species donates an electron pair to form a new bond. Generally, anions are stronger nucleophiles than their neutral counterparts. OH⁻ is an anion and a strong nucleophile. H₂O and NH₃ are both neutral molecules with lone pairs, making them weaker nucleophiles than OH⁻. CH₃COOH is a carboxylic acid and acts as a proton donor (acid), not a nucleophile. Therefore, the hydroxide ion, OH⁻, would be the most effective nucleophile among the choices.