Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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R3.4.1 Nucleophile
Nucleophile
A reactant that forms a new covalent bond to its reaction partner (the electrophile) by donating BOTH bonding electrons. A nucleophile is electron-rich and has an electron pair available to donate, usually a lone pair. Nucleophiles may be negatively charged (OH⁻, CN⁻, Cl⁻, Br⁻) or neutral molecules with a lone pair (H₂O, NH₃); a charge is not required.
Students often think A nucleophile must carry a negative charge, so neutral molecules cannot be nucleophiles. In fact Yes. A nucleophile needs an electron pair to donate, not a charge. The O atom of H₂O and the N atom of NH₃ each have lone pairs they can donate to an electron-deficient carbon atom.
Students often think The names are swapped: an electrophile is the electron-rich species that donates electrons, and a nucleophile is the species that receives them. In fact The nucleophile donates the pair and the electrophile accepts it. A nucleophile is electron-rich; an electrophile is electron-poor.
R3.4.2 Nucleophilic substitution
Nucleophilic substitution
A reaction in which a nucleophile donates an electron pair to form a new bond to a carbon atom as another bond to that carbon breaks, releasing a leaving group. For a halogenoalkane: R–X + Nu⁻ → R–Nu + X⁻, e.g. CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH + Br⁻. The nucleophile bonds to the carbon atom, which carries a partial positive charge because the halogen is more electronegative.
Leaving group
The atom or group that departs when a bond breaks during a substitution, taking BOTH electrons of that bond with it. In the substitution of a halogenoalkane the leaving group is the halide ion, X⁻ (Cl⁻, Br⁻ or I⁻).
Students often think An electron-rich species bonds to the halogen atom of its partner, because the halogen is the functional group where the reaction happens. In fact The carbon atom bonded to the halogen. That carbon carries a partial positive charge because the halogen is more electronegative.
Students often think The hydroxide ion always reacts as a Brønsted–Lowry base, removing H⁺ from whatever it meets. In fact No. OH⁻ acts as a nucleophile: it donates a lone pair to the carbon bonded to Br, forming propan-1-ol, and Br⁻ leaves.
R3.4.3 Heterolytic fission
Heterolytic fission
Breaking of a covalent bond in which both bonding electrons remain with one of the two fragments. When a neutral molecule breaks this way, the fragment that takes the pair becomes negatively charged and the other becomes positively charged, so two ions form: H–Br → H⁺ + Br⁻. The more electronegative atom usually takes the pair.
Homolytic fission
Breaking of a covalent bond in which each fragment keeps one of the two bonding electrons, forming two radicals (e.g. Br–Br → 2Br•, as in radical substitution under UV light). It is the contrast to heterolytic fission and is shown with fish-hook (single-headed) arrows.
Curly arrow
A double-headed arrow showing the movement of an ELECTRON PAIR, never of an atom or ion. It starts at the source of the pair (a lone pair or a bond) and ends where the pair goes (an atom, to form a lone pair or a new bond). For H–Cl → H⁺ + Cl⁻ the arrow starts at the H–Cl bond and ends on Cl.
Students often think When any covalent bond breaks, the pair splits evenly, one electron to each atom, so neutral atoms or radicals are always formed. In fact No. That is homolytic fission. In heterolytic fission one fragment keeps both electrons, so ions form rather than radicals.
Students often think The fragment that takes the bonding pair becomes positively charged (and the one that loses its share becomes negative). In fact Negative. It gains an extra electron compared with its share in the bond, so it becomes an anion (when the bond breaks in a neutral molecule); the other fragment becomes a cation.
R3.4.4 Electrophile
Electrophile
A reactant that forms a new covalent bond to its reaction partner (the nucleophile) by ACCEPTING both bonding electrons from that partner. Electrophiles are electron-deficient; they may be positively charged (H⁺, NO₂⁺) or neutral molecules (HBr, Br₂). A positive charge is not required.
Students often think An electrophile must carry a positive charge, so neutral molecules cannot be electrophiles, and a non-polar molecule such as Br₂, with no δ+ end, cannot be one either. In fact Yes. An electrophile needs to be able to accept an electron pair, not to carry a charge. HBr (through its δ+ H atom) and Br₂ both accept an electron pair from the C=C bond of an alkene.
R3.4.5 Electrophilic addition
Electrophilic addition
The typical reaction of alkenes. The C=C double bond is a region of high electron density, so it attracts electrophiles; the reagent adds across the double bond, which becomes a single bond, and no by-product is formed (all the atoms of the reagent end up in the addition product). The C=C bond is STRONGER overall than a C–C single bond; its reactivity comes from the accessible, electron-rich second bond, not from weakness.
Reactions of alkenes with water, halogens and hydrogen halides
Water (as steam, with an acid catalyst such as phosphoric acid) adds H and OH to form an alcohol: CH₂=CH₂ + H₂O → CH₃CH₂OH. A halogen adds one halogen atom to each carbon to form a dihalogenoalkane: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br; the orange-brown colour of bromine disappears, which is used as a test for C=C. A hydrogen halide adds H and X to form a halogenoalkane: CH₂=CH₂ + HBr → CH₃CH₂Br.
Students often think Alkenes are reactive because the C=C double bond is weaker than a C–C single bond. In fact No. A C=C bond is stronger than a C–C single bond (its bond enthalpy is higher), though less than twice as strong. Alkenes are reactive because the double bond is a region of high electron density, not because it is weak.
Students often think Atoms react in order to complete their octets, so the carbon atoms of C=C react because they need more electrons for a full outer shell. In fact No. Each carbon atom in an alkene already has eight outer electrons. Alkenes react with electrophiles because the C=C double bond is a region of high electron density.
R3.4.6 Lewis acid and Lewis base HL
Lewis acid and Lewis base
A Lewis acid is an electron-pair acceptor; a Lewis base is an electron-pair donor. The definitions need no hydrogen: BF₃ and AlCl₃ are Lewis acids because B and Al have an incomplete valence shell, and H⁺ is a Lewis acid when it accepts a lone pair from H₂O to form H₃O⁺. In organic chemistry the C=C bond of an alkene acts as a Lewis base toward H⁺.
Students often think An acid must contain hydrogen and transfer H⁺, so a species without hydrogen, or a reaction with no H⁺ transfer, cannot involve an acid or a base. In fact Yes, a Lewis acid. The Lewis definition is based on accepting an electron pair, not on donating H⁺. BF₃ and AlCl₃ accept lone pairs because B and Al have incomplete valence shells.
Students often think A Lewis acid donates an electron pair, just as a Brønsted–Lowry acid donates a proton. In fact It accepts an electron pair. The Lewis base donates the pair.
R3.4.7 Coordination bond HL
Coordination bond
A covalent bond in which both shared electrons come from the same atom (a Lewis base donating a lone pair to a Lewis acid). Once formed it is a covalent bond like any other: in NH₄⁺ all four N–H bonds are identical. In the Lewis formula of H₃N–BF₃ the N–B bond is a coordination bond, formed from the nitrogen lone pair.
Nucleophiles as Lewis bases, electrophiles as Lewis acids
Every nucleophile donates an electron pair, so it is a Lewis base; every electrophile accepts an electron pair, so it is a Lewis acid. When OH⁻ attacks CH₃CH₂Br, OH⁻ is both the nucleophile and the Lewis base; when an alkene reacts with H⁺, H⁺ is both the electrophile and the Lewis acid.
Students often think If a Lewis formula shows a positive charge on one atom and a negative charge on the next, the bond between them is ionic. In fact No. The formal charges are a bookkeeping device showing that N has donated a lone pair and B has accepted it. The N–B bond is a covalent coordination bond.
R3.4.8 Ligand and complex ion HL
Ligand and complex ion
A ligand is a molecule or ion with a lone pair that it donates to a transition element cation, forming a coordination bond. Ligands may be neutral (H₂O, NH₃) or negative (Cl⁻, CN⁻). A complex ion is the central cation together with its ligands, written in square brackets, e.g. [Cu(H₂O)₆]²⁺, [Fe(CN)₆]³⁻, [CoCl₄]²⁻.
Charge on a complex ion
The charge on a complex ion is the sum of the charge on the metal ion and the charges on all the ligands. Neutral ligands add nothing. Examples: Fe³⁺ + 6CN⁻ gives (+3) + 6(−1) = −3, [Fe(CN)₆]³⁻; Cu²⁺ + 4NH₃ + 2H₂O gives +2, [Cu(NH₃)₄(H₂O)₂]²⁺; Co²⁺ + 4Cl⁻ gives −2, [CoCl₄]²⁻.
Students often think A bond between a metal (or metal ion) and a non-metal is always ionic, so ligands and species such as AlCl₃ are held by ionic bonds. In fact No. Each ligand donates a lone pair to the metal ion, forming a coordination bond, which is a covalent bond.
Students often think The charge on a complex ion is the charge on the central metal ion; the ligands do not change it. In fact Only when all the ligands are neutral. The complex charge is the metal ion charge plus the sum of the ligand charges.
R3.4.9 SN2 mechanism HL
SN2 mechanism
Bimolecular nucleophilic substitution, typical of primary halogenoalkanes. It is concerted: in ONE step the nucleophile's lone pair forms the new bond to carbon while the C–X bond breaks, through a transition state (not an intermediate). The nucleophile attacks from the side opposite the halogen (backside attack). Rate = k[halogenoalkane][nucleophile]. The '2' refers to the two species in the rate-determining step, not to the number of steps.
SN1 mechanism
Unimolecular nucleophilic substitution, typical of tertiary halogenoalkanes. It has TWO steps: (1) slow heterolytic fission of C–X, forming a carbocation and X⁻; (2) fast attack on the carbocation by the nucleophile. Rate = k[halogenoalkane]. Tertiary halogenoalkanes react this way because the tertiary carbocation is stabilized by three alkyl groups and the alkyl groups hinder backside attack. Secondary halogenoalkanes can react by both SN1 and SN2.
Stereospecific nature of SN2 (inversion of configuration)
Because the nucleophile attacks from the side opposite the leaving group, the other three groups on the carbon turn inside out, like an umbrella in the wind. When the carbon is a chiral centre, one enantiomer of the halogenoalkane gives one enantiomer of the product, with the configuration inverted. SN2 is therefore stereospecific.
Carbocation
An ion with a positively charged carbon atom that has only three bonds and six outer electrons, e.g. (CH₃)₃C⁺. It is an intermediate in SN1 reactions and in electrophilic addition. It is planar around the positive carbon. Carbocations are classed as primary, secondary or tertiary by the number of alkyl groups on the positive carbon.
Students often think SN1 is a one-step reaction and SN2 is a two-step reaction, as the numbers in the names indicate. In fact No. They give the molecularity of the rate-determining step: one species in SN1, two in SN2. SN1 has two steps; SN2 has one.
Students often think Substitution is a direct swap: the incoming group moves into the exact place of the group that leaves, in a single event with no change in arrangement. In fact No. In SN2 the nucleophile attacks from the side opposite the leaving group, and the configuration is inverted. In benzene the electrophile first forms a positive intermediate before H⁺ is lost.
R3.4.10 Effect of the halogen on the rate of substitution HL
Effect of the halogen on the rate of substitution
For otherwise identical halogenoalkanes, the rate of substitution increases in the order chloroalkane < bromoalkane < iodoalkane. The C–X bond enthalpy decreases from C–Cl to C–Br to C–I, so the C–I bond is broken most easily and iodide is the best leaving group. Bond strength, not bond polarity, is the deciding factor: C–Cl is the most polar of the three but reacts slowest.
Students often think The more polar the C–X bond, the more δ+ the carbon and the faster the substitution, so chloroalkanes react fastest. In fact No. The rate is controlled mainly by how easily the C–X bond breaks. C–I has the lowest bond enthalpy, so iodoalkanes react fastest and chloroalkanes slowest.
Students often think A larger halogen atom gets in the way of the nucleophile, so iodoalkanes react slowest and chloroalkanes fastest. In fact No. Iodoalkanes react fastest of the chloro-, bromo- and iodoalkanes, even though iodine is the largest atom, because the C–I bond is the weakest.
R3.4.11 Mechanism of addition of a halogen HL
Mechanism of addition of a halogen
As Br₂ approaches the electron-rich C=C bond, the Br–Br electron pair is repelled, inducing a dipole (Brδ+–Brδ−). A curly arrow goes from the C=C bond to the Brδ+ atom and one from the Br–Br bond to the Brδ− atom, forming a carbocation and Br⁻. Br⁻ then donates a lone pair to the positive carbon (curly arrow from Br⁻ to C⁺), giving CH₂BrCH₂Br. (The IB treatment shows a carbocation intermediate; a cyclic bromonium ion is a more accurate description, beyond the guide.)
Mechanism of addition of a hydrogen halide
H–Br is polar, Hδ+–Brδ−. A curly arrow goes from the C=C bond to the H atom and one from the H–Br bond to the Br atom, giving a carbocation (CH₃CH₂⁺ from ethene) and Br⁻. Br⁻ then donates a lone pair to the positive carbon, giving CH₃CH₂Br.
Mechanism of acid-catalysed addition of water
The C=C bond donates its electron pair to H⁺ (from the acid catalyst, present as H₃O⁺), forming a carbocation. A lone pair on the O atom of H₂O forms a bond to the positive carbon, and the resulting protonated alcohol loses H⁺, which regenerates the catalyst: CH₂=CH₂ + H₂O → CH₃CH₂OH.
Students often think A catalyst that takes part in a mechanism is used up: the H⁺ added in the first step stays in the product, so the acid is consumed. In fact No. H⁺ is used in the first step (it bonds to the alkene, forming a carbocation) and an H⁺ is released in the last step (from the protonated alcohol), so the catalyst is regenerated.
R3.4.12 Carbocation stability HL
Carbocation stability
Stability order: tertiary > secondary > primary. Alkyl groups are electron-releasing (positive inductive effect): each alkyl group pushes electron density toward the positive carbon, spreading its charge. The more alkyl groups on the positive carbon, the more stable the carbocation and the more readily it forms.
Major product with an unsymmetrical alkene
When HX or H₂O adds to an unsymmetrical alkene, H adds to the carbon atom that already has MORE hydrogen atoms, so that the more stable (more substituted) carbocation forms; X or OH then bonds to that carbon. CH₃CH=CH₂ + HBr gives mainly 2-bromopropane via CH₃CH⁺CH₃; (CH₃)₂C=CH₂ + H₂O gives mainly 2-methylpropan-2-ol via (CH₃)₃C⁺. This outcome is known as Markovnikov's rule.
Students often think Alkyl groups withdraw electrons, so the more alkyl groups on the positive carbon, the less stable the carbocation; primary carbocations are the most stable. In fact Tertiary > secondary > primary. Alkyl groups release electron density toward the positive carbon (positive inductive effect), spreading the charge.
Students often think The H atom adds to the carbon atom of C=C that has fewer hydrogen atoms. In fact The end carbon, CH₂, which already has more H atoms. This gives the more stable secondary carbocation, CH₃CH⁺CH₃, so Br bonds to the middle carbon: 2-bromopropane.
R3.4.13 Electrophilic substitution of benzene HL
Electrophilic substitution of benzene
Benzene reacts with electrophiles by substitution, not addition, because substitution keeps the stable delocalized ring of π electrons in the product. Mechanism with a charged electrophile E⁺: a curly arrow from the delocalized π electrons to E⁺ forms a C–E bond and a positively charged intermediate in which the delocalization is partly broken; a curly arrow from the C–H bond into the ring then removes H⁺ and restores the delocalized ring: C₆H₆ + E⁺ → C₆H₅E + H⁺.
Students often think Benzene has C=C double bonds, so it reacts with electrophiles by addition in the same way as an alkene. In fact No. Benzene undergoes electrophilic substitution: an H atom is replaced and the delocalized ring is kept. Addition would destroy the delocalization, which is energetically unfavourable.
Students often think Benzene does not react with electrophiles, as it is saturated like a cycloalkane. In fact No. Benzene is unsaturated and does react with electrophiles, but by substitution rather than addition; for example it is nitrated by NO₂⁺.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement describes how a nucleophile forms a new bond to its reaction partner?
Answer and reasoning
It donates both electrons of the new covalent bond. — A nucleophile is an electron-pair donor: it supplies both electrons of the new bond, usually from a lone pair, and the electrophile supplies none.
It accepts the two electrons of the new covalent bond. — Accepting the pair is what an electrophile does. 'Nucleophile' means nucleus-seeking: it is the electron-rich partner and donates the pair.
It gives one electron, and its partner gives the other. — That is how a covalent bond is pictured in dot-and-cross diagrams such as H₂, but in a nucleophile–electrophile reaction the nucleophile supplies both electrons of the new bond.
It transfers electrons, forming an ionic bond. — 'Donate' here means supply for sharing. The pair ends up shared between the two atoms, so the new bond is covalent (e.g. the C–O bond in an alcohol), not ionic.
2 1-bromopropane is warmed with aqueous sodium hydroxide. Which equation represents the reaction that occurs?
Answer and reasoning
CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂⁻ + HOBr — This has OH⁻ bonding to the bromine atom. Br is the δ− end of the C–Br bond; an electron-rich nucleophile is attracted to the δ+ carbon, so the C–O bond forms and Br⁻ leaves.
CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH + Br⁻ — OH⁻ donates a lone pair to the δ+ carbon bonded to Br, and the C–Br bonding pair leaves with Br as the bromide ion. The product is propan-1-ol.
CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CHBr⁻ + H₂O — This treats OH⁻ as a base removing H⁺ from carbon. With a halogenoalkane, OH⁻ acts as a nucleophile: it donates its lone pair to the carbon bonded to Br, substituting for the halogen.
CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH⁻ + Br — This splits the C–Br bonding pair one electron to each atom, so Br leaves as an atom. In substitution the C–Br bond breaks heterolytically: Br takes both electrons and leaves as Br⁻, and the alcohol is neutral.
3 Which statement describes the heterolytic fission of a covalent bond?
Answer and reasoning
One bonding electron goes to each fragment, so radicals form. — An even split of the bonding pair is homolytic fission, which gives radicals. Heterolytic fission gives both electrons to one fragment, forming ions.
Both bonding electrons go to one fragment, so two ions form. — In heterolytic fission one fragment keeps the bonding pair and becomes an anion; the other becomes a cation, e.g. H–Br → H⁺ + Br⁻.
Both bonding electrons go to one fragment, which becomes positive. — A fragment that takes both electrons gains negative charge, so it becomes the anion. The fragment left without the pair becomes the cation.
It is the breaking of a bond that joins two different elements. — 'Hetero-' describes the two different fragments formed (a cation and an anion), not the atoms in the bond. A bond between different atoms can break either way: C–Cl breaks heterolytically when a nucleophile substitutes for Cl, but homolytically in UV light, giving radicals.
4 Ethene reacts with bromine:
CH₂=CH₂ + Br₂ → CH₂BrCH₂Br
Which reactant acts as the electrophile?
Answer and reasoning
Br₂, as it accepts an electron pair from the C=C bond — Br₂ accepts the electron pair donated by the electron-rich C=C bond, forming a C–Br bond. A neutral molecule can be an electrophile.
Ethene, as its C=C bond is rich in electrons — It is true that the C=C bond is electron-rich, but that makes ethene the electron-pair DONOR, the nucleophile. The electrophile is the electron-pair acceptor, Br₂.
Neither, as neither reactant carries a positive charge — An electrophile need not be charged. Br₂ accepts an electron pair from the C=C bond, so it is the electrophile, even though it is neutral.
Ethene, as its C atoms need electrons for an octet — Each carbon in C=C already has an octet. The C=C bond is electron-rich and donates a pair; Br₂ accepts it, so Br₂ is the electrophile.
5 Aluminium chloride reacts with chloride ions:
AlCl₃ + Cl⁻ → AlCl₄⁻
Which statement identifies the Lewis acid in this reaction? HL
Answer and reasoning
Cl⁻, as it donates an electron pair to AlCl₃ — Donating the electron pair makes Cl⁻ the Lewis BASE. In Lewis theory the acid is the acceptor, AlCl₃.
AlCl₃, as it accepts an electron pair from Cl⁻ — Al in AlCl₃ has only six outer electrons and accepts a lone pair from Cl⁻, forming a coordination bond in AlCl₄⁻. An electron-pair acceptor is a Lewis acid.
Neither, as no H⁺ ion is transferred at all — That is the Brønsted–Lowry test. Lewis theory needs no hydrogen: AlCl₃ accepts an electron pair, so it is a Lewis acid.
AlCl₃, as Al forms an ionic bond with Cl⁻ — AlCl₃ is the Lewis acid, but not because an ionic bond forms. The Cl⁻ lone pair is shared with Al in a covalent coordination bond.
6 In the reaction CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻, which statement about OH⁻ is correct? HL
Answer and reasoning
It is a nucleophile, and so it is a Lewis acid. — A Lewis acid accepts an electron pair. OH⁻ donates one, so as a nucleophile it is a Lewis BASE; electrophiles are the Lewis acids.
It is an electrophile, as it is rich in electrons. — Being electron-rich makes OH⁻ a nucleophile, the electron-pair donor. An electrophile is electron-deficient and accepts a pair.
It is a nucleophile but no base, as it gains no H⁺. — Gaining H⁺ is the Brønsted–Lowry test. In Lewis terms OH⁻ is a base because it donates an electron pair, which it does here.
It is a nucleophile, and so it is also a Lewis base. — OH⁻ donates a lone pair to the carbon atom. Every nucleophile is an electron-pair donor, so it is also a Lewis base.
7 Which statement describes the bonding between H₂O and Cu²⁺ in the complex ion [Cu(H₂O)₆]²⁺? HL
Answer and reasoning
Cu²⁺ donates an electron pair to each H₂O, forming a coordination bond. — The metal cation is the electron-pair acceptor (Lewis acid). The donors are the ligands, each giving a lone pair from oxygen.
Each H₂O donates a lone pair to Cu²⁺, forming a coordination bond. — Each water ligand donates a lone pair from its oxygen atom to Cu²⁺. Six coordination bonds form, and the ligands act as Lewis bases.
Each H₂O is held to Cu²⁺ by an ionic bond, as Cu is a metal. — 'Metal + non-metal = ionic' does not apply here. Each H₂O donates a lone pair that is shared with Cu²⁺, forming a covalent coordination bond.
Each H₂O and the Cu²⁺ give one electron to form a covalent bond. — In a coordination bond both electrons come from one atom. Here the oxygen of each water supplies both; Cu²⁺ supplies none.
8 1-chlorobutane, 1-bromobutane and 1-iodobutane are each warmed with aqueous sodium hydroxide under identical conditions. Which reacts fastest, and why? HL
Answer and reasoning
1-chlorobutane, as its C–Cl bond is the most polar — C–Cl is the most polar, but polarity is not what controls the rate here. The C–Cl bond is the strongest and the hardest to break, so 1-chlorobutane is the slowest.
1-iodobutane, as its C–I bond is the weakest — The C–X bond must break for substitution to occur. Bond enthalpy falls from C–Cl to C–Br to C–I, so C–I breaks most easily and 1-iodobutane reacts fastest.
1-chlorobutane, as the small Cl blocks attack least — The size of the halogen is not what controls this rate. C–I is the weakest C–X bond and breaks most easily, so 1-iodobutane reacts fastest even though iodine is the largest.
None; all react equally fast, as OH⁻ attacks C — OH⁻ does attack carbon, but the C–X bond must also break, and its strength depends on the halogen. The rates differ: iodo > bromo > chloro.
9 Ethene reacts with hydrogen bromide by electrophilic addition. Which describes the curly arrows in the first step of the mechanism? HL
Answer and reasoning
From the H atom of HBr to the C=C bond, and from Br to the H–Br bond — These arrows start at the electron-poor atoms. A curly arrow starts at the electron pair: at the C=C bond and at the H–Br bond.
From the C=C bond to the Br of HBr, and from the H–Br bond to the H — Br is the δ− end of H–Br. The electron-rich C=C bond is attracted to the δ+ H, and the H–Br pair goes to the more electronegative Br, forming Br⁻.
From the C=C bond to the H of HBr, and from the H–Br bond to Br — The electron-rich C=C bond donates a pair to the δ+ H atom, forming a C–H bond, while the H–Br bonding pair moves onto Br. This gives CH₃CH₂⁺ and Br⁻; Br⁻ then bonds to the carbocation.
Fish-hook arrows split H–Br into H• and Br•, which add to C=C — Electrophilic addition is not a radical reaction. The H–Br bond breaks heterolytically, Br taking the pair, while the C=C pair forms the C–H bond.
10 2-methylpropene, (CH₃)₂C=CH₂, reacts with water in the presence of an acid catalyst. What is the major organic product? HL
Answer and reasoning
2-methylpropan-1-ol, as H adds to the C with fewer H — H⁺ adds to the carbon with MORE H atoms (CH₂), so that the tertiary carbocation forms. OH then ends up on the central carbon: 2-methylpropan-2-ol.
2-methylpropane-1,2-diol, as OH adds to each carbon — Water supplies one H and one OH, not two OH groups. H adds to one carbon and OH to the other, giving 2-methylpropan-2-ol as the major product.
2-methylpropan-2-ol, via the tertiary carbocation — H⁺ adds to the CH₂ end, giving the tertiary carbocation (CH₃)₃C⁺, the most stable possible. Water bonds to it and loses H⁺, giving mainly (CH₃)₃COH.
A 1:1 mixture of the two possible alcohols — The two possible carbocations differ greatly in stability: tertiary vs primary. The tertiary one forms far faster, so 2-methylpropan-2-ol is the major product.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
27 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 2-bromo-2-methylpropane reacts with water:
(CH₃)₃CBr + H₂O → (CH₃)₃COH + HBr
Which reactant acts as the nucleophile, and why?
Answer and reasoning
Neither, as neither reactant carries a negative charge — A nucleophile needs an electron pair to donate, not a negative charge. Both reactants are neutral, but a lone pair on the O atom of H₂O forms the new C–O bond, so water is the nucleophile.
(CH₃)₃CBr, as its C atom accepts a pair of electrons — Accepting an electron pair is the role of the electrophile. The carbon of (CH₃)₃CBr is the electron-deficient site that is attacked; the donor, H₂O, is the nucleophile.
H₂O, as its O atom and the C atom each give one electron — H₂O is the nucleophile, but not for this reason. The new C–O bond is made from a lone pair on oxygen; the carbon supplies no electrons.
H₂O, as a lone pair on its O atom forms the new bond to carbon — The O atom of water has lone pairs; one is donated to the electron-deficient carbon atom, forming the C–O bond of the alcohol. A neutral molecule can be a nucleophile.
2 In the reaction CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻ a new C–O bond forms. Where do the two electrons of the new C–O bond come from?
Answer and reasoning
One from the O atom and one from the C atom — Dot-and-cross diagrams show each atom giving one electron, but here the nucleophile supplies both. The carbon's electrons in the C–Br bond leave with Br⁻.
Both from the C–Br bond, which becomes the new C–O bond — The C–Br pair is not handed over to O: it moves onto Br, which leaves as Br⁻ carrying both electrons. The new C–O pair is a lone pair donated by the O of OH⁻.
Both from a lone pair on the O atom of the OH⁻ ion — OH⁻ is the nucleophile: a lone pair on its oxygen is donated to the δ+ carbon and becomes the C–O bonding pair. The C–Br pair leaves with Br⁻.
Both from the C atom, which donates them to the O — This makes the halogenoalkane carbon the donor. The carbon is δ+ and electron-deficient, so it accepts the pair; the electron-rich OH⁻ donates it.
3 Hydrogen chloride undergoes heterolytic fission to form H⁺ and Cl⁻. Which describes the curly arrow for this change?
Answer and reasoning
It starts at the H–Cl bond and ends on the H atom. — If the pair went to H, H would become H⁻ and Cl would become Cl⁺. The product H⁺ shows that H has lost its share of the pair, so the arrow ends on Cl.
It starts on the Cl atom and points back to the H–Cl bond. — An arrow starting on Cl and ending on the bond would show electrons moving INTO the bond, which forms a bond rather than breaking one. Start at the source of the pair: the H–Cl bond.
It starts on the H atom and shows the H⁺ ion moving away. — A curly arrow never shows an atom or ion moving; it shows an electron pair. The pair in the H–Cl bond moves to Cl, and H⁺ is what is left behind.
It starts at the H–Cl bond and ends on the Cl atom. — A curly arrow shows an electron pair moving from its source to its destination. The H–Cl bonding pair moves onto Cl, which becomes Cl⁻, leaving H⁺.
4 Ammonia reacts with hydrogen ions:
NH₃ + H⁺ → NH₄⁺
Which species acts as the electrophile?
Answer and reasoning
NH₃, as its nitrogen atom carries a lone pair of electrons — Having a lone pair to donate makes NH₃ the nucleophile (the electron-pair donor). The electrophile is the acceptor, H⁺.
Neither, as N and H each supply one electron to the bond — H⁺ has no electrons, so it cannot supply one. Both electrons of the new N–H bond come from the nitrogen lone pair; H⁺ accepts them and is the electrophile.
H⁺, as it accepts the lone pair on the nitrogen atom — H⁺ has no electrons and accepts the lone pair of nitrogen to form the new N–H bond. A positively charged species that accepts a pair is an electrophile.
H⁺, as it forms an ionic bond with the nitrogen atom of NH₃ — H⁺ is the electrophile, but the bond it forms is covalent: the nitrogen lone pair is shared between N and H. NH₄⁺ has four identical covalent N–H bonds.
5 Why are alkenes susceptible to attack by electrophiles?
Answer and reasoning
The C=C double bond is weaker than a C–C single bond. — C=C is stronger than C–C (it has a higher bond enthalpy). Alkenes react because of the high electron density of the double bond, not because it is weak.
The C=C double bond is a region of high electron density. — The double bond concentrates electron density between and around the two carbon atoms, and this electron-rich region attracts electron-pair acceptors.
The carbon atoms of C=C are deficient in electrons. — An electrophile seeks electrons, so it attacks an electron-RICH site. The carbon atoms of C=C are surrounded by high electron density, not a deficit.
The C=C carbon atoms need more electrons for an octet. — Each carbon in C=C shares four electron pairs and already has an octet. The reason is the high electron density of the double bond.
6 Ethene is bubbled through a solution of bromine in an organic solvent at room temperature, in the dark. Which statement is correct?
Answer and reasoning
The bromine loses its colour as CH₂BrCH₂Br forms by addition. — Br₂ adds across the C=C bond: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br. The orange-brown colour of bromine disappears; this is the test for C=C.
The bromine loses its colour as CH₂=CHBr forms by substitution. — Replacing an H atom and keeping C=C is how alkanes react (with UV light). Alkenes react by addition, and the product 1,2-dibromoethane has no C=C bond.
No reaction occurs, as Br₂ has no positive charge. — An electrophile need not be positive. Neutral Br₂ accepts an electron pair from the C=C bond, so the reaction occurs and the colour disappears.
No reaction occurs, as UV light is needed to split Br₂. — UV light is needed for radical substitution of alkanes. Alkenes react with Br₂ rapidly in the dark, which is why bromine is used to test for C=C.
7 But-2-ene, CH₃CH=CHCH₃, undergoes hydration when heated with steam in the presence of an acid catalyst. Which statement is correct?
Answer and reasoning
Water adds across the C=C bond, forming butane-2,3-diol. — A diol would need an OH group on both carbons, but water supplies only one OH. One carbon receives H and the other OH, giving butan-2-ol.
An OH group replaces an H atom, forming but-2-en-2-ol. — Alkenes react by addition, not substitution. The C=C bond becomes a single bond as H and OH add across it, giving butan-2-ol.
H atoms from water add across C=C, forming butane. — Adding hydrogen to form an alkane is hydrogenation (H₂ with a catalyst). Hydration adds both parts of water, H and OH, giving butan-2-ol.
Water adds across the C=C bond, forming butan-2-ol. — H adds to one carbon of C=C and OH to the other: CH₃CH=CHCH₃ + H₂O → CH₃CH(OH)CH₂CH₃, butan-2-ol.
8 Ethene is mixed with hydrogen bromide gas at room temperature. Which statement is correct?
Answer and reasoning
HBr adds across the C=C bond, forming CH₂BrCH₂Br. — HBr has only one Br atom, so it cannot put Br on both carbons. One carbon gains H and the other gains Br, giving CH₃CH₂Br.
A Br atom replaces one H atom, forming CH₂=CHBr. — Substitution that keeps C=C is not how alkenes react. HBr adds across the double bond, giving CH₃CH₂Br with no C=C bond.
HBr adds across the C=C bond, forming CH₃CH₂Br. — H adds to one carbon and Br to the other: CH₂=CH₂ + HBr → CH₃CH₂Br, bromoethane.
No reaction occurs, as the HBr molecule is neutral. — A neutral molecule can be an electrophile. HBr accepts an electron pair from the C=C bond at its δ+ H atom, and bromoethane forms.
9 In the first step of the acid-catalysed addition of water to ethene, H⁺ bonds to one carbon atom of the C=C bond, forming CH₃CH₂⁺. Which species acts as the Lewis base in this step? HL
Answer and reasoning
Neither, as ethene has no lone pair to donate — A Lewis base needs an available electron pair, not necessarily a lone pair. The electron pair of the C=C second bond is donated to H⁺.
H⁺, as it accepts the electron pair of the C=C bond — Accepting an electron pair makes H⁺ the Lewis ACID. The Lewis base is the donor, ethene.
Ethene, as its C=C electrons form the new C–H bond — The electron pair of the second bond of C=C is donated to H⁺ and becomes the new C–H bond. The electron-pair donor, ethene, is the Lewis base; H⁺ is the Lewis acid.
Ethene, as it gives two electrons to H⁺ in an ionic bond — Ethene is the Lewis base, but the bond formed is covalent: the donated pair is shared between C and H in the new C–H bond.
10 Ammonia reacts with boron trifluoride to form H₃NBF₃. The Lewis formula of the product is often drawn with a positive formal charge on N and a negative formal charge on B. Which statement about the new N–B bond is correct? HL
Answer and reasoning
It is a coordination bond; both electrons came from N — N in NH₃ has a lone pair and B in BF₃ has only six outer electrons. N donates its lone pair to B, forming a coordination bond; the formal charges record this donation.
It is a coordination bond; both of its electrons came from B — BF₃ is the Lewis acid, the acceptor. B has no lone pair to donate; the pair comes from the nitrogen of NH₃, the Lewis base.
It is a covalent bond; N and B each gave an electron — Each atom giving one electron is the dot-and-cross picture of an ordinary covalent bond. Here both electrons come from the N lone pair, so it is a coordination bond.
It is an ionic bond, as N carries + and B carries − — The + and − are formal charges in one molecule, not the charges of separate ions. N and B share the donated pair, so the bond is covalent (coordination).
12 2-bromo-2-methylpropane, (CH₃)₃CBr, reacts with aqueous hydroxide ions by the SN1 mechanism. Which describes the first step? HL
Answer and reasoning
The C–Br bond breaks heterolytically, forming (CH₃)₃C⁺ and Br⁻. — The slow first step of SN1 is heterolysis of C–Br: Br takes the bonding pair (curly arrow from the C–Br bond to Br), giving a tertiary carbocation and Br⁻. OH⁻ attacks the carbocation in the second, fast step.
OH⁻ attacks the C atom as the C–Br bond breaks, in a single step. — A single concerted step is the SN2 mechanism. The '1' in SN1 refers to one species in the rate-determining step; SN1 has two steps, starting with formation of the carbocation.
The C–Br bond breaks homolytically, giving radicals (CH₃)₃C• and Br•. — Substitution by a nucleophile involves heterolytic fission. Br, the more electronegative atom, takes both bonding electrons and leaves as Br⁻, forming a carbocation, not radicals.
The C–Br bond breaks heterolytically, forming (CH₃)₃C⁻ and Br⁺ ions. — Br takes the bonding pair, and gaining the pair makes it negative: Br⁻. The carbon, having lost its share, becomes positive: (CH₃)₃C⁺.
13 A single enantiomer of 2-bromobutane, CH₃CHBrCH₂CH₃, reacts with aqueous hydroxide ions by the SN2 mechanism. Which statement about this reaction is correct? HL
Answer and reasoning
OH⁻ takes the exact place of Br, so the configuration is retained. — Substitution is not a same-side swap. The nucleophile must approach from the back, away from Br, so the configuration is inverted.
A carbocation forms first, so a racemic mixture is produced. — A carbocation forms in SN1, not SN2. The '2' means two species in the rate-determining step; SN2 is one concerted step with no carbocation, so a single inverted enantiomer forms.
OH⁻ attacks opposite Br, so the configuration is inverted. — In SN2 the nucleophile approaches the δ+ carbon from the side opposite the leaving group. As Br⁻ leaves, the other three groups flip over, so a single enantiomer of butan-2-ol forms with inverted configuration: SN2 is stereospecific.
A stable five-bonded intermediate forms, then Br⁻ leaves. — The five-coordinate species is a transition state, an energy maximum, not an intermediate that can exist. SN2 is a single step.
14 Which statement about the mechanisms of nucleophilic substitution of halogenoalkanes is correct? HL
Answer and reasoning
Secondary ones react by SN1 and SN2, as both paths are feasible. — Primary halogenoalkanes react by SN2 and tertiary by SN1. A secondary carbocation is of intermediate stability and backside attack is only partly hindered, so both mechanisms occur.
Primary ones react by SN1, as their carbocations are most stable. — Primary carbocations are the LEAST stable, because alkyl groups release electrons and a primary carbon has only one. Primary halogenoalkanes react by SN2.
Tertiary ones react by SN1, which takes place in one step. — Tertiary halogenoalkanes do react by SN1, but SN1 has two steps: formation of the carbocation, then attack by the nucleophile. The '1' is not a count of steps.
Primary ones react via a stable intermediate with five bonds to C. — Primary halogenoalkanes react by SN2, which has a transition state (an energy maximum), not a stable intermediate. The reaction is one step.
15 Under identical conditions, which reacts faster with aqueous hydroxide ions, 1-bromobutane or 1-chlorobutane, and why? HL
Answer and reasoning
1-chlorobutane, as the C–Cl bond is more polar than C–Br — C–Cl is more polar, but the rate is controlled by how easily the C–X bond breaks. C–Cl is stronger, so 1-chlorobutane reacts more slowly.
1-chlorobutane, as the smaller Cl atom blocks attack less — The nucleophile attacks from the side opposite the halogen, so the halogen's size does not block it. The weaker C–Br bond makes 1-bromobutane faster.
Neither; both react at the same rate, as OH⁻ attacks C — OH⁻ does attack carbon, but the C–X bond must also break, and its strength depends on the halogen. C–Br is weaker than C–Cl, so 1-bromobutane reacts faster.
1-bromobutane, as the C–Br bond is weaker than the C–Cl bond — The C–Br bond enthalpy is lower than that of C–Cl, so it breaks more easily and bromide is the better leaving group. 1-bromobutane reacts faster.
16 Ethene reacts with bromine by electrophilic addition. How does the non-polar Br₂ molecule come to act as an electrophile? HL
Answer and reasoning
Br₂ first ionizes into Br⁺ and Br⁻, and Br⁺ accepts the C=C electron pair. — Br₂ does not need to ionize first: a neutral molecule can be an electrophile. The Br–Br bond is polarized by the approaching C=C bond and breaks as the C–Br bond forms.
UV light splits Br₂ into Br atoms, and each Br atom adds to a carbon. — UV light is needed for radical substitution of alkanes. Alkenes react with Br₂ in the dark by electrophilic addition, via an induced dipole and heterolytic fission.
One Br atom donates an electron pair to C=C, forming a C–Br bond. — Br₂ is the electrophile, so it accepts the pair; the C=C bond is the donor. The curly arrow goes from the C=C bond to the Brδ+ atom.
The electron-rich C=C induces a dipole in Br₂, and Brδ+ accepts the C=C pair. — As Br₂ approaches, the C=C electron density repels the Br–Br bonding pair, making the near Br δ+. A curly arrow from C=C to Brδ+ and one from Br–Br to Brδ− give a carbocation and Br⁻, which then bonds to it.
17 In the acid-catalysed addition of water to ethene, the first step forms the carbocation CH₃CH₂⁺. Which describes what happens next? HL
Answer and reasoning
A lone pair on the O of H₂O bonds to C⁺, then H⁺ is lost, regenerating the catalyst. — Water is the nucleophile: an O lone pair forms the C–O bond, giving CH₃CH₂OH₂⁺. This loses H⁺ (curly arrow from the O–H bond to O), giving ethanol and regenerating the H⁺ catalyst.
A lone pair on the O of H₂O bonds to C⁺, and the H⁺ stays, so the acid is used up. — The first part is right, but the catalyst is not consumed. CH₃CH₂OH₂⁺ loses H⁺ to give ethanol, so the H⁺ used in the first step is regenerated.
The C⁺ ion donates an electron pair to the O atom of H₂O, forming a C–O bond. — The carbocation has an empty orbital and no pair to donate; it is the electrophile. The O atom of H₂O donates a lone pair to the positive carbon.
H₂O first ionizes to OH⁻, as only a negative ion can bond to the C⁺ ion. — A neutral molecule with a lone pair can be a nucleophile. In the acidic mixture H₂O itself bonds to C⁺; OH⁻ is not needed.
18 Propene, CH₃CH=CH₂, reacts with hydrogen bromide. What is the major product, and why? HL
Answer and reasoning
1-bromopropane, as H adds to the carbon that has fewer H atoms — The rule is reversed. H adds to the carbon with MORE H atoms (the CH₂ end) because that gives the more stable secondary carbocation, so Br ends up on C2.
2-bromopropane, as it forms via the more stable secondary carbocation — H adds to the CH₂ end, giving the secondary carbocation CH₃CH⁺CH₃, stabilized by two electron-releasing alkyl groups. Br⁻ bonds to it, giving mainly 2-bromopropane.
1-bromopropane, as it forms via the more stable primary carbocation — Primary carbocations are the least stable: alkyl groups release electrons, and a primary carbon has only one. The secondary carbocation forms, giving 2-bromopropane.
Equal amounts of both, as H adds to either carbon atom with equal chance — The two directions give carbocations of different stability. The secondary carbocation forms much faster, so 2-bromopropane is the major product.
19 Which statement about the reaction of benzene with electrophiles is correct? HL
Answer and reasoning
It undergoes addition across a C=C bond, in the same way as an alkene. — Benzene has a delocalized ring, not three separate C=C bonds. Addition would destroy the delocalization, so benzene reacts by substitution.
It does not react at all, as its ring is saturated, like a cycloalkane. — Benzene is unsaturated and reacts with strong electrophiles such as NO₂⁺, by substitution. Its stability changes the type of reaction, not whether it reacts.
It undergoes substitution in a single step, with no intermediate ion formed. — Substitution is not a direct swap. E⁺ first bonds to the ring, forming a positive intermediate; H⁺ is then lost to restore the delocalized ring.
It undergoes substitution; the product keeps the delocalized ring. — Benzene's delocalized π ring is especially stable. Substitution replaces an H atom and restores the ring, whereas addition would destroy it.
20 Benzene reacts with a charged electrophile, E⁺. Which sequence describes the mechanism? HL
Answer and reasoning
Ring π electrons bond to E⁺, forming a cation; loss of H⁺ restores the ring. — A curly arrow from the delocalized ring to E⁺ forms the C–E bond and a positive intermediate. A curly arrow from the C–H bond into the ring then releases H⁺ and restores delocalization: C₆H₆ + E⁺ → C₆H₅E + H⁺.
Ring π electrons bond to E⁺, forming a cation; a nucleophile then adds to it. — Adding a nucleophile completes an alkene-style addition and would leave the ring non-aromatic. Benzene's intermediate loses H⁺ instead, restoring the stable ring.
E⁺ donates an electron pair to the ring, forming an anion; H⁻ is then lost. — E⁺ is the electrophile, so it accepts an electron pair from the ring. The intermediate is a cation, and it loses H⁺, not H⁻.
E⁺ swaps directly with an H atom in one step; the ring stays delocalized. — Substitution is not a one-step swap. E⁺ first bonds to the ring, forming a cation in which delocalization is partly broken; H⁺ is then lost.
21 1-bromobutane, CH₃CH₂CH₂CH₂Br, and 2-bromo-2-methylpropane, (CH₃)₃CBr, are each warmed with aqueous hydroxide ions. Which statement about their mechanisms is correct? HL
Answer and reasoning
1-bromobutane reacts by SN1 via a primary carbocation; (CH₃)₃CBr by SN2 in a single step — Primary carbocations are the least stable: alkyl groups release electrons, and a primary carbon has only one. So 1-bromobutane reacts by one-step SN2, and (CH₃)₃CBr, which forms the stable tertiary carbocation, reacts by SN1.
1-bromobutane reacts by SN2 in two separate steps; (CH₃)₃CBr by SN1 in a single step — The mechanisms are matched to the right compounds, but the step counts are reversed. The 2 in SN2 means two species in the rate-determining step of a one-step reaction; SN1 has two steps, carbocation formation and then attack by OH⁻.
1-bromobutane reacts by SN2 in a single step; (CH₃)₃CBr by SN1 via a carbocation — The primary halogenoalkane reacts by SN2: OH⁻ attacks the δ+ carbon from the side opposite Br as the C–Br bond breaks, in one concerted step through a transition state. The tertiary one reacts by SN1: its three methyl groups hinder attack from the back, and C–Br breaks first to give the stable tertiary carbocation (CH₃)₃C⁺, which OH⁻ then attacks.
1-bromobutane reacts by SN2 via a stable intermediate; (CH₃)₃CBr by SN1 via a carbocation — The tertiary part is right, but the five-coordinate species in SN2 is a transition state, an energy maximum, not an intermediate that can exist. SN2 of 1-bromobutane is a single step.
22 Water can undergo heterolytic fission of one O–H bond to form H⁺ and OH⁻. The diagram shows four attempts, P to S, to draw this change with a curly arrow. Which drawing is correct?
Answer and reasoning
Drawing Q — This arrow sends the bonding pair to H, which would form H⁻ and OH⁺. Gaining the pair makes a fragment negative, not positive, so to form OH⁻ the pair must go to the OH group.
Drawing P — A curly arrow starts where the electron pair is (the H–OH bond) and ends where it goes. Both bonding electrons go to oxygen, so OH takes the pair and becomes OH⁻, leaving H with no electrons: H⁺.
Drawing R — Each fish-hook arrow moves ONE electron, so R shows homolytic fission: one electron to each fragment, giving the radicals H• and •OH, not ions. To form H⁺ and OH⁻ both electrons must go to OH, shown by a single double-barbed arrow from the bond to OH.
Drawing S — A curly arrow is drawn from the electrons, not from the atom that receives them. It starts at the H–OH bonding pair and ends on the OH group; it is not drawn from OH back toward the bond.
23 The diagram shows the reactants in the reaction BF₃ + F⁻ → BF₄⁻, with a curly arrow. BF₃ is drawn as a displayed formula; all four lone pairs of F⁻ are shown. Which statement about the reaction is correct? HL
Answer and reasoning
F⁻ is the Lewis acid, since it is the species that donates the electron pair — A Lewis acid ACCEPTS an electron pair. F⁻ donates the pair shown by the arrow, so it is the Lewis base; BF₃, whose B atom accepts the pair, is the Lewis acid.
Neither species is an acid or a base, since no H⁺ ion is transferred — Lewis theory does not require hydrogen. Any electron-pair donor is a Lewis base and any acceptor a Lewis acid, so BF₃ (acceptor) and F⁻ (donor) are a Lewis acid–base pair.
B and F each supply one electron to the new bond, as in the other three B–F bonds — The arrow starts at a lone pair on F⁻, so both electrons of the new bond come from F⁻; B contributes none. Once formed, this coordination bond is the same as the other B–F bonds, but its electrons came from one atom.
F⁻ is the Lewis base, since it supplies both electrons of the coordination bond — The arrow starts at a lone pair on F⁻ and ends on B, which has only six electrons and an empty orbital. F⁻ donates the pair (Lewis base), BF₃ accepts it (Lewis acid), and a bond in which both electrons come from one atom is a coordination bond.
24 The diagram shows a complex ion of cobalt with its ligands, drawn inside square brackets with the overall charge of the ion outside. What is the charge on the cobalt ion in this complex? HL
Answer and reasoning
1+ — This takes the overall charge outside the brackets as the charge on the metal. The two Cl⁻ ligands contribute −2, so the cobalt ion must be +3 for the complex to be +1.
3+ — Count the ligand charges from the diagram: four neutral NH₃ and two Cl⁻ give −2 in total. For the overall charge to be +1 the metal must be +3: (+3) + 2(−1) + 4(0) = +1.
7+ — This counts all six ligands as −1: +1 + 6 = +7. Donating a lone pair does not give a ligand a charge; NH₃ is neutral, and only the two Cl⁻ ligands carry −1, so the metal is +3.
1− — This adds the size of the Cl⁻ charges to the metal charge, x + 2 = +1, giving x = −1. Charges are combined with their signs: x + 2(−1) = +1, so x = +3.
Working Overall charge = charge on metal ion + sum of ligand charges. The diagram shows four NH₃ ligands (each 0), two Cl⁻ ligands (each −1) and an overall charge of +1: +1 = x + 4(0) + 2(−1), so x = +1 + 2 = +3. The ion is [CoCl₂(NH₃)₄]⁺ containing Co³⁺.
25 A primary halogenoalkane and a tertiary halogenoalkane each react with aqueous hydroxide ions. The diagram shows the energy profiles, P and Q, of the two reactions. Which statement is correct? HL
Answer and reasoning
Q is for the tertiary halogenoalkane, as its dip is the carbocation intermediate — Two peaks mean two steps with an intermediate between them. A tertiary halogenoalkane reacts by SN1: slow heterolysis of C–X gives a carbocation (the dip, high in energy but a real species), then OH⁻ attacks it. The one-peak profile P is the concerted SN2 reaction of a primary halogenoalkane.
P is for the tertiary halogenoalkane, as its single peak shows a one-step mechanism — A tertiary halogenoalkane reacts by SN1, and the '1' means one species in the rate-determining step, not one step. SN1 has two steps and so two peaks (profile Q); the single peak of P is the concerted SN2 reaction of the primary halogenoalkane.
Q is for the primary halogenoalkane, as its carbocation is the more stable one — Alkyl groups release electron density, so a primary carbocation is the LEAST stable and does not form. Primary halogenoalkanes react in one concerted step (profile P); tertiary ones form a tertiary carbocation (profile Q).
P is for the primary halogenoalkane, as its peak is the intermediate of the reaction — P is indeed the one-step SN2 profile of the primary halogenoalkane, but its peak is a transition state, not an intermediate. The species at the top has partial bonds to both OH and the halogen, exists only momentarily and cannot be isolated. An intermediate sits in a dip between two peaks, like the carbocation in Q.
26 But-1-ene, CH₃CH₂CH=CH₂, reacts with hydrogen bromide. The diagram shows skeletal formulas of the two carbocations, P and Q, that could form when H⁺ adds to the C=C bond. Which statement is correct? HL
Answer and reasoning
Q forms faster, as its positive carbon has only one electron-withdrawing alkyl group — Alkyl groups release electron density to a positive carbon; they do not withdraw it. More alkyl groups mean a more stable carbocation, so the secondary carbocation P is favoured over the primary Q.
Q forms, as the H atom adds to the carbon atom of C=C that has fewer H atoms — The rule runs the other way: H adds to the carbon with MORE hydrogen atoms (the CH₂ end), so that the positive charge sits on the more substituted carbon. That gives P, the secondary carbocation.
P forms faster, as its positive carbon is stabilized by two electron-releasing alkyl groups — In P the + is on a carbon bonded to two other carbons (a secondary carbocation, CH₃CH₂CH⁺CH₃); in Q it is on an end carbon bonded to one (primary). Alkyl groups release electron density towards the positive carbon, so P is more stable, forms faster, and gives the major product 2-bromobutane.
P and Q form in equal amounts, since H⁺ can add to either carbon of C=C — The two directions of addition give carbocations of different stability. The secondary carbocation P has a much lower activation energy, so it forms far faster and 2-bromobutane is the major product.
27 The diagram shows the intermediate formed when benzene reacts with an electrophile, E⁺, together with the curly arrow for the step that follows. What happens in the step shown? HL
Answer and reasoning
A nucleophile adds to the positive carbon, giving an addition product — Benzene does not react like an alkene here. Adding a nucleophile would leave the ring permanently non-aromatic; instead the C–H pair returns to the ring, H⁺ is lost, and the stable delocalized ring is regained (substitution).
The C–H bonding pair moves into the ring, restoring delocalization, and H⁺ is lost — The arrow starts at the C–H bond and ends in the ring. Both electrons of that bond re-enter the π system, so the six-carbon delocalization is restored, the ring loses its positive charge, and the hydrogen leaves without its electrons as H⁺: C₆H₆ + E⁺ → C₆H₅E + H⁺.
The C–H bond breaks homolytically, releasing a hydrogen atom, H• — The arrow is a double-barbed curly arrow, so it moves a PAIR of electrons: heterolytic fission. Both electrons of the C–H bond go into the ring and the hydrogen leaves as H⁺, not as a radical.
The arrow shows the hydrogen atom moving to another carbon atom of the ring — Curly arrows show electrons, not atoms, moving. This arrow moves the C–H bonding pair into the ring; the hydrogen does not move around the ring, it is lost as H⁺.
That was your twenty minutes. Real practice on R3.4 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
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