Why a 7 in IB Chemistry Reactivity 3.4 depends on drawing the lone pair before the arrow
A senior IB Chemistry tutor's reading of Reactivity 3.4: how curly-arrow mechanism answers are actually marked, and the scoring habits that separate a 6 from a 7.
Reactivity 3.4 sits at the heart of organic chemistry in the IB Chemistry Diploma. It is the sub-topic that asks candidates to explain, with a curly arrow, why one species attacks another, why a leaving group departs, and why an intermediate has the charge it does. Most candidates who land at a 5 or a 6 in IB Chemistry HL do not lose marks here because their factual recall is weak. They lose marks because the examiner cannot read the story they are trying to tell on the page. This article reads Reactivity 3.4 the way the marking scheme reads it: as a scoring discipline, not as a list of named reactions.
What Reactivity 3.4 actually asks you to demonstrate
Reactivity 3.4 belongs to the Reactivity 3 strand, which the IB guide frames around electron-pair sharing reactions. Where Reactivity 3.1, 3.2 and 3.3 build up the descriptive language of acids, bases, redox and organic families, Reactivity 3.4 demands something harder: a working model of how a covalent bond is made and broken, drawn live on the page in front of the examiner. The assessment statement language is precise on purpose. Candidates are expected to show a movement of an electron pair from a donor site to an acceptor site, to identify that donor and acceptor by name, and to justify both identification and movement with reference to electronegativity, charge, or orbital overlap.
In practice on Paper 2, this looks like three or four short-answer parts clustered around a single stimulus. You might be given an alkene and asked why a Br₂ molecule approaches one face rather than the other, or a haloalkane and asked why the rate changes when you swap chloride for iodide. The marks are not in the name of the mechanism. The marks live in the curly arrow, the lone pair drawn before the arrow, the formal charge written after the arrow, and the one-sentence justification that frames the whole diagram. In my experience reading IB Chemistry HL mark schemes, the descriptors on Reactivity 3.4 are unusually generous about the words a candidate uses, but unusually strict about the order in which those words appear on the page. If the lone pair is missing, if the arrow starts in empty space, or if the intermediate carries a charge that contradicts the geometry just drawn, the examiner will not give the band 4 mark even when the surrounding prose is fluent.
This is why a 7 in Reactivity 3.4 depends less on knowing the name of every mechanism and more on internalising a small set of drawing and writing rules. The next sections break those rules open.
Electrophile, nucleophile, and the donor–acceptor test the rubric quietly applies
The single most important sentence a candidate can write in Reactivity 3.4 is also the simplest: the electron pair moves from a region of high electron density to a region of low electron density. From that sentence, every other judgement in the sub-topic flows. An electrophile is a species with a low-electron-density site that accepts the pair. A nucleophile is a species with a high-electron-density site that donates the pair. The arrow always travels from donor to acceptor. When candidates state this explicitly in a Paper 2 answer, they nearly always pick up the band 3 or band 4 reasoning marks; when they assume the examiner can read it from a diagram, they often lose one mark band for a missing justification.
For most IB Chemistry HL candidates, the practical translation is a two-step test. Step one: identify the polar bond in the substrate and the atom that carries the partial positive charge. Step two: identify a lone pair, a pi bond, or a polarised sigma bond on the attacking species that can move. If both identifications are made and written, the rubric will accept the arrow. If only one identification is made, the rubric will often cap the answer at band 2 even if the arrow is drawn correctly, because the rubric is testing reasoning as much as it is testing drawing.
The trap candidates fall into is the adjective trap. Saying that HBr is an electrophile without saying which atom of HBr is electrophilic and why loses the mark. Saying that the alkene is a nucleophile without locating the pi electrons as the donor site loses the mark. The IB examiner is not testing whether you can label a molecule. The examiner is testing whether you can justify the label. This is the most common difference between a script that lands at 6 and one that lands at 7 in Reactivity 3.4: not fewer errors, but better-stated justifications for the errors that remain.
How curly arrows are actually marked on Paper 2
A curly arrow in IB Chemistry HL is a piece of evidence, not a piece of decoration. The arrow tells the examiner which bond you think is breaking, which bond you think is forming, and which way the electrons are moving. Three drawing rules are scored directly, and a fourth is scored indirectly through the justification prose.
The first rule is the tail: the tail of a curly arrow must start on an electron pair, never in empty space, and never on a positive charge. If the substrate has a lone pair and the candidate draws the tail in the middle of the carbon–carbon bond, the examiner treats this as evidence that the candidate does not understand what is being donated. The second rule is the head: the head must land on the atom that receives the electrons, or on the new bond being formed, but it must not stop short in the middle of the two atoms. The third rule is the pairing: the breaking arrow and the forming arrow for a single mechanistic step should be drawn as a pair, ideally with a clear geometric relationship on the page that shows the two events are simultaneous.
The fourth rule, scored through the justification rather than the diagram, is the timing. A mechanism answer that walks the examiner through step one, then step two, then step three, with arrows drawn in that same order, is far easier to mark than a single diagram in which three arrows are drawn at once. In my experience this is the single highest-leverage change a candidate can make in the four to six weeks before Paper 2: rewriting their mechanism notes so that each step is a separate panel, with the intermediate drawn between panels, and a half-sentence beneath each panel explaining why that particular pair of electrons moved. The examiner does not need long prose. They need evidence that the candidate knows which electron pair is moving at each stage.
A common error worth naming is the double-headed arrow for an electron pair. In some textbooks the curly arrow has a single barb to denote a pair of electrons and a double barb to denote a single electron. The IB guide is explicit: candidates are scored on the two-electron curly arrow. Drawing a single-barbed arrow by reflex, especially when revising from a non-IB source, can quietly cap a candidate at band 2 even when the underlying chemistry is correct.
Three mechanism families that dominate Reactivity 3.4 questions
Although the IB guide is deliberately broad, the mark schemes over the last decade cluster around three mechanism families. Candidates who can draw all three fluently, with their intermediates and their justifications, will pick up the bulk of the marks available in the sub-topic.
The first family is electrophilic addition to a carbon–carbon double bond. The substrate is an alkene; the attacking species is an electrophile such as Br₂, HBr, or a proton from an acid catalyst. The arrow sequence is two arrows in the first step: one from the pi bond to the electrophile, one from the bond within the electrophile to the more electronegative atom of the leaving group. The intermediate is a carbocation. The justification sentence must explain why the alkene is the donor and the electrophile is the acceptor, and must explicitly say the carbocation is formed because the pi electrons have been removed. Candidates who skip the carbocation and draw a direct product lose the intermediate mark and a justification mark with it.
The second family is nucleophilic substitution at a saturated carbon. The substrate is a haloalkane, an alcohol under acid catalysis, or a related species with a leaving group. The attacking species is a nucleophile such as OH⁻, CN⁻, NH₃, or water. The arrow sequence is again two arrows in the first step: one from the lone pair on the nucleophile to the carbon, one from the carbon–leaving-group bond to the leaving group. The intermediate is, in the concerted SN2 case, a transition state rather than a discrete species, and the IB guide accepts a single transition-state diagram with partial bonds drawn. The justification sentence must locate the donor lone pair and explain why the carbon is electrophilic: usually a sentence about polarisation of the C–X bond, with X named.
The third family is electrophilic substitution on a benzene ring. The substrate is benzene or a substituted benzene; the attacking species is an electrophile generated in situ, often a nitronium ion, a sigma complex equivalent, or a halogen in the presence of a Lewis acid catalyst. The arrow sequence in the first step is one arrow from the ring to the electrophile, producing a carbocation intermediate in which the aromatic pi system is temporarily broken. The second step is a second arrow from a C–H bond back into the ring to restore aromaticity, with a third arrow kicking out the proton. This is the family where the timing rule matters most, because the three arrows must be drawn in the correct order or the aromatic system does not get restored on the page.
A short tactical aside on memorisation: candidates often try to memorise these three families as three separate stories. In practice, they are the same story told three times. A high-electron-density site attacks a low-electron-density site. A bond breaks to balance the electrons. An intermediate forms. The second step restores the original electron count. If a candidate can hold that skeleton in mind, the named mechanism follows from the substrate given in the question.
How to draw lone pairs and formal charges so the examiner awards band 4
Lone pairs and formal charges are the two annotations that turn a curly-arrow diagram from a sketch into a scored answer. The IB mark scheme consistently allocates one mark per panel of a mechanism for a correct lone pair and one mark per panel for a correct formal charge. A four-step mechanism with all annotations correct therefore carries up to eight annotation marks before the justification marks are even counted.
The first drawing habit is to put lone pairs on the page before the arrow. A candidate who draws the attacking species with a clear lone pair, then draws the arrow starting from that lone pair, will signal to the examiner that the donor is identified. A candidate who draws the arrow first and the lone pair afterwards, or who never draws the lone pair at all, will lose at least one band even if the arrow itself is correct. The second habit is to write the formal charge on the intermediate as soon as it forms. The intermediate in nucleophilic substitution on a neutral haloalkane is a positively charged species because the nucleophile has brought a bond in but the leaving group has not yet left in the SN1 case. Drawing a + sign on the carbon, or on the atom that now carries the extra bond, is the rubric's way of checking that the candidate can count electrons.
Common pitfalls and how to avoid them:
- Drawing the arrow from a positive charge. A positive charge is an electron deficiency, not an electron source. If the tail of the arrow touches a +, the rubric treats it as an error of model, not of neatness.
- Forgetting the lone pair on the nucleophile. OH⁻ has three lone pairs, CN⁻ has one, NH₃ has one. Drawing the molecule without a lone pair is treated as evidence the candidate does not know which pair is being donated.
- Writing the wrong sign on the intermediate. If the candidate adds a bond to a neutral species and breaks no bond in the same step, the intermediate is positive. If the candidate breaks a bond and forms no new one, the intermediate is negative. The sign follows the electron count, not the candidate's intuition.
- Drawing the arrow head in the middle of two atoms. The arrow should land on the new bond or on the receiving atom, not in the empty space between them. A head that stops halfway reads as an unfinished mechanism.
The practical revision exercise is to take three or four past-paper mechanism stimuli and redraw them three times each: once with the lone pairs, once with the formal charges, once with both. The first redraw takes the longest. The third is two to three times faster, and the candidates who reach that speed are the ones who can sustain a four-step mechanism under timed Paper 2 conditions.
What separates a 5 from a 7 in Reactivity 3.4 on Paper 2
The score band on Reactivity 3.4 is not a knowledge band. It is a writing band. The descriptors in the IB guide are written in terms of the candidate's ability to communicate a mechanistic argument, not in terms of how many named reactions they can list. The implications for revision are large. Memorising a reaction map has a small payoff on this sub-topic. Practising the drawing and justification habits has a large payoff.
In my experience the most common reason a strong candidate lands at 6 instead of 7 on Reactivity 3.4 is the missing half-sentence. The arrows are correct, the intermediates are drawn, the formal charges are present, but the candidate has not stated, in words, why the donor is a donor and why the acceptor is an acceptor. The rubric will award the drawing marks but not the reasoning marks, and the script sits in band 3 when the chemistry in it would otherwise justify band 4. A second common reason is the time pressure on Paper 2, which forces candidates to compress a three-step mechanism into a single diagram. The compression usually costs the intermediate mark, because the intermediate is the part the candidate runs out of time to redraw cleanly.
For most candidates, the highest-leverage preparation strategy is not a content review but a writing drill. Pick one mechanism, draw it under timed conditions, then mark it against the IB mark scheme word for word. Repeat with a different mechanism, then a third. After three or four iterations the candidate knows exactly which sentences the rubric is hunting for, and the band 4 descriptors start to feel like a checklist rather than a mystery.
Reactivity 3.4 on Paper 1: the same content, the harder format
Paper 1 is multiple choice, but it is not a soft option for Reactivity 3.4. The IB guide reserves a small but consistent share of Paper 1 for mechanism questions phrased as four-option items. The stems typically present a substrate and an attacking species, then ask which intermediate forms, which arrow sequence is correct, or which statement about a step in the mechanism is true. The distractors are drawn from the same error families described above: arrows from positive charges, lone pairs drawn on the wrong atom, intermediates with the wrong charge. A candidate who has practised the drawing habits described in the previous sections will read the four options in a Paper 1 mechanism question and recognise the distractors within a few seconds. A candidate who has only memorised the names of the mechanisms will find Paper 1 mechanism items slow and uncertain.
The practical implication is that Paper 1 preparation for Reactivity 3.4 should be drawing preparation, not reading preparation. Candidates who can redraw the three mechanism families in under five minutes per family, with all annotations, will be the candidates who handle the Paper 1 items confidently. Candidates who cannot redraw them under time pressure will end up guessing on the mechanism items, and four or five mechanism items in a 40-question paper is enough to swing a final grade boundary by a full band.
Paper 1 versus Paper 2: how the two formats score the same sub-topic differently
The table below summarises the scoring cultures of the two papers as they apply to Reactivity 3.4. The chemistry is identical. The reading the examiner does of the candidate's work is not.
| Dimension | Paper 1 | Paper 2 |
|---|---|---|
| Answer mode | Selecting one of four pre-drawn mechanism diagrams or statements | Drawing the mechanism, naming the intermediate, and writing a one-sentence justification |
| Mark allocation per item | Single mark, no partial credit, no annotation marks available | Three to five marks per panel, including dedicated marks for lone pairs, formal charges, arrow tails and arrow heads |
| Time budget per mark | Around 90 seconds per mark across the whole paper | Around 150 to 200 seconds per mark on a mechanism panel |
| What the examiner rewards | Recognition of a correct arrow sequence among four distractors | Construction of a correct arrow sequence with annotations and justification |
| Typical ceiling for unprepared candidates | Two or three of the four mechanism items answered by elimination only | Band 2 to band 3 ceiling, with annotation marks lost even when the arrows are correct |
The contrast is the point. A candidate who treats Reactivity 3.4 as a reading task will plateau on Paper 1 and fall further behind on Paper 2. A candidate who treats it as a drawing and writing task will gain on both papers at once, because the recognition skill that Paper 1 tests is downstream of the construction skill that Paper 2 tests. This is one of the few sub-topics in the IB Chemistry syllabus where Paper 2 preparation is strictly better preparation for Paper 1 than Paper 1 preparation is for Paper 2.
A 14-day preparation plan for Reactivity 3.4
The preparation plan below is the one I would assign to a candidate sitting IB Chemistry HL with a 7 target who has not yet done a serious pass at Reactivity 3.4. It is not a content plan. It is a writing and drawing plan, because that is what the sub-topic scores on.
- Days 1 to 3: redraw the three mechanism families from memory. Electrophilic addition, nucleophilic substitution, electrophilic substitution. Draw each twice: once with the intermediate named, once with the intermediate's formal charge calculated. Do not use notes.
- Days 4 to 6: annotate three past-paper mechanism stimuli. Use real IB Paper 2 stimuli. For each, list every lone pair, every formal charge, every arrow tail and every arrow head. Mark your own list against the mark scheme. The aim is to internalise the rubric's checklist.
- Days 7 to 9: write justification sentences. For each of the three mechanism families, write a one-sentence justification for why the donor is a donor and a one-sentence justification for why the acceptor is an acceptor. Re-write them in different orders until they feel natural.
- Days 10 to 12: timed Paper 2 panels. Draw a four-step mechanism under timed conditions, with all annotations, in under ten minutes. Repeat until the speed is consistent. This is the speed you will need on the actual paper.
- Days 13 to 14: Paper 1 mechanism items. Work through Paper 1 mechanism items from past papers. For each, identify which distractor corresponds to which error family above. The aim is recognition speed, not new learning.
The plan assumes around 45 minutes of focused work per day. For most candidates that is enough to lift the Reactivity 3.4 score by one full band on Paper 2 and to make the Paper 1 mechanism items near-automatic. For a candidate with less time, the highest-leverage subset is days 1 to 3 plus days 10 to 12: redraw the three families from memory, then redraw them under timed conditions. Those two habits, more than any content review, decide the band on this sub-topic.
How Reactivity 3.4 connects to the rest of the Reactivity 3 strand
Reactivity 3.4 is the mechanism sub-topic, but it is not standalone. It depends on the language of Reactivity 3.1, 3.2 and 3.3 in the same way a paragraph depends on the sentences around it. A curly-arrow diagram in 3.4 is meaningless if the candidate cannot name the acid, base, oxidising agent or reducing agent that the substrate is reacting with. The mechanism is the explanation, not the content. Candidates who arrive at 3.4 having read 3.1, 3.2 and 3.3 as separate silos will find 3.4 difficult; candidates who arrive having read the three earlier sub-topics as a connected language of donor, acceptor, polarisation and charge will find 3.4 the natural extension of that language.
The practical consequence is that Reactivity 3.4 revision should not start in isolation. The first ten minutes of any 3.4 study session should be a quick review of the named reactions in 3.1, 3.2 and 3.3 that the day's mechanism work will reference. Without that warm-up, candidates tend to draw correct arrows on top of misnamed reactions, which loses the same justification marks as a missing half-sentence would.
Conclusion and next steps for the IB Chemistry candidate
Reactivity 3.4 is a writing and drawing sub-topic in disguise. The chemistry is not hard; the rubric is. A 7 in IB Chemistry HL on this sub-topic depends on three habits: redraw the mechanism, not memorise it; annotate the mechanism with lone pairs and formal charges before the arrow; and write the half-sentence that justifies the donor–acceptor assignment. Candidates who build those habits over a two-week stretch will find that Paper 1 mechanism items become recognisable on sight and Paper 2 mechanism panels become a predictable, scoreable routine rather than a stressful improvisation. The single most useful next step is to take a blank page, redraw the three mechanism families from memory with every annotation, and mark yourself against the rubric. That exercise, repeated three times across a fortnight, is the difference between a 6 and a 7 on this sub-topic for most IB Chemistry HL candidates.
IB Courses' IB Chemistry HL Paper 2 mechanism clinic works through Reactivity 3.4 electron-pair sharing reactions panel by panel, redrawing each of the three mechanism families against the rubric until the arrow, the lone pair, the formal charge and the justification sentence become a single habit. Candidates who book the clinic arrive at Paper 2 with a 7-shaped preparation plan, not a hope.