Organic Chemistry (HL extension) — IB Chemistry HL HL Study Guide

For: IB Chemistry HL candidates sitting IB Chemistry HL.

Covers: Reaction mechanisms (SN1, SN2, electrophilic addition), stereochemistry (chirality, R/S assignment), spectroscopic structure determination (NMR, MS, IR), and multi-step organic synthetic pathways.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Chemistry HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Organic Chemistry (HL Extension)?

Organic Chemistry (HL Extension) is the advanced study of carbon-based compound structure, reactivity, synthesis, and identification, building on SL content to prioritize mechanistic reasoning, 3D stereochemical analysis, and quantitative spectroscopic characterization rather than rote memorization of reaction facts. It makes up 20% of your final HL Chemistry exam mark, appearing across Paper 1 (multiple choice), Paper 2 (structured response), and Paper 3 (option papers), with 5–8 mark long-answer questions on spectroscopy and synthetic pathways being very common.

2. Reaction mechanisms — SN1, SN2, electrophilic addition

A reaction mechanism is a step-by-step description of bond breaking and formation during a reaction, where curly arrows represent the movement of electron pairs (always start arrows at a lone pair or covalent bond, not an atom, for full marks).

SN2 (Bimolecular Nucleophilic Substitution)

This concerted (single-step) mechanism occurs for primary halogenoalkanes, with a rate equation: $$\text{rate}=k[\text{halogenoalkane}][\text{nucleophile}]$$ The nucleophile attacks the electrophilic carbon bonded to the leaving group from the backside, forming a high-energy transition state with partial bonds to both the nucleophile and leaving group, before the leaving group is displaced. This causes inversion of configuration (Walden inversion) at the reactive carbon. Worked example: Reaction of bromoethane with aqueous ${\text{OH}}^{-}$: The lone pair on oxygen of ${\text{OH}}^{-}$ attacks the C1 carbon, the C-Br bond breaks to form ${\text{Br}}^{-}$, and the product is ethanol with inverted stereochemistry if the starting material was chiral.

SN1 (Unimolecular Nucleophilic Substitution)

This two-step mechanism occurs for tertiary halogenoalkanes, as alkyl groups stabilize the intermediate carbocation via the positive inductive effect. The rate equation only includes the halogenoalkane (the first, rate-determining step is leaving group departure): $$\text{rate}=k[\text{halogenoalkane}]$$ The planar carbocation intermediate can be attacked by the nucleophile from either face, forming a racemic mixture of enantiomers. Worked example: Reaction of 2-bromo-2-methylpropane with ${\text{OH}}^{-}$: First, the C-Br bond breaks to form a tertiary $(C{H}_3{)}_3{C}^{+}$ carbocation, then ${\text{OH}}^{-}$ attacks either face of the planar ion to form equal amounts of R and S 2-methylpropan-2-ol.

Electrophilic Addition

This mechanism describes reactions of alkenes, where the electron-rich C=C π bond acts as a nucleophile to attack an electrophile, following Markovnikov’s rule: the electrophile adds to the carbon with more hydrogen substituents, forming the more stable (more substituted) carbocation intermediate. Worked example: Reaction of propene with HBr: The π bond electrons attack the H of HBr, which adds to the terminal C (2 H atoms) to form a secondary carbocation on the central C, then ${\text{Br}}^{-}$ attacks the carbocation to form 2-bromopropane as the major product.

3. Stereochemistry — chirality, R/S

Stereochemistry studies the 3D arrangement of atoms in molecules, a key HL topic that explains differences in reactivity and biological activity of molecules with identical structural formulas.

Chirality

A molecule is chiral if it cannot be superimposed on its mirror image, almost always due to a chiral center: an $s{p}^3$-hybridized carbon bonded to 4 distinct groups. The two non-superimposable mirror images are called enantiomers, which have identical physical properties except they rotate plane-polarized light in opposite directions, and identical chemical properties except when reacting with other chiral molecules (e.g. enzyme active sites). A 50:50 mixture of enantiomers is a racemic mixture, which does not rotate plane-polarized light as the two rotations cancel out.

R/S Configuration Assignment

Follow these steps to assign absolute configuration to a chiral center:

1. Assign priority to the 4 attached groups: highest atomic number = highest priority (1), lowest = 4. If the first bonded atom is identical, compare the next bonded atoms to break the tie.
2. Rotate the molecule so the lowest priority group (almost always H) is pointing away from you, behind the plane of the page.
3. Trace the path from priority 1 → 2 → 3: clockwise = R (rectus) configuration, counterclockwise = S (sinister) configuration. Worked example: For 2-chlorobutane, the chiral center on C2 has groups: $\text{Cl}$ (1, Z=17), ${\text{C}}_2{\text{H}}_5$ (2, bonded to C,H,H vs ${\text{CH}}_3$ bonded to H,H,H), ${\text{CH}}_3$ (3), $\text{H}$ (4). With H pointing back, the path 1→2→3 is counterclockwise, so the configuration is S-2-chlorobutane. Exam shortcut: If the lowest priority group is pointing towards you, reverse your final assignment (clockwise = S, counter = R) to save time.

4. Spectroscopy for structure determination — NMR, MS, IR

These three complementary techniques are used together to deduce the full structure of unknown organic compounds, with 5–8 mark deduction questions guaranteed on Paper 2.

IR Spectroscopy

Measures absorption of infrared radiation by covalent bonds, with each bond type having a characteristic wavenumber (cm⁻¹) and peak shape:

• O-H (alcohol): broad, 3200–3600 cm⁻¹
• O-H (carboxylic acid): very broad, 2500–3300 cm⁻¹ (overlaps with C-H peaks)
• C=O (carbonyl): sharp, 1680–1750 cm⁻¹
• C=C (alkene): sharp, 1620–1680 cm⁻¹ Tip: Peak shape is more important than exact wavenumber for identifying O-H groups.

Mass Spectrometry (MS)

Ionizes molecules and separates fragments by mass-to-charge ($m/z$) ratio:

• The molecular ion peak ($M^{+}$) gives the molar mass of the compound.
• The $M+1$ peak (from 1.1% natural abundance of ${}^{13}$C) is used to calculate the number of carbon atoms: $$\text{Number of C atoms}=\frac{\text{Height of }M+1\text{ peak}}{\text{Height of }{M}^{+}\text{ peak}}\times 91$$
• Fragment peaks indicate functional groups: e.g. $m/z=43$ is either $C_3{H}_7^{+}$ or $C{H}_{3}C{O}^{+}$.

¹H NMR Spectroscopy

Gives information about hydrogen environments in a molecule:

1. Chemical shift (δ, ppm): Indicates the type of H environment: 0–2 ppm = alkane H, 2–5 ppm = alcohol/halogenoalkane H, 7–9 ppm = arene H, 10–12 ppm = carboxylic acid H.
2. Integration trace: Gives the relative number of H atoms in each environment.
3. Spin-spin splitting: Follows the $n+1$ rule: if a H has $n$ non-equivalent adjacent H atoms, it splits into $n+1$ peaks. Worked example: Ethanol ¹H NMR has 3 peaks: δ=1.2 (triplet, 3H, $C{H}_3$ adjacent to $C{H}_2$, $n=2$), δ=3.7 (quartet, 2H, $C{H}_2$ adjacent to $C{H}_3$, $n=3$), δ=2.5 (singlet, 1H, $OH$, no adjacent H).

5. Synthetic pathways

You will be asked to design 2–3 step syntheses of a target molecule from a given starting material, with marks awarded for correct reagents, conditions, and intermediate structures.

Core Rules for Pathway Design

1. Use only high-yield, syllabus-listed reactions, avoiding steps that produce multiple side products.
2. Always list both reagents and conditions for every step (1 mark is allocated for each per step).
3. Account for stereochemistry if specified (e.g. if a single enantiomer is required, use SN2 not SN1). Worked example: Synthesize propanoic acid from propene:
4. Step 1: Electrophilic addition of HBr (no UV, room temperature) to propene, forming 2-bromopropane (major product, Markovnikov addition).
5. Step 2: Elimination with alcoholic KOH (heat under reflux) to form propene? No, alternative correct pathway: 1. Hydroboration-oxidation of propene to form propan-1-ol, 2. Oxidation with excess acidified $K_{2}C{r}_2{O}_7$ (heat under reflux) to form propanoic acid.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Drawing curly arrows starting at an atom instead of a lone pair or bond in mechanism questions. Why students do it: Rushing, forgetting arrows represent electron movement. Correct move: Draw lone pairs on nucleophiles first, then start arrows at the center of the lone pair or covalent bond, ending at the atom receiving electrons.
• Wrong move: Assigning R/S configuration without adjusting for the lowest priority group pointing forward. Why students do it: Trying to save time. Correct move: If H is pointing towards you, reverse your final assignment (clockwise = S, counter = R) instead of skipping the adjustment.
• Wrong move: Confusing SN1 and SN2 rate equations and product stereochemistry. Why students do it: Mixing up unimolecular/bimolecular labels. Correct move: Remember "1" = first order, only halogenoalkane in rate equation, racemic product; "2" = second order, both reactants in rate equation, inverted product.
• Wrong move: Forgetting that UV light with HBr causes anti-Markovnikov addition to alkenes. Why students do it: Mixing up addition and free radical substitution conditions. Correct move: Use HBr at room temp, no UV for Markovnikov addition; UV is only for free radical substitution of alkanes.
• Wrong move: Forgetting to write "excess" oxidizing agent when oxidizing primary alcohols to carboxylic acids. Why students do it: Rushing to write reagents. Correct move: Always specify excess acidified $K_{2}C{r}_2{O}_7$ and heat under reflux for full oxidation to carboxylic acid; distill off product to stop at aldehyde.

7. Practice Questions (IB Chemistry HL Style)

Question 1

(a) State the mechanism type for the reaction of 2-bromo-2-methylbutane with aqueous sodium hydroxide. (1 mark) (b) Write the rate equation for this reaction, and explain why the product is a racemic mixture. (3 marks)

Worked Solution

(a) SN1 (unimolecular nucleophilic substitution) (1 mark) (b) Rate = $k[(C{H}_3{)}_{2}CBrC{H}_{2}C{H}_3]$ (1 mark). The rate-determining step is formation of a planar tertiary carbocation intermediate when $B{r}^{-}$ leaves (1 mark). The $O{H}^{-}$ nucleophile can attack either face of the planar carbocation with equal probability, forming equal amounts of R and S enantiomers (a racemic mixture) (1 mark).

Question 2

An unknown compound has a molecular ion peak at $m/z=88$, an IR absorption at 1715 cm⁻¹, and the following ¹H NMR spectrum: δ = 1.1 (triplet, 3H), δ = 2.0 (singlet, 3H), δ = 4.1 (quartet, 2H). Deduce the full structure, showing your reasoning. (4 marks)

Worked Solution

1. $M^{+}=88$ gives molar mass 88 g/mol, IR peak at 1715 cm⁻¹ indicates a C=O carbonyl group (1 mark).
2. NMR analysis: δ=1.1 triplet 3H = $C{H}_3$ adjacent to $C{H}_2$ (n=2, so 3 peaks); δ=4.1 quartet 2H = $C{H}_2$ adjacent to $C{H}_3$ (n=3, so 4 peaks) and bonded to O (high chemical shift); δ=2.0 singlet 3H = $C{H}_3$ adjacent to C=O (no adjacent H, so singlet) (1 mark).
3. Assemble fragments: $C{H}_{3}CO-$ (mass 43) and $-OC{H}_{2}C{H}_3$ (mass 45) sum to 88 g/mol (1 mark).
4. Final structure: ethyl ethanoate, $C{H}_{3}COOC{H}_{2}C{H}_3$ (1 mark).

Question 3

Design a 2-step synthetic pathway to convert 1-bromopropane to propanoic acid, including reagents and conditions for each step. (3 marks)

Worked Solution

1. Step 1: Convert 1-bromopropane to propan-1-ol. Reagents: aqueous NaOH; conditions: heat under reflux (1 mark reagents, 1 mark conditions).
2. Step 2: Oxidize propan-1-ol to propanoic acid. Reagents: excess acidified potassium dichromate(VI); conditions: heat under reflux (1 mark reagents + conditions, total 3 marks).

8. Quick Reference Cheatsheet

9. What's Next

This organic chemistry HL extension content is a prerequisite for all optional IB Chemistry topics: Medicinal Chemistry requires understanding of chiral drug activity, Materials Chemistry relies on polymerization mechanism knowledge, and Energy Chemistry uses synthetic pathway logic for biofuel production. It also forms the basis of most chemistry extended essay topics, and 20-25% of Paper 2 marks are allocated to these HL extension concepts.

If you struggle with any of the mechanism drawing, stereochemical assignment, or spectroscopic deduction steps, you can ask Ollie for step-by-step walkthroughs of additional practice problems, or request custom quizzes tailored to your weak areas. Head to [the homepage](/] to access more IB Chemistry HL study resources, past paper question banks, and personalized feedback on your answers.