Difficulty: Advanced | Prerequisites: All prior material (nomenclature, stereochemistry, SN1/SN2/E1/E2 reactivity). Familiarity with Lewis structures and curved-arrow notation is essential.
Drawing reaction mechanisms with curly arrows is how organic chemists explain why a reaction produces the products it does. It is not just bookkeeping: the arrows show electron flow, and getting them right is the difference between understanding a reaction and memorising it. Spectroscopy (NMR and IR) is the complementary skill. Once you have predicted a product, spectroscopy is how you confirm it in the lab. This section ties together mechanism drawing (specifically E1 with acid catalysis) and structure determination from NMR and IR data, both of which appeared as high-point-value questions on the exam.
E1 mechanisms proceed in two main stages: protonation of the leaving group, ionisation to form a carbocation, then deprotonation by a base (often solvent) to form the alkene. Every curly arrow starts from an electron-rich site and points to an electron-poor site. For spectroscopy, 1H NMR tells you how many distinct proton environments exist, how many protons are in each, and what neighbours they have (splitting). IR tells you what functional groups are present. Together they let you work out a structure from a molecular formula.
Curly arrow (curved arrow)
A notation showing the movement of an electron pair from a source (lone pair or bond) to a destination (an atom or a bond). The tail starts where the electrons are; the head points to where they go.
Carbocation
A carbon atom bearing a positive charge due to having only six electrons in its valence shell. Carbocations are intermediates in SN1 and E1 reactions. Tertiary carbocations are more stable than secondary, which are more stable than primary.
Protonation
The addition of a proton (H+) to an atom, typically a lone pair on oxygen, nitrogen, or a pi bond. In acid-catalysed reactions, protonation is often the first step.
Deprotonation
Removal of a proton by a base. In E1, the base (often solvent or a weak base) removes a proton adjacent to the carbocation to form the alkene.
1H NMR (proton nuclear magnetic resonance)
A spectroscopic technique that reveals the number of distinct hydrogen environments in a molecule, how many hydrogens are in each environment (integration), and what neighbours each hydrogen has (splitting pattern). In simple terms, it is a map of where the hydrogens sit on your molecule.
Chemical shift (ppm)
The position of an NMR signal on the x-axis, measured in parts per million. More deshielded protons (near electronegative atoms or pi bonds) appear further downfield (higher ppm).
Integration
The area under an NMR peak, proportional to the number of hydrogens producing that signal. A peak integrating for 6H likely corresponds to two equivalent methyl groups.
IR spectroscopy (infrared)
A technique that identifies functional groups by the frequencies at which bonds absorb infrared light. Key absorptions: broad O-H stretch near 3200 to 3400 cm-1 (alcohols), sharp C=C stretch near 1640 cm-1 (alkenes), C=O stretch near 1710 to 1750 cm-1.
Chirality centre (stereogenic centre, stereocentre)
A carbon bonded to four different groups. Its spatial arrangement is described as R or S using CIP priority rules.
The exam presented a compound with a hydroxyl leaving group, treated with a strong acid (sulfonic acid) in water with heat, giving an alkene product via E1.
Step 1, protonation of the leaving group: a lone pair on the hydroxyl oxygen attacks the acidic proton (H3O+ or R-SO3H). Draw the arrow from the oxygen lone pair to the H. This converts the poor leaving group (OH) into a good one (OH2+).
Step 2, ionisation (loss of the leaving group): the C-O bond breaks, with both electrons going to the oxygen. Draw the arrow from the C-O bond to the oxygen. This forms a carbocation intermediate and releases water.
Step 3, deprotonation: a base (water, in this case) removes a proton from a carbon adjacent to the carbocation. Draw one arrow from the C-H bond to form the new C=C pi bond, and another from the base's lone pair to the proton. This gives the alkene product.
Key points for earning full marks:
Every arrow must start from electrons (a lone pair or a bond) and point to where those electrons go.
Show all intermediates. Each step should produce a clearly drawn intermediate with correct charges.
The exam awarded 2 points per arrow and 2 points per intermediate.
Use H3O+ as the proton source when told to.
Count the number of signals: each signal corresponds to a distinct proton environment.
Read the integration: the ratio of peak areas gives the ratio of protons (e.g. 6H : 3H : 1H : 1H).
Note chemical shifts: protons near electronegative atoms (O, Cl, Br) or on sp2 carbons resonate further downfield.
0 to 1 ppm: CH3 groups on saturated carbon, far from electronegative atoms
1 to 2 ppm: CH2 and CH on saturated carbon
2 to 4 ppm: protons on carbon adjacent to electronegative atoms or carbonyl groups
3.5 to 4 ppm: protons on carbon bonded to O, N, or halogens
4.5 to 6.5 ppm: vinyl protons (C=C-H)
Note splitting patterns: a proton split into a doublet has one neighbouring non-equivalent proton; a septet has six neighbours, and so on (n+1 rule).
Exam example: compound A (C5H11Br) showed four signals: 6H (two equivalent methyls, a doublet around 0.9 ppm), 3H (one methyl), 1H (a methine), and 1H (a methine further downfield near 3.8 ppm, the one attached to the carbon bearing Br). This pattern matches 2-bromo-3-methylbutane.
A broad absorption near 3200 to 3400 cm-1 indicates an O-H stretch (alcohol). This is broad because of hydrogen bonding.
A sharp absorption near 1640 cm-1 indicates a C=C stretch (alkene).
A strong absorption near 1710 to 1750 cm-1 indicates a C=O stretch (carbonyl).
Exam example: compound B showed a broad peak near 3400 cm-1, confirming it is an alcohol. Compound C showed a peak near 1640 cm-1, confirming it contains a C=C double bond (an alkene).
The exam gave compound A (C5H11Br) and asked students to determine the structures of A, B, and C formed when A is treated with NaOH (aq).
A: 2-bromo-3-methylbutane. The NMR data (four signals, 6H doublet for two equivalent methyls, 1H downfield for CH-Br) confirms this structure. The stereochemistry at the carbon bearing Br was determined by dashes and wedges.
B: an alcohol with S configuration at the chirality centre. NaOH acts as a nucleophile (OH-) in an SN2 reaction, giving inversion of configuration. The broad IR peak near 3400 cm-1 confirms the alcohol. If A had R configuration, B has S (inversion).
C: an alkene. NaOH can also act as a base, promoting E2 elimination. The IR peak near 1640 cm-1 and the 1H NMR quartet downfield of 5.6 ppm (a vinyl proton with three neighbours) confirm an alkene. The specific structure is 3-methylbut-1-ene or 2-methylbut-2-ene, depending on which proton is removed.
Students draw curly arrows pointing the wrong way: from the electrophile to the nucleophile. Arrows always show electron movement, so they always point from the electron-rich species to the electron-poor species.
Forgetting to show intermediates. Each elementary step should produce a complete intermediate with all charges shown. Jumping straight from starting material to product loses most of the marks.
Confusing integration with splitting in NMR. Integration tells you how many protons produce a signal. Splitting tells you how many neighbouring protons that signal's protons have. They are independent pieces of information.
Assuming that a broad OH peak in IR means water. It can, but in the context of an organic product, a broad peak near 3400 cm-1 almost always means an alcohol.
The mechanism question (Q6) was worth 17 points, the highest-value single question on the exam. Getting the arrows and intermediates right is critical.
The spectroscopy question (Q7) was worth 14 points. It requires combining NMR, IR, molecular formula, and reaction knowledge to work backwards from data to structure.
Expect to identify the type of mechanism (E1, E2, SN1, SN2) before drawing it. The exam presented this as a separate sub-question worth 3 points.
NMR integration ratios are a fast way to narrow down structures. A 6H signal near 1 ppm almost certainly means two equivalent methyl groups.
True or false: In a curly arrow mechanism, the arrow tail starts at the nucleophile or electron source. (True)
Fill in the blank: In an E1 mechanism, the first step is ______ of the leaving group (if it is an alcohol), and the second step is ______ to form the carbocation. (protonation; ionisation / loss of the leaving group)
True or false: A broad IR absorption near 3400 cm-1 indicates a carbonyl group. (False, it indicates an O-H stretch, typically an alcohol.)
Fill in the blank: In 1H NMR, a signal integrating for 6H near 0.9 ppm most likely corresponds to ______. (two equivalent methyl groups)
True or false: SN2 on a chiral centre gives retention of configuration. (False, it gives inversion.)
Q: An alcohol is heated with a strong acid in water, producing an alkene. What type of mechanism is this, and what are the key steps?
A: E1. Step 1: protonation of the OH by the acid to form OH2+. Step 2: loss of water to form a carbocation. Step 3: a base (water) removes a proton adjacent to the carbocation, forming the C=C double bond.
Q: A compound with molecular formula C5H11Br shows four signals in its 1H NMR: 6H near 0.9 ppm, 3H near 1.5 ppm, 1H near 1.9 ppm, and 1H near 3.8 ppm. Propose a structure.
A: 2-bromo-3-methylbutane. The 6H signal is two equivalent methyl groups (the isopropyl CH3s), the 3H is the methyl on the carbon bearing Br, the 1H near 1.9 ppm is the CH of the isopropyl group, and the 1H near 3.8 ppm is the CH bearing bromine (downfield due to Br).
Q: When the compound above is treated with aqueous NaOH, two products form. One shows a broad IR peak near 3400 cm-1 and the other shows a peak near 1640 cm-1. Identify the functional group in each product.
A: The product with the broad 3400 cm-1 peak is an alcohol (O-H stretch). The product with the 1640 cm-1 peak is an alkene (C=C stretch).
Q: The alcohol product has S configuration. If the starting bromide had R configuration, what mechanism produced the alcohol?
A: SN2. Backside attack by hydroxide inverts R to S.
Q: How many curly arrows are needed for the ionisation step in an E1 mechanism, and what do they show?
A: One arrow, from the C-leaving group bond to the leaving group (e.g. from the C-O bond to the oxygen of water). This shows the bond breaking heterolytically, with both electrons going to the leaving group.
Mechanism drawing is the thread that runs through all of organic chemistry. The E1 mechanism here connects directly to SN1 (they share the carbocation intermediate) and to acid-catalysed reactions you will encounter in Organic Chemistry II (hydration, dehydration, esterification). Spectroscopy connects to every lab course: you will use NMR and IR to confirm products for the rest of your chemistry career. The structure elucidation approach (molecular formula, degrees of unsaturation, NMR, IR) also applies to mass spectrometry problems in later courses.
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