Source: Exam 4 Key, University of Minnesota Twin Cities
Difficulty: Intermediate to Advanced | Prerequisites: Alkene/alkyne reactions, stereochemistry, IR and NMR basics Tags: reaction mechanism, arrow pushing, synthesis, retrosynthesis, IR spectroscopy, mass spectrometry, 1H NMR, structure determination, heat of hydrogenation, selectivity, polymerisation
Mechanism questions require you to draw every arrow, every intermediate, and the correct electron flow for each step. Synthesis questions ask you to chain known reactions together to reach a target molecule, often requiring you to think backwards (retrosynthesis). Spectroscopy problems use mass spectrometry (molecular formula), IR (functional groups), and ¹H NMR (hydrogen environments) to deduce an unknown structure. Miscellaneous topics round out the exam: heats of hydrogenation, syn vs anti selectivity, polymerisation regiochemistry, and identifying intermediates in specific reaction pathways.
Curved-arrow (electron-pushing) mechanism
A notation system showing how electron pairs move during bond-breaking and bond-forming events. Each arrow starts at the electron source (lone pair or bond) and points to the electron sink (electrophilic atom or bond). In simple terms, arrows show where the electrons go in each step of a reaction.
Retrosynthesis
Working backwards from the target molecule to identify what starting materials and reagents are needed. You mentally "undo" the last reaction to figure out what intermediate came before, then repeat. Think of it as solving a maze from the finish back to the start.
Mass spectrometry (MS)
A technique that measures the mass-to-charge ratio (m/z) of ions. The molecular ion peak (M⁺) gives the molecular weight of the compound. In simple terms, it tells you how heavy the molecule is.
Infrared spectroscopy (IR)
Measures absorption of infrared light by molecular bonds. Different functional groups absorb at characteristic wavenumbers (cm⁻¹). Key absorptions: O-H broad ~3200-3600, C=O sharp ~1700-1750, C=C ~1650, sp C-H ~3300. Think of it as a fingerprint for functional groups.
¹H NMR spectroscopy
Measures the magnetic environments of hydrogen atoms. Chemical shift (ppm) indicates the type of hydrogen; integration gives the relative number; splitting pattern reveals neighbouring hydrogens. In simple terms, it tells you how many types of hydrogens you have, how many of each, and what is next to them.
Heat of hydrogenation
The enthalpy change when an alkene (or alkyne) is completely hydrogenated. Less stable alkenes release more energy upon hydrogenation, so they have a higher (more negative) heat of hydrogenation. Think of it as a direct measure of how much energy the double bond stores. The less stable the alkene, the more energy it releases when reduced.
Enol
A tautomer of a carbonyl compound in which an OH group is attached to a C=C double bond. Enols spontaneously tautomerise to the more stable keto form in solution.
Tautomerism (keto-enol)
An equilibrium between a keto form (C=O with C-H on the adjacent carbon) and an enol form (C=C-OH). In most cases, the keto form dominates. This is relevant in alkyne hydration, where the initial product is an enol that rearranges to a ketone.
This mechanism was tested explicitly on the exam. The substrate is a trisubstituted alkene reacting with Cl₂ in water.
Step 1: Formation of the chloronium ion
The pi electrons of the alkene attack one of the chlorine atoms in Cl₂. This displaces the other chlorine as Cl⁻ and forms a three-membered chloronium ion (cyclic intermediate with a positively charged chlorine bridging the two carbons of the former double bond).
Arrow: from the C=C pi bond to the Cl of Cl₂.
Step 2: Nucleophilic attack by water
Water attacks the more substituted carbon of the chloronium ion from the opposite face (anti to the chlorine bridge). This opens the ring and places the oxygen on the more substituted carbon.
Arrow: from a lone pair on oxygen of H₂O to the more substituted carbon.
Step 3: Deprotonation
The oxonium ion (positively charged oxygen) loses a proton to the solvent (water or Cl⁻), giving the neutral chlorohydrin product plus HCl.
Arrow: from the O-H bond to the base (Cl⁻ or another water molecule).
Final products: the chlorohydrin (with Cl and OH on adjacent carbons, anti to each other) + HCl.
Always show lone pairs where they are the source of an arrow.
Each step should have a clearly drawn intermediate, with formal charges labelled.
Arrows go from nucleophile to electrophile (electron-rich to electron-poor).
For a three-step mechanism, you need at least two intermediates shown between the starting materials and products.
If a proton transfer is needed, show it as its own step (do not combine it with another bond-forming event in the same arrow).
The exam asked: convert methylenecyclohexane to a cyclohexane with an OH on the less substituted carbon (anti-Markovnikov placement, + enantiomer).
Route 1: HCl, then elimination, then hydroboration-oxidation
Step 1: Add HCl to the alkene (Markovnikov, Cl goes to more substituted carbon).
Step 2: Eliminate with NaOEt/EtOH (E2, forms a new alkene with the double bond now inside the ring).
Step 3: Hydroboration-oxidation (1. BH₃, 2. H₂O₂/NaOH) gives anti-Markovnikov, syn addition of OH.
Result: OH on the less substituted carbon, + enantiomer.
Route 2: Hydration, then elimination, then hydroboration-oxidation
Step 1: Acid-catalysed hydration (H₂O/H₂SO₄) places OH on the more substituted carbon (Markovnikov).
Step 2: Dehydrate with H₂SO₄/heat to form the endocyclic alkene.
Step 3: Hydroboration-oxidation gives the anti-Markovnikov alcohol.
Start from the target product and ask: what reaction, as the last step, installs this functional group with this regiochemistry and stereochemistry?
Anti-Markovnikov OH with syn stereochemistry? Last step must be hydroboration-oxidation.
That means the penultimate intermediate is an alkene with the double bond positioned so BH₃ gives the correct regiochemistry.
If the available starting material has the double bond in the wrong place, insert an elimination step to move it.
The exam also tests your ability to fill in missing reagents, starting materials, or products in a multi-step scheme.
Geminal dihalide to alkyne: treat with NaNH₂ (two equivalents, or excess), then H₂O. Eliminates both HX to form the alkyne.
Alkyne to ketone: H₂O/H₂SO₄/HgSO₄ (Markovnikov hydration of the alkyne gives an enol, which tautomerises to the ketone).
Terminal alkyne to internal alkyne: 1. NaNH₂ (deprotonate), 2. CH₃I (alkylate via SN2).
Internal alkyne to cis-alkene: H₂ with Lindlar catalyst.
Alkene to epoxide: m-CPBA.
Three compounds (A, B, C) are connected by reactions, and you must deduce each structure from spectroscopic data.
Compound A: molecular weight 54 (MS)
MW 54 with only C and H: C₄H₆ (4 carbons, 6 hydrogens, degree of unsaturation = 2).
Two degrees of unsaturation could be one triple bond, or two double bonds, or a ring + a double bond.
Treatment with Li/NH₃ (dissolving metal reduction) converts A to B (MW 56). That is a gain of 2 mass units, consistent with adding H₂ across a triple bond to give an alkene.
Conclusion: A is 1,3-butadiene? Actually, MW 54 for C₄H₆ and dissolving metal reduction giving MW 56 (C₄H₈) points to a conjugated diene or a cyclobutene. But given the context (dissolving metal reduction of alkynes to trans-alkenes), A is most likely 1-butyne or 2-butyne. C₄H₆ = butyne. Li/NH₃ on 2-butyne would give trans-2-butene (MW 56).
A = 2-butyne (an internal alkyne, since dissolving metal reduction is used on internal alkynes to give trans-alkenes).
Compound B: molecular weight 56, IR peaks at 3090, 2970, 2930, 1650 cm⁻¹
MW 56 = C₄H₈ (one degree of unsaturation).
IR: 3090 cm⁻¹ (=C-H stretch, sp² C-H), 2970 and 2930 cm⁻¹ (sp³ C-H), 1650 cm⁻¹ (C=C stretch).
This confirms B is an alkene: trans-2-butene.
Compound C: molecular weight 44, IR peak at 1740 cm⁻¹, ¹H NMR shows 1H at ~9 ppm and 3H at ~2 ppm
Ozonolysis of B (trans-2-butene) cleaves the C=C. Since both carbons of the double bond are internal (each bearing one H and one CH₃), ozonolysis gives two equivalents of the same product.
MW 44 = C₂H₄O (acetaldehyde, CH₃CHO).
IR: 1740 cm⁻¹ = carbonyl (C=O) stretch, consistent with an aldehyde.
¹H NMR: 1H at ~9 ppm = aldehyde proton (CHO); 3H at ~2 ppm = the CH₃ group adjacent to the carbonyl.
C = acetaldehyde (ethanal, CH₃CHO).
Start with the molecular formula from MS. Calculate the degree of unsaturation (DoU = (2C + 2 + N - H - X) / 2).
Use IR to identify functional groups: broad O-H (~3200-3600), sharp C=O (~1700-1750), C=C (~1600-1680), sp C-H (~3300), sp² C-H (~3050-3100).
Use ¹H NMR chemical shifts to assign hydrogen types: aldehyde H ~9-10 ppm, aromatic H ~6.5-8 ppm, vinyl H ~4.5-6.5 ppm, allylic/adjacent-to-C=O ~2-2.5 ppm, alkyl H ~0.5-2 ppm.
Integration tells you the ratio of hydrogens. Splitting pattern (n+1 rule) tells you how many neighbouring Hs each signal has.
To convert an alkene with a terminal double bond to a geminal dichloride (both Cl atoms on the same carbon), use excess HCl. Two sequential Markovnikov additions place both chlorines on the more substituted carbon.
Do not confuse with Cl₂ (gives vicinal dihalide, anti addition), Cl₂/light (radical substitution, not addition), NaCl/DMF (Finkelstein-type, not relevant here), or SOCl₂ (converts alcohols to alkyl chlorides).
The least stable alkene has the highest heat of hydrogenation, because it releases the most energy when converted to the alkane.
Stability ranking of alkenes (most to least stable): tetrasubstituted > trisubstituted > disubstituted (trans > cis) > monosubstituted > unsubstituted.
A strained alkene (such as one with a trans double bond in a small ring, or an exocyclic double bond on a strained system) is particularly unstable and will have an unusually high heat of hydrogenation.
When a vinyl monomer like methyl acrylate (CH₂=CH-COOMe) undergoes radical polymerisation, the radical adds to the less hindered CH₂ end (tail) of each monomer, because this forms the more stable radical on the substituted carbon (head). The result is a head-to-tail polymer with alternating substituted and unsubstituted carbons along the backbone.
Of the common reactions covered in this course, hydrogenation of alkenes (H₂/Pd, Pt, or Ni) is the classic syn-selective reaction. Both hydrogen atoms deliver to the same face of the double bond on the metal surface.
Hydroboration is also syn-selective. Epoxidation preserves the alkene geometry (also considered a syn process).
Reactions that are anti-selective: halogenation (via halonium ion) and addition of halogens to alkynes.
Reactions that are not stereoselective (or give mixtures): acid-catalysed hydration, HX addition (no peroxides), and radical HBr addition.
Alkyne hydration with H₂O/H₂SO₄/HgSO₄ proceeds through a vinyl cation or a mercurinium-ion-like intermediate (depending on the level of detail expected). The key intermediate to recognise is the enol (vinyl alcohol), which then tautomerises to the ketone.
On the exam, the correct intermediate choice was the enol form (E in the multiple choice), recognisable as an OH group on a carbon that is still part of a C=C double bond.
Students draw mechanism arrows from the electrophile to the nucleophile. Arrows always go from the electron source (nucleophile, lone pair, pi bond) to the electron sink (electrophile).
Students skip intermediates in mechanisms. Each elementary step forms one intermediate that must be drawn with correct charges before the next step begins.
Students forget that the enol is the intermediate in alkyne hydration, not the final product. The enol tautomerises to the ketone (or aldehyde) spontaneously.
Students confuse "highest heat of hydrogenation" with "most stable alkene." The opposite is true: the least stable alkene has the highest heat of hydrogenation because it releases the most energy upon reduction.
⚠️ Mechanism problems are worth significant marks and are graded step by step. Missing one arrow or one intermediate costs points even if the final product is correct.
⚠️ Synthesis questions often have more than one valid route. Full credit requires that every step is a known, reliable reaction with correct reagents. Partial routes with vague reagents lose marks.
⚠️ Spectroscopy problems test your ability to work systematically: molecular formula first (MS), functional groups next (IR), then structural detail (NMR). Skipping steps leads to wrong structures.
⚠️ The heat of hydrogenation question tests conceptual understanding, not calculation. Remember: least stable = highest heat of hydrogenation.
⚠️ Know which reactions are syn-selective, which are anti-selective, and which give a mixture. This is a common multiple-choice topic.
True or False: In a curved-arrow mechanism, arrows point from the electrophile to the nucleophile. False. Arrows go from electron source (nucleophile) to electron sink (electrophile).
Fill in the blank: The intermediate in alkyne hydration (H₂O/H₂SO₄/HgSO₄) that rearranges to the final ketone product is called an ______. Enol.
True or False: The alkene with the highest heat of hydrogenation is the most stable alkene. False. It is the least stable.
Fill in the blank: To form a geminal dichloride from a terminal alkene, use ______. Excess HCl.
True or False: Hydrogenation of alkenes (H₂/Pd) is an anti-selective reaction. False. It is syn-selective.
Q: Draw the complete mechanism for the reaction of cyclohexene with Cl₂ in water (chlorohydrin formation). Show all intermediates, arrows, and charges.
A: Step 1: Pi electrons of cyclohexene attack Cl₂, forming a chloronium ion and displacing Cl⁻. Step 2: H₂O attacks the more substituted carbon of the chloronium ion from the anti face, forming an oxonium ion. Step 3: Cl⁻ (or another H₂O) deprotonates the oxonium ion, yielding the trans-chlorohydrin + HCl.
Q: Propose a synthesis to convert methylenecyclohexane to an alcohol with OH on the less substituted carbon of the ring (anti-Markovnikov placement).
A: One valid route: (1) HCl adds to give a tertiary chloride (Markovnikov). (2) NaOEt/EtOH eliminates to form 1-methylcyclohexene (endocyclic alkene). (3) Hydroboration-oxidation (BH₃, then H₂O₂/NaOH) places OH anti-Markovnikov with syn addition. Product: trans-2-methylcyclohexanol + enantiomer.
Q: Compound A (MW 54) is treated with Li/NH₃ to give compound B (MW 56, IR shows C=C at 1650 cm⁻¹). Ozonolysis of B gives only compound C (MW 44, IR shows C=O at 1740 cm⁻¹, NMR: 1H at 9 ppm, 3H at 2 ppm). Identify A, B, and C.
A: A = 2-butyne (C₄H₆, internal alkyne). B = trans-2-butene (C₄H₈, from dissolving metal reduction). C = acetaldehyde (CH₃CHO, MW 44, aldehyde C=O at 1740, aldehyde H at 9 ppm, CH₃ at 2 ppm). Ozonolysis of a symmetrical internal alkene gives two equivalents of the same aldehyde.
Q: Among five alkenes of varying substitution, which has the highest heat of hydrogenation, and why?
A: The least substituted (or most strained) alkene has the highest heat of hydrogenation. It is the highest in energy, so it releases the most energy when fully reduced to the alkane. On the exam, the answer was the terminal, unsubstituted alkene (E).
Mechanism drawing is the skill that ties everything together. Every reaction you learn in organic chemistry has a mechanism, and exams increasingly test whether you understand the mechanism rather than just the product.
Synthesis problems connect every reaction in the course into a toolbox. The more reactions you know cold (reagents, regiochemistry, stereochemistry, limitations), the more synthesis routes you can construct.
Spectroscopy reappears throughout Organic Chemistry II (carbonyl chemistry, aromatic compounds) and is essential in lab courses. The systematic approach here (MS, then IR, then NMR) is the same method you will use for every unknown.
curved-arrow mechanism, electron-pushing arrows, nucleophile, electrophile, chlorohydrin mechanism, halonium ion, synthesis, retrosynthesis, multi-step synthesis, hydroboration-oxidation synthesis, elimination reaction, NaOEt EtOH, mass spectrometry, molecular ion, degree of unsaturation, IR spectroscopy, carbonyl stretch, 1H NMR, chemical shift, aldehyde proton, integration, splitting pattern, heat of hydrogenation, alkene stability, syn selective, anti selective, radical polymerisation, head-to-tail polymer, enol, keto-enol tautomerism, alkyne hydration, HgSO4, organic chemistry I, UMN ochem exam 4