Difficulty: Advanced | Prerequisites: Alkene/alkyne reactions, nomenclature, stability
This set of notes covers three exam question types: drawing detailed stepwise mechanisms with curved arrows, designing multi-step syntheses from a given starting material to a target product, and deducing molecular structures from IR and ¹H NMR spectral data. These are the highest-point questions on the exam and require you to combine everything from nomenclature, stability, and reactions into applied problem-solving.
Homolytic cleavage
A bond breaks so that each atom keeps one electron from the shared pair, producing two radicals. Shown with fishhook (single-barbed) arrows. This is how the peroxide O-O bond breaks in the initiation step of a radical mechanism.
Heterolytic cleavage
A bond breaks so that one atom keeps both electrons, producing a cation and an anion. Shown with standard (double-barbed) curved arrows. This is the arrow type used in electrophilic addition mechanisms like HCl addition.
Carbocation
A positively charged, sp² hybridised carbon with an empty p orbital. Formed as an intermediate in electrophilic additions (e.g. HCl or H₂O/H₂SO₄ addition to alkenes). Stability: methyl < 1° < 2° < 3°.
¹H NMR (proton nuclear magnetic resonance)
A spectroscopic technique that reveals the number of distinct hydrogen environments in a molecule, their electronic surroundings (chemical shift in ppm), how many neighbouring hydrogens they have (splitting pattern), and their relative abundance (integration). In simple terms, it is a map of where the hydrogens are and what is next to them.
Chemical shift (δ, ppm)
The position of an NMR signal on the x-axis, measured in parts per million relative to TMS. Electropositive environments (alkyl C-H) appear at 0 to 2 ppm. Vinyl C-H appears at 5 to 7 ppm. Aldehyde C-H at 9 to 10 ppm.
Splitting (multiplicity)
The pattern of an NMR signal, determined by the n+1 rule: a hydrogen with n equivalent neighbouring hydrogens splits into n+1 peaks. A singlet (s) means zero neighbours, a doublet (d) means one neighbour, a triplet (t) means two, and so on.
Integration
The relative area under each NMR peak, which corresponds to the relative number of hydrogens producing that signal. A signal integrating for 9H (singlet) is a strong indicator of a tert-butyl group, (CH₃)₃C-.
IR spectroscopy (infrared)
Measures which frequencies of infrared light a molecule absorbs, corresponding to bond vibrations. Key absorptions for this exam: Csp²-H stretch near 3000 to 3100 cm⁻¹ (alkene), C=C stretch near 1600 to 1680 cm⁻¹, Csp-H stretch near 3300 cm⁻¹ (terminal alkyne), C≡C stretch near 2100 to 2260 cm⁻¹.
Vicinal dihalide
A compound with two halogen atoms on adjacent (vicinal) carbons. Formed by addition of X₂ to an alkene.
Double elimination
Two successive E2 eliminations, typically using a strong base like NaNH₂. A vicinal dihalide undergoes first elimination to give a vinyl halide, then second elimination to give an alkyne.
This was tested as Question 5a on the practice exam. The mechanism has three phases:
Initiation
The peroxide (ROOR) undergoes homolytic cleavage of the weak O-O bond when heated or exposed to light, generating two alkoxy radicals (RO·).
Each step uses fishhook (single-barbed) arrows.
The alkoxy radical then abstracts a hydrogen from HBr, producing ROH and a bromine radical (Br·).
Propagation (two repeating steps)
Step 1: The Br· radical adds to the less substituted carbon of the alkene. This forms a new C-Br bond and places the radical on the more substituted carbon (the more stable radical). Draw the fishhook arrow from Br· to the terminal carbon, and a second fishhook from the pi bond to the internal carbon.
Step 2: The carbon radical abstracts a hydrogen from another molecule of HBr, forming the C-H bond and regenerating a Br· radical to continue the chain.
The product has anti-Markovnikov regiochemistry (Br on the less substituted carbon).
Termination
Any two radicals combine to form a covalent bond. Examples: Br· + Br· → Br₂, or two carbon radicals coupling, or Br· + carbon radical → product.
Termination steps are not productive and consume radicals to end the chain.
This was tested as Question 5b. The mechanism has two steps:
Step 1: Protonation of the alkene
The pi electrons of the alkene act as the nucleophile and attack the electrophilic H of HCl.
Draw a curved arrow from the C=C pi bond to the H of HCl, and a second curved arrow from the H-Cl bond to Cl (heterolytic cleavage).
The proton adds to the less substituted carbon (Markovnikov's rule), forming the more stable carbocation on the more substituted carbon.
Use standard double-barbed curved arrows throughout.
Step 2: Nucleophilic attack
The chloride anion (Cl⁻) attacks the carbocation.
Draw a curved arrow from a lone pair on Cl⁻ to the positively charged carbon.
Product: a chloroalkane with Markovnikov regiochemistry (Cl on the more substituted carbon).
If the carbocation is planar and not adjacent to a ring or other constraint, Cl⁻ can attack from either face, giving a racemic mixture.
Question 4 on the practice exam asks you to convert a starting material into a target product using any reagents, showing all isolable intermediates.
Compare the starting material and the target. Identify what has changed: new functional groups, new bonds, changes in carbon skeleton length, changes in stereochemistry.
Work backwards (retrosynthetic analysis): ask "what reaction could produce the target?" and then "what starting material would I need for that reaction?"
Chain the steps forward in your final answer, showing each reagent set over or under an arrow, and drawing the intermediate product at each stage.
The practice exam shows conversion of 2-methylpropene to a longer-chain alkene. A likely approach:
Identify that the carbon skeleton needs to grow. Alkynes can be extended via acetylide alkylation.
Convert the alkene to a suitable intermediate (e.g. a halide via HBr addition), then use that halide to alkylate an acetylide, then reduce the resulting alkyne to the desired alkene geometry.
The practice exam shows propyne converted to a secondary alcohol (+ enantiomer). A possible route:
Extend the chain: deprotonate propyne with NaNH₂ to form the propynide anion, then alkylate with an appropriate alkyl halide.
Convert the internal alkyne to an alkene (Lindlar for cis, Na/NH₃ for trans, depending on what the target needs).
Hydrate the alkene (H₂O/H₂SO₄ for Markovnikov alcohol) to install the OH at the correct position.
If the product is racemic, confirm that the mechanism goes through a planar intermediate (carbocation) allowing attack from both faces.
Terminal alkyne → acetylide: NaNH₂
Acetylide + R-X → internal alkyne (SN2 alkylation, works best with primary alkyl halides and methyl halides)
Internal alkyne → cis alkene: H₂, Lindlar catalyst
Internal alkyne → trans alkene: Na, NH₃
Alkene → alcohol (Markovnikov): H₂O, H₂SO₄
Alkene → alcohol (anti-Markovnikov): hydroboration-oxidation (BH₃ then H₂O₂/NaOH)
Alkene → alkyl halide (Markovnikov): HBr or HCl
Alkene → alkyl halide (anti-Markovnikov): HBr, ROOR
Question 6 on the practice exam is a worked structure-determination problem. Here is a walkthrough of the reasoning.
Compound A has molecular formula C₇H₁₄ and Z stereochemistry.
IR of A shows peaks near 1500 cm⁻¹ and 3100 cm⁻¹. The 3000 to 3100 cm⁻¹ region corresponds to Csp²-H stretches (alkene C-H). The ~1500 cm⁻¹ peak is in the fingerprint region but is consistent with C=C stretching.
IHD for C₇H₁₄: (2(7) + 2 - 14) / 2 = 1. One degree of unsaturation, so one double bond or one ring. The IR confirms an alkene.
9H singlet (around 1 ppm): a tert-butyl group, (CH₃)₃C-. Nine equivalent methyl hydrogens with no adjacent C-H neighbours give a singlet.
3H doublet (around 1.5 ppm): a -CH₃ group split into a doublet by one neighbouring hydrogen. Likely a =CH-CH₃ unit.
1H multiplet (around 5 ppm): a vinyl hydrogen (on the C=C) that is coupled to several neighbours.
1H doublet (around 5.5 ppm): another vinyl hydrogen split by one neighbour.
Putting it together: C₇H₁₄ with one C=C, Z configuration, a tert-butyl group, and a methyl group on the alkene. The structure is (Z)-4,4-dimethyl-2-pentene: (CH₃)₃C-CH=CH-CH₃ with the two larger groups (tert-butyl and methyl) on the same side (Z).
A is treated with Cl₂ → B.
Cl₂ adds across the double bond (anti addition via a chloronium ion) to give a vicinal dichloride.
B is the anti-addition product of Cl₂ to (Z)-4,4-dimethyl-2-pentene. Two new stereocentres form at C2 and C3. Because the starting alkene is Z (cis), and Cl₂ adds anti, the product has a specific relative stereochemistry. Since the two carbons are not equivalent, expect a racemic mixture (+ enantiomer).
B is treated with 2 equiv. NaNH₂ → C.
Two equivalents of a strong base on a vicinal dihalide means double elimination, producing an alkyne.
The two Cl atoms and two adjacent H atoms are removed to form a triple bond.
C is 4,4-dimethyl-2-pentyne: (CH₃)₃C-C≡C-CH₃.
9H singlet: the tert-butyl group, (CH₃)₃C-. Same as in A.
3H singlet (around 1.8 ppm): a methyl group attached to the triple bond, -C≡C-CH₃. No adjacent C-H neighbours (the adjacent carbon is part of the triple bond), so it appears as a singlet.
Two signals, both singlets, matching a simple symmetrical internal alkyne with a tert-butyl on one end and a methyl on the other. This confirms C = 4,4-dimethyl-2-pentyne.
0 to 2 ppm: alkyl C-H (sp³)
2 to 3 ppm: allylic, benzylic, or alpha-to-carbonyl C-H
3 to 5 ppm: C-H next to O, N, or halogen
5 to 7 ppm: vinyl C-H (sp²)
6.5 to 8.5 ppm: aromatic C-H
9 to 10 ppm: aldehyde C-H
3200 to 3600 cm⁻¹: O-H or N-H
3000 to 3100 cm⁻¹: Csp²-H (alkene)
2850 to 3000 cm⁻¹: Csp³-H
~3300 cm⁻¹: Csp-H (terminal alkyne, sharp)
2100 to 2300 cm⁻¹: C≡C or C≡N
1600 to 1850 cm⁻¹: C=O or C=C
Students often draw ionic (double-barbed) curved arrows in radical mechanisms and vice versa. Radical steps use fishhook (single-barbed) arrows. Ionic steps use full curved arrows. Mixing them up will cost you marks even if the products are correct.
A frequent error in synthesis problems is skipping intermediates. The exam instructions say to show all isolable intermediates. If you go from an alkyne to an alcohol in one arrow without showing the alkene intermediate, you lose points.
In spectral analysis, students sometimes guess a structure that fits the molecular formula but ignore the NMR data. Every signal in the NMR must be accounted for. If your proposed structure has six distinct hydrogen environments but the NMR shows only four signals, your structure is wrong.
Students often confuse the n+1 rule with the number of peaks. A doublet means one neighbouring hydrogen (n=1, so n+1=2 peaks), not two.
⚠️ Question 5 (Mechanisms) is worth 22 points. You must draw every arrow, every intermediate, and every phase (initiation, propagation, termination for radicals; or each discrete step for ionic mechanisms). Missing a phase or using the wrong arrow type loses significant marks.
⚠️ Question 4 (Synthesis) is worth 20 points. Show reagents over each arrow and draw each isolable intermediate. Do not combine steps. Do not forget to specify stereochemistry when the question calls for it.
⚠️ Question 6 (Spectral Analysis) is worth 9 points but is often the question students find hardest. Start by calculating IHD, then match each NMR signal to a structural fragment, and piece them together. Verify your answer against every piece of data before writing it in the box.
⚠️ A 9H singlet at ~1 ppm almost always means a tert-butyl group. Learn to spot this immediately.
True or false: In a radical mechanism, curved arrows should be double-barbed. (False, use single-barbed fishhook arrows)
Fill in the blank: The three phases of a radical chain mechanism are ____, ____, and ____. (Initiation, propagation, termination)
True or false: A 9H singlet in ¹H NMR near 1 ppm indicates a tert-butyl group. (True)
Fill in the blank: Two equivalents of NaNH₂ on a vicinal dihalide produce an ____. (Alkyne, via double elimination)
True or false: In the electrophilic addition of HCl to an alkene, the proton adds to the more substituted carbon. (False, the proton adds to the less substituted carbon to form the more stable carbocation on the more substituted carbon)
Q: Write out the full radical chain mechanism for the reaction of 1-methylcyclohexene with HBr in the presence of ROOR. What is the product, and what is its regiochemistry?
A: Initiation: ROOR undergoes homolytic cleavage to give 2 RO·. Then RO· abstracts H from HBr to give ROH + Br·. Propagation step 1: Br· adds to the less hindered carbon of the double bond (C2 in this case, the =CH₂ end, or the less substituted position), forming a tertiary radical at C1 (the more substituted carbon). Propagation step 2: the carbon radical abstracts H from HBr, forming the C-H bond and regenerating Br·. Termination: any two radicals combine. The product is 2-bromo-1-methylcyclohexane (anti-Markovnikov, Br on the less substituted carbon).
Q: Draw the step-by-step mechanism for the addition of HCl to 2-methylpropene (isobutylene). What product forms?
A: Step 1: the pi electrons attack H of HCl, forming a tertiary carbocation at C2 (the more substituted carbon) and releasing Cl⁻. Step 2: Cl⁻ attacks the carbocation. Product: 2-chloro-2-methylpropane (tert-butyl chloride). Markovnikov regiochemistry.
Q: An unknown compound has molecular formula C₅H₁₀ and shows a 3H singlet at 1.7 ppm, a 3H singlet at 1.7 ppm, and a 1H triplet at 5.1 ppm, plus a 3H triplet at 1.0 ppm and a 2H quartet at 2.0 ppm in the ¹H NMR. What is the structure?
A: C₅H₁₀ gives IHD = 1 (one double bond or ring). The signal at 5.1 ppm (1H, triplet) is a vinyl hydrogen. Two singlets at 1.7 ppm (3H each) suggest two methyl groups on the double bond with no adjacent C-H. A 2H quartet + 3H triplet indicates a -CH₂CH₃ group. Putting it together: 2-methyl-2-pentene, or more precisely, the structure is CH₃CH₂-CH=C(CH₃)₂ is not quite right (that would be C₆). For C₅H₁₀, consider 2-methyl-2-butene: (CH₃)₂C=CH-CH₃. Check the signals: this does not perfectly match the splitting described. This illustrates why you must count every hydrogen against the formula. The correct approach is always: calculate IHD, list fragments from each NMR signal, assemble, and verify.
Q: Propose a synthesis of 3-hexyne from 1-propyne.
A: Step 1: NaNH₂, deprotonates the terminal alkyne to give the propynide anion (CH₃-C≡C⁻ Na⁺). Step 2: add 1-bromopropane (CH₃CH₂CH₂Br) for SN2 alkylation. The acetylide attacks the primary alkyl halide to form CH₃-C≡C-CH₂CH₂CH₃, which is 3-hexyne. Two steps, one intermediate (the sodium acetylide).
Mechanisms tie directly to the reaction products covered in the second set of notes. If you can draw the mechanism, you can predict the product, and vice versa. The radical mechanism explains why HBr/ROOR gives anti-Markovnikov products (the radical forms at the more stable, more substituted position). The ionic mechanism explains why HCl gives Markovnikov products (the carbocation forms at the more substituted position).
Spectral analysis connects to every other topic: you need to know what functional groups are present (IR), how many distinct hydrogen environments exist (NMR), and what reactions could interconvert the compounds described. Question 6 on this exam is a chain of deductions: identify A from its NMR and formula, predict B from A + Cl₂, predict C from B + NaNH₂, and verify C against its own NMR.
Radical mechanism, electrophilic addition, curved arrows, fishhook arrows, homolytic cleavage, heterolytic cleavage, initiation, propagation, termination, carbocation, synthesis, retrosynthesis, acetylide alkylation, NaNH2, double elimination, NMR, proton NMR, 1H NMR, chemical shift, splitting, integration, singlet, doublet, triplet, multiplet, tert-butyl, IR spectroscopy, Csp2-H, degree of unsaturation, IHD, spectral analysis, structure determination, organic chemistry I, CHEM 2301, Exam 4