Great — I converted a standard Chapter 3 on alkenes & alkynes into a compact, exam-focused set of 40 high-quality flashcards optimized for active recall and spaced repetition. Each card is numbered, shows a single focused prompt on the Front, and a concise, exam-ready Back with explanations, examples, mnemonics, and a difficulty/priority marker. After the cards you'll find a short, practical study strategy guide (optimal schedule, grouping, and tips). Flashcards — Alkenes & Alkynes (40) Front: Define an alkene. Back: An alkene is a hydrocarbon containing at least one carbon–carbon double bond (C=C). General formula for a single double bond: CnH2n (acyclic, no rings). Double bond is planar (sp²) and restricts rotation. Difficulty: Easy. Priority: High. Front: Define an alkyne. Back: An alkyne is a hydrocarbon containing at least one carbon–carbon triple bond (C≡C). General formula for a single triple bond: CnH2n−2 (acyclic, no rings). Triple bond carbons are linear (sp). Difficulty: Easy. Priority: High. Front: What hybridizations do C atoms in C–C single, double, and triple bonds have? Back: Single (C–C) → sp³; Double (C=C) → sp²; Triple (C≡C) → sp. Mnemonic: “3, 2, 1” (more bonds → fewer orbital lobes). Difficulty: Easy. Priority: High. Front: Describe geometry and bond angles for sp² and sp carbons. Back: sp²: trigonal planar, ~120° angles; sp: linear, 180°. Difficulty: Easy. Priority: High. Front: How do you name a simple alkene (IUPAC rules — priority points)? Back: Longest chain containing the double bond; number chain from end nearest double bond; give position of double bond (e.g., 2-butene); use –ene suffix; include E/Z if needed. Difficulty: Medium. Priority: High. Front: What is cis/trans vs E/Z stereochemistry for alkenes? Back: cis/trans used when two identical substituents on each double-bond carbon — cis (same side), trans (opposite). E/Z uses Cahn–Ingold–Prelog: highest-priority groups—Z (zusammen, together), E (entgegen, opposite). Difficulty: Medium. Priority: High. Front: How do you assign E/Z configuration step-by-step? Back: 1) For each double-bond carbon, rank substituents by atomic number (CIP). 2) Determine whether highest-priority groups are on same side (Z) or opposite (E). If tie, compare next atoms. Difficulty: Medium. Priority: High. Front: What is Markovnikov's rule? Back: In the addition of HX to an unsymmetrical alkene, H attaches to the carbon with more Hs (less substituted) and X attaches to the more substituted carbon (the one that forms the more stable carbocation). Mnemonic: “H goes to H-rich.” Difficulty: Medium. Priority: High. Front: What is anti-Markovnikov addition and when does it occur? Back: Anti-Markovnikov: the nucleophile (e.g., Br in HBr) adds to the less substituted carbon. Occurs with radical conditions (peroxides) for HBr and with hydroboration-oxidation (BH3 → OH adds anti-Markovnikov). Difficulty: Medium. Priority: High. Front: Describe the mechanism and regiochemistry of hydroboration-oxidation of alkenes. Back: Concerted syn addition of BH3 across double bond (anti-Markovnikov). Oxidation (H2O2, OH−) replaces B with OH, giving alcohol at less substituted carbon. Syn stereochemistry (both H and OH same face). Difficulty: Medium. Priority: High. Front: Describe electrophilic addition to alkenes (general mechanism). Back: Step 1: π bond attacks electrophile → carbocation or cyclic intermediate. Step 2: nucleophile attacks carbocation → product. Carbocation stability (3° > 2° > 1°) controls regiochemistry and possible rearrangements. Difficulty: High. Priority: High. Front: What is carbocation rearrangement and why does it matter in additions? Back: After carbocation forms, a hydride or alkyl shift may occur to form a more stable carbocation, changing product regiochemistry (e.g., unexpected Markovnikov product). Common mistake: forgetting possible rearrangements when predicting products. Difficulty: High. Priority: High. Front: What is oxymercuration-demercuration and why is it used? Back: Oxymercuration (Hg(OAc)₂, H₂O) followed by NaBH₄ gives Markovnikov alcohol without carbocation rearrangement. Mercury forms bridged mercurinium ion preventing rearrangement. Difficulty: Medium. Priority: High. Front: Explain hydrogenation of alkenes: catalysts, stereochemistry, and thermodynamics. Back: H₂ with metal catalyst (Pd, Pt, Ni) adds syn to double bond producing saturated alkane. Reaction is exothermic; used to measure stability (heat of hydrogenation). Difficulty: Medium. Priority: High. Front: How does Lindlar catalyst vs dissolving-metal reduction affect alkynes? Back: Lindlar (Pd/CaCO₃ poisoned) hydrogenates alkynes to cis-alkenes (partial, syn). Dissolving-metal (Na, NH₃(l)) produces trans-alkenes via radical mechanism. Difficulty: Medium. Priority: High. Front: What is catalytic hydrogenation of alkynes to alkanes (full reduction)? Back: H₂ with Pd/Pt/Ni in excess hydrogen reduces alkynes fully to alkanes (no stereocontrol). To stop at alkene, use Lindlar (cis) or Na/NH₃ (trans). Difficulty: Easy. Priority: Medium. Front: How do you synthesize internal and terminal alkynes from dihalides? Back: Sequential elimination (2× E2 using strong base, e.g., NaNH₂) on a vicinal or geminal dihalide yields an alkyne. For terminal alkynes, final step uses a proton source; strong base required for deprotonation if alkylation desired. Difficulty: Medium. Priority: High. Front: How are terminal alkynes alkylated? (key C–C bond-forming method) Back: Deprotonate terminal alkyne with strong base (NaNH₂ or n-BuLi) to form acetylide anion; then perform SN2 with primary (or methyl) alkyl halide to form new C–C bond. Avoid secondary/tertiary halides. Difficulty: High. Priority: High. Front: Compare acidity: alkane vs alkene vs alkyne (give approximate pKa values). Back: Alkane pKa ~50, Alkene pKa ~44, Terminal alkyne pKa ~25. Terminal alkynes are significantly more acidic due to sp hybridization (greater s-character). Difficulty: Medium. Priority: High. Front: What IR and ¹H NMR signals indicate an alkene or alkyne? Back: IR: C=C ~1600–1680 cm⁻¹; =C–H stretch ~3020 cm⁻¹. Alkyne: C≡C ~2100–2260 cm⁻¹ (weak), ≡C–H stretch ~3300 cm⁻¹ (sharp) for terminal alkynes. ¹H NMR: vinylic H appear ~4.5–6.5 ppm; terminal ≡C–H ~2.5–3.0 ppm. Difficulty: Medium. Priority: Medium. Front: What is halogenation of alkenes and its stereochemistry? Back: Addition of X₂ (Br₂, Cl₂) across C=C proceeds by anti addition via a cyclic halonium ion intermediate → vicinal dihalide with anti stereochemistry. In polar solvent, halohydrin can form. Difficulty: Medium. Priority: High. Front: How does halohydrin formation occur and what is regiochemistry? Back: Reaction with X₂ in H₂O: a halonium ion forms; H₂O attacks the more substituted carbon (more stable carbocation character) → halohydrin (OH on more substituted carbon, X on less). Anti stereochemistry. Difficulty: Medium. Priority: Medium. Front: Describe epoxidation of alkenes and stereochemical outcome. Back: Peracid (mCPBA) transfers an O atom to the π bond in a concerted syn process producing an epoxide. Stereospecific: substituents that are syn in alkene remain syn in epoxide. Difficulty: Medium. Priority: Medium. Front: How do you convert an epoxide to a trans-diol? (two common methods) Back: 1) Acid-catalyzed ring opening (H₃O⁺) gives trans diol after hydrolysis (anti opening). 2) Base-catalyzed opening with nucleophile (e.g., OH⁻) also gives trans relationship (but regiochemistry depends on substitution). Difficulty: Medium. Priority: Medium. Front: Describe syn and anti dihydroxylation methods for alkenes. Back: Syn: OsO₄ (or KMnO₄ cold, mild) gives syn-1,2-diol. Anti: epoxidation followed by acidic ring opening yields trans (anti) diol. Difficulty: Medium. Priority: Medium. Front: What is ozonolysis of an alkene and what products result? Back: O₃, then reductive workup (Zn or (CH₃)₂S) cleaves C=C to give carbonyl compounds: aldehydes or ketones (if alkene carbon has H → aldehyde; if not → ketone). Oxidative workup (H₂O₂) converts aldehydes to carboxylic acids. Difficulty: Medium. Priority: High. Front: Heat of hydrogenation and alkene stability—what's the relationship? Back: Lower (less exothermic) heat of hydrogenation → more stable alkene. E (trans) is typically more stable than Z (cis); more substituted alkenes are more stable (tetrasubstituted > trisubstituted > disubstituted > monosubstituted). Difficulty: Medium. Priority: High. Front: Predict major product: elimination (E2) from a secondary alkyl bromide with a strong base (e.g., t-BuOK). Zaitsev or Hofmann? Explain. Back: t-BuOK (bulky base) favors Hoffman (less substituted) alkene due to steric hindrance. Small, strong base (e.g., NaOEt) favors Zaitsev (more substituted). Difficulty: Medium. Priority: High. Front: What conditions favor E1 vs E2 elimination mechanisms? Back: E1: weak base, polar protic solvent, tertiary substrate, heat — proceeds via carbocation (may rearrange). E2: strong base, usually primary/secondary/tertiary, concerted, stereospecific (antiperiplanar H required). Difficulty: Medium. Priority: High. Front: How is syn vs anti addition determined experimentally? Give two reaction examples (one syn, one anti). Back: Syn example: hydrogenation (H₂/Pd) — both H add same face. Anti example: Br₂ addition — anti via halonium ion. Use stereochemistry of product to infer mechanism. Difficulty: Medium. Priority: Medium. Front: What is allylic bromination (NBS) and what radical selectivity does it show? Back: NBS in presence of light or peroxides selectively brominates allylic C–H (resonance-stabilized radical) rather than adding across double bond. Radical substitution gives allylic bromide; product is often a mixture of stereoisomers. Difficulty: Medium. Priority: Medium. Front: For conjugated dienes, what are 1,2 and 1,4 additions and what determines product ratio? Back: 1,2 adds across adjacent carbons (kinetic product; forms faster, at low temp). 1,4 (conjugate) gives more stable product (thermodynamic product; favored at higher temp). Reaction conditions (temp, solvent) and kinetics vs thermodynamics determine ratio. Difficulty: Hard. Priority: Medium. Front: What is conjugation and how does it stabilize alkenes/dienes? Back: Conjugation = alternating π and σ bonds allowing delocalization of π electrons over adjacent p-orbitals. Delocalization lowers overall energy → increased stability and changed reactivity (e.g., resonance-stabilized allylic carbocations). Difficulty: Medium. Priority: High. Front: How do alkynes differ from alkenes in electrophilic addition reactivity? Back: Alkynes have two π bonds; they undergo addition twice. Terminal alkynes are less reactive toward electrophiles than alkenes but can be hydrated or halogenated; addition often proceeds through vinyl carbocations (less stable) so different regio/stereochemistry and catalysts (e.g., Hg²⁺ for hydration). Difficulty: Medium. Priority: Medium. Front: Describe hydration of alkynes (Markovnikov) and catalysts used. Back: Acid-catalyzed hydration with Hg²⁺ (HgSO₄, H₂SO₄) yields a ketone (via enol tautomerization) following Markovnikov regiochemistry. For terminal alkyne, gives methyl ketone. Difficulty: Medium. Priority: High. Front: Describe hydroboration-oxidation of alkynes and what product it gives for terminal alkynes. Back: Hydroboration (R₂BH) then H₂O₂/OH⁻ on internal alkyne gives enol → tautomerizes to aldehyde (anti-Markovnikov) if using sterically bulky borane and terminal alkyne protocol (weaker conditions). For terminal alkynes, hydroboration-oxidation gives an aldehyde (anti-Markovnikov). Difficulty: Hard. Priority: Medium. Front: What is the acidity test to identify a terminal alkyne in a lab? Back: Terminal alkyne reacts with strong base (NaNH₂, NaH, n-BuLi) to form a soluble acetylide anion (often precipitate or reacts with electrophile). Terminal alkynes deprotonate; internal alkynes do not under mild conditions. Difficulty: Easy. Priority: Medium. Front: Name and describe a method for cyclopropanation of alkenes (product and stereochemistry). Back: Simmons–Smith (CH₂I₂ + Zn(Cu)) or carbenoid/metal-carbene (e.g., CH₂N₂ + Cu) adds a CH₂ across C=C to form a cyclopropane. Addition is concerted and stereospecific (cis alkenes → cis cyclopropane). Difficulty: Medium. Priority: Low. Front: Common mistake: predicting addition product without considering stereochemistry and mechanism — give an example and short fix. Back: Example: predicting Markovnikov HBr addition but forgetting peroxides → anti-Markovnikov product instead. Fix: always note reagents/conditions (radical vs ionic), stereochemical course (syn/anti), and possibility of rearrangement. Difficulty: Easy. Priority: High. Front: Give three quick heuristics to choose a reaction for converting alkene → alcohol (marking when each is best). Back: (1) Hydroboration-oxidation (BH₃ → H₂O₂/OH⁻): anti-Markovnikov, syn, no rearrangement. (2) Oxymercuration-demercuration (Hg(OAc)₂ / NaBH₄): Markovnikov, no rearrangement. (3) Acid-catalyzed hydration (H₃O⁺): Markovnikov, may rearrange. Choose based on desired regiochemistry and rearrangement tolerance. Difficulty: Medium. Priority: High. Study Strategy Guide (brief & practical) 1) Priority & Scheduling (spaced repetition): Label colors in your SRS: High priority (cards 1–3,6–9,11–14,17–19,26–29,34–36,40) — review daily until mastered. Medium priority — review every 2–3 days. Low priority — review weekly. Suggested initial schedule (first 10 days): Day 1 — all 40; Day 2 — review High + missed cards; Day 4 — all High + Medium; Day 7 — all cards; Day 14 — full review; then space by mastery (SRS algorithm recommended). 2) Grouping for efficient sessions (three 15–20 minute blocks): Block A — Foundations & structure (cards 1–7,19,20) Block B — Mechanisms & major additions (8–16,21–26,30) Block C — Alkynes, advanced, problems & exceptions (17–19,31–40) Group cards so each session mixes concept types (definition, mechanism, application) to strengthen retrieval cues. 3) Active-recall and variation tips: Shuffle frequently: each card is self-contained — test blind recall, then check back. Use the cover-and-recall approach: read front, answer aloud and write short mechanism or drawing, then verify back. For mechanism cards, always draw stepwise curly arrows on paper — drawing enforces deeper encoding. For stereochemistry/regiochemistry, draw starting materials plus reagents — then predict product and compare. 4) Error management (learning from mistakes): Keep an “error log” with 3 columns: Wrong prediction, Why wrong (mechanism/reagent oversight), Correct reasoning. Review these prior to each study session. Pay special attention to rearrangements and radical vs ionic conditions (common exam trap). 5) Memory aids included in cards: Use the included mnemonics (e.g., “H goes to H-rich” for Markovnikov) and mental imagery: visualize BH₃ approaching less substituted face for hydroboration, picture bridged halonium ring for anti attack, picture linear carbon when thinking sp hybrid alkyne acidity. 6) Practice & exam simulation: After mastering cards, simulate timed problems: pick 6 random cards and write mechanism for each in 30–60 seconds; then check and correct. For synthesis questions, practice multi-step planning: e.g., convert alkene to aldehyde — pick hydroboration-oxidation then oxidation steps and write reagents. Would you like these exported in a specific flashcard format (Anki CSV, Quizlet import, or spaced-repetition-ready CSV)? I can produce a downloadable file formatted for Anki (front/back, tags, difficulty) or a printable PDF with 2-up index card layout — tell me which and I’ll generate it now.