📍 Chapter Overview
Haloalkanes — Complete Mind Map
Topics Covered: Classification (1°, 2°, 3°, allylic, benzylic, vinylic) · Preparation · Physical Properties · SN2 Mechanism (Walden inversion, polar aprotic solvent) · SN1 Mechanism (racemization, carbocation intermediate, polar protic) · E2 Elimination (anti-periplanar H, Zaitsev) · E1 Elimination · SN vs E Competition Rules · Industrial Uses
🤖 AI Prompt — Chapter Mind Map:
Deep indigo-violet themed chemistry mind map on dark background. Central node: "HALOALKANES — R-X" (bold white). Bond shown: Cδ+–Xδ− in red with polarity arrows. Seven branches: (1) "Classification" — 1° RX, 2° RX, 3° RX, allylic (CH2=CH-CH2X), benzylic (Ph-CH2X), vinylic (CH2=CHX); (2) "Preparation" — from ROH (HX, SOCl2, PBr3), from alkene (HX), Finkelstein (NaI/acetone), Swarts (AgF); (3) "SN2 Mechanism" — one-step, back attack, inversion shown with arrows, rate = k[RX][Nu], CH3 best; (4) "SN1 Mechanism" — two-step, carbocation intermediate (trigonal planar), racemization, rate = k[RX], 3° best; (5) "E2" — anti-periplanar H required, single step, Zaitsev product; (6) "E1" — same first step as SN1 (carbocation), then base removes beta-H; (7) "SN vs E" — conditions box: temperature, nucleophile strength, solvent type. Color code: SN2 in blue, SN1 in red, E2 in green. High resolution, educational.
1. Classification of Haloalkanes
1° Alkyl Halide: –X on a carbon bonded to 1 other carbon. (e.g. CH₃CH₂Br — 1-bromoethane)
2° Alkyl Halide: –X on a carbon bonded to 2 other carbons. (e.g. (CH₃)₂CHBr)
3° Alkyl Halide: –X on a carbon bonded to 3 other carbons. (e.g. (CH₃)₃CBr)
Allylic halide: –X on carbon adjacent to C=C. (e.g. CH₂=CHCH₂Br). Very reactive in SN1 (allylic carbocation stabilized by resonance).
Benzylic halide: –X on carbon adjacent to benzene ring. (e.g. C₆H₅CH₂Br). Very reactive in SN1 (benzylic carbocation stabilized by 4 resonance structures).
Vinylic halide: –X directly on the C=C carbon. (e.g. CH₂=CHCl). Very UNreactive in SN (covered in Module 10).
2. Preparation of Alkyl Halides
From Alcohols + HX: R–OH + HX → R–X + H₂O (reactivity: HI > HBr > HCl)
From Alcohols + SOCl₂: R–OH + SOCl₂ → R–Cl + SO₂ + HCl (neat product easily isolated — SO₂ and HCl gas escape)
From Alcohols + PBr₃: 3R–OH + PBr₃ → 3R–Br + H₃PO₃
From Alkenes + HX: (Markovnikov addition — see Module 06)
Finkelstein Reaction: R–Cl + NaI (in acetone) → R–I + NaCl↓ (NaCl is insoluble in acetone → drives equilibrium forward)
Swarts Reaction: R–Cl + AgF → R–F + AgCl (preparation of alkyl fluorides)
3. SN2 — Nucleophilic Substitution Bimolecular
SN2 Mechanism — Back-Attack, Trigonal Bipyramidal TS, Walden Inversion
Draw a detailed, step-by-step SN2 mechanism diagram for the reaction of (R)-2-bromobutane with NaOH (HO⁻ as nucleophile) on a white background. The diagram should show: STARTING MATERIAL: (R)-2-bromobutane drawn in 3D perspective (wedge-dash notation) — central C2 with Br going back (dashed wedge), OH group approaching from the front-back of opposite face, H up (wedge), CH₃ left (line), C₂H₅ right (line). Label: "(R)-2-bromobutane". Large bold curved arrow from the HO⁻ lone pair attacking the C2 carbon from the BACK (the face OPPOSITE to Br). TRANSITION STATE: Draw a dotted-bond transition state — a trigonal bipyramidal arrangement where HO is partially bonded (dotted line) to C2 from one axial position, Br is partially leaving (dotted line) from the other axial position, and the three remaining groups (H, CH₃, C₂H₅) are in the equatorial plane (flat/perpendicular to the HO---C---Br axis). Label: "Transition State [‡] — trigonal bipyramidal — CONCERTED (one step)". Large bold curved arrow from Br leaving the C2. PRODUCT: (S)-2-butanol drawn in 3D — the configuration has INVERTED (Walden inversion). The HO is now where Br was and the three groups have flipped (umbrella flip). Label: "(S)-butan-2-ol — inversion of configuration (Walden inversion)". Add notes: "Rate = k[RX][Nu] — bimolecular, 2nd order. Best for: CH₃X and 1° RX. Polar aprotic solvent (DMSO, acetone) enhances nucleophilicity." White background, clean bold curved arrows, all 3D tetrahedral centers shown properly with wedge-dash notation. Chemistry textbook quality.
Key features of SN2:
- One step (concerted): bond forming and bond breaking happen simultaneously
- Rate = k[RX][Nu] — second order overall
- Inversion of configuration (Walden inversion) at the chiral centre
- Reactivity order: CH₃X > 1° > 2° >> 3° (steric hindrance blocks back attack)
- Best solvent: Polar aprotic (DMSO, DMF, acetone) — these dissolve the nucleophile but don't cage it with H-bonds, so Nu remains free and reactive
- No rearrangements (no carbocation intermediate)
4. SN1 — Nucleophilic Substitution Unimolecular
SN1 Mechanism — Carbocation Intermediate, Racemization
Draw a detailed two-step SN1 mechanism diagram for (R)-3-bromo-3-methylhexane reacting with water (H₂O solvent) on a white background. Starting material: the R compound drawn in 3D with a tertiary carbon center — Br attached as a dashed wedge, three carbon groups in plane. STEP 1 (Slow, Rate-Determining): Show Br leaving with both electrons (full curved arrow from C-Br bond to Br). Product: a CARBOCATION (planar, sp2 carbon) drawn as a flat triangle — the central carbon with 3 groups at 120°, empty p orbital shown as dashed dumbbell above and below the plane. Label: "Tertiary carbocation — PLANAR, sp2, empty p orbital. SLOW step. Rate = k[RX]". Show resonance stabilization by the three alkyl groups (arrows showing hyperconjugation optional). STEP 2 (Fast): Show H₂O (nucleophile) attacking from BOTH FACES of the flat carbocation: Top face attack → gives (R) product. Bottom face attack → gives (S) product. Draw both products side by side with their wedge-dash structures and configuration labels. Label: "FAST step — equal probability from both faces → RACEMIZATION (50% R + 50% S)". Below: show the final racemic product as ±-3-methyl-3-hexanol. Add notes: "Rate = k[RX] — unimolecular, 1st order. Best for: 3° RX. Polar protic solvent (H₂O, ROH) stabilizes the carbocation and the leaving group by solvation. Rearrangements possible (1,2-H shift, 1,2-methyl shift to more stable carbocation)." White background, textbook chemistry quality, all steps clearly labeled.
| Feature | SN2 | SN1 |
|---|
| Steps | One (concerted) | Two (via carbocation) |
| Rate law | k[RX][Nu] — 2nd order | k[RX] — 1st order |
| Best substrate | CH₃X, 1° RX | 3° RX (benzylic, allylic) |
| Stereochemistry | Inversion (Walden) | Racemization |
| Best solvent | Polar aprotic (DMSO, DMF) | Polar protic (H₂O, ROH) |
| Rearrangements | None | Possible (carbocation rearranges) |
| Nucleophile needed | Strong Nu required | Even weak Nu works |
5. Elimination Reactions — E2 and E1
E2 Elimination Mechanism — Anti-Periplanar Requirement (Newman Projection)
Draw the E2 elimination mechanism for 2-bromobutane reacting with KOH in ethanol on white background. Two sections: LEFT SECTION "Anti-periplanar requirement": Draw a Newman projection looking down the C2-C3 bond of 2-bromobutane. Front carbon (C2) has: Br (pointing left), H (pointing up-right), and CH₃ (pointing down-right). Back carbon (C3) has: H (pointing directly up, opposite to the H being abstracted), CH₃ (down-left), and CH₂CH₃ (middle... use appropriate arrangement). The key: show that the BASE (KOH) can only abstract the β-H that is ANTI-PERIPLANAR (180° dihedral angle) to the leaving group Br. Highlight the anti-periplanar H and Br in red/blue. Label: "Anti-periplanar — H and Br on opposite sides, 180° apart — required for E2". RIGHT SECTION "Mechanism": Draw the 2-bromobutane molecule (CH₃-CHBr-CH₂-CH₃) with KOH base. Show THREE curved arrows simultaneously: (1) KO⁻ lone pair → beta H (abstracting H). (2) C-H bond electrons → forming the pi bond. (3) C-Br bond electrons → leaving as Br⁻. Label: "CONCERTED — all arrows in one step". Products: Show primarily but-2-ene (CH₃CH=CHCH₃) labeled "Major product — Zaitsev's Rule (more substituted alkene)" and but-1-ene (CH₂=CHCH₂CH₃) labeled "Minor product". Add box: "Zaitsev's Rule: More substituted alkene = more stable = major E2 product". White background, textbook chemistry quality.
E1 Elimination:
- Two steps: same as SN1 first step (carbocation forms) → then base removes β-H → alkene
- Rate = k[RX] — unimolecular
- Best for 3° RX, polar protic solvents, heat
- Competes with SN1 under same conditions. Increasing temperature favours elimination over substitution.
- Zaitsev's Rule applies: more substituted alkene is major product.
6. SN vs E — Competition Rules
7. Important Haloalkane Compounds
Chloroform (CHCl₃): Trichloromethane. Formerly used as anaesthetic (toxic). Now used as solvent. Must be stored in dark bottles with added ~0.5% ethanol (prevents oxidation to phosgene COCl₂ — a deadly war gas). If stored improperly: CHCl₃ + O₂ → COCl₂ + HCl
Carbon Tetrachloride (CCl₄): Non-flammable, used in fire extinguishers (now limited — toxic). Dry cleaning solvent. Hepatotoxic (toxic to liver).
Iodoform (CHI₃): Yellow crystalline solid, antiseptic smell. Iodoform test: compound + I₂ + NaOH → yellow precipitate of CHI₃. Positive for: methyl ketones (CH₃COR), ethanol, secondary alcohols with CH₃CHOH– group, acetaldehyde.
Freons (CFCs — Chlorofluorocarbons): CCl₂F₂ (Freon-12), etc. Formerly in refrigerants, aerosols. NOW BANNED (cause ozone depletion — Cl• radicals in stratosphere catalytically destroy O₃).
DDT (Dichlorodiphenyltrichloroethane): First synthetic organic insecticide (used to control malaria). NOW BANNED — persistent in environment, bioaccumulation in food chain, carcinogenic.
Ex 1 M: (R)-2-bromobutane reacts with NaI in acetone. What is the product configuration?
Solution: NaI in acetone = Finkelstein condition (SN2). SN2 → Walden inversion. (R) → (S)-2-iodobutane.
Ex 2 M: 2-bromo-2-methylbutane + KOH/alcohol (heat) → major product?
Solution: 3° RX + strong base + alcohol solvent + heat → E2 (elimination). Zaitsev's Rule → most substituted alkene is major. The alkene with MORE substitution at C=C: 2-methylbut-2-ene (CH₃C(CH₃)=CHCH₃, trisubstituted) vs 2-methylbut-1-ene (CH₂=C(CH₃)CH₂CH₃, disubstituted). Major: 2-methylbut-2-ene (E2, Zaitsev).
Ex 3 H: Why does CHCl₃ need to be stored with ethanol?
Solution: CHCl₃ oxidizes slowly in air and light: CHCl₃ + O₂ → COCl₂ (phosgene) + HCl. Phosgene was used as a chemical weapon in WWI and is extremely toxic (causes pulmonary oedema). Ethanol present reacts with any COCl₂ formed (COCl₂ + 2 C₂H₅OH → (C₂H₅O)₂CO + 2HCl) and prevents its accumulation.
E Q1. Arrange in SN2 reactivity: (CH₃)₃CBr, C₂H₅Br, CH₃Br, (CH₃)₂CHBr.
E Q2. What is the product of Finkelstein reaction of 1-chlorobutane? What role does acetone play?
M Q3. When (±)-2-bromobutane undergoes SN2 with NaOH, what is the optical activity of the product? Explain.
M Q4. Compare: CH₃Cl + NaOH in DMSO vs NaOH in water. Which gives faster SN2 and why?
H Q5. A compound C₄H₉Br undergoes SN1 and gives a racemic mixture. What is the compound? Identify the carbocation intermediate and explain why racemization occurs.
H Q6. Explain with a diagram why E2 of (2R,3S)-2-bromo-3-deuterio-butane gives only the (Z)-but-2-ene (not E). Use anti-periplanar requirement.
- Adding heat and expecting SN product — heat ALWAYS FAVORS ELIMINATION over substitution.
- Applying SN2 to tertiary halides — 3° RX cannot undergo SN2 (too much steric hindrance blocks back-attack). Use SN1 or E2 for 3° compounds.
- Saying polar protic solvent promotes SN2 — WRONG! Polar protic solvents (H₂O, ROH) cage the nucleophile with H-bonds → reduce its nucleophilicity → HINDERS SN2. Use polar APROTIC solvents for SN2.
- Forgetting Walden inversion — every SN2 INVERTS configuration. If starting material is R, product is S, and vice versa.
- Confusing Zaitsev (more substituted, major) with Hofmann (less substituted, with bulky base) — in exams, unless the base is specified as bulky (t-BuOK), use Zaitsev's Rule.
| Reaction | Best RX | Best Conditions | Stereochemistry |
|---|
| SN2 | CH₃X, 1° | Strong Nu, polar aprotic, low T | Inversion (Walden) |
| SN1 | 3° (allylic, benzylic) | Polar protic, weak/any Nu, low T | Racemization |
| E2 | 2°, 3° | Strong (bulky) base, alcoholic KOH, high T | Anti-periplanar; trans/E alkene (Zaitsev) |
| E1 | 3° | Polar protic, weak base, high T | Zaitsev product (most substituted) |
