VKC Notes · Elimination reactions · Civil Services P-II · IFoS P-II

2.9 Elimination Reactions: E1, E2, E1cB, Saytzeff and Hofmann Orientation, Syn Eliminations

★★★ 2021–2025: 19 parts, 150 marks (Civil Services 9 parts, 85 marks, one of them 20 marks; IFoS 10 parts, 65 marks). Asked every year: a Newman projection or a chair, Saytzeff against Hofmann, or a syn pyrolytic elimination.

Syllabus lines. Civil Services P-II: “Elimination reactions: E1, E2 and E1cb mechanisms; orientation in E2 reactions (Saytzeff and Hoffmann); pyrolytic syn elimination (acetate pyrolysis, Chugaev and Cope eliminations).” IFoS P-II: the same E1, E2, E1cb mechanisms with orientation and pyrolytic syn eliminations. Substitution: 2.8; carbocations and carbanions: 2.3; isotope effects: 2.7; conformations and Newman projections: 2.10.

Two-page short notes for revision All VKC notes

Download PDF PLAN

PDF edition, print-ready. It opens with the VKC plan, together with every locked answer, note and test on the site.

CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

2.9.1 The E2 mechanism ★★★

(2.9.1)

One step: the base takes the β-H, the C–H bond pair becomes the π bond and the leaving group goes with the C–X pair. Second order; a large primary isotope effect (k_H/k_D ≈ 7, 2.7.5).SN
Geometry: the C–H and C–X bonds must be anti-periplanar (dihedral 180°), so the filled σ(C–H) overlaps the empty σ*(C–X) in one plane. Syn-periplanar (0°) is possible but slower; a gauche H cannot be removed by E2.SN
2-Bromobutane + hot alcoholic KOH: two conformers put an H anti to Br on C3; the one with the methyls anti gives (E)-but-2-ene (major), the one with the methyls gauche gives (Z)-but-2-ene; but-1-ene is minor with this small base. The olefinic H–H coupling tells them apart: ³J ≈ 15–16 Hz trans, ≈ 10 Hz cis IFoS 2021 Q2(d)(i) 5m.SN

Fig. 2.9.1 E2: one step, H and Br anti.

Draw it in 60 s: EtO⁻ lone pair to the β-H; C–H bond arrow into the C–C bond; C–Br bond arrow to Br; (E)-but-2-ene.

Fig. 2.9.2 Newman projections of 2-bromobutane: why (E) is major.

Draw it in 60 s: circle = back carbon; front Br down, back H up (both red); methyls anti → E, gauche → Z.

2.9.2 Which alkene: Saytzeff or Hofmann ★★★ 10 marks four times

Saytzeff (Zaitsev) rule: with a small base and a good leaving group (Br, I, OTs) the more substituted, more stable alkene is major: the E2 TS already has partial C=C character and hyperconjugation stabilises it. 2-Chloro-2,3-dimethylpentane + NaOEt: 2,3-dimethylpent-2-ene (tetrasubstituted, major) and 2,3-dimethylpent-1-ene (minor) CSE 2022 Q2(b)(iii) 5m.SN
Rates with NaOEt/EtOH, 55 °C: tert-butyl bromide > isopropyl > n-propyl > ethyl bromide (II > III > IV > I): more β-H and a more substituted, more stable alkene in the TS IFoS 2023 Q4(b) 5m.SN
Conjugation beats substitution: CH₂=CHCH₂CH(Br)CH(CH₃)₂ + MeO⁻ gives the conjugated diene CH₂=CHCH=CHCH(CH₃)₂ (major) over the more substituted but isolated CH₂=CHCH₂CH=C(CH₃)₂ IFoS 2025 Q1(d) 5m.SN
Hofmann rule: the least substituted alkene is major when the leaving group is a bulky, poor one (N⁺Me₃, S⁺Me₂, F) or the base is bulky (t-BuO⁻). Reasons: (i) in the anti conformer that leads to the internal alkene the alkyl group must sit gauche to the large N⁺Me₃ (strain); a CH₃ group always has an anti H without strain; (ii) a poor leaving group makes the TS carbanion-like (E1cB-like), so the most acidic, least hindered H (on CH₃) goes CSE 2023 Q3(c) 10m IFoS 2024 Q2(c)(i) 10m.SN

Fig. 2.9.3 Hofmann rule: sec-butyltrimethylammonium ion.

Draw it in 60 s: two Newman projections with N⁺Me₃ up (front, red) and the anti H down (back, red): along C2–C3 the CH₃ is gauche to N⁺Me₃; along C2–C1 three H, no strain.

Hofmann elimination in practice: amine → exhaustive methylation (excess CH₃I) → quaternary iodide → moist Ag₂O (R₄N⁺HO⁻) → heat (100–200 °C) → alkene + R₃N. Sec-butyltrimethylammonium hydroxide: but-1-ene 95 %, but-2-ene 5 %. Ethyl(n-propyl)dimethylammonium hydroxide: the ethyl group's CH₃ hydrogens go: ethene + N,N-dimethylpropan-1-amine (propene minor); methyl groups have no β-H IFoS 2021 Q2(c)(ii) 5m.SN

2.9.3 Anti-periplanar in rings and diastereomers ★★★ 20 marks in 2021

Cyclohexanes: anti-periplanar means trans-diaxial. The leaving group must be axial and the H on the next carbon axial on the opposite face; an equatorial leaving group cannot do E2 until the ring flips.SN
cis- and trans-1-chloro-2-methylcyclohexane CSE 2021 Q2(a) 20m: in the cis isomer the stable chair has Cl axial (Me equatorial) with axial H on both C2 and C6: fast E2, and the Saytzeff product 1-methylcyclohexene is major (with some 3-methylcyclohexene). The trans isomer must flip to the chair with Cl and Me both axial; there the only anti H is on C6: slow, and only 3-methylcyclohexene forms (non-Saytzeff).SN

Fig. 2.9.4 Trans-diaxial E2 in 1-chloro-2-methylcyclohexanes.

Draw it in 60 s: chair; Cl axial up at C1; axial H down at C2 and C6 (red); cis: Me equatorial at C2; trans: Me axial at C2, so only C6–H is red.

Neomenthyl and menthyl chlorides CSE 2025 Q1(c) 10m: in neomenthyl chloride the stable chair (i-Pr and CH₃ equatorial) already has Cl axial, with axial H anti on C2 (next to i-Pr) and C6: fast; the Saytzeff product 3-menthene (≈ 75 %) with 2-menthene (≈ 25 %). Menthyl chloride must flip to the all-axial chair; only the C6 H is anti: about 200 times slower and only 2-menthene.SN
meso-2,3-Dibromobutane CSE 2021 Q3(c)(i) 5m IFoS 2022 Q2(a) 10m: with I⁻ (attacks Br; both Br leave anti, Br₂ loss) → (E)-but-2-ene; with EtO⁻ (takes the H anti to Br, HBr loss) → (E)-2-bromobut-2-ene (the two CH₃ cis). Both are stereospecific: the (±)-dibromide gives (Z)-but-2-ene and (Z)-2-bromobut-2-ene.SN

Fig. 2.9.5 meso-2,3-dibromobutane: two anti eliminations.

Draw it in 60 s: left: Br front up, Br back down (red), methyls anti → E-alkene; right: rotate the back carbon 120°; front H and back Br anti (red); methyls gauche → bromo-alkene with CH₃ cis.

Drawing chairs and Newman projections, R/S and E/Z: 2.10 Stereochemistry.

2.9.4 E1 and rearrangement ★★★

Two steps: slow ionisation to the carbocation (as in S_N1), then fast loss of a β-H to a weak base (solvent). First order; no isotope effect; no anti requirement (the cation is planar), so the most stable (Saytzeff) alkene and the E-isomer dominate. Heat and weak nucleophiles favour E1 over S_N1.SN
Lewis-acid dehydration: 2-cyclohexylpropan-2-ol + BF₃·OEt₂: O binds BF₃, the C–O bond ionises to the tertiary cation; loss of the ring H gives A = isopropylidenecyclohexane (tetrasubstituted, major); loss of a methyl H gives B = isopropenylcyclohexane (disubstituted), though statistics (6 H against 1 H) favour B CSE 2021 Q1(c) 10m. A 1,2-H shift into the ring gives 1-isopropylcyclohexene, and ring expansion is also possible IFoS 2023 Q2(c) 10m.SN

Fig. 2.9.6 E1 with BF₃: tertiary cation, two alkenes.

Draw it in 60 s: O lone pair to B; C–O arrow to O; 3° cation; ring C–H arrow into the C–C⁺ bond → A; methyl H → B.

Rearrangement first: CH₃CH(Cl)CH(CH₃)CH(CH₃)₂ in hot ethanol: the 2° cation takes a hydride from the next carbon (1,2-H shift) to become tertiary; loss of H⁺ gives the tetrasubstituted 2,3-dimethylpent-2-ene IFoS 2024 Q2(b)(iii) 5m. Methyl shifts: 2.8.6; carbocations: 2.3.SN

Part of VKC Basic

Read all 12 sections of this topic, its short notes and the PDF with VKC Basic

VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.

See what is included Already a member? Sign in

Solved PYQs, short notes and books: vkcscience.comTelegram: t.me/upsc_chemistry_vkc