(a) (i) Consider the following reactions : PhCH₂CH₂Br + EtO⁻ → PhCH=CH₂ (Rate = k_H) and PhCD₂CH₂Br + EtO⁻ → PhCD=CH₂ (Rate =…

2. (a) (i) Mechanism from k_H/k_D = 7.1

• What the number means: replacing the two benzylic β-hydrogens by deuterium slows the reaction 7.1-fold. This is a primary kinetic isotope effect, so the C–H(D) bond must be breaking in the rate-determining step. Secondary isotope effects (bond not broken) are only about 1.1–1.4.
• Why C–D reacts more slowly: the heavier D gives a lower stretching frequency and a lower zero-point energy. In the transition state this stretch becomes the reaction coordinate, so the zero-point difference (about 4.8 kJ mol⁻¹) is lost and Eₐ(D) > Eₐ(H). The theoretical maximum near room temperature is about 7, so 7.1 shows the proton is roughly half-transferred in a fairly symmetrical transition state.
• Mechanism: E2 (bimolecular, concerted). EtO⁻ removes a β-H while the C–Br bond breaks and the C=C forms in a single step. Rate = k[PhCH₂CH₂Br][EtO⁻]. The H and Br are anti-periplanar (dihedral angle 180°), and the product is styrene, PhCH=CH₂.
• E1 is ruled out: a primary bromide ionises poorly, and in E1 the slow step is C–Br ionisation with C–H cleavage afterwards, which would give k_H/k_D ≈ 1–1.5.
• E1cB is ruled out: reversible E1cB would exchange H for D when the reaction is run in EtOD, and this is not observed. Irreversible E1cB needs a much more acidic β-H and a poor leaving group, whereas Br⁻ leaves easily. The benzylic β-H gives the E2 transition state some carbanion character, but no free carbanion forms.
• Saunders and Edison (1960) measured k_H/k_D = 7.11 for this substrate with ethoxide, which is the classic evidence for E2.

(a) (ii) 1,2- versus 1,4-addition of HBr to buta-1,3-diene

• H⁺ adds to C-1 and gives a resonance-stabilised allylic cation: CH₃–C⁺H–CH=CH₂ ↔ CH₃–CH=CH–C⁺H₂. Br⁻ can then bond at C-2 (1,2-product, 3-bromobut-1-ene) or at C-4 (1,4-product, 1-bromobut-2-ene).

At −80 °C the reaction is under kinetic control. Br⁻ attacks C-2 faster for two reasons:

• C-2 is secondary and carries the larger share of the positive charge.
• Br⁻, released next to C-1/C-2 in the ion pair, is closest to C-2 (proximity effect).
• The lower Eₐ makes the 1,2-product form faster. At this temperature the addition is effectively irreversible, so the product ratio equals the ratio of rates (typically about 80 : 20, 1,2 : 1,4).

At −45 °C and above the reaction shifts to thermodynamic control. The allylic C–Br bonds re-ionise to the same allylic cation, so the two products equilibrate.

• The 1,4-product has an internal, disubstituted (mainly E) C=C with more hyperconjugation, so it is lower in free energy and predominates (about 15 : 85 near +40 °C).
• Supporting evidence: warming the pure 1,2-product gives the same equilibrium mixture.
• Summary: the 1,2-product forms faster and the 1,4-product is more stable. Low temperature freezes the rate ratio; higher temperature lets the system reach equilibrium.

(b) (i) Knoevenagel condensation

• X = diethyl benzylidenemalonate, PhCH=C(CO₂Et)₂ (diethyl 2-benzylidenepropanedioate). The reaction is the Knoevenagel condensation.
• Step 1 – enolate formation: pyridine (pKₐ of pyH⁺ ≈ 5.2) removes a proton from the active methylene group (pKₐ ≈ 13). Only a small equilibrium concentration of enolate is needed; its charge is delocalised over both ester carbonyls. A trace of piperidine is often added as a more effective catalyst.
• Step 2 – C–C bond formation: the enolate carbon adds to the carbonyl carbon of benzaldehyde, giving an alkoxide.
• Step 3 – proton transfer: pyH⁺ protonates the alkoxide, giving the aldol-type β-hydroxy diester PhCH(OH)CH(CO₂Et)₂.
• Step 4 – dehydration by E1cB: the remaining C–H, flanked by two esters, is removed to give a stabilised carbanion, which expels OH⁻. A poor leaving group can depart here because the anion is so well stabilised.
• Driving force: the new C=C is conjugated with the phenyl ring and both C=O groups.
• Benzaldehyde has no α-H, so it cannot undergo self-condensation. (With free malonic acid in pyridine, the Doebner modification, decarboxylation gives cinnamic acid; the diethyl ester stops at the diester.)

(b) (ii) Chugaev elimination

• Formation of X: NaOH converts the tertiary alcohol to its alkoxide. The alkoxide adds to CS₂ to give the sodium xanthate RO–C(=S)S⁻ Na⁺, which CH₃I S-methylates. X = (CH₃)₂CH–C(CH₃)₂–O–C(=S)–SCH₃, the S-methyl xanthate (O-2,3-dimethylbutan-2-yl S-methyl dithiocarbonate).
• Elimination: heating X to about 150–200 °C gives the alkene plus COS and CH₃SH. This pyrolysis is the Chugaev elimination (an Ei reaction).

Why it is a syn-elimination:

• Six-membered cyclic transition state: the thiocarbonyl sulfur itself abstracts the β-H intramolecularly. The ring is H–Cβ–Cα–O–C=S, so the β-H and the C–O bond must lie on the same side, syn-periplanar with a dihedral angle near 0°.
• Kinetics: the reaction is first order and needs no external base.
• Stereospecificity: diastereomeric (erythro and threo) xanthates give different E/Z alkenes, each matching cis removal of H and OC(S)SCH₃. In cyclohexyl xanthates, only a β-H cis to the xanthate group is removed. An E2 reaction would instead need a trans-diaxial H.
• No carbocation: because no carbocation forms, there is no rearrangement. This is useful for a tertiary alcohol, which acid would dehydrate through a cation.
• Regiochemistry: syn-pyrolytic eliminations follow the number of available cis β-H atoms. Besides the tetrasubstituted alkene shown in the question, (CH₃)₂C=C(CH₃)₂ (from the single methine H), the six methyl H atoms also give some CH₂=C(CH₃)–CH(CH₃)₂.

(c) (i) Products

(A) Fischer indole synthesis

Mechanism:

• Phenylhydrazine condenses with the ketone to give the phenylhydrazone.
• Acid tautomerises it to the ene-hydrazine.
• A [3,3]-sigmatropic rearrangement forms the new C–C bond and breaks the N–N bond.
• Rearomatisation and cyclisation give an aminal, which loses NH₃ to form the indole ring.
• Regiochemistry: in a 3-substituted cyclohexanone, the ene-hydrazine can form towards C-2 or C-6. The C-6 side, away from the bulky quaternary substituent, is favoured.
• Major product: ethyl 2-cyano-2-(2,3,4,9-tetrahydro-1H-carbazol-2-yl)butanoate, a 2-substituted tetrahydrocarbazole.
• Minor product: the 4-substituted isomer.
• The nitrile and ester survive AcOH/heat.

(B) Skraup synthesis

Mechanism:

• Hot H₂SO₄ dehydrates glycerol to acrolein.
• The aromatic NH₂ adds 1,4 (Michael addition) to the acrolein.
• The aldehyde, activated by acid, closes the ring onto the ring carbon ortho to nitrogen.
• Dehydration gives a dihydropyridine ring, and oxidation aromatises it. The classical oxidant is nitrobenzene; here hot H₂SO₄/air serves.
• FeSO₄ and H₃BO₃ moderate the notoriously violent reaction.
• Regiochemistry: of the two positions ortho to NH₂ (C-7 and C-9), C-7 is para to the ring oxygen and less hindered. Ring closure there gives the linear pyrido-fused azaxanthone: a new pyridine ring fused across C-7/C-8, with the ring N attached at the former C-8 amino position. Angular closure at C-9 is a possible minor pathway.

(C) Hofmann exhaustive methylation and elimination

• Excess CH₃I converts N-methylazepane into N,N-dimethylazepanium iodide.
• Moist Ag₂O precipitates AgI and gives the quaternary ammonium hydroxide.
• On heating, OH⁻ removes a ring β-H in an E2 reaction. Both β-carbons are ring CH₂ groups, and the elimination follows the Hofmann rule (least substituted alkene). Because the leaving nitrogen is part of the ring, the ring opens.
• Product: CH₂=CH–(CH₂)₄–N(CH₃)₂, N,N-dimethylhex-5-en-1-amine, plus H₂O.
• A second methylation–elimination cycle would give hexa-1,5-diene and N(CH₃)₃.

(c) (ii) Reformatsky route to CH₃CH₂CH=C(CH₃)COOH (2-methylpent-2-enoic acid)

• Disconnection: α,β-unsaturated acid ⇐ β-hydroxy ester ⇐ aldehyde + α-bromo ester. C-3 to C-5 come from propanal; C-1, C-2 and the 2-methyl group come from ethyl 2-bromopropanoate.
• Step 1: activated zinc inserts into the C–Br bond of CH₃CH(Br)CO₂Et in dry ether or benzene, giving the BrZn ester enolate (the Reformatsky reagent).
• Step 2: the zinc enolate adds to CH₃CH₂CHO through a Zn-chelated six-membered transition state. Aqueous acid work-up gives ethyl 3-hydroxy-2-methylpentanoate, CH₃CH₂CH(OH)CH(CH₃)CO₂Et.
• Step 3: dehydration with H⁺/heat (or POCl₃/pyridine) gives the conjugated ester CH₃CH₂CH=C(CH₃)CO₂Et, mainly the E isomer.
• Step 4: saponification (NaOH/H₂O) followed by H₃O⁺ gives CH₃CH₂CH=C(CH₃)COOH.
• Why zinc: the zinc enolate is less reactive than Grignard or lithium enolates. It does not attack the ester group, so self-condensation is avoided and it adds cleanly to the aldehyde.