(a) Calculate the value of λ_max in the following compounds using Woodward-Fieser rules : three tricyclic…

(a)

All three compounds are tricyclic ketones. In each, the C=O is conjugated with two C=C bonds (C=C–C=C–C=O), so each is a linearly conjugated dienone. The Woodward–Fieser rules for α,β-unsaturated ketones (ethanol solvent) are used with these values:

• Base value, six-membered ring (or acyclic) enone: 215
• Double bond extending the conjugation: +30
• Homoannular diene component: +39
• Alkyl group or ring residue at α / β / γ and higher: +10 / +12 / +18
• Each exocyclic C=C: +5

The atoms are labelled O=C–Cα=Cβ–Cγ=Cδ.

Compound A. The two C=C (Cα=Cβ and Cγ=Cδ) both lie in the central ring, so the diene is homoannular.

• Ring residues: one at α (the ring-fusion CH), one at β, one at γ (both from the lower ring) and one at δ (the CH₂ of the central ring).
• Exocyclic double bonds: Cα=Cβ is exocyclic to the ketone ring and to the lower ring (2), and Cγ=Cδ is exocyclic to the lower ring (1), a total of 3.

λmax = 215 + 30 + 39 + 10 (α) + 12 (β) + 18 (γ) + 18 (δ) + 3 × 5 = 357 nm

Compound B. Cα=Cβ is the fusion bond between the ketone ring and the central ring, and Cγ=Cδ is in the central ring, so the diene is homoannular (central ring).

• Ring residues: α (the ring-fusion CH of the lower ring), β (the CH₂ of the ketone ring), γ has none (it is CH), δ (the ring-fusion CH).
• No C=C is exocyclic to any ring, because both atoms of each double bond lie in the same rings.

λmax = 215 + 30 + 39 + 10 (α) + 12 (β) + 18 (δ) = 324 nm

Compound C. Both C=C lie in the left-hand ring, so the diene is homoannular.

• Ring residues: α (the ring-fusion CH); β and γ have none; δ (the CH₂ of the left ring).
• Exocyclic: Cα=Cβ is exocyclic to the ketone ring (1).

λmax = 215 + 30 + 39 + 10 (α) + 18 (δ) + 5 = 317 nm

• A: 357 nm
• B: 324 nm
• C: 317 nm

Compound A absorbs at the longest wavelength because it is the most heavily substituted and has three exocyclic double bonds. All three lie far above a simple cyclohexenone (about 215–250 nm) because of the extra conjugated C=C and its homoannular geometry.

(b)

The starting material is tert-butyl 3-methylidene-4-(2-oxoethyl)piperidine-1-carboxylate. It is an N-Boc piperidine carrying an exocyclic C=CH₂ at C-3 and a –CH₂CHO side chain at C-4.

X: LiAlH₄, THF, 0 °C. At 0 °C and with a short reaction time, hydride reduces the most electrophilic group, the aldehyde, to a primary alcohol. The Boc carbamate is much less electrophilic and survives: it is reduced to N-CH₃ only under forcing conditions such as reflux. The isolated C=C is inert to LiAlH₄. X = tert-butyl 4-(2-hydroxyethyl)-3-methylidenepiperidine-1-carboxylate.

Y: BH₃·THF, then H₂O₂/HO⁻. Hydroboration of the 1,1-disubstituted exocyclic alkene is anti-Markovnikov. Boron adds to the unsubstituted terminal CH₂, the less hindered carbon, while the partial positive charge of the four-centre transition state develops on the more substituted C-3, which receives the H. The addition is a syn, four-centre process. Oxidation with alkaline H₂O₂ replaces C–B by C–OH with retention. The result is a primary CH₂OH on C-3, and C-3 becomes a new stereocentre, formed as a mixture of cis/trans diastereomers relative to C-4. Y = tert-butyl 4-(2-hydroxyethyl)-3-(hydroxymethyl)piperidine-1-carboxylate (a 1,5-diol).

Z: CrO₃/H₂SO₄ (Jones reagent). In aqueous acid Jones reagent oxidises primary alcohols through the aldehyde hydrate all the way to carboxylic acids. Both primary OH groups are oxidised. Z = 1-(tert-butoxycarbonyl)-4-(carboxymethyl)piperidine-3-carboxylic acid, a dicarboxylic acid.

Because Y is a 1,5-diol, some intramolecular lactol formation and oxidation to a fused δ-lactone can compete during the Jones step. Short reaction times at 0 °C also limit acid-promoted loss of the Boc group.

(c)(i)

C₁₀H₁₂O₂ has (2×10 + 2 − 12)/2 = 5 degrees of unsaturation: a benzene ring (4) and a C=O (1). The band at 1730 cm⁻¹ is the C=O stretch of a saturated (non-conjugated) ester. In both spectra δ 7.3 (5H, s) shows a monosubstituted phenyl group.

Compound (1): 2-phenylethyl acetate, CH₃COO–CH₂CH₂–C₆H₅.

• δ 2.0 (3H, s) is CH₃–C(=O)–O, an acetate methyl with no neighbours.
• δ 4.3 (2H, t) is –O–CH₂–, deshielded by the ester oxygen and split by the adjacent CH₂.
• δ 2.93 (2H, t) is Ph–CH₂–, benzylic and split by the O–CH₂.
• The two mutually coupled triplets show an isolated –CH₂–CH₂– unit between O and Ph.

Compound (2): ethyl phenylacetate, C₆H₅–CH₂–COO–CH₂CH₃.

• δ 1.23 (3H, t) and δ 4.13 (2H, q) form an O-ethyl group (O–CH₂ deshielded, J ≈ 7 Hz).
• δ 3.72 (2H, s) is a CH₂ flanked by the phenyl ring and the C=O. It has no neighbouring H, so it is a singlet, and it is doubly deshielded.

The two compounds are therefore metamers (both esters) of each other: one is the ester of an alcohol carrying the phenyl group, and the other is the ester of an acid carrying the phenyl group.

(c)(ii)

Compound I: 5-phenylhexan-2-one (M = 176). The m/z 58 ion comes from a McLafferty rearrangement.

• The ketone has a γ-hydrogen: C-5, the benzylic CH bearing CH₃ and Ph, is γ to the C=O.
• In the molecular ion the carbonyl oxygen radical abstracts this γ-H through a six-membered cyclic transition state.
• The Cα–Cβ (C3–C4) bond then breaks. Neutral α-methylstyrene, CH₂=C(CH₃)Ph (118), is expelled, and the charge and radical stay on the enol fragment.

Fragment: [CH₂=C(OH)CH₃]⁺•, the acetone enol radical cation, C₃H₆O⁺•, m/z = 36 + 6 + 16 = 58. The ion mass is 176 − 118 = 58. The ion has even mass and is odd-electron, which is diagnostic of a rearrangement, and it is stabilised by resonance between the enol oxygen and the C=C (CH₂=C(O⁺•H)CH₃ ↔ •CH₂–C(=O⁺H)CH₃). Simple α-cleavage would give odd masses (43, CH₃CO⁺), not 58.

Compound II: propylbenzene (M = 120). The m/z 92 ion comes from a McLafferty-type rearrangement with the ring as acceptor.

• A γ-H of the terminal CH₃ is transferred through a six-membered transition state to an ortho ring carbon.
• The benzylic Cα–Cβ bond breaks, and ethylene (28) is expelled.
• The ion left is [C₇H₈]⁺•, the methylenecyclohexadiene radical cation (it can isomerise to the toluene radical cation), m/z = 120 − 28 = 92.

The base peak of propylbenzene, m/z 91 (tropylium, C₇H₇⁺), arises instead by simple benzylic cleavage with loss of C₂H₅•. The even-mass m/z 92 peak is the mark of the hydrogen-transfer rearrangement, which is possible only because the side chain carries a γ-hydrogen.