(a) (i) Among the following compounds I and II, which has more carbonyl stretching frequency in IR spectra? Explain: I =…

(a)(i) Which ketone has the higher ν(C=O)?

Compound II (1-acetyl-2-methylcyclohexene) has the higher carbonyl frequency.

Both ketones are formally α,β-unsaturated. In I (1-acetylcyclohexene) the acetyl group can lie in the plane of the ring C=C, so the C=O is fully conjugated. Conjugation gives the C=O some single-bond character (C=C–C=O ↔ ⁺C–C=C–O⁻), and it absorbs near 1685 cm⁻¹. In II the methyl group at C-2 clashes with the acetyl group (allylic 1,3-strain), which twists the C=O out of the plane of the C=C. This steric inhibition of resonance reduces the conjugation, restores C=O double-bond character, and moves the band to higher wavenumber, towards the 1700–1715 cm⁻¹ of a saturated ketone.

(a)(ii) Acid-catalysed dehydration of 1,2-dimethylcyclohexan-1-ol

H⁺ protonates the OH, water leaves, and the tertiary carbocation forms at C-1. Loss of a β-proton from each of the three different β-positions gives three alkenes:

• A, 1,2-dimethylcyclohexene (from C-2 H): tetrasubstituted, the Zaitsev (major) product.
• B, 1,6-dimethylcyclohexene (from C-6 H): trisubstituted ring alkene.
• C, 2-methyl-1-methylidenecyclohexane (from the C-1 methyl): exocyclic 1,1-disubstituted alkene, the minor product.

IR distinguishes them:

• A has no vinylic H, so there is no =C–H stretch above 3000 cm⁻¹. The almost symmetrical tetrasubstituted C=C stretch near 1670 cm⁻¹ is very weak or absent.
• B shows a =C–H stretch near 3020 cm⁻¹, a medium C=C near 1670 cm⁻¹, and a trisubstituted =C–H out-of-plane bend near 800–840 cm⁻¹.
• C shows a =C–H stretch near 3080 cm⁻¹, C=C near 1650 cm⁻¹, and a strong =CH₂ out-of-plane wag near 890 cm⁻¹, which is diagnostic for the exocyclic methylene.

None of them shows the broad O–H band of the starting alcohol near 3400 cm⁻¹.

(b)(i) Woodward–Fieser λmax

Compound (1). The conjugated system is C3=C4–C5=C6–C7=C8–C9=C11, a linear tetraene. Ring B contains two of its double bonds (C5=C6 and C7=C8), so the homoannular base value is used.

• Homoannular diene base: 253 nm
• Two extra conjugated C=C (C3=C4 and C9=C11): 2 × 30 = 60
• Ring residues on the conjugated carbons (C-2 on C3, C-10 on C5, C-14 on C8, C-10 on C9, C-12 on C11): 5 × 5 = 25
• Exocyclic C=C (C5=C6 to ring A, C7=C8 to ring C, C9=C11 to ring B): 3 × 5 = 15
• λmax (calc) = 253 + 60 + 25 + 15 = 353 nm

Compound (2). The conjugated system is CH₂=CH–C1=C8a–C4a=C5, a triene whose ring double bonds lie in different rings, so it is heteroannular.

• Heteroannular (transoid) diene base: 214 nm
• Extra conjugated C=C (the vinyl group): 30
• Ring residues (C-2 on C1, C-8 on C8a, C-4 on C4a, C-6 on C5): 4 × 5 = 20
• Exocyclic C=C (C1=C8a and C4a=C5): 2 × 5 = 10
• λmax (calc) = 214 + 30 + 20 + 10 = 274 nm

(b)(ii) Butadiene vs hexatriene

1,3,5-Hexatriene has three conjugated C=C bonds (6 π electrons), while butadiene has two (4 π electrons). Each added conjugated double bond raises the HOMO and lowers the LUMO, so the HOMO–LUMO gap shrinks (Hückel: 1.24β for butadiene, 0.89β for hexatriene). A smaller ΔE means absorption at a longer wavelength. Butadiene therefore absorbs at about 217 nm and hexatriene at about 258 nm.

(c)(i) McLafferty rearrangement and butyl butyrate

McLafferty rearrangement: in the radical cation of a carbonyl compound that has a γ-hydrogen, the γ-H moves to the carbonyl oxygen through a six-membered transition state and the β,γ bond breaks. A neutral alkene is expelled, leaving an enol radical cation (or, if the charge stays on the alkene fragment, an alkene radical cation).

Butyl butyrate is CH₃CH₂CH₂CO–O–CH₂CH₂CH₂CH₃, C₈H₁₆O₂, M = 144.

• m/z 101: [M − C₃H₇]⁺, loss of a propyl radical (43). α-Cleavage next to the carbonyl gives the butoxycarbonyl cation, ⁺O≡C–OC₄H₉. Cleavage in the butyl chain, giving C₃H₇COOCH₂⁺, has the same mass.
• m/z 73: C₄H₉O⁺, the butoxy-derived oxonium ion, from cleavage of the acyl–O bond with the charge on the alkoxy side. It is the partner of the 71 ion.
• m/z 71: the butanoyl acylium ion, CH₃CH₂CH₂C≡O⁺, from α-cleavage with loss of •OC₄H₉ (73). It is usually a strong peak.
• m/z 56: C₄H₈⁺•, the butene radical cation from a McLafferty-type transfer of an H from the butyl chain to the carbonyl oxygen. The complementary neutral is butanoic acid (88).

(c)(ii)(1) C₂H₂BrCl showing two doublets, J = 16 Hz

Three isomers are possible: cis- and trans-1-bromo-2-chloroethene, and 1-bromo-1-chloroethene. Each has two non-equivalent vinyl protons, so each gives two doublets. The coupling constant decides between them. Trans vicinal coupling across a C=C is about 12–18 Hz, cis vicinal coupling is about 6–12 Hz (smaller still with electronegative halogens), and geminal =CH₂ coupling is only about 0–3 Hz. J = 16 Hz identifies trans-1-bromo-2-chloroethene, BrHC=CHCl (E). The cis isomer and 1-bromo-1-chloroethene are the other possible structures.

(c)(ii)(2) Assigning the UV data to A and B

Both are dienones of the type C=C–C=C–C=O, and Woodward rules for enones fit them.

• B, 4-(cyclohex-1-enyl)but-3-en-2-one: base 215 + extended conjugation 30 + γ ring residue 18 + δ ring residue 18 = 281 nm.
• A, β-ionone: base 215 + extended conjugation 30 + γ ring residue 18 + δ ring residue and δ methyl 2 × 18 = 299 nm.

So A absorbs at 296 nm and B at 281 nm. The extra alkyl groups on the ring double bond of A cause the bathochromic shift. The lower ε of A (10 700 against 20 800) fits as well: the gem-dimethyl group and the ring methyl twist the side chain out of the ring plane, which weakens the π→π* band and keeps λmax slightly below the calculated 299 nm.