(a)(i) For the following compound : (i) Identify the site of initial ionization under EI conditions. (ii) Draw the structure of…

(a)(i) Mass spectrum of doxepin

The compound is doxepin, a dibenz[b,e]oxepine with a 3-(dimethylamino)propylidene side chain, C₁₉H₂₁NO, M = 279.

Site of ionisation: the electron is removed from the non-bonding lone pair on the tertiary amine nitrogen, which is the most loosely held electron pair in the molecule (lower ionisation energy than the aromatic π system or the ether oxygen). The molecular ion is best written with the radical cation on N.

Ion at m/z 58: the N-centred radical cation undergoes α-cleavage, in which the C–C bond next to the N–CH₂ carbon breaks and the unpaired electron on N pairs to form C=N⁺. The ion is the dimethyl(methylene)iminium ion, CH₂=N⁺(CH₃)₂, C₃H₈N⁺, m/z = 36 + 8 + 14 = 58. It is resonance-stabilised (CH₂=N⁺Me₂ ↔ ⁺CH₂–NMe₂) and is the base peak.

Metastable ion: for M⁺• (m₁ = 279) → 58⁺ (m₂ = 58) in the field-free region, m* = m₂²/m₁ = 58²/279 = 3364/279 = 12.06. A broad peak therefore appears near m/z 12.1.

(a)(ii) Ketone C₇H₁₂O with λmax 249 nm

C₇H₁₂O has (2×7 + 2 − 12)/2 = 2 degrees of unsaturation, which are one C=O and one C=C with no ring, so an absorption at 249 nm means an acyclic α,β-unsaturated ketone. By the Woodward–Fieser rules: base value 215 nm, one α-alkyl group +10 nm, two β-alkyl groups 2 × 12 = +24 nm, total 249 nm.

The ketone is 3,4-dimethylpent-3-en-2-one, CH₃–CO–C(CH₃)=C(CH₃)₂.

(b)(i) Microwave rotational spectra and rotational constants

(A) A molecule shows a pure rotational spectrum only if it has a permanent dipole moment, because the rotating dipole interacts with the radiation field. HCl and BrF are polar heteronuclear diatomics, and H₂O is a bent polar molecule, so these three are microwave active. H₂ is homonuclear (μ = 0) and CH₄ is a tetrahedral spherical top (μ = 0), so these are inactive.

(B) For a rigid rotor E_J = BJ(J + 1), and the first line (J = 0 → 1) lies at 2B.

¹²C¹⁶O: B = 3.84235/2 = 1.92118 cm⁻¹ (about 57.60 GHz).

¹³C¹⁶O: B = 3.67337/2 = 1.83669 cm⁻¹ (about 55.06 GHz).

As a check, both values give r = 1.131 Å from B = h/(8π²cμr²), so the bond length is unchanged by isotopic substitution. B falls for ¹³C¹⁶O only because its reduced mass is larger.

(b)(ii) Splitting of the lettered protons

(A) 1,1-Dichloroethene: Hₐ and H_b are geminal and chemically equivalent, so no splitting is observed and the signal is a singlet (²J is about 0–3 Hz in magnitude).

(B) 1,1-Dichloro-2-methylidenecyclopentane: Hₐ and H_b are the geminal =CH₂ protons, non-equivalent (one cis to CCl₂, one cis to the ring CH₂). ²J is small, about 0–3 Hz (typically about 2 Hz), because the H–C–H angle is near 120°, with additional allylic couplings of about 1–2 Hz to the ring CH₂.

(C) 3-Methoxycyclohexene: Hₐ and H_b are vicinal, cis protons on the ring C=C. Their dihedral angle is near 0°, and ³J(cis) is about 10 Hz (cyclohexene range about 9–11 Hz).

(c)(i) Keto–enol composition of acetylacetone

Integration height is proportional to the number of protons. Keto, CH₃CO–CH₂–COCH₃, has 2 H at the CH₂ signal, so 10 mm ÷ 2 = 5 units per molecule. Enol, CH₃C(OH)=CH–COCH₃, has 1 H at the =CH signal, so 22 mm ÷ 1 = 22 units. Total = 5 + 22 = 27 units.

% enol = 22/27 × 100 = 81.5%, and % keto = 5/27 × 100 = 18.5%.

The enol predominates because its C=C is conjugated with C=O and a strong intramolecular O–H···O hydrogen bond closes a six-membered chelate ring (the enolic OH appears near δ 15).

(c)(ii) Distinguishing the isomeric amides by mass spectrometry

Both are C₅H₁₁NO with M⁺• = 101 (odd, in line with the nitrogen rule). Compound I is N,N-dimethylpropanamide and II is N-methylbutanamide.

Compound II has a γ-hydrogen on its butanoyl chain, so a McLafferty rearrangement with loss of C₂H₄ gives an ion at m/z 73. Compound I has no γ-hydrogen on the acyl side and shows no such ion. α-Cleavage also differs: I loses C₂H₅• to give (CH₃)₂N–C≡O⁺ at m/z 72 and gives the acylium C₂H₅CO⁺ at m/z 57, while II loses C₃H₇• to give CH₃NH–C≡O⁺ at m/z 58 and gives the acylium C₃H₇CO⁺ at m/z 71. The set 73, 71 and 58 identifies II, and the pair 72 and 57 without 73 identifies I.

(c)(iii) Fundamental vibrations of CO₂

CO₂ is linear (N = 3), so it has 3N − 5 = 4 normal modes: the symmetric stretch, the antisymmetric stretch and a doubly degenerate bend. The bend counts twice but has one frequency, so three fundamental frequencies are expected.

A vibration is IR-active only if it changes the dipole moment. The symmetric stretch keeps the centre of symmetry and the dipole stays zero, so it is IR-inactive (Raman-active). The antisymmetric stretch (about 2349 cm⁻¹) and the bend (about 667 cm⁻¹) produce an oscillating dipole, so they are IR-active. The IR spectrum therefore shows two fundamental bands, consistent with the rule of mutual exclusion for a centrosymmetric molecule.