(a) (i) Calculate the rotational constant of a diatomic molecule if the moment of inertia is 13·97 × 10⁻⁴⁷ kg m². (Given …

(a) Three spectroscopic parts (15 marks)

(i) Rotational constant (5 marks). The rotational constant in wavenumber units is B = h / (8π²cI).

B = 6.626 × 10⁻³⁴ / (8 × 9.8696 × 2.998 × 10⁸ × 13.97 × 10⁻⁴⁷)

The denominator is 78.957 × 2.998 × 10⁸ × 13.97 × 10⁻⁴⁷ = 3.307 × 10⁻³⁶, so

B = 6.626 × 10⁻³⁴ / 3.307 × 10⁻³⁶ = 200.4 m⁻¹ = 2.00 cm⁻¹.

In other units this is h²/8π²I = 3.98 × 10⁻²³ J, or 6.01 × 10¹⁰ Hz.

(ii) Order of vibrational frequency (5 marks). The frequency is ν̃ = (1/2πc)√(k/μ), where k is the force constant and μ the reduced mass. The increasing order is NO < CO < HI < HF.

• NO: ν̃ about 1904 cm⁻¹ (k about 1600 N/m, bond order 2.5, and a large μ of about 7.5 u).
• CO: about 2143 cm⁻¹ (k about 1860 N/m, a triple bond, μ about 6.9 u). The stiffer bond gives a higher value than NO.
• HI: about 2309 cm⁻¹ (k about 314 N/m, but μ is only about 1.0 u because the iodine is heavy and the hydrogen moves almost alone).
• HF: about 4138 cm⁻¹ (k about 966 N/m, a very strong H-F bond, and μ about 0.96 u).

The two hydrides are above the two heavy diatomics because of their small reduced mass, and HF is the highest because its bond is much stiffer than HI.

(iii) C=O and O-H frequencies in two chelated carbonyl compounds (5 marks). Pentane-2,4-dione exists as a keto-enol mixture. The keto form shows two coupled C=O stretches at about 1725 and 1707 cm⁻¹ and no O-H. In the enol form, the OH forms a strong intramolecular hydrogen bond to the second carbonyl, in a six-membered chelate ring with a conjugated O=C-C=C-OH unit. The C=O stretch is lowered to about 1600-1640 cm⁻¹ (mixed with C=C, and weakened by conjugation and hydrogen bonding), and the O-H stretch appears as a very broad, weak band centred near 2700 cm⁻¹ (spread over 2500-3200), far below the free-OH position at about 3600. The bands do not change on dilution because the bond is intramolecular.

2-Hydroxypropiophenone (taken here as the ortho-hydroxy ketone, 1-(2-hydroxyphenyl)propan-1-one, in which the phenolic OH sits next to the C=O): the same chelation occurs, through a six-membered ring. The C=O stretch falls from about 1685 cm⁻¹ in propiophenone to about 1630-1640 cm⁻¹, and the O-H stretch is broad at about 3000-3200 cm⁻¹ and independent of concentration. (If the alpha-hydroxy ketone, 2-hydroxy-1-phenylpropan-1-one, is meant, the five-membered hydrogen bond is weak: C=O about 1685-1690 cm⁻¹ and O-H about 3450 cm⁻¹.)

(b)(i) ¹H NMR of furfuraldehyde (5 marks)

Furfuraldehyde has four protons: the aldehyde H and three ring protons, H3, H4 and H5. Spin-spin coupling between neighbouring protons splits each signal. The ring couplings are J₃,₄ about 3.6 Hz, J₄,₅ about 1.7 Hz and a small four-bond coupling J₃,₅ about 0.8 Hz (the coupling to the aldehyde H is also about 0.8 Hz).

• CHO, δ about 9.65: a narrow doublet or singlet (small coupling to H3).
• H5 (next to O), δ about 7.70: a doublet of doublets, J₄,₅ = 1.7 and J₃,₅ = 0.8 Hz; it is the most deshielded ring proton because it is α to oxygen.
• H3 (next to the CHO-bearing carbon), δ about 7.26: a doublet, J₃,₄ = 3.6 Hz (with small additional splitting); it is deshielded by the carbonyl.
• H4, δ about 6.62: a doublet of doublets, J₃,₄ = 3.6 and J₄,₅ = 1.7 Hz.

J₃,₄ (the β-β coupling) is the largest ring coupling; the α-β couplings such as J₄,₅ are small in furan (about 1.7 Hz), a characteristic of five-membered heteroaromatic rings, so the three ring protons are easily assigned from their splittings.

(b)(ii) Distinctions by ¹H NMR (5 marks)

(A) Anthracene and phenanthrene. Anthracene is highly symmetric (D₂h) and has only three sets of protons: H9/H10 (2H, a singlet, δ about 8.4), H1/H4/H5/H8 (4H, δ about 8.0) and H2/H3/H6/H7 (4H, δ about 7.45), in the ratio 2:4:4. Phenanthrene (C₂ᵥ) has five sets, each of 2H: H4/H5 (δ about 8.7), H1/H8 (7.9), H9/H10 (a singlet, about 7.75), H3/H6 (about 7.65) and H2/H7 (about 7.6). Thus five signals of equal intensity, including the lowest-field one (H4/H5, the bay-region protons deshielded by the neighbouring ring), distinguish phenanthrene from anthracene's three signals at 2:4:4.

(B) Pure ethanol and ethanol with acid. In carefully purified ethanol the OH proton exchanges slowly and couples to the CH₂ protons: OH is a triplet (J about 5 Hz), the CH₂ is a doublet of quartets (eight lines) and CH₃ is a triplet. A trace of acid (or base) catalyses fast intermolecular proton exchange. If the exchange rate is much greater than the coupling (in Hz), the OH proton sees an average of the spin states of the CH₂ protons and the coupling is lost: OH becomes a broad singlet, CH₂ a simple quartet and CH₃ a triplet. This exchange averaging is why the OH signal in most samples of ethanol is a singlet.

(b)(iii) 2-Nitroacetophenone and 3-nitroacetophenone (5 marks)

Considering only ortho coupling (J about 8 Hz):

2-Nitroacetophenone has four adjacent aromatic protons (H3, H4, H5, H6) in a row. H3 (next to NO₂) is a doublet at δ about 8.1; H4 and H5 are each a triplet (each has two ortho neighbours), at about 7.7 and 7.6; H6 (next to the acyl group) is a doublet at about 7.4. Pattern: d, t, t, d. The acetyl CH₃ is at δ about 2.55, slightly lower than in the meta isomer because the ortho nitro group twists the acetyl out of the ring plane and it is less deshielded.

3-Nitroacetophenone has three adjacent protons and one isolated proton. H2, between the two substituents, has no ortho neighbour and is a singlet (a fine meta splitting is ignored here) at the lowest field, δ about 8.75; H4 (next to NO₂) is a doublet at about 8.4; H6 (next to acyl) is a doublet at about 8.25; and H5 is a triplet at about 7.7. Pattern: s, d, d, t. The CH₃ is at δ about 2.68.

The meta isomer therefore has a singlet near 8.75 and only one triplet; the ortho isomer has two triplets, no singlet, and nothing above 8.1.

(c)(i) Conjugation in 3-buten-2-one and acetone (5 marks)

Acetone (isolated C=O): n → π* at about 279 nm (ε about 15, weak because it is symmetry-forbidden), and π → π* at about 187-190 nm (ε about 900).

3-Buten-2-one (methyl vinyl ketone, conjugated C=C-C=O): n → π* at about 324 nm (ε about 25) and π → π* at about 219 nm (ε about 3,600).

The reason is that conjugation of C=C with C=O produces four π molecular orbitals (like butadiene), with a raised HOMO and a lowered LUMO, so the π → π* gap narrows. The band moves by about 30 nm to longer wavelength (bathochromic shift) and is several times more intense (hyperchromic), because the transition is now more allowed (a larger transition dipole over the extended system). The non-bonding orbital n on oxygen is almost unaffected, while the π* level is lowered, so the n → π* band also moves to longer wavelength, by about 45 nm, and its intensity increases only slightly because it remains a forbidden transition.

(c)(ii) Charge-transfer complexes (5 marks)

A charge-transfer (CT) complex forms between an electron-rich donor (D) and an electron-poor acceptor (A). In the ground state there is only a weak association, but light can transfer an electron from the HOMO of D to the LUMO of A. This gives a new, broad, intense absorption band, the CT band, that belongs to neither component and usually lies in the visible region, so the complex is coloured. The energy is given by Mulliken's relation, hν_CT ≈ I_D - E_A - C, so a donor of lower ionisation energy or an acceptor of higher electron affinity moves the band to longer wavelength. Its intensity increases with the concentration of the complex (Benesi-Hildebrand analysis gives K and ε), and it is solvent-dependent.

(A) Picric acid and anthracene. Picric acid (2,4,6-trinitrophenol) is the π-acceptor, since three nitro groups make the ring electron-poor. Anthracene, a large polycyclic aromatic hydrocarbon with a high-lying π HOMO, is the π-donor. They stack face to face and form a red-orange 1:1 complex (a π-π CT complex), whose CT band appears in the visible region (at about 450 nm) where neither component absorbs.

(B) Tetracyanoethylene (TCNE) and aniline. TCNE is a very strong π-acceptor (four CN groups, very low LUMO); aniline is the donor through its nitrogen lone pair conjugated with the ring, a stronger donor than benzene. The solution becomes strongly coloured, with a CT band well into the visible region (far to the red of the benzene-TCNE complex at 384 nm). Aniline is so electron-rich that it can also undergo an irreversible substitution with TCNE, so spectra must be recorded quickly.

(c)(iii) Mass spectrum of ethyl 4-chlorobenzoate (5 marks)

The formula is C₉H₉ClO₂. Chlorine has two isotopes, ³⁵Cl and ³⁷Cl, in a ratio of about 3:1, so the molecular ion appears as a pair: M⁺• at m/z 184 (³⁵Cl) and M+2 at m/z 186 (³⁷Cl), with relative intensity about 3:1.

McLafferty rearrangement. The ester has γ-hydrogens on the ethyl group, in a six-membered arrangement O=C-O-CH₂-CH₃. The carbonyl oxygen abstracts a hydrogen from the CH₃ through a six-membered transition state, and ethene (28) is lost. The rearranged ion is the radical cation of 4-chlorobenzoic acid, [4-Cl-C₆H₄-C(OH)₂]⁺• ↔ 4-ClC₆H₄-COOH⁺•, at m/z 156 (³⁵Cl) and 158 (³⁷Cl), again 3:1.

Other prominent ions are m/z 139/141, 4-ClC₆H₄-C≡O⁺, from loss of •OCH₂CH₃ (the base peak, formed by α-cleavage), and m/z 111/113, the chlorophenyl cation, from loss of CO.

(c)(iv) Mass spectrum of anisole (5 marks)

Anisole (methoxybenzene, C₆H₅-O-CH₃) has M⁺• at m/z 108, the radical cation C₆H₅-O⁺•-CH₃. Its fragments are as follows.

• m/z 93, C₆H₅O⁺ (the phenoxy cation): loss of •CH₃ (15) from M⁺• by cleavage of the O-CH₃ bond.
• m/z 78, C₆H₆⁺• (the benzene radical cation): loss of formaldehyde, CH₂=O (30), from M⁺• after transfer of a hydrogen from the methyl group to the ring.
• m/z 77, C₆H₅⁺ (the phenyl cation): loss of the methoxy radical, •OCH₃ (31), by cleavage of the C-O bond.
• m/z 65, C₅H₅⁺ (the cyclopentadienyl cation): loss of CO (28) from the m/z 93 phenoxy ion, with ring contraction.

Pathway: M⁺• (108) → 93 → 65; M⁺• → 78; M⁺• → 77 (which may lose C₂H₂ to give m/z 51).