Question 13 of 13: Mass Spectrometry, IR, and NMR Structure Elucidation

Question 13: Mass Spectrometry, IR, and NMR Structure Elucidation (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) Dopamine's molecular formula by exact mass

Using exact isotope masses (C 12.0000, H 1.007825, N 14.003074, O 15.994915):

$$m(\text{C}_8\text{H}_{11}\text{NO}_2) = 8(12.0000)+11(1.007825)+14.003074+2(15.994915) = 153.0790$$ $$m(\text{C}_7\text{H}_{11}\text{N}_3\text{O}) = 7(12.0000)+11(1.007825)+3(14.003074)+15.994915 = 153.0902$$

Comparing both to the measured 153.0680: |153.0790 – 153.0680| = 0.0110 vs. |153.0902 – 153.0680| = 0.0222 — C₈H₁₁NO₂ is the closer match by a factor of two, so C₈H₁₁NO₂ is the correct formula (this also matches dopamine's real, known structure: a catechol ring with two phenolic OH's and a simple primary-amine ethylamine chain — a single nitrogen, exactly as C₈H₁₁NO₂ requires, whereas C₇H₁₁N₃O would demand three separate nitrogens that have no place in dopamine's simple monoamine structure).

b) Distinguishing morphine, heroin, and oxycodone by IR

The three narcotics differ exactly in which O–H and C=O groups are present, which IR reads directly:

Morphine converts to heroin by acetylating both O–H's, so the O–H stretch vanishes and two ester carbonyls appear — that single presence/absence pair already separates morphine from heroin. Oxycodone keeps one O–H (so it still shows the O–H stretch, like morphine) but replaces morphine's allylic alcohol with a ketone, so it uniquely shows both an O–H stretch and a carbonyl band, with the carbonyl frequency itself (≈1710, ketone) distinguishing it from heroin's higher-frequency ester carbonyls (≈1740–1770) were the two ever confused. The three compounds are therefore each uniquely identified by the (O–H present?, C=O present?, C=O frequency class) triple.

c) Structure elucidation from spectral data

Each part is solved the same way: use the IR to fix the functional-group class (or rule classes out), then use the NMR splitting pattern (n+1 rule) and integration to fix the carbon skeleton and symmetry.

i) C₄H₈Br₂, IR shows only C–H (no O–H/C=O); NMR: two singlets only (6H, 2H) — a highly symmetric molecule with no adjacent-H coupling anywhere. A quaternary carbon bearing two equivalent methyls, one Br, and a –CH₂Br arm fits perfectly: neither the gem-dimethyl (6H, no neighbouring H since the carbon they're on has none) nor the CH₂Br (2H, likewise flanked only by that same H-less carbon) has anything to couple to, so both appear as clean singlets, and the CH₂Br is the more downfield one (3.86, next to two electronegative-ish substituents through the adjacent C) — 1,2-dibromo-2-methylpropane, (CH₃)₂CBr–CH₂Br.

ii) C₃H₆Br₂, IR shows only C–H; NMR: quintet + triplet, no OH/ester. A symmetric 1,3-disubstituted propane: the central CH₂ is flanked by 4 equivalent H's (2 on each terminal CH₂Br) ⇒ quintet (n+1=5); each terminal CH₂Br is flanked by only the central CH₂'s 2 H's ⇒ triplet (n+1=3), and the two terminal CH₂Br's (4H total) are chemically equivalent by the molecule's own mirror symmetry — 1,3-dibromopropane, BrCH₂CH₂CH₂Br.

iii) C₅H₁₀O₂, IR 1740 (ester C=O); NMR: two triplet/quartet ethyl patterns, one quartet shifted far downfield (4.72, O–CH₂). Two independent ethyl groups — one attached to the carbonyl carbon (CH₂ at 2.30, alpha to C=O; its CH₃ at 1.15) and one attached through the ester oxygen (OCH₂ at 4.72, strongly deshielded; its CH₃ at 1.24) — ethyl propanoate, CH₃CH₂C(=O)OCH₂CH₃.

iv) C₆H₁₄O, IR 3600–3200 (O–H present); NMR: triplet (6H)/quartet (4H) ethyl pattern + a lone methyl singlet (3H) + an exchangeable 1H singlet (OH). Two equivalent ethyl groups (triplet 6H/quartet 4H) plus one more methyl (singlet, no neighbouring H) all converging on one quaternary, OH-bearing carbon — 3-methylpentan-3-ol, (CH₃CH₂)₂C(OH)CH₃. The IR O–H stretch (present here, absent in part v) is what rules out the isomeric ether in the first place.

v) C₆H₁₄O, IR shows only C–H (no O–H) — must be an ether, not an alcohol, despite being an O-isomer of part (iv); NMR: one doublet (12H) and one septet (2H), ratio 12:2 = 6:1. Two equivalent isopropyl groups joined through an ether oxygen: each isopropyl's two methyls (12H total, all four equivalent by the molecule's symmetry) couple only to their own CH ⇒ doublet; each isopropyl CH (2H total) couples to 6 equivalent methyl H's ⇒ septet (n+1=7), strongly deshielded (3.60) by the adjacent ether oxygen — diisopropyl ether, (CH₃)₂CH–O–CH(CH₃)₂.