Question 9 of 13: Structure Elucidation from MS/IR/NMR

Question 9: Structure Elucidation from MS/IR/NMR (20 marks)

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.

Given. M⁺ = 86 (2-butanone, MW 72, plus one CH₃ group, +14, regardless of which α-carbon is alkylated — both possible mono-methylation products are C₅H₁₀O, MW 86). NMR integrations 6H : 1H : 3H with, respectively, a doublet-type simple multiplet (6H), a complex many-line multiplet (the small 1H peak), and (from the IR) a strong sharp C=O near 1710–1715 cm⁻¹ with no O–H.

Find. Which of the two possible mono-methylation regiochemistries (at the C1 methyl vs. the C3 methylene of 2-butanone) is actually observed, and the structure of W.

Approach. 2-Butanone has two different α-carbons: C1 (a methyl, 3H) and C3 (a methylene, 2H). Deprotonating either one and alkylating with CH₃I still gives a compound of formula C₅H₁₀O (MW 86) — the mass spectrum alone cannot distinguish the two regiochemical outcomes, so the NMR symmetry/splitting pattern is the decisive evidence.

1. Alkylation at C1 (the methyl) would give 3-pentanone. CH₃CH₂–CO–CH₂CH₃ is perfectly symmetric: its two ethyl groups are chemically equivalent, so the ¹H NMR would show only two signals total (a 4H quartet near 2.4 ppm + a 6H triplet near 1.0 ppm). The given spectrum instead shows three distinct signals (3H : 1H : 6H) — this rules out 3-pentanone immediately.
2. Alkylation at C3 (the methylene) gives 3-methyl-2-butanone (methyl isopropyl ketone), CH₃–CO–CH(CH₃)₂. This unsymmetrical ketone has exactly three different proton environments, matching the spectrum perfectly:
   • the acetyl CH₃ (3H) has no neighbouring C–H's, so it is an isolated singlet near 2.1–2.2 ppm;
   • the single isopropyl methine C–H (1H) is coupled to six equivalent neighbouring protons, giving the classic complex septet (seven lines) near 2.5–2.7 ppm — exactly the small, many-lined peak shown;
   • the two equivalent isopropyl methyls (6H) are coupled only to that one methine proton, giving a simple doublet near 1.0 ppm.
3. IR confirms a ketone, not an ester/acid/alcohol. A single strong, sharp C=O near 1710–1715 cm⁻¹ with no O–H anywhere is exactly consistent with a simple dialkyl ketone (an ester would sit higher, ≈1735–1750; an acid would show the broad 2500–3300 O–H shelf).

Mechanistic note. This regiochemistry (alkylation at the more substituted α-carbon) is the thermodynamic enolate outcome — the more substituted enolate is the more stable (more highly substituted) alkene-like species and predominates when the enolisation step is reversible/at equilibrium, which "a strong base" at ordinary temperature (as opposed to careful kinetic LDA/−78°C conditions) permits.