Question 9 of 13

Question 9 (9/13)

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.

Approach. Combine the reaction type (enolate alkylation, so W must be a methylated ketone one CH₂ unit heavier than 2-butanone) with the three spectra to nail down exactly which methylated ketone: the molecular ion fixes the formula, the IR confirms the ketone stays a ketone (no new O–H), and the ¹H integration pattern (1H : 3H : 6H, with the 6H peak a doublet) fixes the connectivity uniquely.

Formula from M⁺ = 86. 2-Butanone is C₄H₈O (MW 72). Deprotonating an alpha C–H and alkylating with CH₃I adds exactly one CH₂ unit (+14), giving C₅H₁₀O, MW 86 — consistent with the stated molecular ion.

IR. A single strong carbonyl stretch near 1710–1715 cm⁻¹ persists (a saturated dialkyl ketone) and there is no new O–H or C≡N/C≡C band — W is still a simple ketone, just heavier.

NMR. 2-Butanone has two inequivalent alpha positions: the methyl ketone's CH₃ (3H) and the ethyl's CH₂ (2H, more substituted/more stabilised enolate under thermodynamic, strong-base conditions). Alkylating at the more substituted (thermodynamic) enolate carbon — the CH₂ — installs the new methyl there, converting –CH₂CH₃ into –CH(CH₃)CH₃ (isopropyl) while leaving the original ketone CH₃ untouched. The product, 3-methyl-2-butanone, (CH₃)₂CH–CO–CH₃, has exactly the three ¹H environments the spectrum shows: a 6H doublet ≈1.0 ppm (the two equivalent isopropyl methyls, split by the adjacent CH), a 3H singlet ≈2.1 ppm (the methyl ketone, no neighbouring H), and a 1H septet/multiplet ≈2.6 ppm (the single isopropyl methine, coupled to all six adjacent methyl H's) — total 6+3+1 = 10H, matching C₅H₁₀O exactly.