Methanol containing the oxygen isotope ¹⁸O is allowed to react, in separate reactions, with each of the acid chlorides shown in Fig. P25.29a–c, and each of the resulting compounds A, C, and E is treated with one equivalent of sodium hydroxide. This treatment regenerates methanol in all three cases, along with hydrolysis products B, D, and F, respectively. Identify the compounds A–F and account for the distribution of the isotope in all three cases. Specifically, why is isotopically substituted methanol formed in reaction (b) but not in reactions (a) and (c)?

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(a) (b) * CH3OH+ Ph- 0 = 180 0 = 180 ő - CH₂OH+H3C-C-Cl acetyl chloride * CH,OH+ 0 = 180 Figure P25.29 -Cl benzenesulfonyl chloride i-Pro-P-Cl i-Pro pyridine pyridine pyridine A E C NaOH NaOH NaOH. CH3OH + B (contains no isotope) CH₂OH + D CH3OH + F (contains no ¹80 isotope)