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Unformatted text preview: 4 H 8 is probable, suggesting a butyl group. From this information one can infer the CH 3 CO (C 2 H 3 O), CH 3 COO and -C 4 H 9 subunits pointing to butyl acetate. Correct answer: n-Butyl acetate Problem 3: The m/z 156 peak satisfies the molecular ion criteria. The (A + 1)/A ratio shows a very small number of A+1 type elements, pointing to C 2 . The rest must be A-type elements. The m/z 29 ion points to C 2 H 5 , which is corroborated by the m/z 127 ion. The high relative abundance of M indicates a substituent of a low ionization energy. The absence of fragments between m/z 127 and 29 is a clue to the presence of a heavy substituent (iodine). Correct answer: Iodoethane...
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This note was uploaded on 03/04/2010 for the course CHEM chem 520 taught by Professor Turecek during the Winter '10 term at University of Washington.
- Winter '10