homework_HWK1_Key - 4 H 8 is probable, suggesting a butyl...

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Homework 1 Solutions CHEM 520 Problem 1: The isotope pattern at m/z 196-200 indicates the presence of 6 C and 2 Cl. The m/z 196 ion satisfies the criteria for a molecular ion. The 4 r + db equivalents and the high relative abundance (= stability) of the molecular ion indicate a benzene derivative. Correct answer: Dichlorobenzene Problem 2: The highest-mass peak is at m/z 101. However, m/z 87 would be loss of 14 Da, which is illogical. The spectrum does not show a molecular ion. The (A +1)/A ratio for m/z 44/43 indicates 2C, hence C 2 H 3 O or C 2 F for the ion. Both are EE + ions. The (A +1)/A ratio for m/z 62/61 indicates 1-2 C, hence C 2 H 5 O 2 or HCO 3 (EE+ ions). The (A +1)/A ratio for m/z 74/73 indicates 3C, hence C 3 H 6 O 2 , an OE ion. The m/z 56 ion has a maximum of 5C, hence C
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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.

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