homework_HWK2_Key - CHEM 520 - Analytical Mass Spectrometry...

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CHEM 520 - Analytical Mass Spectrometry Homework #2 Solutions Unknown #1: The A+1/A ratio for m/z 174/173 = 11.5% indicates a maximum of 10 C with the possibility for another A+1 element (N). The A+2/A ratio for m/z 175/173 = 0.44% excludes abundant A+2 elements (Si, S, Cl, Br), although O remains a possibility. The accurate mass fits best C 10 H 7 NO 2 (173.0477, OE). The other peaks are identified by elemental composition as C 10 H 7 (123.0548, EE), C 9 H 7 (115.0548, EE), and NO 2 (45.9929, EE). The m/z 173 ion is an OE and passes the tests for M. Ring+db = (23-7)/2 = 8. Since the presence of an NO 2 group (1 r+db) is indicated, the carbon skeleton must have 7 r + db, which is indicative of naphthalene or azulene (C 10 H 8 ). The final distinction among the possible isomers cannot be based on the mass spectrum alone, but should use the auxiliary info. The m.p. identifies the unknown as 2- nitronaphthalene . Unknown #2:
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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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