exam_iii--2009_answers

exam_iii--2009_answers - CHEM 302 Organic Chemistry I Exam...

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CHEM 302 Organic Chemistry I Exam III 12-December-2009 Answers 1) (18 pts) Compound A , C 9 H 16 , was found to be optically active. On catalytic reduction over a palladium catalyst, 2 equivalents of hydrogen were absorbed to yield compound B . Ozonolysis of A gave two compounds. One was identified as acetaldehyde, CH 3 CHO; the other compound, C , was an optically active dialdehyde, C 5 H 8 O 2 . By formulating the reactions involved, determine the structures of A , B , and C . Here is the reaction scheme, which should help you get the structures: A C 9 H 16 2H 2 Pd B (C 9 H 20 ) O 3 C C 5 H 6 O 2 (dialdehyde) + H 3 C H O We see from C and acetaldehyde that we need 2 acetaldehydes to make up the necessary formula. Also, since C is a dialdehyde, both ends must be –CHO and it must be optically active. This leaves as the only possible structure for C : O O H H Given this structure, both A and B follow. (Note that A can have either the Z or E conformation around the double bond; ozonolysis destroys any stereochemistry there): H H H H A B
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2) (20 pts) Provide all necessary reagents (i-v) and missing reaction products (A-E) in the following sequences: a) 1) NaNH 2 /NH 3 2) CH 3 I A H 2 Pd/CaCO
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exam_iii--2009_answers - CHEM 302 Organic Chemistry I Exam...

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