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Exam 2 Solutions--M2016

1 Chemistry 2302 Monday, July 11, 2016 Exam 2 Answer Key Exam 2 Mean: 60 Exam 2 Median: 60 Exam 2 St. Dev.: 19 2 1. (12 pts) Each of the reactions below is drawn with two possible reaction conditions. If only one of the two reaction conditions would generate the given molecule as the major product, circle those conditions. If both sets of conditions would accomplish the reaction, circle BOTH . If neither set of reaction conditions would succeed, circle NEITHER . Circle one answer only. LiAlH4 and NaBH4 are both reducing agents, and add hydride ( H- ) to C=O bonds. LiAlH4 is a strong reducing agent, and will reduce almost anything (including the aldehyde here) to an alcohol. NaBH4 is choosier, and will only reduce ketones and aldehydes but that s exactly what we ve got here. So both work. Both of these reactions illustrate addition of an alkylmetal to a three-carbon electrophile.

equivalent to a C=O containing functional group, and our amide needs two to make an amine. So regardless of whether LiAlH(OtBu)3 adds or not, it certainly won’t make ... 4 points for each protection/deprotection cycle. 1 point for using correct reagents in each protection/deprotection; 2 points for drawing correct molecules on the right.

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