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Unformatted text preview: Chemistry 14D Winter 2012 Midterm Exam 2 Solutions Page 1 Statistics : High score, average and low score will be posted on the course web site after exam grading is complete. A note about exam keys : The answers presented here may be significantly longer than expected from a student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as well. To see the projected course grade cutoffs, consult the grading scale on the Chem 14D course web page. 1. (a) CH 3 1. BH 3 2. NaOH, HOOH H CH 3 OH Anti-Markovnikov addition and syn addition. (b) Ph CH 3 H 2 Pt Ph CH 3 H H Neither Markovnikov or anti-Markovnikov, and syn addition. 2. (a) B and C . These are derived from the more stable carbocation intermediate . (b) B and D . The new CH and CBr bonds are formed on the same face of the alkene . (c) None of these. Markovnikov's rule is irrelevant when the pi bond has no attached hydrogens . (d) D and E . In the presence of a peroxide, HBr addition to an alkene or alkyne follows a radical mechanism,...
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- Winter '08