14D_w07_e02_key

14D_w07_e02_key - C h e m is t r y 1 4 D W in t e r 2 0 0 7...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Chemistry 14D Winter 2007 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. The exam is ready to be picked up when these numbers are posted. 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. Same mechanism for either enantiomer. H OH 2 OH 2 O H H OH 2 OH 2. (c) The S and R stereoisomers are produced in equal amounts. 3. Both 3-hexene and 2-hexene give 3-hexanol, when reacted with H 2 SO 4 /H 2 O. However 2-hexene is a less efficient source of 3-hexanol because it also yields a significant amount of 2-hexanol. 4. H 2 SO 4 H 2 O OH 5. Compared to the reaction of H 2 SO 4 /H 2 O with 3-hexene, this reaction is faster because the carbocation intermediate is more stable. 6.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

14D_w07_e02_key - C h e m is t r y 1 4 D W in t e r 2 0 0 7...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online