Final - Practice C

Final - Practice C - Organic Chemistry Chem 38 Practice...

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Organic Chemistry: Chem 38 Practice Exam 4 C There are 35 questions on this exam. Check that you have done all of the problems and filled in the first 35 bubbles on the scantron. Most questions are worth 4 points; there are several 1 and 2-point questions clearly labeled in the text. Instructions Answer sheet 1) You need to clearly write your name and fill in your student number, section number and test form on the scantron. 2) White = test form A Yellow = test form B 3) Section one =12:20 pm class Section two = 3:35 pm class 4) Use a #2 pencil Exam policy 1) Molecular models are allowed, calculators are not allowed. 2) The back of the test can be used for scratch paper. 3) Relevant tables, including the periodic table, are attached the end of this exam. 4) You cannot take this copy of the exam with you. You may copy your answers onto the provided page. The answer key will be posted on the web after the exam. Hints 1) As you read the question, underline or circle key words to highlight them for yourself. 2) Some questions have more than one acceptable answer. Pick the best one for full credit. Partial credit will be given for other acceptable answers. 3) There is no penalty for guessing. 1
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1 . (4 pts) One can exchange 127 I with 128 I in some iodo compounds. When the reaction is run on optically active ( S)- 2-iodobutane, the rate of racemization (i.e. the rate of loss of optical activity) is found to be exactly twice the initial rate of exchange ( 128 I for 127 I). Which reaction mechanism is most consistent with this observation ? NOTE: Optical rotation does not depend on the isotope present, i.e. ( S) -2- iodobutane with 127 I has the same rotation as ( S)- 2-iodobutane with 128 I. 127 S -2-Iodobutane + CH 3 CH 2 CHCH 3 I 128 I a) S N 1 b) S N 2 c) S N 1- partially through a tight ion pair d) E1 e) E2 f) between S N 1 and S N 2 g) elimination followed by substitution h) not enough information is provided to decide 2 . (4 pts) Addition of Br 2 to cis -2-hexene gives racemic products, even though attack of thebromide ion on the unsymmetrical bromonium ion intermediate is not equally likely at both carbons. The reasons for the racemic products are: a) The reaction is reversible, after a while both enantiomers are produced in equal amounts b) The product is a meso compound c) The bromonium ion formed has a plane of symmetry d) Two enantiomeric bromonium ions are formed in equal amounts e) The Br
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Final - Practice C - Organic Chemistry Chem 38 Practice...

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