ps9 - CH 3 H OH 1 3 4 2 (a) Explain why the two reactions...

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Chemistry 3590 Honors Organic Chemistry I PROBLEM SET #9 FALL, 2009 Please answer problems 31, 36, 37, 45, 53 and 56 in Chapter 9 of Bruice and the following questions: (1) For each of the two compounds named below, draw the structure of the product resulting from an S N 2 reaction using exactly one equivalent of NaOH in methanol. Indicate the (R) or (S) designation, where appropriate. (R)-1-iodo-2-methylbutane (S)-1-chloro-3-iodo-2-methylpropane (2) Both of the reactions shown below exhibit first-order rate laws, indicative of an S N 1 mechanism. CH 3 H 3 C CH 3 Cl water-acetone CH 3 H 3 C CH 3 OH CH 3 H Cl water-acetone
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Unformatted text preview: CH 3 H OH 1 3 4 2 (a) Explain why the two reactions proceed at similar rates even though chloride 1 has a tertiary α-carbon, whereas chloride 3 has a secondary α-carbon. (b) Suppose you ran the second reaction using the pure (R)-enantiomer of 3 . What stereoisomer(s) of the product 4 would be formed? REVIEW PROBLEM (3) The alkene ( E )-2-deuterio-3-methyl-2-pentene (C 6 H 11 D) undergoes reaction with HCl to form C 6 H 12 DCl. Indicate what product(s) are formed. If any are stereoisomers, be sure to indicate which ones are enantiomers and which are diastereomers....
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This note was uploaded on 11/10/2009 for the course CHEM 1570-001 taught by Professor Tadhgbegley during the Fall '08 term at Cornell University (Engineering School).

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