Wiley PLUS C edugen.wileyplus.com main uni Return to Blackboard Wiley PLUS Brown
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Wiley PLUS C edugen.wileyplus.com main uni Return to Blackboard Wiley PLUS Brown & Poon, Introduction to Organic Chemistry, 5e Help l Sistem Announcements (1 Unread) Assignment Open Assignment NEXT PRINTER VERSION BACK ASSIGNMENT RESOURCES HW t6 2 Problem 7.10 Problem 7.28 Problem 7.26 Problem 7.24 Bond (2) haloalkane or halocycloalkane: Problem 7.34 Problem 7.50 Edit Problem 7.12 Bond (2) metal alkoxide: Review Results by Study Objective which combination gives the higher yield of ether? 1 License Agreement l Privacy Policy I 2000-2016 John Wiley & Sons Inc. All Rights Reserved. A Division of John Wiley & Sons Inc. 4.18 7:15 PM a e 9 wiley PLUS- G 9 Chegg Study IG 9 hemistry quest. A Sticky Notes A 4x HE Search the web and Windows 6/6/2016Explanation / Answer
I don´t understand well what is the right way to answer the question on the webpage that you use, however, I will explain how to get the higher yield in this reaction.
The Williamson Ether synthesis (alkyl halide + metal alkoxide) is an SN2 reaction. Remember that since the SN2 reaction proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the “big barrier” for the SN2 reaction is steric hindrance. The rate of the SN2 reaction was highest for methyl halides, then primary, then secondary, then tertiary (which essentially don’t happen at all). The same pattern exists for the Williamson Ether reaction.
The higher yield is obtained when we use a methyl halide with the alkoxide (RO–) in addition to the alcohol (ROH) in the reaction. The reaction of RO– with an alkyl halide is always going to be much faster than the reaction of ROH because of the higher electron density on the nucleophile (oxygen). Remerber the conjugate base is always a better nucleophile.
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