P Predentknight Full Member 10+ Year Member 15+ Year Member Joined Feb 12, 2008 Messages 87 Reaction score 0 Points 0 Pre-Dental May 30, 2008 #1 Advertisement - Members don't see this ad Take a look at step 2 in the synthesis problem. I understand that it is an E2 elimination but I do not understand why it would form the less substituted alkene? Anybody with destroyer who understands this problem, please explain.
Advertisement - Members don't see this ad Take a look at step 2 in the synthesis problem. I understand that it is an E2 elimination but I do not understand why it would form the less substituted alkene? Anybody with destroyer who understands this problem, please explain.
drtoothy Full Member 10+ Year Member 15+ Year Member Joined May 14, 2007 Messages 72 Reaction score 0 Points 0 Dental Student May 30, 2008 #2 tBuO-K+ is used -- any bulky base won't sterically be able to form the most substituted product. therefore, the primary halide forms. Upvote 0 Downvote
tBuO-K+ is used -- any bulky base won't sterically be able to form the most substituted product. therefore, the primary halide forms.
