NAOH strong base goes E2 with more substituted

[Ibraiz]

Answer is E.
I doubts about answer choice B as this is a strong base (not a bulky base) thus it will form an Zaitsev elimination product (double bond with more substituted carbon). Answer choice B is anti-Zaitsev forming a double bond with less substituted. Is that a mistake or am I not thinking right?



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[MSUstudent]

Answer is E.
I doubts about answer choice B as this is a strong base (not a bulky base) thus it will form an Zaitsev elimination product (double bond with more substituted carbon). Answer choice B is anti-Zaitsev forming a double bond with less substituted. Is that a mistake or am I not thinking right?



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I understand why B is correct because the Hoffman product will form because the (CH3)3 group will prevent a double bond to form the zaitsev's prdt.

For A, under aprotic conditions the Br will leave and make -OH a good Nucleophile and have a backside attack (SN2 like fashion).

I hope this helps.

 

[Ibraiz]

Thanks MSUstudent, I didn't even think about that. I kept looking at (CH3)3C without realizing that there is no H attached directly to this C thus no elimination could take place with this carbon. Now that you mentioned, it clicked.