How does the orientation of hydrogen affect E2 elimination?

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The discussion focuses on the orientation of hydrogen in E2 elimination reactions, specifically identifying which hydrogen is anti-periplanar to the leaving group. Participants emphasize the importance of understanding the mechanism of E2 elimination to determine the correct orientation of the 2-hydrogen relative to the 1-bromine. Clarification is sought on how these orientations affect the final product's structure. The conversation highlights the critical relationship between hydrogen positioning and the success of the elimination process. Understanding these concepts is essential for mastering E2 reactions in organic chemistry.
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Please see the attached picture. I'm not sure which hydrogen would be anti-periplanar. Any help would be appreciated. Thanks!
 

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You're really close. Think about the mechanism of the E2 elimination reaction. Which way does the 2-hydrogen need to be oriented relative to the 1-bromine for the elimination to occur? What does that do to the orientations of the groups in the final product?
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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