Most Stable Isomer: Ortho, Meta, or Para Benzene?

  • Thread starter Thread starter skull
  • Start date Start date
  • Tags Tags
    Stable
AI Thread Summary
The stability of ortho, meta, and para isomers of benzene depends on the substituents attached to the benzene ring. In the case of ortho-di-chlorobenzene, meta, and para isomers can be compared for stability based on steric hindrance and electronic effects. Larger substituents, like bromine, create more steric strain when positioned close together, making the ortho isomer less stable. The para isomer generally exhibits the highest stability due to minimized steric interactions and optimal distance between substituents. Understanding these factors is crucial for determining the most stable isomer in substituted benzene compounds.
skull
Messages
22
Reaction score
0
Which of the following is most stable. Ortho, meta, or para benzene? Also, why is this the case.
Thank you in advance
 
Physics news on Phys.org
your question makes no sense seing as benzen has no substituants to compare stability. And no hydrogens don't count as substituants
 
you may want to check the question just incase you may have miss-spelt the compound.

For example ortho-di-chlorobenzene can be compared to meta or para dichlorobenze in terms of stabilites
 
OMG sorry about that! I forgot to include the branch names. Assume the branches are bromine
 
well then that makes things a lot more clearer...know assuming you know how the orth meta and para positions look like on a molecule and the trend for increasing ionic radius you can further deduce that bromine is a fairly large molecule thus it wants its space otherwise if forced to close to another large molecule it would make it unstable.

From this i think you can figure out which orientation is the most stable
 
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...
Back
Top