I have a question to the solution regarding the product formed in a reaction with chlorobenzene with the use of (i) $\ce{CH3CH2Cl}$ and (ii)$ \ce{AlCl3}$. Since the -Cl substituent is ortho- para-directing shouldn't the answer then be a mixture of 1-chloro-4-ethylbenzene and 1-chloro-2-ethylbenzene? - I suppose that the para-orientation is the major product. But how is that determined?

Friedels-Craft alkylation

  • $\begingroup$ chemistry.stackexchange.com/questions/47088/… $\endgroup$ – Mithoron Jul 31 '18 at 21:02
  • $\begingroup$ Thanks. The answers in the given link are very helpful. Though i still dont understand, why the mayor formed product is the para-orientation instead of ortho. Is the para vs ortho only determined experimentially? Because I dont know which orientation to choose in other excersises that include reactions with benzenes that have an ortho- and para-directing substituent. $\endgroup$ – Eryk Jul 31 '18 at 21:17
  • 1
    $\begingroup$ Chlorine is $\sigma$ acceptor, thus para > orto. But in general orto/para orientation it poorly controlled and separation step is usually needed. Notable exceptions are reactions with sterically hindered orientants and and cases when both aromatic compound and active particle coordinate to the same Lewis acid. (acilation of phenols with TiCl4, for example) $\endgroup$ – permeakra Jul 31 '18 at 21:58

In general para isomera are thermodynamically more stable than ortho isomers (except in cases when ther is hydrogen bonding in ortho isomer). There are several reasons:- 1. Steric hindrance makes para isomer more stable. 2.Benzene substituents rotate and change thier plane due to steric hindrance. It stops delecolaization of lone pair of electrons (if present in substituents) causing elevation in energy. Both proper orientation (steric hindrance) and appropriate energy (temperature) is required for formation of product. Para isomer satisfies these conditions readily and hence are major product.


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