# Ortho vs para alkylation of chlorobenzene

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?

• chemistry.stackexchange.com/questions/47088/… – Mithoron Jul 31 '18 at 21:02
• 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. – Eryk Jul 31 '18 at 21:17
• 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) – permeakra Jul 31 '18 at 21:58