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?
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.