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Why is the para position preferred more than ortho in the electrophilic substitution reactions of haloarenes?

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  • $\begingroup$ It is true already in the case of halogenation, nitration, sulphonation, fiedel crafts alkylation and fiedel crafts acylation. $\endgroup$
    – user103292
    Commented Mar 1, 2016 at 12:25
  • $\begingroup$ Klaus Warzecha is right of course.Well,if you aren't still satisfied then you could think of steric hinderance (which is less in para isomer than ortho).But as Klaus Warzecha mentioned,it is completely dependent on reaction condition. $\endgroup$ Commented Mar 1, 2016 at 16:25
  • $\begingroup$ But in my ncert textbook it's written that para position is preferred.... $\endgroup$
    – user103292
    Commented Mar 1, 2016 at 16:30
  • $\begingroup$ yup i am also studying from ncert book itself.then let me ask why do we get salicylic acid in kolbe and riemmer an tieman reaction?both give ortho product instead of para.(its salycilaldehyde in the latter one.doesn't matter its still ortho product only) $\endgroup$ Commented Mar 1, 2016 at 16:37
  • $\begingroup$ Okay. I'm not doubting your answer but what's the reason behind it and is what's written in the text book wrong? $\endgroup$
    – user103292
    Commented Mar 1, 2016 at 16:39

1 Answer 1

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Is it really? Always?

In an older article (J. Chem. Educ., 1988, 65, 367), the authors described an experiment that demonstrates the outcome of the alkylation of chlorobenzene (1) with 2-chloropropane (2) under kinetic control.

They first added 25 mL of 1 and 2 mL of nitromethane to 1.2 g anhydrous aluminium chloride, and stirred until the aluminium chloride was fully dissolved. To the cooled mixture (3-5 °C), a solution of 3 mL 2 in 20 mL of 1 was dropwise added within 10 min, while keeping the temperature of the reaction mixture below 10 °C. The reaction was stirred for another hour, allowed to come to room temperature.

After workup, gas chromatographic analysis showed that the o/m/p ratio for the resulting isopropyl chlorobenzenes was 63/5/32.

To cut the long story short, under the conditions describe above, the ortho isomer prevailed.

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    $\begingroup$ This says the ortho and para positions are equally reactive, and the ortho dominates because there are two potential reaction sides leading to the same product. $\endgroup$
    – Lighthart
    Commented Feb 29, 2016 at 18:57
  • $\begingroup$ @Lighthart Yes, two for the price of one :) $\endgroup$ Commented Feb 29, 2016 at 20:02

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