Destroyer Ochem # 15

Zerconia2921

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    Can someone explain problem number 15 in OCHEM for destroyer.

    I understand the activation and deactivation. What I have trouble in is understanding why it adds para instead of ortho when it can add to both. How do you figure out which one will be the major product?
     

    eatkabab

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      I actually noticed this as well and went hunting for the answer. basically, if you have to choose between an activating and deactivating group being the controller, the activating group always wins. amines are activating, carbonyls are deactivating.

      I believe even if you have a methyl and a fluorine, the methyl wins. both methyl and fluorine are activating, but the fluorine is inductively deactivating, therefore not as strong of an activator as methyl. but I'm not 100% sure on that one. I'd like some clarification under this extreme case. also, if you've gotta choose between two deactivating groups, I believe you go with the most deactivating (like a nitro over a methoxy group).


      of course activating groups add ortho/para, with para leading to the lower energy product (99% of the time both thermal and kinetic). and as for why only one Cl adds, well thats just how the reaction works... this should clarify things for you a little:

      http://en.wikipedia.org/wiki/Electrophilic_halogenation
       
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      PreDent2009

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        Eatkabab, you are right in saying that the methyl group will direct over the fluorine or any other halogen. Even though halogens are o/p directing, they are technically still inductive-withdrawers that provide resonance stabalization. In the case of two deactivating groups, this is basically like a battle between two evils, there is no true director that will win. In the case of an activator vs. deactivator, or activator vs. activator, the most activating group will win and direct the electrophile. O->OH>OR etc.. Also, a methoxy group is an activating group to clarify, since the lone pair on O can donate its e- towards the ring for resonance.
         

        Orgodox

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          Can someone explain problem number 15 in OCHEM for destroyer.

          I understand the activation and deactivation. What I have trouble in is understanding why it adds para instead of ortho when it can add to both. How do you figure out which one will be the major product?

          The para is the major product here because of steric factors. These are not present in the para position making it the major product and the ortho minor.
           

          doc3232

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            Ok, so I just took a test on this.
            Advantage of Para: it is farther from the group.
            Advantage of ortho: there are 2 ortho positions.

            Lets say the group attacking is Chlorine or some other small group, then it adds like 70% ortho (about double the para position).

            I hope I helped.
             

            Orgodox

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              Lets say the group attacking is Chlorine or some other small group, then it adds like 70% ortho (about double the para position).

              Sterics are not limited to what is being added but ALSO what is already there. (Even if you are adding chlorine which is small, to tert-butylbenzene it will go mostly para because of sterics)
              In this case there is another ring attached which makes addition to the ortho position minor.
               
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