Question about bases and electron donating/withdrawing groups...

FROGGBUSTER

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Jul 22, 2010
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    Please see attached pic.

    All parts of the question have the substituents para to each other. My question is, how would you answer the question of which is the most basic if they were all META to each other?

    If the methoxy group (part A) is meta to the -NH2, it can't donate its electrons next to the -NH2, so would we just look at the inductive withdrawing power of each substituent to determine the answer (look for LEAST inductively withdrawing???)???
     

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    whiteshadodw

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      Please see attached pic.

      All parts of the question have the substituents para to each other. My question is, how would you answer the question of which is the most basic if they were all META to each other?

      If the methoxy group (part A) is meta to the -NH2, it can't donate its electrons next to the -NH2, so would we just look at the inductive withdrawing power of each substituent to determine the answer (look for LEAST inductively withdrawing???)???

      well...or you could recognize that O-R (where R is H, or alkyl) groups are electron donating. in fact, anything with a lone pair of electrons that is directly attached to the ring is electron donating. double bonds are electron withdrawing, and NO2 is strongly withdrawing (the strongest usually).
       

      FROGGBUSTER

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      Jul 22, 2010
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        well...or you could recognize that O-R (where R is H, or alkyl) groups are electron donating. in fact, anything with a lone pair of electrons that is directly attached to the ring is electron donating. double bonds are electron withdrawing, and NO2 is strongly withdrawing (the strongest usually).

        Sorry I'm not sure I follow.

        If the substituents were meta to each other, -OR can't donate its electrons next to the -NH3+ that results after protonation. Therefore, we should only be looking at inductive effects instead of any resonance stabilization.

        This was my original proposal and biophilia okay'd it.

        -OR is electron-donating thru the pi framework but withdrawing thru the sigma framework, is it not?
         

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        WikiPremed
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          It's really best to take a chemical thermodynamics, or, in other words, spontaneity and equilibrium approach to questions of relative acidity or basicity. The equilibrium constant is determined by the free energy change, which for this type of comparison is going to get pretty much down to internal energy change comparison - neglecting solvent entropy effects which can be important sometimes. In other words, look at the stability of the conjugate acid. A strong base has a stable conjugate acid in aqueous solution.

          The original question is interesting because you have to understand that the ortho-para directors are so because they can donate into the ring and distribute negative charge around in the intermediate in electrophilic aromatic substitution, making the already resonance stable intermediate formed in that reaction even more stable by feeding electrons into the ring, which in the case of the acid base comparison can lead to greater stability for a conjugate acid in the para position. This property is also going to lead to an ability to increase the basicity of substituents ortho or para, so it's kind of a curve ball way to get at some fundamental organic learning goals. Although it would have been too hard if the original question had an amine as well as the methoxy substituent to choose from regarding para substituents.

          I think the question of meta substituents is too advanced for the MCAT. On the one hand there is a proximal partial negative charge to the ammonium group in the conjugate acid which will be strongest for the methoxy substituent, but its own oxygen substituent is the most electron withdrawing by induction. My own feeling is that the partial negative from one bond away on both sides, which would probably only be a minor effect, translated through sigma bonds from next door, not nearly as strong an effect as if the arrangement were para, but unimpeded on one side, though impeded by the electron-withdrawing induction of the oxygen's electronegativity itself translated through the adjacent carbon, would outweigh those inductive effects and carry the day, so I would guess the answer to still be methoxy. If the counter tendency of the electronegativity of the substituent itself actually outweighs the vicinal partial negative charges from resonance then the answer would probably have to be the aldyl group.
           
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