Lecture outline Directing effects of substituents Substituents already attached to an aromatic ring influence the preferred site of attachment of an incoming electrophile. e.g., nitration of toluene 3 2 2 S 4 + + 2 2 63% 3% 34% Reaction at the meta position is clearly disfavored compared to ortho and para. There is about twice as much ortho as para product. What does this tell you about the relative reactivities at the ortho vs para positions? Because ortho and para substitution products are favored over meta, is called an ortho,para-director it "directs" incoming electrophiles to the ortho and para positions. In addition, nitration of toluene is 23 times as fast as nitration of benzene, so is also called an activator. Activating and directing effects of substituents must originate from their effects on a for the rate determining of the electrophile on the aromatic ring. The transition state for this mechanistic step lies closer in energy, and therefore structure (ammond postulate), to the intermediate arenium ion. So activating and directing effects are due primarily to the variation in stability of the intermediate cation; the structure and charge distribution of the reactant are less important. (complete the resonance structures below by carefully adding the double bonds and charges) para of +
ortho of + for simplicity, let's just draw the actual structure for this one... meta of + ow does the charge placement affect the stabilities of these three cations? More stable cations are formed faster, so rxn at o and p positions is faster than rxn at m ("o, p-director"), and faster than rxn with benzene ("activator") Because the effect is on the stability of the intermediate, a methyl group would more properly be called a "cationic intermediate stabilizer", rather than an "activating, o,p-director", i.e., it doesn't really "activate" the ring of the reactant or "direct" the + to the o and p positions, as once thought. Unfortunately, the misleading terminology is permanently embedded in the jargon associated with this reaction. In the space at left, sketch a rxn coordinate diagram that shows the three competing rxn pathways, leading from reactants to the three possible intermediate cations
R and R 2 groups are powerful activating, o,p-directors (complete the resonance structures below) R R R R R an analogous all-octet structure results from para... draw it... but not from meta (why?) Does the m-methoxy stabilize or destabilize this intermediate? and are electronegative, so they re e -withdrawing inductively (via σ-bonds), but their available lone pairs in conjugation with the π-system of the ring make them strongly e - donating by resonance this is clearly the dominant effect when these atoms are adjacent to a an electron deficient site of the arenium ion. In contrast, 2 is a powerful deactivating meta-director para ortho is also bad...
... meta is bad, but not as bad as o and p Because an 2 meta to the point of withdraws e -density inductively from the positively-charged ring of the intermediate, an + s nitrobenzene slower than it s benzene. Thus 2 is deactivating. The nitro group is also meta-directing, not because meta is any good but because the alternative ortho and para s are even worse. What does the rxn coordinate diagram for these three look like? S 3, 3 +, C, and carbonyl groups are also deactivating, m-directors. Draw their structures and explain why. alogens are a special case they re weakly "deactivating" and "o,p-directing" In addition to the three resonance structures you completed above, a fourth all-octet structure is possible.
Summary Fill in the table below with the substituents we've already encountered, plus, CR, CR, CR, C 2 R, CR 2, C. In general, the most powerful activators are the lonepair donors; the most powerful deactivators are substituents with full or substantial ( S 3 ) positive charges on the atom attached to the ring. activating o,p-directors Strongly act. Weakly act. deactivating m-directors Strongly deact. Weakly deact. (very weakly) deactivating o,p-directors Warning: Do not memorize this! You can figure out the important aspects of this table from your knowledge of the structures, resonance, and inductive effects.