Electrophilic Aromatic Substitution: Direction

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Electrophilic Aromatic Substitution: Direction or each of the following species, show the most likely site(s) for electrophilic aromatic substitution, and predict whether the molecule reacts with electrophiles more quickly or more slowly than benzene itself. 2 more quickly more slowly (The para position is much less sterically hindered; EAS will occur predominantly at that postiion over the ortho positions.) more slowly more slowly C 3 C 3 C 3 more quickly (Two EDGs in competition; the alkyl group is a weak EDG by hyperconjugation, while the methoxy group is a strong EDG by resonance -- its directing effect "wins.") 2 (much!) more slowly (Two EWGs in competition; the ester group is a weaker EWG than the nitro group, so the new substituent would go meta to the nitro.)

Electrophilic Aromatic Substitution: More Direction 1. There are two possible positions for substitution on the aromatic compound called pyrrole. Predict which of these two positions an incoming electrophile will be directed to, and justify your answer with resonance structures. pyrrole 3 2 S 4 2 2-nitropyrrole or 2 3-nitropyrrole Incoming electrophiles are directed predominantly to the 2-position. This is because the arenium cation resulting from attack at this position has two resonance structures to stabilize it, while the arenium cation intermediate from attack at the 3-position only has one. Attack at 2-position: 2 2 2 Attack at 3-position: 2 2 2. Electrophilic aromatic substitution on anilines (aminobenzenes) typically proceeds quite poorly (surprisingly slow reaction rate relative to benzene) and gives an unexpectedly high yield of the meta product. iefly explain this observation, basing your argument on the representative example below. 2 3 2 2 2 2 S 4 2 2 10% 2 40% Electrophilic aromatic substitutions typically require strong Lewis or mineral acids to be effective. owever, amines are basic and thus easily protonated under these conditions! So there is a large proportion of arylammonium ion in the mixture, not just the aniline. As you learned in lecture, arylammonium is avery strong deactivator and meta director, hence the reaction is slow and you see a large amound of meta product! 2 3 2 S 4 50% activator, o/p-director deactivator, m-director

Electrophilic Aromatic Substitution: Applications of Direction 1. Before the advent of modern spectroscopy Korner's absolute method was used to determine whether a disubstituted benzene derivative was the ortho, meta, or para isomer. Korner's method involves adding a third group (usually a nitro group) and determining how many isomers are formed. or example, when o-xylene is nitrated, two isomers are formed: 2 3 2 2 S 4 o-xylene (a) ow many isomers (including any minor isomers) are formed by nitration of m-xylene? Draw them. 3 2 2 m-xylene 2 S 4 2 Three isomers are formed. (b) ow many isomers (including any minor isomers) are formed by nitration of p-xylene? Draw them. 3 2 2 S 4 p-xylene ne isomer is formed. (c) A chemist isolated an aromatic compound with molecular formula C 6 4 2. e carefully nitrated this compound and purified three isomers of formula C 6 3 2 2 (including any minor isomers). Propose structures for the original compound and the three nitrated derivatives. 3 2 2 2 S 4 2 m-dibromobenzene

Electrophilic Aromatic Substitution: Reagents Each of the following transformations can be carried out in one or two steps. ill in the reagents required for each step. 1., Al 3 2. 2, Al 3 1. 3, 2 S 4 2. Sn 0, 2 1., Al 3 2. ab 4, Me C 3 1., Al 3 C 3 S 3 2. S 3, 2 S 4 1., Al 3 2. X

Electrophilic Aromatic Substitution: Mechanisms Provide a complete curved-arrow mechanism for the following transformation. Show resonance structures for the arenium intermediate to make it clear why the substitution takes place at the indicated position on the ring. Al Al Al 4 This dissociation is favorable because tetrachloroaluminate is a good leaving group, and tert-butyl carbocation is quite stable. Al 3 :sol Positive charge can be stabilized by donation from the high-energy lone pair. The best curved arrow mechanism shows formation of the best resonance structure immediately; the carbocation intermediates are shown here only for illustration of the various resonance structures possible.

Electrophilic Aromatic Substitution: More Mechanisms 1. The following reaction produces a mixture of two products as shown below. 2 C 3 C 3 C 3 Compound A Compound B a) Provide a complete curved-arrow mechanism to account for the formation of Compound A. : C 3 C 3 C 3 b) Provide a complete curved-arrow mechanism to account for the formation of Compound B. You will want to keep in mind the mechanism for electrophilic aromatic substitution. ow might the methoxy group play an important role in this reaction? : C 3 C 3 C 3 C 3 ote that the third step has the three-membered ring being opened at the less-substituted side; this is the only way we can produce Compound B. If the bromide ion opens the threemembered ring at the more substituted side in this mechanism, Compound A results. In fact, some (if not all!) of the Compound A formed in this reaction is probably via that mechanism.

Electrophilic Aromatic Substitution: Even More Mechanisms 1. Under special conditions, 2,4,6-tribromophenol, shown below, can undergo a fourth bromination. The reaction does not occur at the meta position, as might be expected, but instead occurs at the para position to give a nonaromatic product. 2 e 3 2,4,6-tribromophenol (a) Provide a complete curved-arrow mechanism for this transformation. e 3 3 e solv (b) The product in (a) can be used for bromination of other aromatic compounds, such as the aniline shown below. Provide a complete curved-arrow mechanism for this transformation. C 3 C 3 Me 2 Me 2 C 3 C 3 Me 2 Me 2 C 3 Me 2

Aryl Diazonium Salts Propose a synthetic pathway for each of the following targets starting from benzene or toluene. Each synthesis will require the use of an aryl diazonium salt as a key step. a) 3 2 Sn 2 2 S 4 a 2, 2 CuC C b) 2 3 Al 3 2 S 4 2 Sn, a 2 KI 2 2 I c) 3 2 2 2 2 S 4 Al 3 Sn, 2 2 2 a 2 d) C 3 3 C 3 2 (xs) C 3 Sn C 3 2 S 4 e 3 C 3 C 3 2 3 P 2 2 a 2, 2 2

ucleophilic Aromatic Substitution: Mechanisms Provide a complete curved-arrow mechanism for the following transformation. You must draw the most favorable resonance structure for all intermediates. : 2 K 2 2 2 2 2 The best curved-arrow mechanism shows formation of the most stable resonance structure in a single step, so you need't show the carbanion resonance structure here! Provide a complete curved-arrow mechanism for the following transformation. t Bu: K I K t Bu I t Bu

ucleophilic Aromatic Substitution: Direction 1. or each pair shown below, choose the species that will react more quickly with sodium methoxide (ame), and explain your reasoning. Resonance structures will be necessary. a) 3 C vs. 2 3 C more quickly 2 The first species reacts more slowly because nucleophilic attack results in a resonance structure where the negative charge is localizd adjacent to an electron-donating group -- this is a repulsive, and thus destabilizing, interaction that slows the reaction relative to that of the other molecule. u 3 C 2 Destabilizing! b) vs. C 2 C more quickly 2 C u 2 C stabilizing resonance u 2 The first species reacts more quickly because nucleophilic attack results in a resonance structure where the negative charge can be delocalized into the π system of the electron-withdrawing cyano group, stabilizing the intermediate and accelerating the reaction relative to the other species. 2. Explain the unusual regiochemistry observed in the products of the following reaction. : C 3 K a C 3 b C 3 C 3 C 3 b a : After the benzyne intermediate has formed by the pathway shown above, the incoming nucleophile can attack at either end of the "triple bond", resulting in a near 50:50 mixture of products.

ucleophilic Aromatic Substitution: More Mechanisms 1. Sanger's reagent, 1-fluoro-2,4-dinitrobenzene, has been used to determine the - terminal amino acid of polypeptide chains. Its use was pioneered by rederick Sanger, who used it to determine the -terminal amino acid residue in each of the four polypeptide chains of insulin. Sanger's reagent reacts with the -terminal amino acid residue of a polypeptide as shown below: 2 2 polypeptide 2 2 2 (a) Provide a complete curved-arrow mechanism for the reaction shown above. 2 polypeptide 2 2 polypeptide 2 2 polypeptide 2 2 polypeptide 2 polypeptide (b) After the reaction above is complete, the polypeptide is hydrolyzed to release free amino acids. The - terminal amino acid is easily identified, as it is the only one which is bound to Sanger's reagent. In fact, it is usually identified using paper chromatography, since it appears yellow while all the other amino acids are colorless. Provide an M-based explanation for why this first amino acid acquires a yellow color. This is a challenging question! Consider the structure of the -terminal amino acid bound to Sanger's reagent: We have extensive conjugation: five double bonds are in conjugation with each other. As a result, the M-LUM gap is relatively small, which means the amino acid absorbs low-frequency and long-wavelength light. The wavelength absorbed is long enough to be in the visible (rather than ultraviolet) light region: the amino acid absorbs violet light, and thus appears yellow (which is the complementary color of violet).

ucleophilic Aromatic Substitution: Even More Mechanisms 1. During the Vietnam War, U.S. military forces utilized a mixture of defoliant chemicals called Agent range as an herbicide to reduce the dense jungle foliage so that enemy guerilla forces could not use it for cover. Tragically, Agent range caused nearly a million deaths, disabilities, and birth defects; it was contaminated with 2,4,5-trichlorophenol, which can dimerize under basic (and indeed, physiologic) conditions to form notoriously toxic compound known as dioxins. An example of this dimerization is shown below; provide a complete curved-arrow mechanism for this transformation. a 2 2,4,5-trichlorophenol dioxin This set of arrows should actually be two steps; we just ran out of space!