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2014 Purple Book Answers Week 4.Pdf Chem S-20ab Purple Book Solutions Week 4 - Page 1 of 38 Aromaticity 1. Given the following structures, circle any that are aromatic. Explain why the other species are not aromatic. 4n electrons not a continuous antiaromatic cyclic system. NH not a continuous 4n electrons cyclic system. antiaromatic 2. In the following molecule, which nitrogen atom is more basic (i.e. more willing to accept a proton)? Why? The other one has its lone pair as part of CH3 the 6 electron aromatic system. Protonating that lone pair would destroy the aromaticity, N which would be bad: H CH3 this one is more basic N N not a continuous cyclic system. N 3. When the following molecule is oxidized, the expected ketone is not formed, but phenol is formed instead. Describe what is going on here, and explain this unusual behavior. OH O OH CrO3 oxidation phenol Tautomerizes in order to achieve aromaticity. Chem S-20ab Purple Book Solutions Week 4 - Page 2 of 38 Electrophilic Aromatic Substitution: Mechanisms 1. Provide a curved-arrow mechanism for the following transformation. Show resonance structures for the cationic intermediate which make it clear why the substitution goes where it does. O O Cl HN CH3 HN CH3 AlCl3 catalyst Cl AlCl3 Cl AlCl3 O O O O HN CH3 HN CH3 HN CH3 HN CH3 H H H want to put + charge next to :Solvent electron-donating nitrogen 2. Provide a curved-arrow mechanism for the following transformation. H+ + O: H2SO4 catalyst H OH :Sol :Solvent OH2 H OH H+ OH OH Chem S-20ab Purple Book Solutions Week 4 - Page 3 of 38 Electrophilic Aromatic Substitution: Direction 1. For each of the following species, show the most likely site(s) for electrophilic aromatic substitution, and predict whether the molecule is more reactive than benzene or less reactive than benzene. NO2 more reactive less reactive Strong deactivator Cl O Note: Imagine that these two groups are separate. The fact that they are connected is irrelevant. Weak activator less reactive less reactive CH3 O OCH3 O CH3 NO2 more reactive less reactive Chem S-20ab Purple Book Solutions Week 4 - Page 4 of 38 Electrophilic Aromatic Substitution: Reagents 1. Each of the following transformations can be carried out in one or two steps. Fill in the reagents required for each step. If a second step is not needed, please put an "X" through the second box. ,AlCl 1. Cl 3 Note: Don't try to get fancy with F-C alkylation. You can use it 2. ONLY for an isopropyl or tert-butyl group! O O ,AlCl 1. Cl 3 Cl Cl2 ,AlCl3 2. O OH ,AlCl 1. Cl 3 NaBH4,EtOH 2. CH3 O CH3 SO H ,AlCl 3 1. Cl 3 SO3 ,H2SO4 2. O O O Cl ,AlCl 1. Cl O 3 O 2. Chem S-20ab Purple Book Solutions Week 4 - Page 5 of 38 Electrophilic Aromatic Substitution: Mechanism 1. Provide a complete curved-arrow mechanism for the following transformation. CH3 CH3 H2SO4 H3C OH CH3 H+ (aromatic resonance) CH CH3 3 H3C OH2 CH3 :Solvent H H3C H3C CH3 CH3 Chem S-20ab Purple Book Solutions Week 4 - Page 6 of 38 More Mechanisms 1. The following reaction produces a mixture of the two products A and B: Br Br Br Br Br2 + OCH3 OCH3 OCH3 AB a) Provide a complete curved-arrow mechanism that leads to the product A. Br – Br: Br :Br—Br Br OCH OCH3 3 OCH3 b) Provide a complete curved-arrow mechanism that leads to the product B. (Hint: The lone pairs of the – OCH3 play an important role in the mechanism. Also, consider the mechanism for electrophilic aromatic substitution.) :Br—Br Br Br:– Br Br Br OCH OCH 3 3 OCH3 OCH3 Chem S-20ab Purple Book Solutions Week 4 - Page 7 of 38 Chemistry S-20ab Nucleophilic Aromatic Substitution: Mechanisms Week 4 Provide a complete curved-arrow mechanism for the following transformation. You must draw the most favorable resonance structure for all intermediates. :OH O N Cl O N OH 2 KOH 2 O2N O2N OH OH O2N O2N Cl Cl O O N N O O Provide a complete curved-arrow mechanism for the following transformation. tBuO: H OK I O KOtBu I tBuO H O O Chem S-20ab Purple Book Solutions Week 4 - Page 8 of 38 Nucleophilic Aromatic Substitution: Direction 1. For each pair, choose the species which will react faster with NaOCH3, and explain your reasoning. (Resonance structures will be helpful!) – CH3O: Cl Cl NO2 NO2 vs. H3C faster CH3 H3CO Cl In this resonance structure, the NO2 negative charge is next to the electron-donating CH3 group. This is bad. So the other one is faster; it never has a negative charge next to CH3 the methyl. – CH3O: Cl Cl NO 2 NO2 vs. NC faster CN H3CO Cl In this resonance structure, the NO negative charge is next to the 2 electron-withdrawing CN group. This is good. So this one will react faster. CN 2. Explain the unusual product mixture observed in the following reaction. CH3 CH3 CH3 Cl NaOH, heat OH + This is the "benzyne" mechanism! H OH –:OH Cl CH3 –:OH can attack either carbon. Chem S-20ab Purple Book Solutions Week 4 - Page 9 of 38 Nucleophilic Aromatic Substitution: Reactivity 1. Explain the following observations. Writing chemical equations and resonance structures will be more useful than long sentences! a) Although most hydrocarbons are fairly nonpolar, the following hydrocarbon has a significant dipole moment. (As part of your explanation, please indicate the direction of the dipole moment: which end is positive and which end is negative.) resonance This resonance structure is particularly stable. The left half is an aromatic ring with 2 pi-electrons. The right half is an aromatic ring with 6 pi-electrons. This resonance contributor gives an overall dipole moment to the molecule, with the left side as the positive end and the right side as the negative end. b) Sodium ethoxide will substitute for chloride in one of the substrates below but shows no reaction with the other, similar substrate. Cl OEt Cl NaOEt NaOEt no but reaction This substrate has no EtO:– special mechanism for stabilizing the Cl EtO Cl anionic intermediate. Notice the aromatic ring with 6 pi-electrons. This stabilizes the anionic intermediate in this nucleophilic aromatic substitution reaction, even without the presence of an electron withdrawing group. Chem S-20ab Purple Book Solutions Week 4 - Page 10 of 38 Carbonyl Reactivity 1. A site on a molecule will be electrophilic if it has a partial positive charge and a low-lying LUMO. Explain, using resonance and molecular orbital theory, why the carbonyl carbon is electrophilic. From MO point of view, O O we know that a * antibonding orbital will already be a good LUMO, and the electronegative oxygen makesitevenlower,soitisan Resonance shows partial excellent LUMO. electrophilic carbon positive charge on carbon (even though this is not a particularly good res. structure). 2. Using resonance and molecular orbital theory, rationalize the reactivity of the following carbonyl analogs towards nucleophilic attack: More reactive Less reactive + + OH NR2 O NR OH NR2 O NR These resonance structures become increasingy worse from left to right, so there's less + charge on carbon. From an MO perspective, the two factors changing here are electronegativity and charge. We know that a more electronegative atom (like O) makes the energy of the LUMO lower, in turn making the molecule more electrophilic. Positive charge has the same effect. It makes electrons more strongly attracted, lowering the energy of the LUMO. 3. The electrophilicity of a carbonyl compound is affected by both inductive and resonance effects of neighboring groups. Rationalze the reactivity of the following carbonyl derivatives: More reactive Less reactive O O O O O O – Cl H R OR NR2 O electron C-H bonds withdrawing No resonance * by induction or induction donate to lone pairs are electron donating (hyperconjugation) O O O OR NHR2 O increasingly important alternate res. structures The energy of the * LUMO is affected by the electronegativity of the neighboring group (which is why the Cl lowers the energy of the LUMO), but also by electron donation from neighboring lone pairs (which is why the ester, amide, and carboxylate raise the energy of the LUMO and make it a poorer acceptor.) Chem S-20ab Purple Book Solutions Week 4 - Page 11 of 38 Carbonyl Addition: Mechanisms 1. Provide curved-arrow mechanisms for the following transformations: O NaBH4 H C H H C OH 3 EtOH 3 H H OEt BH3 O H H3C H O OH MgBr + H /H2O H3C H H3C H+ CH2 O H3C H H3C H+ O + H2O, H HO OH H3C H H3C H HO H O OH H H3C H H3C H H2O :OH2 + O HO OH H2O, H H CH3 H CH3 H2O H O OH H 2 O H H CH3 Chem S-20ab Purple Book Solutions Week 4 - Page 12 of 38 Reactivity of Carbonyl Addition Reactions 1. Explain the observed ranking of equilibrium constants for carbonyl hydration in the following substrates: O H2O HO OH [hydrate] K= [carbonyl] R R' R R' Carbonyl Compound K O 1x106 electron-withdrawing groups (F's) destabilize the highly electron- F C CF 3 3 deficient C=O. (The hydrate is more O electron rich than the carbonyl.) 3x104 F3C H O 2x103 H H O alkyl groups (CH3)aremore 1 electron-donating than H, and H3C H thus stabilize the electron- deficient C=O O 1x10–3 H3C CH3 O 9x10–6 resonance with aromatic CH3 ring stabilizes pi-bonding in C=O 2.
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