Unformatted text preview: Practice Problem 6‐33 – Answer Write a stepwise mechanism for each of the following reactions. Use curved arrows to show the flow of electrons in each step. Practice Problem 6‐34 – Answer Aromatic iodination can be carried out with a number of reagents, including iodine monochloride, ICl. What is the direction of polarization of ICl? Propose a mechanism for the iodination of benzene with ICl. Practice Problem 6‐35 – Answer The carbocation electrophile in a Friedel‐Crafts reaction can be generated in ways other than by reaction of an alkyl chloride with AlCl3. For example, reaction of benzene with 2‐methylpropene in the presence of H3PO4 yields tert‐butylbenzene. Propose a mechanism for this reaction. Practice Problem 6‐36 – Answer Pyridine undergoes electrophilic aromatic substitution preferentially at the 3 position as illustrated by the synthesis of 3‐nitropyridine. Under these acidic conditions, the species undergoing nitration is not pyridine but its conjugate acid. Write resonance contributing structures for the intermediate formed by attack of NO2+ at the 2, 3, and 4 positions of the conjugate acid of pyridine. From examination of these intermediates, offer an explanation for preferential nitration at the 3 position. When nitration occurs at the 2 or the 4 position, the intermediate has a contributing resonance structure which places a +2 charge on nitrogen. This unfavorable resonance structure does not exist when nitration occurs at the 3 position, so the intermediate that forms in this reaction is much more stable. Practice Problem 6‐37 – Answer The sulfonation of an aromatic ring with SO3 and H2SO4 is a reversible reaction. That is, heating benzenesulfonic acid with H2SO4 yields benzene. Show the mechanism of the desulfonation reaction. What is the electrophile? A proton in sulfuric acid acts as the electrophile in the first step of this reaction mechanism: Practice Problem 6‐38 – Answer Arrange the compounds in each set in order of decreasing reactivity (fastest to slowest) toward electrophilic aromatic substitution. [1 = fastest; 3 = slowest] Practice Problem 6‐39 – Answer The following molecules each contain two aromatic rings. Which ring in each undergoes electrophilic aromatic substitution more readily? Draw the major product formed on nitration. In each case, the ring with an activating substituent directly attached will undergo electrophilic aromatic substitution more readily. For steric reasons, the nitro group will end up in the para position in the major product. Practice Problem 6‐40 – Answer Propose an explanation for the fact that the trifluoromethyl group is almost exclusively meta directing. The intermediate that forms if NO2+ reacts at the meta position is shown below: The intermediate that forms if NO2+ reacts at the para position is shown below: If NO2+ reacts at the para position, the intermediate has a very unfavorable resonance structure which places the positive charge on the carbon atom bonded directly to the very electron‐deficient carbon atom in the –CF3 group. A similar situation occurs if NO2+ reacts at the ortho position. Since the intermediate that forms upon reaction at the meta position does not have this very unstable contributing resonance structure, it forms more readily, and the substituent therefore has a meta directing effect. Practice Problem 6‐41 – Answer Suggest a reason why the nitroso group, –N=O is ortho‐para directing whereas the nitro group, –NO2, is meta directing. The nitrogen atom in –N=O has a lone pair. As we discussed in class, all substituents in which the atom directly bonded to the ring has at least one lone pair are ortho‐para directing, because the intermediate formed in these reaction pathways is more stable than the intermediate that would form upon reaction at the meta position. Refer to your notes for the details as to why. The nitrogen atom in –NO2 has a positive formal charge. See the explanation for Problem 6‐40 to understand why reaction occurs preferentially at the meta position with respect to a nitro group. Practice Problem 6‐42 – Answer + The N,N,N‐trimethylammonium group, –N(CH3)3, is one of the few groups that is a meta‐directing deactivator yet has no electron‐withdrawing resonance effect. Explain. + See the explanation for Problem 6‐40, but replace the –CF3 substituent with –N(CH3)3 ...
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- Summer '08