Solutions to Ex..

### Chapter 1

Solution 1.1

Solution 1.2

Explanation: Recall that nitrogen atoms in amides have partial double bond character (Section 1.6) and are best thought of as sp2-hybridized atoms. In ciprofloxacin, the nitrogen atom in the bicyclic ring is conjugated via a double bond to a carbonyl (C=O) function and to the aromatic ring (as illustrated in the resonance form shown above). Thus, this nitrogen atom contributes its lone pair electrons to an aromatic ring system containing 10 π electrons. The planar arrangement of p orbitals required for aromaticity is possible only if the nitrogen atom is sp2-hybridized.

Solution 1.3

Explanation: The ring systems in examples (a) and (b) possess a planar, contiguous array of p orbitals with 10 and 6 π electrons, respectively, and thus are aromatic. The ring system in (c) is planar and contains 8 π electrons and is thus anti-aromatic. The eight-membered ring in (d) cannot form a planar structure and so is a non-aromatic ring system.

Solution 1.4

Explanation: The correct answer is (b). To construct the relevant Frost circle we first draw the cyclic ring system with one of its vertices pointing down (as illustrated above and in Figure 1.21). The cyclobutadienyl dication has a co-planar array of p orbitals and has only 2 π electrons. These two electrons will fill the single bonding π molecular orbital and produce an aromatic π system (4n +2 where n = 0).

Solution 1.5

Explanation: The allyl anion is more stable since the negative charge is shared by multiple atoms. This can be illustrated with a resonance hybrid as shown above. In molecular orbital terms, the negative charge is delocalized into the π bond by overlap of the p orbitals on the three sp2-hybridized carbon atoms in the allyl anion.

Solution 1.6

Explanation: The cation is aromatic since it is composed of a cyclic, co-planar array of p orbitals containing 2 π electrons (4n + 2 when n = 0).

Solution 1.7

### Chapter 2

Solution 2.14: Shown below ...

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