Ch_15-sample_Q_key_.pdf - Chapter 15 Solutions of the sample questions 1 2a b c d m-nitrotoluene(1-methylethyl)benzene or isopropylbenzene or cumene(not

Ch_15-sample_Q_key_.pdf - Chapter 15 Solutions of the...

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Chapter 15 Solutions of the sample questions 1- 2- a) m -nitrotoluene b) (1-methylethyl)benzene or isopropylbenzene or cumene (not IUPAC) c) 1-((1 R ,2 R )-2-chlorocyclopentyl)benzene d) 1-methyl-2,4-diaminobenzene or 2,4-diaminotoluene 3- a) To meet the criterion of a cyclic, planar, conjugated molecule the oxygen and the N-H must both be sp 2 -hybridized. If this occurs both heteroatoms contribute two electrons to the π system. This results in a total of eight π electrons, which is NOT aromatic according to the Hückel (4n+2) rule. b) The carbocation is sp 2 -hybridized, so if the oxygen is also sp 2 -hybridized, then each atom in the cycle has a conjugated p -orbital. Each double bond contributes two π electrons, for a total of six π electrons in the conjugated system. Therefore, this cation is predicted to be aromatic. c) This compound appears to be cyclic planar, conjugated molecule with 16 π electrons. However, 16 is not a Hückel number, so the compound is probably not planar and is NOT aromatic.
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