Naphthalene resonance forms

All polycyclic aromatic hydrocarbons can be represented by a number of different resonance forms. Naphthalene, for instance, has three. 

Azulene, a beautiful blue hydrocarbon, is an isomer of naphthalene. Is azulene aromatic Draw a second resonance form of azulene in addition to that shown. 

Resolution (enantiomers), 307-309 Resonance, 43-47 acetate ion and, 43 acetone anion and. 45 acyl cations and, 558 allylic carbocations and, 488-489 allylic radical and, 341 arylamines and, 924 benzene and, 44. 521 benzylic carbocation and, 377 benzylic radical and, 578 carbonate ion and. 47 carboxylate ions and, 756-757 enolate ions and, 850 naphthalene and, 532 pentadienyl radical and. 48 phenoxide ions and, 605-606 Resonance effect, 562 Resonance forms, 43... 

The following reasoning was used to eliminate the less probable mechanisms shown in Figure 4. A H atom is added to naphthalene to form an a-radical in reaction 1A and a /3-radical in reaction IB. Both are resonance-stabilized radicals. They can lose either a 2H atom or a H atom to regenerate naphthalene. We have shown a 2H atom lost to form a protium-enriched product in reactions 1A and IB. The fact that we observe a fourfold increase of protium in the a-position of spent naphthalene suggests that reaction IB is faster than reaction 1A and, therefore, is the predominant mechanism. 

This may be rationalized by considering the stability of intermediate addition cations. When the electrophile attacks at C-5 or C-8, the intermediate cation is stabilized by resonance, each having two favourable forms that do not perturb the aromaticity of the pyridinium system. In contrast, for attack at C-6 or C-7 there is only one such resonance form. We used similar reasoning to explain why naphthalene... 

The reaction involves the transfer of an electron from the alkali metal to naphthalene. The radical nature of the anion-radical has been established from electron spin resonance spectroscopy and the carbanion nature by their reaction with carbon dioxide to form the carboxylic acid derivative. The equilibrium in Eq. 5-65 depends on the electron affinity of the hydrocarbon and the donor properties of the solvent. Biphenyl is less useful than naphthalene since its equilibrium is far less toward the anion-radical than for naphthalene. Anthracene is also less useful even though it easily forms the anion-radical. The anthracene anion-radical is too stable to initiate polymerization. Polar solvents are needed to stabilize the anion-radical, primarily via solvation of the cation. Sodium naphthalene is formed quantitatively in tetrahy-drofuran (THF), but dilution with hydrocarbons results in precipitation of sodium and regeneration of naphthalene. For the less electropositive alkaline-earth metals, an even more polar solent than THF [e.g., hexamethylphosphoramide (HMPA)] is needed. 

Draw three resonance forms for naphthalene, showing the positions of the double bonds. 

Table 2 summarizes pK measurements for the simplest protonated aromatic hydrocarbons. The columns to the right and left of the benzenonium ion correspond to benzoannelation of ions subject to protonation at the 2- and 4-positions of the benzene ring, respectively. In the parent ion the two positions correspond to resonance forms (one of which has been rotated through 120° in the table). The naphthalenonium ion 17 is shown as being formed from the 1,4-water adduct (hydrate) of naphthalene. It may also be formed from the isomeric 2,1 hydrate (l,2-dihydro-2-naphthol) with pAR = -6.7 and pAn2o = 13.7. 

For naphthalene (CioH8), there are a number of resonance forms ... 

The regioselectivity of Ar-SE reactions with naphthalene follows from the different stabilities of the Wheland complex intermediate of the 1-attack (Figure 5.10, top) compared with that of the 2-attack (Figure 5.10, bottom). For the Wheland complex with the electrophile at Cl these are five sextet resonance forms. In two of them the aromaticity of one ring is retained. The latter forms are thus considerably more stable than the other three. The Wheland complex with the electrophile at C2 can also be described with five sextet resonance forms. However, only one of them represents an aromatic species. The first Wheland complex is thus more stable than the second. The 1-attack is consequently preferred over the 2-attack. 

Draw pictures of the compounds allene (CHy=C=CH2), naphthalene, and azulene (shown below), representing the 4 double bonds with dumbbellshaped -electrons. Show that the Kekule resonance forms of naphthalene... 

How do you know that all possible resonance forms have been written This is accomplished only by trial and error. If you keep pushing electrons around the naphthalene ring, you will continue to draw structures, but they will be identical to one of the three previously written. 

As was true for benzene with its two equivalent resonance forms, no individual structure is an accurate representation of naphthalene. The true structure of naphthalene is a hybrid of the three resonance forms. 

Stannaaromatic compounds, which contain a tin atom instead of a skeletal carbon atom in aromatic hydrocarbons, have a formal Sn—C double bond in their resonance forms. To date, tin analogs of cy-clopentadienyl anions and dianions, phenanthrene, and naphthalene have been reported. 

The most efficient stabilization of the intermediate carbocation produced by electrophilic attack on naphthalene comes from those resonance forms which retain one fully benzenoid ring. In the case of 1-substitution, two such structures can be drawn, 8 and 9 (plus their Kekule forms). For 2-substitution there is only one structure, 10 (and its Kekule form). The intermediate from attack at the 1-position is the more stable and therefore the 1 -substituted product is favoured. 

It is common practice to use the shorthand notation shown in Fig. 9-34 (a) and (b) to represent benzene and naphthalene, respectively. The circle emphasizes the delocalization of the electrons without the necessity of showing the resonance forms. 

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