Complete ring bromination

Treating imidazole with bromine in a NaOH solution afforded the complete ring bromination product, 2,4,5-tribromoimidazole. 4-Bromoimidazole was then produced in 62% yield upon selective reductive debromination [5]. The same procedure converted 2-methylimidazole and 1-methylimidazole to 4(5)-iodo-2-methylimidazole [6] and 4(5)-bromo-l-methylimidazole [7], respectively. 

The other bromine atom comes from another bromine-containing molecule or ion. This is clearly not a problem in reactions with benzylic species since the benzene ring is not prone to such addition reactions. If the concentration is sufficiently low, there is a low probability that the proper species will be in the vicinity once the intermediate forms. The intermediate in either case reverts to the initial species and the allylic substitution competes successfully. If this is true, it should be possible to brominate an alkene in the allylic position without competition from addition, even in the absence of NBS or a similar compound, if a very low concentration of bromine is used and if the HBr is removed as it is formed so that it is not available to complete the addition step. This has indeed been demonstrated. ... 

Replacement of the fluorines by bromines,4 5 as in the procedure described here, resulted in several major improvements that made the reaction synthetically useful (a) ring (retaliation is completely suppressed, (b) the starting material is either commercially available, as in the present example, or much more easily synthesized than corresponding fluorobromo analogues. 

In the light of the more complete study of ring-halogenated triphenylchloro-methanes in this paper, the free radical hypothesis was back - if it ever was excluded in the previous paper - in the final discussion of the constitution of triphenylmethyl , now with two tautomeric triphenylmethyl radical structures in equilibrium with each other and the Jacobson dimer 1 (Scheme 2). Note that the radical was symbolized by an open valence (a thick line is used here for clarity). The strong results obtained with 3 (Scheme 1) were explained by removal of the quinoid bromine atom from 4 giving a radical 6 which tautomerized to the triphenylmethyl analogue 7. By analogy with the... 

Aminomethylation of cyclododecanone (79) gave 86a which was converted to 1,2-dimethylenecyclododecane (87) via the Wittig condensation, quaternization, and Hofmann elimination. Bromination of 87 to the ( )-dibromide 88 (98 % yield) was the key step in this approach. The routine synthetic sequence modified the side chains of 88 to provide the ( )-dialdehyde 88b whose McMurry ring closure, followed by removal of the newly formed unsaturated center, completed the synthesis. 

This product is then treated with acryloyl chloride. The initial step in this case probably involves the acylation of nitrogen on the enamine conjugate addition then completes the formation of the lactam ring (23-5). Treatment of that product with triethyl silane then reduces the ring unsaturation and cleaves the benzylic nitrogen bond on the auxiliary to yield (23-6) as the optically pure tmns isomer. Displacement of bromine with the mercapto benzthiazole (23-7) completes the synthesis of izonsteride (23-8) [25]. 

The resonance energies of fused systems increase as the number of principal canonical forms increases, as predicted by rule 6 (p. 35).75 Thus, for benzene, naphthalene, anthracene, and phenanthrene, for which we can draw, respectively, two, three, four, and five principal canonical forms, the resonance energies are, respectively, 36, 61, 84, and 92 kcal/mol (152, 255, 351, and 385 kJ/mol), calculated from heat-of-combustion data.76 Note that when phenanthrene, which has a total resonance energy of 92 kcal/mol (385 kJ/mol), loses the 9,10 bond by attack of a reagent such as ozone or bromine, two complete benzene rings remain, each with 36 kcal/mol (152 kJ/mol) that would be lost if benzene was similarly attacked. The fact that anthracene undergoes many reactions across the 9,10 positions can... 

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