Aromaticity Guide Sheet

anti-aromatic compounds unstable. are - cannotform they resonance structures More Aromaticity Why? properly stable structures: more resonance l *I I benzene the #I #y toluene ortho-Xylene IR 2000cm-1 - > overtones 1450-1250cm Rules for ① All resonance ② Find or lone electron pair structures atom ① end must be sources, which electrons ③ more your electron other bonds Resonance structures: sources be the double resonance structure, bond, a spe a so or but the · hybridized no election can have · interruption. spy carbon be an spa or SP a · must · no resonance ④ p contributing The orily the one the :e bond countI double, passing through nitrogen 10 cnitrogenwasone sinceitsonly I doneare [F,T,yy, Aromatic Answer: · or Anti-Aromatic? (18 electrons) Aromatic or more same in the atoms Conjugated Aromatic compounds available on to of i delocalized where n 0, 1,2,3.... = the have electrons Hockel's same rules for an Aromatic compound HOWEVER... ↳ the number electrons is delocalized of 42, where n 0, 1, = conjugated · 4(4) + 2 = pi system 18 elections Heternators (other Ift he double ONLY atoms than bond If heteroatoms bond have no Passing through them, elections then passing them, elections. double count election CorH( the heternators have double bond count one pair π 2,3... follows Hickel's rule · Rule of electrons the number Anti-Aromaticity follows must every single atom. number 4H+2, place without system orbital refers Example stable the structure w/ all Coplanar ③ ADelicalized a typically atoms contributing in one contributing move are structure:All rings is Answer: Aromatic rates on uncarbonishoneti ② compounds Aromatic arranged - #Ox aromaticrematic you'll the anti-aromatic? , Example: in a cyclic compoundneed ① cyclic xy,,z 0yy elections bond is not Aromaticity Huckel's Rule ring or planar Sp-linear/bent carbon-carbon Answer: Anti-aromatic how many elections? Answer:Let the double bond does NOT affectthe # of pielections Spetrigonal stretching vibrations as Example # of #,aromatic either is elections towards MUST stretching bands A combination - well - Example both or bond the of as correct Lewis structures are 3.5.8.5 NMR proton Sp3-tetrahedral 3000-31000m"-Csp" - in Hybridization key Information spectroscopy for aromatic bonds 1700 between ppon ranges and typically peaks appears sheet Guide compound NMR Tips: aromatic π