Chapter 16 Worksheet 01 Aromatic 11 e 2 6 10 14 Keep adding 4 Anti aromatic 11 e 4 8 12 Multiples of 4 An Sp3 atom will have four Sp3 hybrid orbitals and NONE of p orbitals An Sp2 atom will have three Sp2 hybrid orbitals and Q J g p orbital NO P orbital 1 P orbital Example 1 f Sp3 lacks p orbital 0h A sp 2 I sp 2 Sp2 Sp2 To qualify as aromatic or anti aromatic a cyclic compound must have a continuous ring of overlapping p orbitals and must be planar for proper overlap In the above example all the carbons are Sp2 having 1 p orbital each except carbon A which is Sp3 lacking p orbital Since carbon A lacks a p orbital continuous overlap of rest of the p orbitals in a cyclic fashion is not possible making the compound non aromatic When to count the lone pair to determine aromaticity To apply Huckel s rule only the electrons in the p orbitals are counted If the lone pair is in Sp2 or Sp3 hybrid orbital it cannot be counted as 11 e participating in the overlap Example 2 lone pa r n sp hybr d orb tal All the atoms including Nitrogen are Sp2 hybridized having 3 Sp2 hybrid and 1 p orbital each Nitrogen has to use the only p orbital ithas to form pi bond double bond with the adjacent carbon Now that the only p orbital is occupied the lone pair has no choice but to be in 1 of the 3 Sp2 hybrid orbital and thus cannot be counted as 11 e participating in the overlap 3 pi bonds contribute 6 electrons making the compound aromatic www ochemweb com Example 3 N is Sp2 lone pair in p orbital o N Initially you might think that the Nitrogen is Sp3 If N was Sp3 No p orbital lone pair has to be in Sp3 hybrid orbital l cannot be counted as n e only 4 n e plus Sp3 Nitrogen breaking overlap making the compound non aromatic The compound will try to be aromatic to gain the stability by hybridizing the Nitrogen as Sp2 instead of Sp3 By adopting Sp2 hybridization the nitrogen can place its lone pair into a p orbital This p orbital can overlap with the p orbitals from the neighboring Sp2 hybridized carbons in order to join into a larger molecular orbital making it aromatic 6 n e Combining example 2 3 Not counted It is in Sp2 hybrid orbital since p orbital is occupied by pi bond N r N is Sp2 lone pair in p orbital N H l Total 6 n e Aromatic Example 4 Carbon is Sp2 has 1 empty p orbital o All the carbons are Sp2 having 1 p orbital each participating in the delocalization of 6 n e making the compound aromatic www ochemweb com Example 5 8 Carbon MUST be Sp2 due to resonance o Sp2 When the resonance contributors have two different hybridizations the overall hybridization has to be the lower one Sp2 in this case The lone pair is in p orbital counting towards 8 IT e making the compound anti aromatic assuming planarity Example 6 in Sp2 hybrid U Oxygen is Sp2 0 lone pair in p orbital Oxygen is Sp2 instead of Sp3 making this a 6 IT e system Aromatic Example 7 Non aromatic If Oxygen is Sp2 it would be anti aromatic 8 IT e and less stable Would prefer to be non aromatic instead by remaining Sp3 www ochemweb com Classify the following compounds as aromatic A antiaromatic AA or nonaromatic NA Assume all as planar 1 0 Tie 4 p V O iVl 10 ml4st 11 b Sp z 0 rc OV i c D I II r loDf d 1 12 M rev 0 Tie www ochemweb com MA J 0 v H O 13 t C Y r f 2 h i cA OH Q e f 1 14 r o 4 ne 15 1 1 r 5 is sf 2 h Jbr tl NiSSr l I 10 iit 6J b 11 e 9 A lOY JtJJ ft o H I 16 NH2 sp L Cc p f 17 hqS IY dtlp pJek f 1 ifeI 1 b 80 0 oc 1 6J 2 ie 18 C I I C fr I I 19 L I V f IJ i N i f oYhilol I IV b 20 c1 I Sf b71e Yl S r 2 AUC gf1e 21 Sfl r rr DYbH 1 ht J I trt 1 e J CP I E0J 6EJ 23 S LOLtt bCC t J by I 1 1 f r l r ci 6J D t Sf 2 1 r N S TNHI f2 I 24 I ll J r TY I y u www ochemweb com c 25 if boft S rJ r sf l the VI AAC rre N b4 e H W d e f l c W VI fl p I 70 t1IVo APt 26 be j 0 28 29 0 o iI 31 I 1 f3 S M exfJOt 1 J Olr I Sf h h not l f 32 U 1 1 f 6f c lAJ OYbl 0 33 it 4 II e 0 www ochemweb com