Chapter 16 Worksheet 01 Aromatic e 2 6 10 14 Keep adding 4 Anti aromatic 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 one p orbital sp3 atom sp3 sp3 sp3 sp3 NO p orbital sp2 atom sp2 sp2 sp2 p 1 p orbital Example 1 sp3 lacks p orbital A sp sp2 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 H ckel 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 e participating in the overlap Example 2 N lone pair in sp2 hybrid orbital 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 it has 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 e participating in the overlap 3 pi bonds contribute 6 electrons making the compound aromatic www ochemweb com Example 3 H N N is sp2 lone pair in p orbital Initially you might think that the Nitrogen is sp3 If N was sp3 counted as e aromatic No p orbital lone pair has to be in sp3 hybrid orbital cannot be 3 only 4 e plus sp Nitrogen breaking overlap making the compound non 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 e Combining example 2 3 N is sp2 lone pair in p orbital Not counted It is in sp2 hybrid orbital since p orbital is occupied by pi bond Total 6 e N N H Aromatic Example 4 Carbon is sp2 has 1 empty p orbital All the carbons are sp2 having 1 p orbital each participating in the delocalization of 6 e making the compound aromatic www ochemweb com Example 5 Carbon MUST be sp2 due to resonance 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 e making the compound anti aromatic assuming planarity Example 6 in sp2 hybrid Oxygen is sp2 O lone pair in p orbital Oxygen is sp2 instead of sp3 making this a 6 e system Aromatic Example 7 Non aromatic sp3 O If Oxygen is sp2 it would be anti aromatic 8 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 H N 1 S 3 2 N N N H B N N H3C 6 5 4 N CH3 O 8 7 S 10 11 9 S B H 12 www ochemweb com H O N 15 14 13 S NH2 H N O 16 B 17 18 N O H S 20 19 21 B H CH2CH3 23 22 24 O www ochemweb com S N 25 26 27 H CH3 O O 28 N 30 29 N O CH3 O O 31 32 O 33 O N B 34 B N N O 35 36 N N B www ochemweb com