Question

Circle the correct label for each structure. Assume planarity for all structures. Aromatic Aromatic Aromatic Aromatic Aromatic Nonaromatic Nonaromatic Nonaromatic Nonaromatic Nonaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic a o los Aromatic Aromatic Aromatic Aromatic Aromatic Nonaromatic Nonaromatic Nonaromatic Nonaromatic Nonaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic

Answer 1

V v 4n+2 = 6., here, n=1 Aromatic Aromatie. The lone pairs of N are not involved in delocalisation : 4n+2=6 ; 2=1. Sp3 = Non aromatic. Electrons are not delocalised Spa through out the ring. A Aromatic. 4n+2 = 2; n=0 Aromatic. 44+2=61. nal. Non-aromatie. Electront are not delocalized through out the things 2.: 0 ° : Aromatic. 4n+2=6 ;re) one e pair of oxygen involves in delocalisation and other one doesn't ence steneturres! t ole erisoon.. PH 0.0

Antiaromatic. 4n = 4;n=1 romatici 4n+2 =10;n=2. Tedemence structu Sare 480 on, .. in aliomalie. In - asa - 3 8 1 " Aromatici smo 10,032. 8-- -- - Rewo, the structure is similar Aromatic. 4n+2= 2;nso. mesomewe atuwekener SÅ ... Now, this structure is similar. to cyclopropenye catim. 49 + 22 Tesonence A

Note: According to, Huckel rule we've concluded:

If the no. of delocalizable electrons are 4n+2 then the compound is aromatic and if no. of delocalizable electrons are 4n then the compound is anti-aromatic , where n= 0,1,2,3,...

Otherwise, it is non-aromatic.