Question

1. Draw Frost circle diagrams for each compound shown. Assume that all atoms in the ring are sp2 hybridized. Populate the pi electrons in the corresponding MO's. Identify the compound as aromatic or antiaromatic. (5 pts) 2. What is the Hückel number of pi electrons for the molecule shown below? How many pi electron pairs does it have? Would you expect this molecule to be aromatic, antiaromatic or non aromatic? Explain. (3 pts)

Answer 1

CS DATE: PAGE Frost Diagroom Sol 1 ARMO Tefal mo.of selections=2 1L Bмо . - Bro e Bondine molecular orbital ARMO = Antibonding Molecular orbital NB Nonbonding Molecular eskital. >> | Conditions form_exomaticity O conjugation o Planarity * Huckel Rule t 40+2 xe) where =0,1,2.... p o 0+ Conjugation is present. . Dois blanan 460) + 2 = 2 x electrons - 4n+2 - for no Il e As the mole ale satisfying all the three Conditions hence it is Ogrometi'. GOOD WRITE

DATE E - PAGE o conditions for _Antiaromatisity-- ao Conjugation @_--Baplananity 1 4n a electrons where n = o ... The compound is showing conjugation → The compound is planar. I → 461 x electrons nule has been followed for mali Hence this comporind is arti'aromatic Peso Frost Diagrama iliABMO - total no. of electron present in tine

PAGE – x no. of -electrons are Huckel Av on 14 bona 463) + 2 = 14 Pi- electrons pairs = 7 Molecule shows conjugation. Molealle is some planar due to of spa Nitrogen and which is in blane trenta a Also the lo present on Nitrogen is perpendicular to the plane hence not disturbing planarity. Oxygen lone pair (1p is in conjugation Wence in plane. JO So, the molecule is Aromatic