100%(2)2 out of 2 people found this document helpful
This preview shows pages 3–6. Sign up to view the full content.
1.ring is planar 2.all ring atoms are sp2-hybridized (the lone pair on C is in a p-orbital, the H is in an sp2-hybridized orbital) 3.4 πelectrons (4 from πbonds, 2 from O lone pair) = 4n + 2, n = ½ 4 πelectrons = 4n n = 1 Antiaromatic (1 point)
has intentionally blurred sections.
Sign up to view the full version.
Midterm Answers 4 of 115.Circle the most basic atom in imidazole and explain your choice. (5 points)NNHlocatedinap-orbitaland inresonancewiththeπbonds(this is an aromatic ring).located in an sp2-orbitalat90°totheπsystem. Notinvolvedinresonancethereforemoreavailabletoreactwithaproton.111116.The pKa of the indicated proton on the sp3-hybridized carbon in cyclopentadiene is 15, while the pKaof the proton on the allylic sp3-hybridized carbon is 43. Explain why is one proton so much more acidic than the other. (5 points)1HHpKa= 40pKa= 15:BH:BCHH2Cstabilized by resonancewithonlyonedoublebondstabilized by resonancewithbothdoublebondsThis is an aromaticanionAromaticy confers additional stability to the anionMore stable base→stronger conjugate acid11 point bonus for recognizing that this is an aromatic anion11H1