Organic Chemistry II - Midterm 2008

1 ring is planar 2 all ring atoms are sp 2 hybridized

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1. ring is planar 2. all ring atoms are sp 2 -hybridized (the lone pair on C is in a p-orbital, the H is in an sp 2 -hybridized orbital) 3. 4 π electrons (4 from π bonds, 2 from O lone pair) = 4n + 2, n = ½ 4 π electrons = 4n n = 1 Antiaromatic (1 point)
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Midterm Answers 4 of 11 5. Circle the most basic atom in imidazole and explain your choice. (5 points) N N H l o c a t e d i n a p - o r b i t a l and i n r e s o n a n c e w i t h t h e π b o n d s (this is an aromatic ring). located in an s p 2 - o r b i t a l a t 9 0 ° t o t h e π s y s t e m . N o t i n v o l v e d i n r e s o n a n c e t h e r e f o r e m o r e a v a i l a b l e t o r e a c t w i t h a p r o t o n . 1 1 1 1 1 6. The pK a of the indicated proton on the sp 3 -hybridized carbon in cyclopentadiene is 15, while the pK a of the proton on the allylic sp 3 -hybridized carbon is 43. Explain why is one proton so much more acidic than the other. (5 points) 1 H H pK a = 40 pK a = 15 :B H :B CH H 2 C stabilized by r e s o n a n c e w i t h o n l y o n e d o u b l e b o n d stabilized by r e s o n a n c e w i t h b o t h d o u b l e b o n d s This is an a r o m a t i c anion Aromaticy confers additional stability to the anion More stable base stronger conjugate acid 1 1 point bonus for recognizing that this is an aromatic anion 1 1 H 1
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