NH A B

Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter6: Solutions And Colloids
Section: Chapter Questions
Problem 6.82P: 6-82 (Chemical Connections 6C) A solution contains 54 mEq/L ofCI- and 12 mEq/L of HCO3-. If Na+ is...
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Explain why A is aromatic but B is not aromatic (see attached file) 

NH
A
B
Transcribed Image Text:NH A B
Expert Solution
Step 1

Given:

Ring A: 9 membered unsaturated ring with 1 unsubstituted nitrogen atom

Ring B: 9 membered unsaturated ring with 1 acyl substituted nitrogen atom

 

Step 2: Molecule A

Molecule A has 4 double bonds and one nitrogen atom with 1 lone pair, which corresponds to total  2 X4 + 2 = 10 π electrons.

For a molecule to be aromatic, it should obey Huckel's rule of aromaticity.

4n + 2 π electrons, where n is an integer.4n + 2 = 10        4n =10-2        4n  = 8          n   = 84 = 2

 

Hence, Molecule A is aromatic.

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