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EBK GET READY FOR ORGANIC CHEMISTRY
2nd Edition
ISBN: 8220100576379
Author: KARTY
Publisher: PEARSON
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Question
Chapter 14, Problem 14.53P
Interpretation Introduction
Interpretation:
Why the given molecule has quite a large dipole, as indicated in its electrostatic potential map is to be explained.
Concept introduction:
If two unsaturated rings are joined by a conjugated double bond, the electrons of this double bond can move to one of the atoms if it leads to both rings gaining an
These charges, one positive and one negative, are delocalized on the two rings in such a manner as to make both rings aromatic. The charge separation then makes the molecule polar. Due to polarization and delocalization of the electrons over the two rings, the molecule has quite a large dipole moment.
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Chapter 14 Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
Ch. 14 - Prob. 14.1PCh. 14 - Prob. 14.2PCh. 14 - Prob. 14.3PCh. 14 - Prob. 14.4PCh. 14 - Prob. 14.5PCh. 14 - Prob. 14.6PCh. 14 - Prob. 14.7PCh. 14 - Prob. 14.8PCh. 14 - Prob. 14.9PCh. 14 - Prob. 14.10P
Ch. 14 - Prob. 14.11PCh. 14 - Prob. 14.12PCh. 14 - Prob. 14.13PCh. 14 - Prob. 14.14PCh. 14 - Prob. 14.15PCh. 14 - Prob. 14.16PCh. 14 - Prob. 14.17PCh. 14 - Prob. 14.18PCh. 14 - Prob. 14.19PCh. 14 - Prob. 14.20PCh. 14 - Prob. 14.21PCh. 14 - Prob. 14.22PCh. 14 - Prob. 14.23PCh. 14 - Prob. 14.24PCh. 14 - Prob. 14.25PCh. 14 - Prob. 14.26PCh. 14 - Prob. 14.27PCh. 14 - Prob. 14.28PCh. 14 - Prob. 14.29PCh. 14 - Prob. 14.30PCh. 14 - Prob. 14.31PCh. 14 - Prob. 14.32PCh. 14 - Prob. 14.33PCh. 14 - Prob. 14.34PCh. 14 - Prob. 14.35PCh. 14 - Prob. 14.36PCh. 14 - Prob. 14.37PCh. 14 - Prob. 14.38PCh. 14 - Prob. 14.39PCh. 14 - Prob. 14.40PCh. 14 - Prob. 14.41PCh. 14 - Prob. 14.42PCh. 14 - Prob. 14.43PCh. 14 - Prob. 14.44PCh. 14 - Prob. 14.45PCh. 14 - Prob. 14.46PCh. 14 - Prob. 14.47PCh. 14 - Prob. 14.48PCh. 14 - Prob. 14.49PCh. 14 - Prob. 14.50PCh. 14 - Prob. 14.51PCh. 14 - Prob. 14.52PCh. 14 - Prob. 14.53PCh. 14 - Prob. 14.54PCh. 14 - Prob. 14.55PCh. 14 - Prob. 14.56PCh. 14 - Prob. 14.57PCh. 14 - Prob. 14.58PCh. 14 - Prob. 14.59PCh. 14 - Prob. 14.60PCh. 14 - Prob. 14.61PCh. 14 - Prob. 14.62PCh. 14 - Prob. 14.63PCh. 14 - Prob. 14.64PCh. 14 - Prob. 14.65PCh. 14 - Prob. 14.66PCh. 14 - Prob. 14.1YTCh. 14 - Prob. 14.2YTCh. 14 - Prob. 14.3YTCh. 14 - Prob. 14.4YTCh. 14 - Prob. 14.5YTCh. 14 - Prob. 14.6YTCh. 14 - Prob. 14.7YTCh. 14 - Prob. 14.8YTCh. 14 - Prob. 14.9YTCh. 14 - Prob. 14.10YTCh. 14 - Prob. 14.11YTCh. 14 - Prob. 14.12YTCh. 14 - Prob. 14.13YT
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- Among the resonance structures that occur in the structure below, please tell us a resonance structure that does not actually contribute to a resonance hybrid. I want the reason to be detailed.arrow_forwardWhich of the following species (B, C,D) is a valid resonance of A? Use curved arrows to show how A is converted to any valid resonance structure.arrow_forwardRefer to the structure of imidazolium ion in the preceding equation and write a second resonance contributor that obeys the octet rule and has its positive charge on the other nitrogen. Use curved arrows to show how you reorganized the electrons.arrow_forward
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- a) Draw the formal charges to the molecule below. Label atleast two functional groups found in this molecule. Identify and circle the π bonds. b) Draw a curly arrow notation to show how the electrons are redistributed to show a new resonance structure in which the formal charge has moved to a different heteroatom.arrow_forwardDraw all the reasonable resonance structures for the following which delocalizes the negative charge on different atoms. You must use electron pushing arrows notation to show the conversion of a resonance structure into another for full credit.arrow_forwardDraw the major resonance structure for the compound shown; include lone pairs of electrons, formal charges, and condensed hydrogen atoms (located in the More menu). Then draw curved arrows to show how this can be converted to the Lewis structure givenarrow_forward
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