
Interpretation:
The observation that the N,N,N-trimethylammonium group –N(CH3)3, though a meta directing deactivator does not exhibit electron withdrawing resonance effect is to be explained.
Concept introduction:
In planar molecules the overlapping of p orbitals on adjacent carbon atoms takes place leading to the formation of different resonance forms. Resonance is not possible if the molecule is not co-planar.
If the group already present on the benzene ring is electron withdrawing by resonance and/or inductive effect it will deactivate the ring towards electrophiles.
The stability of a resonance hybrid depends on the number of stable individual resonance forms that contribute to the hybrid. More the number, the more stable is the hybrid.
To explain:
The observation that the N,N,N-trimethylammonium group –N(CH3)3, though a meta directing deactivator does not exhibit electron withdrawing resonance effect.

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Chapter 16 Solutions
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- Assign all the carbonsarrow_forward9 7 8 C 9 8 200 190 B 5 A -197.72 9 8 7 15 4 3 0: ང་ 200 190 180 147.52 134.98 170 160 150 140 130 120 110 100 90 90 OH 10 4 3 1 2 -143.04 140. 180 170 160 150 140 130 120 110 100 90 CI 3 5 1 2 141.89 140.07 200 190 180 170 160 150 140 130 120 110 100 ៖- 90 129. 126.25 80 70 60 -60 50 40 10 125.19 -129.21 80 70 3.0 20 20 -8 60 50 10 ppm -20 40 128.31 80 80 70 60 50 40 40 -70.27 3.0 20 10 ppm 00˚0-- 77.17 30 20 20 -45.36 10 ppm -0.00 26.48 22.32 ―30.10 ―-0.00arrow_forwardAssign all the carbonsarrow_forward
- C 5 4 3 CI 2 the Righ B A 5 4 3 The Lich. OH 10 4 5 3 1 LOOP- -147.52 T 77.17 -45.36 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm B -126.25 77.03 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm 200 190 180 170 160 150 140 130 120 110 100 90 80 TO LL <-50.00 70 60 50 40 30 20 10 ppm 45.06 30.18 -26.45 22.36 --0.00 45.07 7.5 1.93 2.05 -30.24 -22.36 C A 7 8 5 ° 4 3 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 8 5 4 3 ཡི་ OH 10 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 5 4 3 2 that th 7 I 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 115 2.21 4.00 1.0 ppm 6.96 2.76 5.01 1.0 ppm 6.30 1.00arrow_forwardCurved arrows were used to generate the significant resonance structure and labeled the most significant contribute. What are the errors in these resonance mechanisms. Draw out the correct resonance mechanisms with an brief explanation.arrow_forwardWhat are the: нсе * Moles of Hice while given: a) 10.0 ml 2.7M ? 6) 10.ome 12M ?arrow_forward
- You are asked to use curved arrows to generate the significant resonance structures for the following series of compounds and to label the most significant contributor. Identify the errors that would occur if you do not expand the Lewis structures or double-check the mechanisms. Also provide the correct answers.arrow_forwardhow to get limiting reactant and % yield based off this data Compound Mass 6) Volume(mL Ben zaphone-5008 ne Acetic Acid 1. Sam L 2-propanot 8.00 Benzopin- a col 030445 Benzopin a Colone 0.06743 Results Compound Melting Point (°c) Benzopin acol 172°c - 175.8 °c Benzoping to lone 1797-180.9arrow_forwardAssign ALL signals for the proton and carbon NMR spectra on the following pages.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning

