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Concept explainers
Interpretation:
- The products of the given reaction have to be predicted.
Concept Introduction:
Resonance Contributor: The appropriate structure with the localized electrons is called a resonance contributor, a resonance structure, or a contributing resonance structure.
Delocalized electrons: The sharing of electrons between two or more atoms known as delocalization of electrons. In order to have delocalized electrons, the system must be planar and have alternative double bonds and single bonds.
Resonance hybrid: The actual structure with delocalized electrons is called a resonance hybrid.
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Answer to Problem 57P
Correct Answer:
The compounds A, B, C, E, F, L, M, N, and O have delocalized electrons.
Explanation of Solution
Reason for correct options:
The above compounds all have delocalized electrons, which undergo electron movement. The resonance structures can be drawn as shown above.
Hence, the correct options are A, B, C, E, F, L, M, N, and O.
Reason for in-correct options:
The remaining compounds do not contain delocalized electrons, conjugated system is absent.
Hence, the in-correct options are D, G, H, I, J, and K.
- The compounds containing delocalized electrons were predicted.
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Chapter 8 Solutions
Organic Chemistry
- 1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward
- 6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forwardNonearrow_forwardQ1: Draw the most stable and the least stable Newman projections about the C2-C3 bond for each of the following isomers (A-C). Are the barriers to rotation identical for enantiomers A and B? How about the diastereomers (A versus C or B versus C)? H Br H Br (S) CH3 (R) CH3 H3C (S) H3C H Br Br H A C enantiomers H Br H Br (R) CH3 H3C (R) (S) CH3 H3C H Br Br H B D identicalarrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning
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