Electrophilic aromatic substitutions proceed in two stages. 1. Attack of the aromatic ring on the electrophile. 2. Regeneration of the stable aromatic system by loss of a positively charged unit, usually a proton. Br Br₂/FeBr3 Draw one resonance structure for the intermediate formed when the electrophile is attacked by benzene in the reaction above.

Electrophilic aromatic substitutions proceed in two stages. 1. Attack of the aromatic ring on the electrophile. 2. Regeneration of the stable aromatic system by loss of a positively charged unit, usually a proton. Br Br₂/FeBr3 Draw one resonance structure for the intermediate formed when the electrophile is attacked by benzene in the reaction above.

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter9: Nucleophilic Substitution And Β-elimination

Section: Chapter Questions

Problem 9.54P

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Electrophilic aromatic substitutions proceed in two stages. 1. Attack of the aromatic ring on the electrophile. 2. Regeneration of the stable aromatic system by loss of a positively charged unit, usually a proton. Br Br₂/FeBr3 Draw one resonance structure for the intermediate formed when the electrophile is attacked by benzene in the reaction above. • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • Do not include lone pairs in your answer. They will not be considered in the grading. (+) ChemDoodleⓇ
A benzene ring alters the reactivity of a neighboring group in the so-called “benzylic” position, similarly to how a double bond alters the reactivity of groups in the “allylic” position. Benzylic cations, anions, and radicals are all more stable than simple alkyl intermediates. a) Use resonance structures to show the delocalization of the positive charge, negative charge, and unpaired electron of the benzyl cation, anion, and radical.
Fill the blank space. Compounds containing a phenol group may work as ANTIOXIDANTS to prevent free radical damage. This is accomplished when a free radical (or UV light) encounters a phenol group, turning the phenol group into a radical. However, contrary to typical radical behavior, the structure of the phenol radical can neutralize (or quench) the unpaired electron. Specifically, the phenol structure neutralizes (or quenches) the unpaired radical electron by doing the following: taking the electron and ---------. The correct name (or abbreviation) of an example compound containing a phenol group with antioxidant properties is: ---------.
Write the mechanism leading to the product(s) for the acid-base reactions below. Label the acid, base, nucleophile, and electrophile. Where possible, label the conjugate base and the conjugate acid.
All electrophilic aromatic substitution reactions occur via a two-step mechanism: addition of the electrophile to form a resonance-stabilized carbocation, followed by deprotonation by a base. In the step shown below, loss of a proton to form the substitution product was drawn using the resonance structure with the + 1 formal charge at the ortho position to the carbon bonded to the electrophile. Redraw the step shown with the other two resonance structures and use curved arrows to show how these other two resonance structures can be converted to the substitution product by removal of a proton with the base shown. Part: 0/2 Part 1 of 2 H :CI-AICI, Draw the mechanism that generates the final substitution product with the +1 formal charge at the para position. Cl HC1 AICI3
Which among the five aromatic molecules will have the least ability to undergo molecularstacking?A. Structure 1B. Structure 2C. Structure 3D. Structure Which aromatic structures contains functional groups which are electron-withdrawing?A. Structures 1, 2 and 4B. Structures 2, 3 and 5C. Structures 2, 4, and 5D. Structures 1, 3, and 4 Which aromatic structures contain functional groups which are electron-donating?A. Structures 1 and 3B. Structures 1 and 4C. Structures 3 and 5D. Structures 3 and 4 In which structure do you expect two products after alkylation?A. Structure 1B. Structure 2C. Structure 3D. Structure 4 In which carbon position will an incoming halogen be located in Structure 3, relative tothe OH group?A. Ortho positionB. Meta positionC. Para positionD. Ortho and para position In which aromatic molecule do you expect to have a carbocation form during resonancestabilization?A. Structures 2, 3 and 4B. Structures 2, 4 and 5C. Structures 2, 3 and 5D. Structures 1, 3 and 4
Compounds containing a phenol group may work as ANTIOXIDANTS to prevent free radical damage. This is accomplished when a free radical (or UV light) encounters a phenol group, turning the phenol group into a radical. However, contrary to typical radical behavior, the structure of the phenol radical can neutralize (or quench) the unpaired electron. Specifically, the phenol structure neutralizes (or quenches) the unpaired radical electron by doing the following: taking the electron and The correct name (or abbreviation) of an example compound (discussed in the lecture videos) containing a phenol group with antioxidant properties is:
As we will learn in Section 13.12, many antioxidants–compounds that prevent unwanted radical oxidation reactions from occurring–are phenols, compounds that contain an OH group bonded directly to a benzene ring. a. Explain why homolysis of the O–H bond in phenol requires considerably less energy than homolysis of the O–H bond in ethanol (362 kJ/mol vs. 438 kJ/mol). b.Why is the C–O bond in phenol shorter than the C–O bond in ethanol?
N,N-diethyl-m-toluamide (DEET) is the active ingredient in many insect repellent preparations. Following is one of the steps in its synthesis. In the box below draw the structure of the product of this reaction. H3C MgBr 1. CO₂ 2. H₂O* product • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • Do not include lone pairs in your answer. They will not be considered in the grading. • Draw the Grignard reagent as a covalent magnesium bromide.
Fill in the table with the partial charges of the atoms. The atoms are numbered with the following scheme with respect to the substituent on the aromatic ring. x c/ 3. | H./ Atom Chlorobenzene Nitrobenzene Aminobenzene C2 Ca Which substituent results in a ring atom with the greatest positive charge (smallest negative charges)? What is the mechanism that causes this pattern of electron distribution on the aromatic ring? Diagram the mechanism. Which substituent results in a ring atom with the least positive charge (largest negative charges)? What is the mechanism that causes this pattern of electron distribution on the aromatic ring? Diagram the mechanism.
Experiment 8: Electrophilic Aromatic Substitutions 1. There are 2 possible ways for CuBr2 to release Br2: i) CuBr2 → CuBr + 1/2 Br2 or ii) CuBr2 → Cu + Br2Look up the colors for CuBr2, CuBr, and Cu. explain which chemical equation best describes how Br2 is formed in this experiment. Note: during the experiment our mixture turned light blue/green.
As we will learn in Section 15.12, many antioxidants-compounds that prevent unwanted radical oxidation reactions from occurring-are phenols, compounds that contain an OH group bonded directly to a benzene ring. a. Explain why homolysis of the O-H bond in phenol requires considerably less energy than homolysis of the O-H bond in ethanol (362 kJ/mol vs. 438 kJ/mol). b. Why is the C-O bond in phenol shorter than the C-O bond in ethanol? -O-H CH,CH2-0-H phenol ethanol