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- Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CH3 CH3 CH3COCI/ AICI3 H₂C. You do not have to consider stereochemistry. Include all valence lone pairs in your answer. • In cases where there is more than one answer, just draw one.Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CH3O + You do not have to consider stereochemistry. • Include all valence lone pairs in your answer. • In cases where there is more than one answer, just draw one. ? [F H3PO4 CH3O OHDraw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. LOCH3 LOCH3 HNO3 / CH3CO₂H O₂N • You do not have to consider stereochemistry. • Include all valence lone pairs in your answer. ● In cases where there is more than one answer, just draw one.
- Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CH3 CH3 CH3CI / AICI3 • H3C • You do not have to consider stereochemistry. Include all valence lone pairs in your answer. In cases where there is more than one answer, just draw one.Draw the least stable resonance form for the intermediate in the following electrophilic substitution reaction. CEN Br₂/FeBra • You do not have to consider stereochemistry. Include all valence lone pairs in your answer. In cases where there is more than one answer, just draw one. Br CEN IConsider this reaction: Br CH;OH Br-Br H3CO The mechanism proceeds through a first cationic intermediate, intermediate 1. Nucleophilic attack leads to intermediate 2, which goes on to form the final product. In cases that involve a negatively charged nucleophile, the attack of the nucleophile leads directly to the product. Br + CH3OH Br Intermediate 1 Intermediate 2 (product) In a similar fashion, draw intermediate 1 and intermediate 2 (final product) for the following reaction. OH + Br2 + HBr Br racemic mixture
- Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. H N. Br2 Br • You do not have to consider stereochemistry. • Include all valence lone pairs in your answer. • In cases where there is more than one answer, just draw one. IZDraw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CH3 Br2 / FeBr3 • You do not have to consider stereochemistry. Include all valence lone pairs in your answer. • In cases where there is more than one answer, just draw one. OO. #[ ] در ChemDoodle CH3 BrDraw the least stable resonance form for the intermediate in the following electrophilic substitution reaction. NO₂ C Br₂ / FeBr3 Br • You do not have to consider stereochemistry. Include all valence lone pairs in your answer. • In cases where there is more than one answer, just draw one. NO₂
- Draw the structure(s) of the major organic product(s) of the following reaction. + N₂ CI CuCN KCN Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. • You do not have to consider stereochemistry. • • Separate multiple products using the + sign from the drop-down menu. Cn ? ChemDoodleⓇ √n [FConsider this reaction: Br CH3OH Br-Br H3CO The mechanism proceeds through a first cationic intermediate, intermediate 1. Nucleophilic attack leads to intermediate 2, which goes on to form the final product. In cases that involve a negatively charged nucleophile, the attack of the nucleophile leads directly to the product. H. Br + CH;OH Br Intermediate 2 (product) Intermediate 1 In a similar fashion, draw intermediate 1 and intermediate 2 (final product) for the following reaction. OH + Br2 + HBr Br racemic mixtureDraw a structural formula for the substitution product of the reaction shown below. H3C H Br Na' acetone • Use the wedge/hash bond tools to indicate stereochemistry where it exists. • If more than one stereoisomer of product is formed, draw both. Separate multiple products using the + sign from the drop-down menu. • Products that are initially formed as ions should be drawn in their neutral forms. C P opy aste [F ChemDoodle