Fill in the boxes with reactants, reagents or products. If enantiomers or diastereomers are made, please draw the stereoisomers.
Answer:-
This is elimination (E-2) reaction of trans-1-bromo-2-methylcyclohexane in the presence of strong base (sodium ethoxide). In this case, hydrogen that is eliminated is not removed from the beta-carbon bonded to the fewest hydrogens. Thus, this elimination does not follow Zaistev's rule. The reason is that the elimination can occur only if the two substituents that are eliminated are in axial positions, the axial positions, the axial hydrogen is removed even though is not bonded to the beta-carbon with the fewest hydrogens.
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Fill in the boxes with reactants, reagents or products. If enantiomers or diastereomers are made, please...
Page 2: Box (5 x 2 marks = 10 marks) Fill in the boxes with reactants, reagents or products. If enantiomers or diastereomers are made, please draw the stereoisomers. 4. PhCH2CH3 PhCOH - 6 NaOE, EtOH
Reactions of Epoxides Fill in the boxes with the appropriate reactants and products to complete the reaction. All reagents are used in excess to assure high yields in respect to the reactant. Indicate stereochemistry using wedges and dashes if applicable. If you get more than one product (c-g. two enantiomers as racemic mixture), draw both polar, agrotic b) pc㎞.peutic c) e)
2. Fill in the missing reactants, reagents & conditions, or products. Don't worry about stereochemistry, unless indicated. 1. LDA, -78c OCH then H3o 0 CO Et H3o C10H100 CH NaOEt (aq) 0 0 NaOH (aq) NaOEt (aq) 0
Q1- Fill in the boxes with the product of the following reactions. Draw only the PREDOMİNANT REGIOISOMER. When a racemic mixture is formed mark the chiral center with an ASTERICK () and write RACEMİC in the box. 1 Eq LDA 0 1 Eq NaOH 1) 0.5 Eq LDA1) Cat NaOH 2) H30 јнзо-, Heat , NaOEt, EtOH 1. NaOEt EtOH Br 1) NaOH, H2O 2) H3O 3) Heat 1) (CH3)zMgBr 2) H30 2 a2- Fill in the boxes with the...
4. Fill-in-the-blanks: draw the missing reactants or products involved in the following single or multi-step reactions. Indicate stereochemistry and draw multiple stereoisomers where applicable. i EtOH EtOH [H2SO4] 1) EtMgBr 2) H20 5. For most ketones, hydrate formation is unfavorable, because the equilibrium favors the ketone rather than the hydrate. However, the equilibrium for hydration of hexafluoroacetone favors the hydrate. Propose a mechanism for this transformation below and provide an explanation for this observation. H2O Но он F3CX CF3 F3C...
Fill in the boxes below with the correct structures of enolate reaction products. NaOEt EtOH 1. NaOH 0 3. heat Michael addition NaOH NaOH heat H+ Crossed Aldol 3 NaOH 1. NaOH CH3 3 Br2 Haloform Fill in the boxes below with the correct structures of enolate reaction products. NaOEt EtOH 1. NaOH 0 3. heat Michael addition NaOH NaOH heat H+ Crossed Aldol 3 NaOH 1. NaOH CH3 3 Br2 Haloform
(10 points) Fill in the boxes with the missing starting material, intermediate products, reagents, and/or products. Then for each reaction draw the reaction mechanism with correct electron flow arrows if appropriate or intermediates and correct formal charges. -OH -NH2 NH (10 points) Use the Gabrielle synthesis to make your starting amine below. Then use this starting amine to synthesize the secondary amine using reductive amination. Draw the mechanisms for both reactions below. Be sure to draw correct intermediates, electron flow...
Draw BOTH products and label them as either enantiomers or diastereomers:
NAME! Please Print! 4. . Fill in the missing reactants, reagents & conditions, or products in the following road map. Show stereochemistry where appropriate. 10 points. CH3 AICI: CH₃ (но CH3 AICI: CH,CI NO2 Brmg then H,07
ochem puzzle 2 Puzzle 2 - Fill in all missing reagents and products in the boxes. Ph Br NBS (PhCO2)2 OH Ph OH DMP 1. CH_MgBr 2. H2O