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- Curved arrows are used to illustrate the flow of electrons. Follow the arrows and draw the intermediate and product in this reaction. Include all lone pairs. Ignore stereochemistry. Ignore inorganic byproducts. H-O :OCH₂ CH₂OH₂* :0: protonation Draw Intermediate CH3OH₂+ protonation CH3OH deprotonation loss of H₂O elimination 1 H₂C CH₂OH nucleophilic addition O-H Draw Intermediate CH3OH n deprotonatio Draw ProductDraw a structural formula for the major product of the reaction shown. -CH₂CH3 Br₂ H₂O • Show product stereochemistry IF the reactant alkene has both carbons of the double bond within a ring. • Do not show stereochemistry in other cases. • If the reaction produces a racemic mixture, just draw one stereoisomer.3) Draw the structure of the major product that results when cyclohexane carbaldehyde is treated with each of the reagents shown. Write NR if no reaction is expected. CH3NH2 (CH3)2NH N(CH3)3 O H cyclohexane carbaldehyde C 1) CH3MgBr 2) H3O+ NaBH4, EtOH H2₂O, NaOH H₂SO4, EtOH (excess)
- Draw the alkene that would react with the reagent given to account for the product formed. ? + HCI My 3 You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one. CH3 CH₂ CHOCH3 TT CI CH3 L ▼ {n [F ? ChemDoodleⓇAcid catalyzed dehydration reaction of 2-methyl-1-butanol produces 2-methyl-2-butene as the major product. Also acid catalyzed dehydration reaction of 3-methyl-1-butanol give the same product as major product. Explain the reason why both of the reaction produce the same product as the major product.Draw the alkene that would react with the reagent given to account for the product formed. ? + + H₂O **** H₂S04 • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one. CH3 CH3 CHCCH3 | | OH CH3 +1
- Draw all stereoisomers formed in the reaction shown. Use wedges and dashes for tetrahedral stereogenic centers, if applicable. OH CH₂CH₂OH HClWhen carbonyl compounds are reduced with a reagent such as LiAlH4 or NaBH4 and a new stereogenic center is formed, what will the composition of the product mixture be? Forms more of one enantiomer than another because of steric reasons around the carbonyl Forms more of one enantiomer than another depending on the temperature of the reaction Forms different products depending on the solvent used Forms a racemic mixture of the two possible enantiomersDraw the alkene that would react with the reagent given to account for the product formed. CH3 HCI CH3 CHCCH, ? + Či CH3 You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one.
- The pictured reaction shows an alkyl bromide being converted into an alkene. Choose all reagents that would produce the pictured alkene as the major product. A) NaOH/H2O B) H2O C) tBuOK/tBuOH D) EtONa/EtOHDraw the structure for an alkene that gives the following reaction product. 1. Hg(OАc)2, Н2О ? -CHCH3 2. NABH4 ОН Ignore alkene stereochemistry. You do not have to explicitly draw H atoms. In cases where there is more than one answer, just draw one. Visited opy aste ChemDoodleDraw the major organic product(s) of the following reaction. ● CH3CH₂-CEC-H ? NaNH, / NH3(I) ChemDoodle You do not have to consider stereochemistry. Separate multiple products using the + sign from the drop-down menu. If no reaction occurs, draw the organic starting material. Ⓡ H₂C H₂C Sn [F CH-CH₂CH₂-Br