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- b) Which of the flowing reactions will have the faster rate and give a better yield? Use drawings of the transition state (show orbitals!) to explain why. Br NaOCH3 CH3OH Br NaOCH3 CH3OHDraw 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 +1Draw 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Ⓡ
- Use the E2 mechanism to explain why when I is mixed with sodium ethoxide (NaOEt) in ethanol, the major product is III, but when II is mixed with sodium ethoxide in ethanol, the major product is IV. H3C CH3 H3C CH3 ..CI CI CH3 I CH3 || H3C. CH3 H3C CH3 E ||| CH3 CH3 IV SANExplain the Carbocation Rearrangements ?Draw the alkene that would react with the reagent given to account for the product formed. ? + H₂O H₂SO4 CH3 CH3 CHCCH3 OH 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. Sn [F ChemDoodle
- CH3 CH3 Br- Br2 .CH3 CH2Cl2 CH3 H3C H3C Br Electrophilic addition of bromine, Brɔ, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermediate known as a bromonium ion. The reaction occurs in an anhydrous solvent such as CH,Cl,. In the second step of the reaction, bromide is the nucleophile and attacks at one of the carbons of the bromonium ion to yield the product. Due to steric clashes, the bromide ion always attacks the carbon from the opposite face of the bromonium ion so that a product with anti stereochemistry is formed. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions CH3 CH3 CH3 CH3 H3C H3C :Br: :Br:Determine the double bond stereochemistry (E or Z) for the following molecules. F. H3C C A CH3 F. H B D BrOrganic Chemistry Indicate which carbocation is the most stable carbocation and which carbocation is the least stable carbocation?Çok Satırlı Metin.
- 2. Explain the selectivity of the following reaction, which produces a single product despite both alkene carbons being equally substituted. H3C CF3 HBr CF3 H3C CF3 H3C Br- H3C only product -H CF3For the reaction run at room temperature, compound I is formed. However, over prolonged reaction time, compound II becomes the major product. Which explains this phenomenon? H. но II O Compound Il is the both the thermodynamic and kinetic product. O Compound I is the kinetic product and compound II is the thermodynamic product. O Compound l is the thermodynamic product and compound Il is the kinetic product. O Compound l is the both the thermodynamic and kinetic product. O Both compounds I and II are the thermodynamic products. O Both compounds I and II are the kinetic products. %Da Draw the structure of the intermediate (I) for the reaction below. Hg(O2CCH3)2 NaBH4 Product CH;OH H3C • Use the wedge/hash bond tools to indicate stereochemistry where it exists. If the reaction produces a racemic mixture, just drawv one stereoisomer.