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- CH3 O O CH3CH₂CHCH₂CH=CHCOH HCI CH3 CHOMO CH3CH₂CHCH₂CH- CHCOH ChemDoodleⓇ CI H CH3 CH3CH₂CHCH₂CH- O CHCOH H CI this product is not observed Account for the fact that this reaction yields only a single product by drawing the carbocation intermediate leading to the observed product. Sty √n [FA student adds NBS to a solution of 1-methylcyclohexene and irradiates the mixture with a sunlamp until all the NBS has reacted. After a careful distillation, the product mixture contains two major products of formula C7H11Br. Draw the products obtained from each free-radical intermediate.Which statement best describes the stereochemistry of the product?
- Draw the major products obtained from the reaction of one equivalent of HBr with the following compounds. For each reaction, indicate the kinetic product and the thermodynamic productFor the following dehydrohalogenation (E2) reaction, draw the major organic product(s), including stereochemistry CI (Cн Сок СH, HII (Cн, СОнDraw the major organic product of the reaction shown below. HO Hoa + HCI • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. /n [F ? ChemDoodleⓇ
- Draw the neutral organic starting material. The Hint: The reaction conditions support electrophilic addition of Br2 to an alkene's C=C double bond, which would normally yield a dibromo product. However, the product has only one bromine atom, with a C–O bond on the adjacent carbon. This fragment is diagnostic for a halohydrin, where an oxygen nucleophile (water or alcohol) reacts with a bromonium intermediate to generate the O–CH2CH2–Br motif. Work backwards to determine what the starting material must look like. There should be an alcohol and an alkene in the neutral organic starting material.Br Brz CH3 CH3 H3C CH2CI2 H3C Br Electrophilic addition of bromine, Br2; 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 Br: :Br: .CH3 H3C H3C CH3 Br:Identify the expected major product of the following electrocyclic reaction. → OI Oll O III OIV OV A ? KYYAK 8 +En ||| + En I || IV + En V
- Consider the series of the trans effect: CO, CN-, C2H4 > PR3, H-, CH3- > C6H5- > NO2-, SCN-, I- > Br- >Cl- > py > NH3 > H20 What would be the major product of the following reaction? Select one:What is the predicted product of the reaction shown? || снобоон CH3COOH III IV VH3C N- H₂NNH₂ H⭑ CH3 H3C IN CH3 Hydrazine reacts with 2,4-pentanedione to yield 3,5-dimethylpyrazole. Including protonations and deprotonations, the reaction takes 12 steps. Write out the mechanism on a sheet of paper and then draw the structure of the product of step 6. • 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. • In an elimination step, include the structure of the leaving group, but draw it in its own sketcher. Separate structures with + signs from the drop-down menu. ? Sn th Previous Next