Draw a reaction with reactant(s) and expected product(s) for each of the 6 alkyl halides in the SN2 reaction. If no product formed for any of the reactions indicates no reaction . In SN2 12% sodium iodide in acetone was used to react with 6 alkyl halids 1-chlorobutane, 2-chlorobutane, Allyl chloride, 2-chloro-2-methylpropane, 1-chloro-2-methylpropane, 2-bromobutane
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Draw a reaction with reactant(s) and expected product(s) for each of the 6 alkyl halides in the SN2 reaction. If no product formed for any of the reactions indicates no reaction .
In SN2 12% sodium iodide in acetone was used to react with 6
1-chlorobutane,
2-chlorobutane,
Allyl chloride,
2-chloro-2-methylpropane,
1-chloro-2-methylpropane,
2-bromobutane
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- Draw a reaction with reactant(s) and expected product(s) for each of the 6 alkyl halides in the SN1 reaction. If no product formed for any of the reactions indicates no reaction . In SN1 1% silver nitrate in ethanol was used to react with 6 alkyl halids 1-chlorobutane, 2-chlorobutane, Allyl chloride, 2-chloro-2-methylpropane, 1-chloro-2-methylpropane, 2-bromobutaneWittig reactions with the following -chloroethers can be used for the synthesis of aldehydes and ketones. (a) Draw the structure of the triphenylphosphonium salt and Wittig reagent formed from each chloroether. (b) Draw the structural formula of the product formed by treating each Wittig reagent with cyclopentanone. Note that the functional group is an enol ether or, alternatively, a vinyl ether. (c) Draw the structural formula of the product formed on acid-catalyzed hydrolysis of each enol ether from part (b).Secondary halides can react via SN1 or SN2 reactions depending on reaction conditions. Draw the starting material and products that would be formed from the reaction of (R)-2-bromobutane with methanol via an SN1 and SN2 reaction.
- rank them according to their reactivity in an Sn2 reaction. isopropyl e-methylbenzenesulfonate, vinyl chloride, tert butyl chlorideButanone undergoes a nucleophilic addition with a Grignard reagent made from 1-bromopropane and magnesium metal in THF solution. The alkoxide formed from the nucleophilic addition is then conveted into the final product by the careful addition of dilute acid. Complete the mechanism by following the instructions provided for each step. Step 1. Nucleophilic Addition in THFGive the products of the following substitution reactions. For every reaction, show electron pairs on both nucleophile and leaving group.
- What is the role of phosphoric acid in the synthesis of cyclohexene? it is an antioxidant that prevents free radical side reactions it is a safe, non-toxic solvent it lowers the boiling point of the reaction mixture (a colligative property of adding phosphoric acid to water) it protonates the hydroxyl of cyclohexanol to make it a better leaving groupWhat reagents would be needed to complete the following reaction? H2 CH3 CH3 H2/Pd AIC13/CH3CI HNO3/H2SO4 CH3CH2OH/H2OSolvolysis of 2-jodo-2-methylpentane in propanoic acid containing potassium propanoate gave three products. Select the three products generated from this reaction. 4-methyl-2-pentene (or 4-methylpent-2-ene) 3-methyl-2-pentanol (or 3-methylpentan-2-ol) 1,1-dimethylbutyl propanoate (or 2-methylpentan-2-yl propionate) 2-methyl-2-pentene (or 2-methylpent-2-ene) 2-methyl-1-pentene (or 2-methylpent-1-ene) 2-methyl-2-pentanol (or 2-methylpentan-2-ol) X 86)
- Draw the major product formed when HBrHBr reacts with the epoxide. Use wedge–dash bonds, including hydrogen atoms at each stereogenic center, to show the stereochemistry of the product.Choose the right reagent or series of reagents from the ones listed below to prepare 2-pentanone from acetylene. NaNH2NaNH2 followed by CH3CH2CH2BrCH3CH2CH2Br, then disiamylborane followed by H2O2H2O2, HO−HO− H2H2, Lindlar followed by H2OH2O, H+H+ NaNH2NaNH2 and CH3CH2CHOCH3CH2CHO NaNH2NaNH2 followed by H2OH2O, H+H+ NaNH2NaNH2 followed by CH3CH2CH2BrCH3CH2CH2Br, then H2OH2O, H2SO4H2SO4, HgSO4 Choose one.The following reaction involves two sequential Heck reactions. Draw structural formu- las for each organopalladium intermediate formed in the sequence and show how the final product is formed. Note from the molecular formula given under each structural formula that this conversion corresponds to a loss of H and I from the starting material. Acetonitrile, CH,CN, is the solvent. 1% mol Pd(OAc), 4% mol Ph,P CH,CN C4H171 C4H16