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Chemistry, 16.06.2020 03:57 funnynunny2903
Electrophilic addition of HBr to alkenes yields a bromoalkane. The reaction begins with an attack on the hydrogen of the electrophilic HBr by the π electrons of the double bond to give a carbocation. This step follows Markovnikov’s rule with the electrophilic H atom adding to the sp2 carbon containing the most hydrogens, leading to the formation of the most stable carbocation (1°<2°<3°). If possible, a 1,2-shift of either a neighboring hydride or methyl group can occur prior to the last step in order to form a more stable carbocation. In the final step of the reaction, nucleophilic bromide adds to the carbocation to give the neutral product. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions
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Chemistry, 22.06.2019 21:50
Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes
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Electrophilic addition of HBr to alkenes yields a bromoalkane. The reaction begins with an attack on...
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