(a) What product(s) are formed when the E isomer of C6H5CH = CHC6H5 is treated with Br2, followed by one equivalent of KOH? Label the resulting alkene(s) as E or Z. (b) What product(s) are formed when the Z isomer of C6H5CH = CHC6H5 is subjected to the same reaction sequence? (c) How are the compounds in parts (a) and (b) related to each other?

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Asked Dec 26, 2019
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(a) What product(s) are formed when the E isomer of C6H5CH = CHC6H5 is treated with Br2, followed by one equivalent of KOH? Label the resulting alkene(s) as E or Z. (b) What product(s) are formed when the Z isomer of C6H5CH = CHC6H5 is subjected to the same reaction sequence? (c) How are the compounds in parts (a) and (b) related to each other?

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Expert Answer

Step 1

a.

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The general steps involved in the addition of halogens to alkenes are stated below: • The first step is the electrophilic attack of halide ion to form a halonium ion. • The second step is the attack of halide ion from back side to open the halonium ion. The product(s) formed by the reaction of E isomer of C6H5CH = CHC&H5 with Br2 is shown below. сн, CHs Н CHs Н (R) н Br Br. Br (S) H- Br Сн сна "Н Н ČHs

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Step 2
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The first step is the bromination of alkene that results in the formation of bromo product. In the next step, dehydrohalogenation takes place in the presence of base. This results in the formation of E alkene because bromine and hydrogen are in antiperiplanarity. The reaction for the formation of E alkene is shown below. ÇÇH5 COHS (E) CH5 (R) H- Br КОН |(S) H. Br Br Н C,H5

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Step 3
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The product(s) formed by the reaction of E isomer of C6H;CH = CHC,H; with Br2, followed by one equivalent of KOH and labeling of resulting alkene as E or Z is shown below. CHs СН, САН, Н Н (R) н Br Br2 Br Br. (S) -H- Br сHн С,Н; H. H. Сан CHs Сан, (R) -H- Br КОН (S) -H- Br Br н CHs

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