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Question

(B) The reaction of $1,2-diphenyl-1-bromoethane$ with a strong base like potassium tert-butoxide $(t-BuOK)$ in the presence of heat $(\Delta)$ undergo

Vedclass · Accepted Answer

$C_6H_5CH=CHC_6H_5$

Vedclass · Answer

(B) The reaction of $1,2-diphenyl-1-bromoethane$ with a strong base like potassium tert-butoxide $(t-BuOK)$ in the presence of heat $(\Delta)$ undergoes a dehydrohalogenation reaction. This is an $E_2$ elimination reaction where the base abstracts a proton from the $\beta$-carbon,leading to the formation of a double bond between the $\alpha$ and $\beta$ carbons. The product formed is $1,2-diphenylethene$ $(C_6H_5CH=CHC_6H_5)$,which is a stable conjugated alkene. The reaction is: $C_6H_5CH(Br)CH_2C_6H_5 + t-BuOK \xrightarrow{\Delta} C_6H_5CH=CHC_6H_5 + t-BuOH + KBr$

The major product obtained in the following reaction is $:$

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