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(A) The reaction of $1-$phenyl$-2-$bromobutane with $NaOMe$ (a strong base) proceeds via an $E2$ elimination mechanism.$1-$phenyl$-2-$bromobutane has

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$(E)-1-$phenylbut$-1-$ene

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(A) The reaction of $1-$phenyl$-2-$bromobutane with $NaOMe$ (a strong base) proceeds via an $E2$ elimination mechanism.$1-$phenyl$-2-$bromobutane has the structure $Ph-CH_2-CH(Br)-CH_2-CH_3$.Elimination of $HBr$ can occur to form either $1-$phenylbut$-1-$ene or $1-$phenylbut$-2-$ene.Although Zaitsev's rule generally predicts the more substituted alkene as the major product,in this specific case,the formation of $(E)-1-$phenylbut$-1-$ene is favored because the double bond is in conjugation with the phenyl ring,which provides significant extra stability.Between the $(E)$ and $(Z)$ isomers of $1-$phenylbut$-1-$ene,the $(E)-$isomer is more stable due to reduced steric hindrance between the phenyl group and the ethyl group.Therefore,the major product is $(E)-1-$phenylbut$-1-$ene.

What is the major product obtained in the reaction of $1-$phenyl$-2-$bromobutane with $NaOMe$?

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