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Question

(A) The reaction of $1$-phenyl-$2$-bromobutane with $NaOMe$ (a strong base) proceeds via an $E_2$ elimination mechanism.In this reaction,the base abst

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The reaction of $1$-phenyl-$2$-bromobutane with $NaOMe$ (a strong base) proceeds via an $E_2$ elimination mechanism.In this reaction,the base abstracts a proton from the $\beta$-carbon to form a double bond.The formation of the conjugated alkene,$(E)-1$-phenylbut-$1$-ene,is favored because the double bond is in conjugation with the phenyl ring,which provides extra stability to the product.Thus,$(E)-1$-phenylbut-$1$-ene is the major product.

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

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