xrightarrow[Delta]{KOH text{ (alc.) }} major | vedclass

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Question

(A) The reaction is a dehydrohalogenation reaction using alcoholic $KOH$ (an elimination reaction,$E2$).The substrate is $1	ext{-phenyl-2-bromo-3-met

Vedclass · Accepted Answer

$1-$phenyl$-3-$methylbut$-1-$ene

Vedclass · Answer

(A) The reaction is a dehydrohalogenation reaction using alcoholic $KOH$ (an elimination reaction,$E2$).The substrate is $1	ext{-phenyl-2-bromo-3-methylbutane}$.Elimination of $HBr$ occurs from the $\beta$-carbon atoms adjacent to the carbon bearing the $Br$ atom.There are two possible $\beta$-hydrogens:$1$. From $C-1$ (benzylic position),which leads to a conjugated alkene $(1	ext{-phenyl-3-methylbut-1-ene})$.$2$. From $C-3$,which leads to a less substituted alkene $(1	ext{-phenyl-3-methylbut-2-ene})$.According to Saytzeff's rule,the more substituted and conjugated alkene is the major product.Therefore,the major product $P$ is $1	ext{-phenyl-3-methylbut-1-ene}$.

$\xrightarrow[\Delta]{KOH \text{ (alc.) }}$ major product $P$
Product $P$ is

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$\xrightarrow[\Delta]{KOH \text{ (alc.) }}$ major product $P$Product $P$ is