The reaction Ph-CH_2-CH(F)-CH_3 xrightarrow[E | vedclass

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(C) The reaction involves the treatment of $1-phenyl-2-fluoropropane$ with $KOH$ in $EtOH$. This is a dehydrohalogenation reaction,which proceeds via

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$Ph-CH=CH-CH_3$

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(C) The reaction involves the treatment of $1-phenyl-2-fluoropropane$ with $KOH$ in $EtOH$. This is a dehydrohalogenation reaction,which proceeds via an $E2$ mechanism to form an alkene. The base $KOH$ in $EtOH$ abstracts a proton from the $\beta$-carbon. There are two possible $\beta$-carbons: $1$. The $\beta$-carbon attached to the phenyl group $(Ph-CH_2-)$,which would lead to $Ph-CH=CH-CH_3$. $2$. The $\beta$-carbon of the methyl group $(-CH_3)$,which would lead to $Ph-CH_2-CH=CH_2$. The formation of $Ph-CH=CH-CH_3$ ($1$-phenylprop$-1-$ene) is favored because the resulting double bond is conjugated with the phenyl ring,making the product more stable. Thus,the major product $X$ is $Ph-CH=CH-CH_3$.

The reaction $Ph-CH_2-CH(F)-CH_3 \xrightarrow[EtOH]{KOH} X$. What is $X$?

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