The starting material is butan-2-ol. The reac | vedclass

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(B) The dehydration of $butan-2-ol$ $(CH_3CH_2CH(OH)CH_3)$ in the presence of $H^+/\Delta$ proceeds via an $E1$ mechanism involving a carbocation inte

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$3, 5$

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(B) The dehydration of $butan-2-ol$ $(CH_3CH_2CH(OH)CH_3)$ in the presence of $H^+/\Delta$ proceeds via an $E1$ mechanism involving a carbocation intermediate.$1$. The carbocation formed is $CH_3CH_2CH^+CH_3$.$2$. Elimination of a proton from the adjacent carbons leads to the formation of three possible alkene products:- $but-1-ene$ $(CH_3CH_2CH=CH_2)$- $cis-but-2-ene$ $(CH_3CH=CHCH_3)$- $trans-but-2-ene$ $(CH_3CH=CHCH_3)$Thus,the number of products $(x) = 3$.$3$. When these three alkenes react with $Br_2/CCl_4$ (electrophilic addition),they form vicinal dibromides:- $but-1-ene$ gives $1,2-dibromobutane$ (one chiral product,a racemic mixture of $R$ and $S$ enantiomers).- $cis-but-2-ene$ gives $meso-2,3-dibromobutane$ (one product).- $trans-but-2-ene$ gives a racemic mixture of $(2R, 3R)$ and $(2S, 3S)-2,3-dibromobutane$ (two enantiomers).$4$. Counting the distinct stereoisomeric products: $1,2-dibromobutane$ ($2$ enantiomers),$meso-2,3-dibromobutane$ ($1$ isomer),and $(2R, 3R)/(2S, 3S)-2,3-dibromobutane$ ($2$ enantiomers). Total products $(y) = 2 + 1 + 2 = 5$.Therefore,$x = 3$ and $y = 5$.

The starting material is $butan-2-ol$. The reaction with $H^+/\Delta$ yields $(x)$ possible products,and the subsequent reaction of these products with $Br_2/CCl_4$ yields $(y)$ possible products. Determine the values of $x$ and $y$.

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