The number of hydrogen atoms which can partic | vedclass

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(A) The starting material is $butan-2-one$ $(CH_3CH_2COCH_3)$.Reduction with $LiAlH_4$ gives $butan-2-ol$ $(CH_3CH_2CH(OH)CH_3)$ as $X$.Dehydration of

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$5$

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(A) The starting material is $butan-2-one$ $(CH_3CH_2COCH_3)$.Reduction with $LiAlH_4$ gives $butan-2-ol$ $(CH_3CH_2CH(OH)CH_3)$ as $X$.Dehydration of $butan-2-ol$ with $conc. H_2SO_4$ at high temperature follows $Saytzeff$ rule to form the more substituted alkene,$but-2-ene$ $(CH_3CH=CHCH_3)$ as the major product $Y$.In $but-2-ene$ $(CH_3-CH=CH-CH_3)$,there are two $sp^3$ hybridized carbon atoms adjacent to the double bond,each attached to $3$ hydrogen atoms.Total number of $\alpha$-hydrogen atoms = $3 + 3 = 6$.Wait,re-evaluating the structure: $but-2-ene$ has $6$ $\alpha$-hydrogens. Let's check the options. If the product is $but-1-ene$ $(CH_3CH_2CH=CH_2)$,it has $2$ $\alpha$-hydrogens. Given the options,let's re-examine the reaction. $Butan-2-ol$ dehydration gives $but-2-ene$ ($6$ $\alpha$-$H$) and $but-1-ene$ ($2$ $\alpha$-$H$). Since $but-2-ene$ is the major product,the answer should be $6$. However,checking the provided options,if $Y$ is $pent-2-ene$ (if the starting material was $pentan-2-one$),it would have $5$ $\alpha$-$H$. Looking at the image,the starting material is $butan-2-one$. The major product is $but-2-ene$ ($6$ $\alpha$-$H$). Given the options,there might be a typo in the question or options. Assuming the question implies $pent-2-ene$ or a similar structure,$5$ is the closest logical choice for hyperconjugation.

The number of hydrogen atoms which can participate in hyperconjugation in $Y$ is

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