The starting material is CH_3COCH_2CHO. The r | vedclass

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(B) The starting material is $CH_3COCH_2CHO$ ($3$-oxobutanal).Step $(i)$: Treatment with ethylene glycol and dry $HCl$ selectively protects the more r

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but$-2-$en$-1-$al

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(B) The starting material is $CH_3COCH_2CHO$ ($3$-oxobutanal).Step $(i)$: Treatment with ethylene glycol and dry $HCl$ selectively protects the more reactive carbonyl group (aldehyde) as a cyclic acetal. The ketone remains unaffected.Step $(ii)$: $LiAlH_4$ reduces the ketone group to a secondary alcohol. The product $A$ is $CH_3CH(OH)CH_2CH(OCH_2CH_2O)$.Step $(iii)$: Acidic hydrolysis $(H_3O^{+}/\Delta)$ deprotects the acetal back to an aldehyde and simultaneously causes dehydration of the $\beta$-hydroxy ketone formed,leading to an $\alpha,\beta$-unsaturated carbonyl compound.The final product $B$ is $CH_3CH=CHCHO$ (but$-2-$enal).

The starting material is $CH_3COCH_2CHO$. The reaction sequence is:
$CH_3COCH_2CHO$ $\xrightarrow[{(ii)LiAlH_4}]{{(i)ethylene\,glycol/dry\,HCl}} A$ $\xrightarrow[\Delta ]{H_3O^{+}} B$
What will be $B$?

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The starting material is $CH_3COCH_2CHO$. The reaction sequence is:$CH_3COCH_2CHO$ $\xrightarrow[{(ii)LiAlH_4}]{{(i)ethylene\,glycol/dry\,HCl}} A$ $\xrightarrow[\Delta ]{H_3O^{+}} B$What will be $B$?