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(A) The reaction described is the Aldol condensation of acetaldehyde $(CH_3CHO)$.Step $1$: Two molecules of acetaldehyde undergo aldol condensation in

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$CH_3CHO$

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(A) The reaction described is the Aldol condensation of acetaldehyde $(CH_3CHO)$.Step $1$: Two molecules of acetaldehyde undergo aldol condensation in the presence of dilute $NaOH$ to form $\beta$-hydroxybutanal.$CH_3CHO + CH_3CHO \xrightarrow{dil. NaOH} CH_3-CH(OH)-CH_2-CHO$ ($\beta$-hydroxybutanal).Step $2$: Upon heating with dilute acid,$\beta$-hydroxybutanal undergoes dehydration to form crotonaldehyde $(CH_3CH=CH-CHO)$.$CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, H^+} CH_3CH=CH-CHO + H_2O$ (Crotonaldehyde).Therefore,the starting compound is $CH_3CHO$.

$A$ compound contains an $\alpha$-hydrogen atom and forms a $\beta$-hydroxy aldehyde in the presence of a dilute alkali. When this product is heated with a dilute acid,it yields unsaturated crotonaldehyde. Identify the compound.

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