CH_3-CHO xrightarrow{dil. OH^{-} / Delta} Maj | vedclass

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Question

(C) The reaction of $CH_3-CHO$ with $dil. OH^{-}$ followed by heating $(\Delta)$ is an Aldol condensation reaction.
Step $1$: Two molecules of acetal

Vedclass · Accepted Answer

$CH_3-CH=CH-CHO$

Vedclass · Answer

(C) The reaction of $CH_3-CHO$ with $dil. OH^{-}$ followed by heating $(\Delta)$ is an Aldol condensation reaction.
Step $1$: Two molecules of acetaldehyde undergo aldol addition in the presence of a dilute base to form $3-hydroxybutanal$ $(CH_3-CH(OH)-CH_2-CHO)$.
Step $2$: Upon heating $(\Delta)$,the aldol product undergoes dehydration (loss of $H_2O$) to form the $\alpha,\beta$-unsaturated aldehyde,$but-2-enal$ $(CH_3-CH=CH-CHO)$,which is commonly known as crotonaldehyde.

$CH_3-CHO \xrightarrow{dil. OH^{-} / \Delta}$ Major product formed in the above reaction is:

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