An organic compound A has the molecular formu | vedclass

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(C) The compound $A$ with formula $C_3H_6O$ gives the iodoform test,so it is propanone $(CH_3COCH_3)$.Propanone undergoes self-aldol condensation in t

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Propanone and $2,6-$dimethyl$-2,5-$heptadien$-4-$one

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(C) The compound $A$ with formula $C_3H_6O$ gives the iodoform test,so it is propanone $(CH_3COCH_3)$.Propanone undergoes self-aldol condensation in the presence of dilute $HCl$ (acid-catalyzed) to form mesityl oxide $(C_6H_{10}O)$ and further condensation leads to phorone $(C_9H_{14}O)$,which is $2,6-$dimethyl$-2,5-$heptadien$-4-$one.Thus,$A$ is propanone and $B$ is $2,6-$dimethyl$-2,5-$heptadien$-4-$one.

An organic compound $A$ has the molecular formula $C_3H_6O$ and it undergoes the iodoform test. When saturated with dilute $HCl$,it gives $B$ with the molecular formula $C_9H_{14}O$. $A$ and $B$ respectively are:

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