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(C) Compound $A$ gives the iodoform test,which means it must contain the $CH_3CO-$ or $CH_3CH(OH)-$ group. Given the molecular formula $C_3H_6O$,compo

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

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(C) Compound $A$ gives the iodoform test,which means it must contain the $CH_3CO-$ or $CH_3CH(OH)-$ group. Given the molecular formula $C_3H_6O$,compound $A$ is propanone $(CH_3COCH_3)$. When propanone is treated with $HCl$ (acid-catalyzed aldol condensation followed by dehydration),it undergoes self-condensation to form phorone $(2,6-dimethyl-2,5-heptadien-4-one)$. The reaction is: $3CH_3COCH_3 \xrightarrow{HCl} (CH_3)_2C=CH-CO-CH=C(CH_3)_2 + 2H_2O$. Thus,$A$ is propanone and $B$ is $2,6-dimethyl-2,5-heptadien-4-one$.

An organic compound $A$ has the molecular formula $C_3H_6O$. It gives the iodoform test. When it is saturated with $HCl$,it gives a compound $B$ with the molecular formula $C_9H_{14}O$. What are $A$ and $B$ respectively?

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