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An organic compound 'A' has the molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$, it undergoes iodoform test. When saturated with dil. $\mathrm{HCl}$ it gives ' $\mathrm{B}$ ' of molecular formula $\mathrm{C}_{9} \mathrm{H}_{14} \mathrm{O}$. A and B respectively are
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Propanone and 2,6-dimethyl-2, 5 -heptadien-4-one
The compound $\mathrm{A}$ with formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ gives iodoform test, it is propanone. It forms a compound B having carbon atoms three times to the number of carbon atoms in propanone, it is 2,6 -dimethyl-2, 5 -heptadien-4-one.
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