An organic compound 'A' has the molecular formula `C_3H_6O`, it undergoes iodoform test. Wheir saturated with dil. HCl it gives 'B' of molecular formula `C_9H_(14)O`. A and B respectively are

Options

A. Propanal and mesitylene
B. Propanone and mesityl oxide
C. Propanone and 2,6-dimethyl-2,5-heptadien-4-one
D. Propanone and mesitylene oxide

Answer

C

Explanation

The compound A with formula `C_3H_6O` 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 (phorone).
`underset((A))(CH_3-underset(CH_3)underset(|)C=O)+CH_3-underset(O)underset(||)C-CH_3+O=underset(CH_3)underset(|)C-CH_3 overset("dil. HCl")to underset("2,6-Dimethyl-2,5-heptadien-4-one")(CH_3-underset(CH_3)underset(|)C=CH-underset(O)underset(||)C-CH=underset(CH_3)underset(|)C-CH_3)`