An organic compound ‘A’ has the molecular formula \[{C_3}{H_6}O\] it undergoes iodoform test. When saturated with HCl it gives ‘B’ of the molecular formula \[{C_9}{H_{14}}O\]. A and B are respectively.
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

Hint: This type of chain questions is easy to solve when we can get the compound from using the molecular formula given, since for compound ‘A’ molecular formula is given as \[{C_3}{H_6}O\] thus we can choose the functional groups having only one oxygen atom and moving forward with them.

Complete answer:
We have to know that the iodoform test confirms for the functional group having ketone in it. Since an iodoform test is confirmed for the methyl ketone thus compound ‘A’ \[{C_3}{H_6}O\] has the only possibility of forming ketone thus we can easily neglect option A as it has aldehyde group present in it.
Propanone has molecular formula': \[{C_3}{H_6}O\] which can be represented as,

Now we need to find compound ‘B’ so when propanone reacts with hydrochloric acid from looking at the molecular formulas given we can infer that compound ‘A’ has three carbon atoms whereas Compound ‘B’ has nine carbon atoms it can be seen that there can be a formation of a trimer, so the possible trimer of propanone is phorone which has an IUPAC Name\[2,6 - dimethyl - 2,5 - heptadien - 4 - one\]. Thus from this information we can easily neglect option B and D, and earlier we have neglected option A.
Thus the correct option for this question is Option (C) which is Compound ‘A’: Propanone and Compound ‘B’: \[2,6 - dimethyl - 2,5 - heptadien - 4 - one\].

Note:
Since iodoform test is confirmed for the methyl ketone thus compound ‘A’ \[{C_3}{H_6}O\] has the only possibility of forming a ketone. Phorone is a trimer of propanone that means it is three times the propanone when we look at molecular formulas it can be easily known.