Formaldehyde can be distinguished from acetaldehyde by the use of

Reaction of iso-propylbenzene with $O_2$ followed by treatment with $H_3O^{+}$ forms phenol and a by-product $P$. Reaction of $P$ with $3$ equivalents of $Cl_2$ gives compound $Q$. Treatment of $Q$ with $Ca(OH)_2$ produces compound $R$ and calcium salt $S$. The correct statement(s) regarding $P, Q, R$ and $S$ is(are):
$(A)$ Reaction of $P$ with $R$ in the presence of $KOH$ gives chloritone.
$(B)$ Reaction of $R$ with $O_2$ in the presence of light gives phosgene gas.
$(C)$ $Q$ reacts with aqueous $NaOH$ to produce $Cl_3CCH_2OH$ and $Cl_3CCOONa$.
$(D)$ $S$ on heating gives $P$.

Assertion: Acetoacetic ester,$CH_3COCH_2COOC_2H_5$,will give iodoform test.
Reason: It does not contain $CH_3CO-$ group.

The reaction of phthalaldehyde with concentrated $NaOH$ produces:

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?

In the following reaction sequence,the correct structures of $E, F$ and $G$ are