Acetaldehyde cannot give:

$A$ carbonyl compound $P$,which gives a positive iodoform test,undergoes reaction with $MeMgBr$ followed by dehydration to give an olefin $Q$. Ozonolysis of $Q$ leads to a dicarbonyl compound $R$,which undergoes intramolecular aldol reaction to give predominantly $S$.
$P$ $\xrightarrow[\substack{2. H^{+}, H_2 O \\ 3. H_2 SO_4, \Delta}]{1. MeMgBr} Q$ $\xrightarrow[2. Zn, H_2 O]{1. O_3} R$ $\xrightarrow[2. \Delta]{1. OH^{-}} S$
$1.$ The structure of the carbonyl compound $P$ is
$2.$ The structures of the products $Q$ and $R$,respectively,are
$3.$ The structure of the product $S$ is
Give the answer for questions $1, 2$ and $3.$

The product obtained from the following sequence of reactions is
$CH_3-C\equiv CH$ $\xrightarrow{HgSO_4, H_2SO_4} A$ $\xrightarrow{NaBH_4} B$

The reaction of phthalaldehyde with concentrated $NaOH$ produces:

Reagents to carry out the above conversion,$P$,$Q$,$R$ respectively are:

The reaction of $C_6H_5CH=CHCHO$ with $LiAlH_4$ gives