$CH_3CHO$ $\xrightarrow[5\,^{\circ}C]{10\%\,NaOH}$ $\xrightarrow[\Delta]{}$ $\xrightarrow[Ni]{H_2} (A)$; Product $(A)$ of the reaction is

Match the example given in Column-$I$ with the name of the reaction in Column-$II$.

Column-$I$ (Example)	Column-$II$ (Reaction)
$(A)$ $CH_3COCl + H_2 \xrightarrow{Pd-C/BaSO_4} CH_3CHO$	$(1)$ Friedel-Crafts acylation
$(B)$ $C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$	$(2)$ $HVZ$ reaction
$(C)$ $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3$	$(3)$ Aldol condensation
$(D)$ $R-CH_2COOH \xrightarrow{Br_2/Red P} R-CH(Br)COOH$	$(4)$ Cannizzaro's reaction
$(E)$ $CH_3-CN \xrightarrow{SnCl_2/HCl, H_2O/H^+} CH_3-CHO$	$(5)$ Rosenmund's reduction
$(F)$ $2CH_3CHO \xrightarrow{NaOH} CH_3-CH=CH-CHO$	$(6)$ Stephen's reaction

Benzaldehyde on heating with concentrated $NaOH$ gives:

Glyoxal $(OHC-CHO)$ on being heated with concentrated $NaOH$ forms:

Compounds $P$ and $R$ upon ozonolysis produce $Q$ and $S$,respectively. The molecular formula of $Q$ and $S$ is $C_8H_8O$. $Q$ undergoes Cannizzaro reaction but not haloform reaction,whereas $S$ undergoes haloform reaction but not Cannizzaro reaction.
$(i)$ $P \xrightarrow[\text{ii) } Zn / H_2O]{\text{i) } O_3 / CH_2Cl_2} \underset{(C_8H_8O)}{Q}$
$(ii)$ $R \xrightarrow[\text{ii) } Zn / H_2O]{\text{i) } O_3 / CH_2Cl_2} \underset{(C_8H_8O)}{S}$
The option$(s)$ with a suitable combination of $P$ and $R$,respectively,is(are):

Two moles of benzaldehyde and one mole of acetone under alkaline conditions using aqueous $NaOH$ after heating gives $x$ as the major product. The number of $\pi$ bonds in the product $x$ is.................