The reagent used in the Wolff-Kishner reduction is

An aldehyde or ketone containing at least one $\alpha-H$ atom reacts with dilute $NaOH$ to form a $\beta-$hydroxy aldehyde or $\beta-$hydroxy ketone. This reaction is known as $—$

The option$(s)$ with the correct sequence of reagents for the conversion of $P$ to $Q$ is(are):
$(A)$ $i$) Lindlar's catalyst,$H_2$; $ii$) $SnCl_2 / HCl$; $iii$) $NaBH_4$; $iv$) $H_3O^{+}$
$(B)$ $i$) Lindlar's catalyst,$H_2$; $ii$) $H_3O^{+}$; $iii$) $SnCl_2 / HCl$; $iv$) $NaBH_4$
$(C)$ $i$) $NaBH_4$; $ii$) $SnCl_2 / HCl$; $iii$) $H_3O^{+}$; $iv$) Lindlar's catalyst,$H_2$
$(D)$ $i$) Lindlar's catalyst,$H_2$; $ii$) $NaBH_4$; $iii$) $SnCl_2 / HCl$; $iv$) $H_3O^{+}$

What will be the compound $X$ in the given reaction?
$MeO-C_6H_4-CHO + (X) \xrightarrow{CH_3COONa, H_3O^+} MeO-C_6H_4-CH=CHCOOH$

Which among the following has the highest boiling point?

Which of the following compounds will give a silver mirror test with ammoniacal silver nitrate (Tollens' reagent)?
$(A)$ Formic acid
$(B)$ Formaldehyde
$(C)$ Benzaldehyde
$(D)$ Acetone
Choose the correct answer from the options given below: