The reaction $CH_3CHO \xrightarrow[Zn(Hg) / \text{Conc. } HCl]{[H]} CH_3CH_3$ is:

In the following reaction:
$\text{Aldehyde} + \text{Alcohol} \xrightarrow{HCl} \text{Acetal}$
$HCHO \qquad ^tBuOH$
$CH_3CHO \qquad MeOH$
The best combination for the formation of acetal is:

$CH_3-C \equiv CH \xrightarrow[dil. H_2SO_4]{Hg^{+2}} (P)$,$CH_3-C \equiv CH \xrightarrow[(2) H_2O_2]{(1) BH_3, THF} (Q)$. $P$ and $Q$ can be differentiated by

Product $(B)$ in this reaction is

An organic compound $CH_3-CH=CH-CH_2-CHO$ is taken in two different containers $A$ and $B$. Sample in $A$ is treated with $H_2 / Ni$ forming new compound $P$. Sample in $B$ is treated with $LiAlH_4$ and hydrolyzed further forming new compound $Q$. Identify $P$ and $Q$.

The major products in the reaction $Br_3CCHO \stackrel{NaOH}{\longrightarrow}$ are $....$