Acetylene gas when passed through $20\% \ H_2SO_4$ at $80 \, ^\circ C$ gives acetaldehyde. The catalyst required for this conversion is:

The reaction $R-C \equiv C-R \xrightarrow{H_2 / \text{Lindlar's catalyst}}$ gives the major product as ..............

In the given reactions,find the sum of $(x + y)$.
$(1)$ $Ph-C(Br)_2-CH_3$ $\xrightarrow{xNaNH_2} (A)$ $\xrightarrow{CH_3I} Ph-C \equiv C-CH_3$
$(2)$ $CH_3-CH_2-C(Br)_2-CH_3 \xrightarrow{yNaNH_2} CH_3-C \equiv C-CH_3$
Find the value of $(x + y)$.

$Ph-C \equiv CH \xrightarrow[MeOH]{MeO^{-}}$ The major product of the reaction is:

$CH \equiv CH$ $\xrightarrow[H_2SO_4]{H_2O/Hg^{2+}} X$ $\xrightarrow{LiAlH_4} Y$ $\xrightarrow{P_4/Br_2} Z$. Here $Z$ is

What will be $X$ in the following reaction?
${C_2}{H_2} \xrightarrow[{HgSO_4 / H_2SO_4, 60^{\circ}C}]{{H_2O}} X \rightleftharpoons CH_3CHO$