Assertion : Acetylene on reacting with sodamide gives sodium acetylide and ammonia.
Reason : $sp$ hybridised carbon atoms of acetylene are considerably electronegative.

Which of the following reactions will yield $2, 2-$dibromopropane?

Consider the following transformations of a compound $P$ $(C_9H_{12})$:
$(i)$ $P \xrightarrow{NaNH_2, C_6H_5COCH_3, H_3O^+/\Delta} R$ (Optically active)
(ii) $P \xrightarrow{Pt/H_2} \text{propylcyclohexane}$
(iii) $P \xrightarrow{X, KMnO_4/H_2SO_4/\Delta} Q$ ($C_8H_{12}O_6$,Optically active acid)
Choose the correct option$(s)$:
$(A)$ $P$ is $3-$cyclohexylprop$-1-$yne
$(B)$ $X$ is $Pd-C/\text{quinoline}/H_2$
$(C)$ $P$ is $1-$cyclohexylprop$-2-$yne
$(D)$ $R$ is $1-$cyclohexyl$-4-$phenylbut$-3-$yn$-2-$ol

Which of the following compounds forms a $Na$-salt upon reaction with $NaNH_2$?

Identify the alkyne in the following sequence of reactions:
Alkyne $\xrightarrow{\text{H}_2, \text{Lindlar's catalyst}} A$ $\xrightarrow{\text{Ozonolysis}} B \text{ (only)}$
Given that $B$ is obtained from $\text{CH}_2=\text{CH}_2$ via the Wacker process.

Identify $Z$ in the following sequence:
${C_3}{H_7}OH$ $\xrightarrow{conc. \ H_2SO_4} X$ $\xrightarrow{Br_2} Y$ $\xrightarrow[alc. \ KOH]{high \ level} Z$