In the given reaction: $NH_2-CH_2-CH_2-NH_2 + C_2H_5OOC-COOC_2H_5 \xrightarrow{\text{Pyridine}} [X]$,$[X]$ will be:

$\text{CH}_3\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{X} \xrightarrow{\text{PCl}_5} \text{Y} \xrightarrow{\text{Alcoholic KOH}} \text{Z}$. Find product '$Z$' in the above reaction.

What will be the product obtained from the following reaction?
$CH_3CH_2C \equiv N + C_2H_5OH + H_2O \xrightarrow{\text{conc. } H_2SO_4}$

Compound $(B)$ is

Choose the correct option$(s)$ for the following reaction sequence. Consider $Q$,$R$ and $S$ as major products.

Consider the following sequence of reactions to give the major product $(X)$: $(i)$ $CH_3Cl$ / anhydrous $AlCl_3$, (ii) $Cl_2$ / $FeCl_3$, (iii) $K_2Cr_2O_7$ / $H_2SO_4$. $P$ g of the major product $(X)$ formed is reacted with $NaHCO_3$ solution to liberate a gas which occupied $11.2 \ dm^3$ at $STP$. $P = \text{ . . . . . . }$ g.