The major product Z formed in the following sequence of reactions is :
$$C_2H_6 \xrightarrow{Cl_2, \text{UV light}} X \text{ (monochlorinated product)} \xrightarrow{NH_3} Y \xrightarrow{(i) NaNO_2, HCl, (ii) H_2O} Z$$
The question asks for the major product (Z) formed from a sequence of reactions starting with ethane ($C_2H_6$).
Ethane ($C_2H_6$) reacts with chlorine ($Cl_2$) under UV light via a free-radical substitution mechanism. This produces monochlorinated ethane ($X$).
Reaction: $C_2H_6 + Cl_2 \xrightarrow{UV \text{ light}} C_2H_5Cl + HCl$
Thus, $X$ is monochlorinated ethane, $C_2H_5Cl$.
Monochlorinated ethane ($X = C_2H_5Cl$) reacts with ammonia ($NH_3$) through nucleophilic substitution (ammonolysis) to form ethylamine ($Y$).
Reaction: $C_2H_5Cl + NH_3 \rightarrow C_2H_5NH_2 + HCl$
Thus, $Y$ is ethylamine, $C_2H_5NH_2$. This is a primary amine.
The primary amine ($Y = C_2H_5NH_2$) is treated with sodium nitrite ($NaNO_2$) and hydrochloric acid ($HCl$) at low temperatures. This generates nitrous acid ($HONO$) in situ.
Reaction: $NaNO_2 + HCl \rightarrow HONO + NaCl$
Primary aliphatic amines react with nitrous acid ($HONO$) to form unstable diazonium salts, which immediately decompose in the aqueous acidic medium to yield alcohols, nitrogen gas, and regenerate the acid.
Reaction: $C_2H_5NH_2 + HONO \xrightarrow{HCl} [C_2H_5N_2]^+Cl^- \xrightarrow{H_2O, \text{heat}} C_2H_5OH + N_2 + HCl$
Therefore, the final product $Z$ is ethanol, $C_2H_5OH$.
The major product Z formed in the sequence is ethanol ($C_2H_5OH$).
Select the reagents that reduce nitriles to primary amines :
A. (i) $LiAlH_4$ ; (ii) $H_2O$
B. $Sn + HCl$
C. $H_2 / Ni$
D. $Na(Hg) / C_2H_5OH$
E. $Br_2 / \text{aq. } NaOH$
Choose the correct answer from the options given below :
The correct IUPAC name of the following compound is :

In the following reaction sequence, X and Z respectively are :
The number of chlorine atoms present in the organic products X and Y of the following reactions, respectively, are :
Two products X and Y are formed in the following reaction sequence.
The suitable method that can be used for the separation of products X and Y is :
| List I (Reactions) | List II (Reagents) |
A.![]() | I. (i) oleum, (ii) $NaOH, \Delta$, (iii) $H^+$ |
| B. $CH_3COOH \rightarrow CH_3CH_2OH$ | II. (i) $O_3$, (ii) $H_2O/H^+$ |
C. $CH_3CH_2CH_2OH \rightarrow $ ![]() | III. (i) $CH_3OH, H^+$, (ii) $H_2$, catalyst |
D. ![]() | IV. (i) conc. $H_2SO_4, \Delta$, (ii) $H_2O$ |
Match List I with List II.
The following two reactions give the same foul smelling product Z.
$$C_2H_5Cl \xrightarrow{X} Z$$

X and Z, respectively, are :
Given below are two statements:
Statement I: The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce 
Statement II: R-CN hydrolyses under acidic medium to produce a compound which on treatment with $\text{SOCl}_2$, followed by the addition of $\text{NH}_3$ gives another compound(x). This compound (x) on treatment with $\text{NaOCl/NaOH}$ gives a product, that on treatment with $\text{CHCl}_3\text{/KOH/}\Delta$ produces R-NC
In the light of the above statements, choose the correct answer from the options given below