Question

The major product Z formed in the following sequence of reactions is:  

$$ {C2H5Cl →[AgNO2] X →[Sn/HCl] Y →[(i)\ NaNO2][(ii)\ HCl,\ H2O] Z} $$

• A. \({C2H5NO2}\)
• B. \({C2H5-N=N-OH}\)
• C. \({C2H5NH2}\)
• D. \({C2H5OH}\)

Hint:

Alkyl halide with \(AgNO_2\) gives nitroalkane, nitroalkane on reduction gives amine, and primary aliphatic amine with nitrous acid gives alcohol.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a three-step organic synthesis starting from Ethane ($C_2H_6$). We must identify the intermediate products X and Y to determine the final product Z. The sequence involves a free-radical substitution, a nucleophilic substitution, and a diazotization reaction.
Step 2: Key Formula or Approach:
The approach involves identifying each reaction type:
Step 1: Halogenation of alkanes ($Cl_2 / h\nu$).
Step 2: Reaction of alkyl halides with ammonia (Ammonolysis).
Step 3: Reaction of primary aliphatic amines with nitrous acid ($NaNO_2/HCl$).
Step 3: Detailed Explanation:

Step 1 ($X$): Ethane reacts with $Cl_2$ in the presence of UV light. This is a free radical substitution where one hydrogen is replaced by chlorine. $C_2H_6 + Cl_2 \rightarrow C_2H_5Cl (X) + HCl$. $X$ is Ethyl chloride.
Step 2 ($Y$): Ethyl chloride reacts with ammonia. The $NH_3$ acts as a nucleophile and displaces the $Cl$ atom. $C_2H_5Cl + NH_3 \rightarrow C_2H_5NH_2 (Y) + HCl$. $Y$ is Ethylamine.
Step 3 ($Z$): Ethylamine (a primary aliphatic amine) reacts with nitrous acid ($HNO_2$ generated in situ). Primary aliphatic amines react with $HNO_2$ to form highly unstable aliphatic diazonium salts ($[C_2H_5N_2^+]Cl^-$). Unlike aromatic diazonium salts, these decompose immediately in water (aqueous medium) to produce nitrogen gas and the corresponding alcohol. $C_2H_5NH_2 \xrightarrow{HNO_2} [C_2H_5N_2^+] \xrightarrow{H_2O} C_2H_5OH (Z) + N_2 \uparrow$.
Step 4: Final Answer:
The final major product $Z$ is Ethanol ($C_2H_5OH$).