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

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

Step 1: Chlorination of ethane gives ethyl chloride (X = C2H5Cl). Step 2: Reaction with ammonia gives ethylamine (Y = C2H5NH2). Step 3: Diazotization of primary aliphatic amine yields unstable diazonium salt which hydrolyzes to ethanol (Z = C2H5OH). Therefore, major product Z is ethanol (C2H5OH). Aliphatic primary amine + nitrous acid (NaNO2/HCl) always decomposes to alcohol + N2 gas. Class 12 Chemistry: Amines

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Solution & Explanation

Related Formula

R-NH₂ NaNO₂ + HCl [R-N₂^+ Cl^-] H₂O R-OH + N₂ + HCl

Core Logic

Step 1: Chlorination of ethane gives ethyl chloride ( $X = C₂H₅Cl$ $X = C_2H_5Cl$). Step 2: Reaction with ammonia gives ethylamine ( $Y = C₂H₅NH₂$ $Y = C_2H_5NH_2$). Step 3: Diazotization of primary aliphatic amine yields unstable diazonium salt which hydrolyzes to ethanol ( $Z = C₂H₅OH$ $Z = C_2H_5OH$).

Step 1: Final Product Identification

Therefore, major product $Z$ $Z$ is ethanol ( $C₂H₅OH$ $C_2H_5OH$).

Pattern Recognition

Aliphatic primary amine + nitrous acid ( $NaNO₂/HCl$ $NaNO_2/HCl$) always decomposes to alcohol + $N₂$ $N_2$ gas.

Chapter Mix

Class 12 Chemistry: Amines

Solution

Related Formula

R-CN Reduction R-CH₂-NH₂

Core Logic

A. $LiAlH₄$ $LiAlH_4$ reduces nitriles to $R-CH₂-NH₂$ $R-CH_2-NH_2$. B. $Sn + HCl$ $Sn + HCl$ reduces nitriles to aldehydes (Stephen reduction intermediate). C. $H₂/Ni$ $H_2/Ni$ catalytically reduces nitriles to $R-CH₂-NH₂$ $R-CH_2-NH_2$. D. $Na(Hg)/C₂H₅OH$ $Na(Hg)/C_2H_5OH$ (Mendius reduction) yields $R-CH₂-NH₂$ $R-CH_2-NH_2$. E. $Br₂/NaOH$ $Br_2/NaOH$ is Hoffmann bromamide degradation for amides, not nitriles.

Step 1: Selection

Reagents A, C, and D perform reduction of nitriles to primary amines.

Pattern Recognition

$LiAlH₄$ $LiAlH_4$, catalytic hydrogenation, and $Na(Hg)/EtOH$ $Na(Hg)/EtOH$ convert $R-CN arrow R-CH₂NH₂$ $R-CN \rightarrow R-CH_2NH_2$.

Chapter Mix

Class 12 Chemistry: Amines

Q89 neet_2024_05_may_morning Carbylamine Reaction and Isocyanides

The following two reactions give the same foul smelling product Z:

C₂H₅Cl X Z

C₂H₅CONH₂ Br₂, NaOH Y [Δ]CHCl₃ / ethanolic KOH Z

X and Z, respectively, are:

A. (1) $X = AgCN; Z = C₂H₅CN$ $X = AgCN; Z = C_2H_5CN$
B. (2) $X = KCN; Z = C₂H₅CN$ $X = KCN; Z = C_2H_5CN$
C. (3) $X = KCN; Z = C₂H₅NC$ $X = KCN; Z = C_2H_5NC$
D. (4) $X = AgCN; Z = C₂H₅NC$ $X = AgCN; Z = C_2H_5NC$

Solution

Related Formula

R-Cl + AgCN arrow R-NC + AgCl

R-NH₂ + CHCl₃ + 3KOH arrow R-NC + 3KCl + 3H₂O

Core Logic

Reaction 1: Ethyl chloride + $AgCN arrow$ $AgCN \rightarrow$ Ethyl isocyanide ( $Z = C₂H₅NC$ $Z = C_2H_5NC$). Reaction 2: Propanamide + $Br₂/NaOH arrow$ $Br_2/NaOH \rightarrow$ Ethylamine ( $Y = C₂H₅NH₂$ $Y = C_2H_5NH_2$). Carbylamine test on $Y$ $Y$ produces foul-smelling ethyl isocyanide ( $Z = C₂H₅NC$ $Z = C_2H_5NC$).

Step 1: Conclusion

$X = AgCN$ $X = AgCN$ and $Z = C₂H₅NC$ $Z = C_2H_5NC$.

Pattern Recognition

Foul-smelling substance = Isocyanide ( $R-NC$ $R-NC$). Covalent $AgCN$ $AgCN$ attacks via nitrogen to yield isocyanide.

Chapter Mix

Class 12 Chemistry: Amines

The major product Z formed in the following sequence of reactions is: $C₂H₆ [UV light]Cl₂ X(monochlorinated Product) NH₃ Y [(ii) H₂O](i) NaNO₂/HCl Z$

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Product Identification

Pattern Recognition

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