Question

Identify the product C in the series
CH₃CN□(→┴(Na/C₂H₅OH) )A(→┴HNO₂) )C(→┴(Cu/573K) )
(a) CH₃COOH (b) CH₃CH₂NHOH
(c) CH₃CONH₂ (d) CH₃CHO

Answer 1

$\huge\underline \mathfrak \blue{Answer}$

(b) CH₃CH₂NHOH

Answer 2

The product 'C' is CH₃CHO (ethanal). Hence, option (d) is correct.

• CH₃CN + Na/C₂H₅OH → CH₃CH₃NH₂ (A)

The reaction of an alkyl cyanide with sodium and alcohol to give an alkyl amine is called the Mendius reaction.

• In Mendius reaction, sodium dissolved in ethanol acts as reducing agent by supplying nascent hydrogen. Therefore, ethyl cyanide is reduced to ethanamine (A) by sodium and alcohol.

• CH₃CH₃NH₂ + HNO₂ → CH₃CH₂OH (B) + N₂ + H₂O

Nitrous acid (HNO₂) is an oxidizing agent that oxidises an amine to alcohol on reacting with it.

• Therefore, the reaction between ethanamine and nitrous acid oxidises ethanamine to ethanol (B) along with nitrogen gas and water molecules as the by-products.

• CH₃CH₂OH + Cu/573K  → CH₃CHO (C) + H₂

The reaction between an alcohol and copper catalyst at 573 K yields the corresponding aldehyde.

• Copper acts an oxidizing agent and oxidises an alcohol to aldehyde by the evolution of hydrogen gas. Therefore, ethanol is oxidised to ethanal (C) by reacting with copper at 573 K.