Identify the compound

Ethyl isocyanide has the structure: $\mathrm{C_2H_5{-}N\equiv C}$

Isocyanides (or carbylamines) have the general formula: $\mathrm{R{-}N\equiv C}$

Recall the hydrolysis reaction of isocyanides in acidic medium

On acidic hydrolysis, isocyanides give a primary amine and formic acid. In acidic medium, the amine exists as its ammonium salt.

General reaction: $\mathrm{R{-}N\equiv C + 2H_2O + HCl \rightarrow R{-}NH_3^+Cl^- + HCOOH}$

So, the products are:

• amine corresponding to $\mathrm{R}$ as its salt
• methanoic acid

Apply to ethyl isocyanide

Here, $\mathrm{R = C_2H_5}$

Therefore: $\mathrm{C_2H_5{-}N\equiv C \xrightarrow[acid]{hydrolysis} C_2H_5NH_3^+X^- + HCOOH}$

This means the products are:

• ethylamine salt
• methanoic acid

Check the options

• A: propanoic acid and ammonium salt — incorrect
• B: ethanoic acid and ammonium salt — incorrect
• C: methylamine salt and ethanoic acid — incorrect
• D: ethylamine salt and methanoic acid — correct

Final answer

The correct option is: $\boxed{\text{D}}$

Comparison with stored correct answer

Stored correct answer = D

My derived answer = D

Hence, they agree.