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Step-by-Step Solution
Step 1: Identify the Possible Hydrolysis Products from the Options
Among the given options, only isocyanides (R–NC) typically hydrolyze to give a primary amine and formic acid. Hence, we focus on the compound propyl isocyanide (CH3CH2CH2NC).
Step 2: Write the Hydrolysis Reaction
On hydrolysis, propyl isocyanide splits into two products:
$ \text{CH}_3\text{CH}_2\text{CH}_2\text{NC} + 2\,\text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{HCOOH} $
So the two hydrolysis products are:
Propylamine ($ \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 $)
Formic acid ($ \text{HCOOH} $)
Step 3: Consider the Reaction of the Amine with NaNO2 and HCl
Primary amines ($ \text{R–NH}_2 $) react with sodium nitrite (NaNO2) and hydrochloric acid (HCl) to form a diazonium salt. At lower temperatures (0–5 °C), the diazonium salt can undergo hydrolysis to form the corresponding alcohol. For propylamine, you get 1-propanol ($ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} $).
$ \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 \xrightarrow[\text{HCl}]{\text{NaNO}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} $
Since this alcohol is 1-propanol, which has the structural formula CH3CH2CH2OH, it does not possess the methyl-substituted carbinol group (–CH(OH)CH3) required for the iodoform test. Hence, it does not give iodoform test.
Step 4: Recognize the Reducing Nature of the Second Product (Formic Acid)
Formic acid ($ \text{HCOOH} $) can be easily oxidized to carbon dioxide ($ \text{CO}_2 $) and hence reduces mild oxidizing agents like Tollens’ reagent and Fehling’s solution. This confirms that the second product is indeed formic acid.
Step 5: Conclude the Parent Compound
Because the hydrolysis leads to a primary amine plus formic acid, and because one of the products obtained from the amine does not give the iodoform test, the original compound must be propyl isocyanide (CH3CH2CH2NC).
Final Answer
CH3CH2CH2NC is the correct compound.
