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(A) The hydrolysis of an alkyl isocyanide $(R-NC)$ yields a primary amine $(R-NH_2)$ and formic acid $(HCOOH)$.Here,the compound is $CH_3CH_2CH_2NC$ (

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$CH_3CH_2CH_2NC$

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(A) The hydrolysis of an alkyl isocyanide $(R-NC)$ yields a primary amine $(R-NH_2)$ and formic acid $(HCOOH)$.Here,the compound is $CH_3CH_2CH_2NC$ (propyl isocyanide).Hydrolysis: $CH_3CH_2CH_2NC + 2H_2O \rightarrow CH_3CH_2CH_2NH_2 + HCOOH$.$1$. $CH_3CH_2CH_2NH_2$ (propylamine) reacts with $NaNO_2/HCl$ to form $CH_3CH_2CH_2OH$ (propan$-1-$ol),which does not give the iodoform test.$2$. $HCOOH$ (formic acid) reduces both Tollens reagent and Fehling's solution.Thus,the compound is $CH_3CH_2CH_2NC$.

On hydrolysis of a "compound",two compounds are obtained. One of which on treatment with sodium nitrite and hydrochloric acid gives a product which does not respond to iodoform test. The second one reduces Tollens reagent and Fehling's solution. The "compound" is

Similar Questions

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