Question

The product of an amine 'X' with benzene sulphonyl chloride produces the product which is insoluble in alkali. The product of 'X' with ethanoyl chloride is:

• A. C$_6$H$_5$NHCOCH$_3$
• B. C$_6$H$_5$N(CH$_3$)COCH$_3$
• C. C$_6$H$_5$N(CH$_3$)CH$_2$CH$_3$
• D. C$_6$H$_5$NHCH$_2$CH$_3$

Hint:

Hinsberg's test is a crucial method for distinguishing between primary, secondary, and tertiary amines based on the solubility of their sulphonamide derivatives in alkali. Acyl chlorides react with amines to form amides.

Answer 1

The Correct Option is B

Step 1: Recall Hinsberg's test.
Benzene sulphonyl chloride (C$_6$H$_5$SO$_2$Cl), also known as Hinsberg's reagent, reacts with amines to form sulphonamides. The solubility of the sulphonamide product in alkali depends on the type of amine (primary, secondary, or tertiary). \begin{itemize} \item Primary amines react to form N-alkyl or N-aryl benzene sulphonamides, which have an acidic hydrogen attached to nitrogen and are soluble in alkali. \item Secondary amines react to form N,N-dialkyl or N-alkyl-N-aryl benzene sulphonamides, which do not have an acidic hydrogen attached to nitrogen and are insoluble in alkali. \item Tertiary amines do not react with benzene sulphonyl chloride. \end{itemize}
Step 2: Determine the type of amine 'X'.
The problem states that the product of amine 'X' with benzene sulphonyl chloride is insoluble in alkali. Based on Hinsberg's test, this indicates that 'X' must be a secondary amine.
Step 3: Analyze the options for the product of 'X' with ethanoyl chloride (CH$_3$COCl).
Ethanoyl chloride is an acyl chloride that reacts with amines via nucleophilic acyl substitution to form amides. Since 'X' is a secondary amine, its reaction with ethanoyl chloride will form an N,N-disubstituted amide. Let's examine the options to see which one is an amide formed from a secondary amine containing a benzene ring. \begin{itemize} \item (1) C$_6$H$_5$NHCOCH$_3$: This is N-phenylacetamide, formed from a primary amine (aniline, C$_6$H$_5$NH$_2$). \item (2) C$_6$H$_5$N(CH$_3$)COCH$_3$: This is N-methyl-N-phenylacetamide, formed from a secondary amine (N-methylaniline, C$_6$H$_5$NHCH$_3$). \item (3) C$_6$H$_5$N(CH$_3$)CH$_2$CH$_3$: This is N-ethyl-N-methylaniline, which is a tertiary amine and not a product of reaction with ethanoyl chloride. \item (4) C$_6$H$_5$NHCH$_2$CH$_3$: This is N-ethylaniline, a secondary amine, but the product with ethanoyl chloride would be N-ethyl-N-phenylacetamide (C$_6$H$_5$N(CH$_2$CH$_3$)COCH$_3$), not this compound. \end{itemize} Since 'X' is a secondary amine containing a phenyl group, the most likely secondary amine is N-methylaniline (C$_6$H$_5$NHCH$_3$), which would react with benzene sulphonyl chloride to give N-methyl-N-phenylbenzenesulfonamide (C$_6$H$_5$SO$_2$N(CH$_3$)C$_6$H$_5$), insoluble in alkali. The reaction of N-methylaniline with ethanoyl chloride would produce N-methyl-N-phenylacetamide (C$_6$H$_5$N(CH$_3$)COCH$_3$). Final Answer: \[ \boxed{\text{C}_6\text{H}_5\text{N(CH}_3\text{)COCH}_3} \]