The amine which can react with ${C_6}{H_5} - S{O_2} - Cl$ to form a product insoluble in alkali shall be
A
Primary amine
B
Secondary amine
C
Tertiary amine
D
Both primary and secondary amines
Medium
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B
Secondary amine
b (b) ${C_6}{H_5}S{O_2}Cl$ is called Hinsberg’s reagent they react with sec amine to form a product in soluble in alkalies. This reaction used to separate ${1^o}$, ${2^o}$ and ${3^o}$ amine from their mixture.
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In the given reaction $\begin{array}{*{20}{c}}
{{C_6}{H_5} - C - C{H_3}} \\
{||} \\
O
\end{array}$ $\xrightarrow{{N{H_2}OH/{H^ \oplus }}}X\xrightarrow{{Na/{C_2}{H_5}OH}}Y$. $Y$will be
Assertion : Aniline does not undergo Friedel- Crafts reaction. Reason : $-NH_2$ group of aniline reacts with $AlCl_3$ (Lewis acid) to give acid-base reaction