How many of the following amine compounds for | vedclass

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(C) Hinsberg's reagent $(C_6H_5SO_2Cl)$ reacts with primary amines $(R-NH_2)$ to form $N$-alkylbenzenesulfonamide,which contains an acidic hydrogen at

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$5$

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(C) Hinsberg's reagent $(C_6H_5SO_2Cl)$ reacts with primary amines $(R-NH_2)$ to form $N$-alkylbenzenesulfonamide,which contains an acidic hydrogen atom attached to the nitrogen. This acidic hydrogen makes the product soluble in aqueous $NaOH$.Secondary amines $(R_2NH)$ form $N,N$-dialkylbenzenesulfonamide,which lacks an acidic hydrogen and is insoluble in $NaOH$.Tertiary amines $(R_3N)$ do not react with Hinsberg's reagent.Let us analyze the given compounds:$1$. Aniline $(C_6H_5NH_2)$: Primary amine,forms soluble product.$2$. $o$-methoxyaniline $(o-CH_3OC_6H_4NH_2)$: Primary amine,forms soluble product.$3$. Ethanamine $(CH_3CH_2NH_2)$: Primary amine,forms soluble product.$4$. $N$-phenyl-$p$-phenylenediamine $(C_6H_5NH-C_6H_4-NH_2)$: Contains a primary amine group,forms soluble product.$5$. Cyclohexanamine $(C_6H_{11}NH_2)$: Primary amine,forms soluble product.Other compounds like amides,ureas,and secondary/tertiary amines do not meet the criteria.There are $5$ such primary amine compounds.

How many of the following amine compounds form a product with Hinsberg's reagent $(C_6H_5SO_2Cl)$ that is soluble in $NaOH$?

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