For the reaction shown, which statement is co | vedclass

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(C) The reaction shown is the Hofmann bromamide degradation of $2$-methylbenzamide using $NaOBr$, which yields $2$-methylaniline (o-toluidine) as Prod

Vedclass · Accepted Answer

It has one isomer of $2^\circ$ amine.

Vedclass · Answer

(C) The reaction shown is the Hofmann bromamide degradation of $2$-methylbenzamide using $NaOBr$, which yields $2$-methylaniline (o-toluidine) as Product "$X$".$2$-methylaniline is a primary aromatic amine $(Ar-NH_2)$.$1$. It reacts with Hinsberg's reagent (benzenesulfonyl chloride) to form a sulfonamide, which is soluble in alkali.$2$. It reacts with nitrous acid $(HNO_2)$ at $0-5^{\circ}C$ to form a diazonium salt, which gives a positive Azo dye test with $\beta$-naphthol.$3$. It is a primary amine, and its $2^\circ$ amine isomer is $N$-methylaniline $(C_6H_5NHCH_3)$.$4$. Primary aromatic amines are generally insoluble in aqueous $NaOH$ because they are basic in nature, not acidic.Therefore, the correct statement is that it has one isomer of $2^\circ$ amine, which is $N$-methylaniline.

For the reaction shown, which statement is correct for Product "$X$"?

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