Compound P is neutral. Q gives effervescence | vedclass

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(B) Compound $P$ has the molecular formula $C_{14}H_{13}ON$. Upon hydrolysis with $HCl$ and $\Delta$,it yields a carboxylic acid $(Q)$ and an amine $(

Vedclass · Accepted Answer

$N$-phenyl$-4-$methylbenzamide

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(B) Compound $P$ has the molecular formula $C_{14}H_{13}ON$. Upon hydrolysis with $HCl$ and $\Delta$,it yields a carboxylic acid $(Q)$ and an amine $(R)$. $Q$ gives effervescence with $NaHCO_3$,indicating it is a carboxylic acid. $R$ reacts with Hinsberg's reagent $(PhSO_2Cl)$ to form a product that is soluble in $NaOH$,which confirms that $R$ is a primary amine ($1^{\circ}$ amine). Based on the molecular formula $C_{14}H_{13}ON$,the compound $P$ is $N$-phenyl$-4-$methylbenzamide $(CH_3-C_6H_4-CONH-C_6H_5)$. Hydrolysis of $N$-phenyl$-4-$methylbenzamide gives $4-$methylbenzoic acid $(Q)$ and aniline $(R)$. Aniline $(R)$ reacts with Hinsberg's reagent to form $N$-phenylbenzenesulfonamide,which is soluble in $NaOH$ due to the acidic hydrogen on the nitrogen atom.

Compound $P$ is neutral. $Q$ gives effervescence with $NaHCO_3$ while $R$ reacts with Hinsberg's reagent to give a solid soluble in $NaOH$. Compound $P$ is:

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