The product (C) of the given reaction sequenc | vedclass

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(C) $1$. Phthalic anhydride reacts with methanol $(MeOH)$ via nucleophilic acyl substitution to open the anhydride ring,forming monomethyl phthalate $

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(C) $1$. Phthalic anhydride reacts with methanol $(MeOH)$ via nucleophilic acyl substitution to open the anhydride ring,forming monomethyl phthalate $(A)$,which is $C_6H_4(COOCH_3)(COOH)$.$2$. Treatment of the carboxylic acid group in $(A)$ with $PCl_3$ converts it into an acid chloride,yielding methyl $2-$(chlorocarbonyl)benzoate $(B)$,which is $C_6H_4(COOCH_3)(COCl)$.$3$. Finally,the reaction of the acid chloride $(B)$ with methylamine $(MeNH_2)$ replaces the chlorine atom with a methylamino group,resulting in the final product $(C)$,which is methyl $2-$(methylcarbamoyl)benzoate,$C_6H_4(COOCH_3)(CONHCH_3)$.

Column-$I$ (Formula)	Column-$II$ (Name)
$(A)$ $CH_3-CH=CHCHO$	$(i)$ $1,3$-diphenylprop-$2$-en-$1$-one
$(B)$ $CH_3-C(CH_3)=CH-COCH_3$	(ii) Acetophenone-$2,4$-dinitrophenylhydrazone
$(C)$ $C_6H_5-CH=CH-COC_6H_5$	(iii) But-$2$-enal
$(D)$ $C_6H_5-C(CH_3)=N-NH-C_6H_3(NO_2)_2$	(iv) $4$-methylpent-$3$-en-$2$-one

The product $(C)$ of the given reaction sequence is:

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