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(A) Step $1$: Acetophenone $(10 \ mol)$ $\xrightarrow{NaOBr, H_3O^+}$ Benzoic acid $(A)$ with $60\%$ yield. Moles of $A = 10 	imes 0.6 = 6 \ mol$.Ste

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$495$

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(A) Step $1$: Acetophenone $(10 \ mol)$ $\xrightarrow{NaOBr, H_3O^+}$ Benzoic acid $(A)$ with $60\%$ yield. Moles of $A = 10 	imes 0.6 = 6 \ mol$.Step $2$: Benzoic acid $(A)$ $\xrightarrow{NH_3, \Delta}$ Benzamide $(B)$ with $50\%$ yield. Moles of $B = 6 	imes 0.5 = 3 \ mol$.Step $3$: Benzamide $(B)$ $\xrightarrow{Br_2/KOH}$ Aniline $(C)$ with $50\%$ yield. Moles of $C = 3 	imes 0.5 = 1.5 \ mol$.Step $4$: Aniline $(C)$ $\xrightarrow{Br_2 (3 \ equiv), AcOH}$ $2,4,6$-Tribromoaniline $(D)$ with $100\%$ yield. Moles of $D = 1.5 	imes 1.0 = 1.5 \ mol$.Molar mass of $D$ $(C_6H_4NBr_3) = (6 	imes 12) + (4 	imes 1) + 14 + (3 	imes 80) = 72 + 4 + 14 + 240 = 330 \ g \ mol^{-1}$.Amount of $D = 1.5 \ mol 	imes 330 \ g \ mol^{-1} = 495 \ g$.

In the following reaction sequence, the amount of $D$ (in $g$) formed from $10$ moles of acetophenone is. . . . . . (Atomic weights in $g \ mol^{-1}: H = 1, C = 12, N = 14, O = 16, Br = 80$. The yield (%) corresponding to the product in each step is given in the parenthesis)

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