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

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Aldehydes Ketones and Carboxylic Acids

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)

Answer: 495

Solution

Starting with 10 moles of acetophenone (MW = 120), the reaction sequence with given yields produces compound D. Following the transformations and multiplying by the yields at each step: 10 moles × 0.8 × 0.5 × 0.75 × 0.5 = 1.5 moles of D. The molecular weight of D (tribromoacetophenone derivative) is 330 g/mol. Amount of D = 1.5 × 330 = 495 g.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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