JEE Advanced · Chemistry · 24. Haloalkanes & Haloarenes
The reaction sequence given below is carried out with 16 moles of \(\mathbf{X}\). The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of \(\mathbf{S}\) produced is ______ .
Use: Atomic mass (in amu) : \(\mathrm{H}=1, \mathrm{C}=\) \(12, \mathrm{O}=16, \mathrm{Br}=80\)
- A 175
- B 185
- C 195
- D 155
Answer & Solution
Correct Answer
(A) 175
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{M}=\left(\mathrm{C}_6 \mathrm{H}_4\right)_2\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \\
& =350
\end{aligned}\)
Weight of product of \(S=350 \times 1 / 2=175 \mathrm{gm}\) Ans.
* WES : Williamson ether synthesis
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