JEE Mains · Chemistry · STD 11 - 9. Hydricarbon
In bromination of Propyne, with Bromine \(1,1,2, \;2-\)tetrabromopropane is obtained in \(27\%\) yield. The amount of \(1,1,2,2\) tetrabromopropane obtained from \(1\; g\) of Bromine in this reaction is \(\dots \;\times10^{-1}\; g\). (Nearest integer) (Molar Mass : Bromine \(=80\; g / mol\) )
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(=\frac{1}{160} \times \frac{1}{2} \times 360 \times 0.27\) \(=0.30375\) \(=3.0375 \times 10^{-1}\) Ans \(=3\)
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