JEE Mains · Chemistry · STD 11 - 9. Hydricarbon
In the following sequence of reactions the maximum number of atoms present in molecule '\(C\)' in one plane is \(A \xrightarrow[ { Cu\; tube }]{\text { Redhot }}\mathrm{B} \xrightarrow[ Anhydrous AlCl_3]{\mathrm{CH}_{3} \mathrm{Cl}(1 \mathrm{eq}} \mathrm{C}\) (\(A\) is a lowest molecular weight alkyne)
- A \(13\)
- B \(15\)
- C \(11\)
- D \(9\)
Answer & Solution
Correct Answer
(A) \(13\)
Step-by-step Solution
Detailed explanation
Total \(13\) atom are present in same plane (\(7\) carbon and \(6\) hydrogen atoms.)
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