Q40 / 155 · WBJEE 2021 · Chemistry

The product (s) in the following sequence of reactions will be $\mathrm{Me}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Me} \frac{\begin{array}{l}\text { 1. } \mathrm{Na} / \mathrm{NH}_{3} \text { (liq.) } \\ \text { ethanol, }-33^{\circ} \mathrm{C}\end{array}}{\text { 2. dil. alkaline } \mathrm{KMnO}_{4}}$ Product (s)

  • A
    Figure
  • Correct answer
    B
    Figure
  • Correct answer
    C
    Figure
  • D
    Figure

Hence, the correct answer is options: 'B', 'C'.

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