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Current Question (ID: 17637)

Question:

$\text{The type of radicals that can be formed as intermediates during}$ $\text{monochlorination of 2-methylpropane is-}$

Options:

1. $\text{Primary and tertiary radicals}$
2. $\text{Two types of primary radicals}$
3. $\text{Primary and secondary radicals}$
4. $\text{Two types of tertiary radical}$

Solution:

$\text{Hint: All different types of hydrogen atoms replaced and form free radical}$ $\text{Step 1:}$ $\text{The structure of 2-methylpropane is as follows:}$ $\text{The substrate contains two types of carbon atoms primary carbon and}$ $\text{tertiary carbon. Hence, two type of free radical are formed.}$ $\text{Make free-radical at both possible carbon of substrate. one carbon is tertiary}$ $\text{radical and other is primary radical}$ $\text{Step 2:}$ $\text{The stability of the free radical will be decided on the basis of the}$ $\text{hyperconjugation electronic effect. On the basis of}$ $\text{hyperconjugation tertiary free radical is more stabilized than primary}$ $\text{radical.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}