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Current Question (ID: 17623)

Question:
$\text{The reaction of HBr with}$ $\begin{array}{c} \text{CH}_3 \\ \text{CH}_3 - \text{C} = \text{CH}_2 \end{array}$ $\text{in the presence of peroxide will give:}$
Options:
  • 1. $\begin{array}{c} \text{Br} \\ \text{CH}_3 - \text{C} - \text{CH}_3 \\ \text{CH}_3 \end{array}$
  • 2. $\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{Br}$
  • 3. $\begin{array}{c} \text{CH}_3 \\ \text{CH}_3 - \text{C} - \text{CH}_2 - \text{Br} \\ \text{H} \end{array}$
  • 4. $\begin{array}{c} \text{CH}_3 \\ \text{CH}_3 - \text{CH}_2 - \text{C} - \text{CH}_3 \\ \text{H} \end{array}$
Solution:
$\text{Hint: Follow anti-Markovnikov rule}$ $\text{When an alkene reacts with HBr in the presence of peroxide then alkyl bromide is obtained as a product. The Br atom is}$ $\text{attached to less substituted carbon of double bond and H is attached to more substituted carbon of double bond. In this}$ $\text{reaction, the Anti-Markovnikov rule is followed.}$ $\text{The reaction is as follows:}$ $\begin{array}{c} \text{CH}_3 - \text{C} = \text{CH}_2 \\ \text{HBr / Peroxide} \\ \text{anti, Markovnikoff addition} \\ \rightarrow \\ \text{CH}_3 - \text{CH} - \text{CH}_2 - \text{Br} \\ \text{CH}_3 \end{array}$

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{
  "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
}