Import Question JSON

Current Question (ID: 17609)

Question:

$\text{The main product A and B in the above mentioned reaction are respectively-}$

Options:

1. $\text{(A) } \text{Br-CH}_2\text{-CH=CH}_2 \quad \text{(B) } \text{Br-CH}_2\text{-CH}_2\text{-CH}_3$
2. $\text{(A) } \text{Br-CH}_2\text{-CH}_2\text{-CH}_3 \quad \text{(B) } \text{Br-CH}_2\text{-CH=CH}_2$
3. $\text{(A) } \text{Br-CH}_2\text{-CH}_2\text{-CH}_3 \quad \text{(B) } \text{Br-CH}_2\text{-CH}_2\text{-CH}_3$
4. $\text{(A) } \text{Br-CH}_2\text{-CH=CH}_2 \quad \text{(B) } \text{Br-CH}_2\text{-CH=CH}_2$

Solution:

$\text{Hint: The first reaction follows the free radical addition reaction.}$ $\text{Step 1:}$ $\text{When alkene reacts with HBr in the presence of peroxide then a free radical}$ $\text{mechanism takes place.}$ $\text{Hydrogen atom is attached to more substituted carbon of double bond.}$ $\text{Step 2:}$ $\text{In the second reaction, the first step is the formation of the carbocation as}$ $\text{an intermediate. Here, rearrangement of}$ $\text{carbocation also takes place because benzyl carbocation is more stable.}$ $\text{Then benzyl radical reacts with bromide ion and}$ $\text{an alkyl bromide is formed as a product.}$ $\text{The reaction is as follows:}$ $\text{The mechanism of the reaction is as follows:}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}