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Current Question (ID: 17633)

Question:

$\text{The major product(s) in the below reaction is-}$ $\text{H}_3\text{C} - \text{CH} = \text{CH}_2 \xrightarrow{\text{HBr}}^{(\text{Ph} - \text{CO} - \text{O})_2} S$

Options:

1. $\text{H}_3\text{C} - \text{CH}_2 - \text{CH}_2\text{Br}$ $1-\text{Bromopropane}$
2. $\text{CH}_3 - \text{CH} - \text{CH}_3 \ | \ \text{Br}$ $2-\text{Bromopropane}$
3. $\text{CH}_3 - \text{CH} = \text{CH}_2$ $\text{Propene}$
4. $\text{None of the above}$

Solution:

$\text{Hint: Follow Anti-Markovnikov rule}$ $\text{Explanation:}$ $\text{Step 1: In the presence of organic peroxides, the addition of HBr to propene follows anti-Markowinkov's rule (or peroxide effect) to form 1-bromopropane (n-propyl bromide).}$ $\text{Step 2: The reaction is as follows:}$ $\text{Ph - C - O - O - C - Ph} \rightarrow \text{2Ph} \cdot + \text{2CO}_2$ $\text{Ph} \cdot + \text{HBr} \rightarrow \text{C}_6\text{H}_6 + \text{Br} \cdot$ $\text{CH}_3 - \text{CH} = \text{CH}_2 + \text{Br} \cdot \rightarrow \text{CH}_3 - \text{CH}_2 - \text{CH}_2\text{Br}$ $\text{1-Bromopropane (more stable)}$

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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
}