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Current Question (ID: 10930)

Question:
$\text{Free radical formation will take place in:}$
Options:
  • 1. $\text{H}_3\text{CO}-\text{OCH}_3 \longrightarrow \text{CH}_3\text{O}^\bullet + \text{CH}_3\text{O}^\bullet$ (Correct)
  • 2. $\text{H}_3\text{C}\text{COCH}_3 + \text{OH}^- \longrightarrow \text{H}_3\text{C}\text{COCH}_2^- + \text{H}_2\text{O}$
  • 3. $\text{(CH}_3\text{)}_3\text{CBr} \longrightarrow \text{(CH}_3\text{)}_3\text{C}^+ + \text{Br}^-$
  • 4. $\text{C}_6\text{H}_6 + \text{E}^+ \longrightarrow \text{C}_6\text{H}_5\text{E} + \text{H}^+$
Solution:
$\text{Hint: Homolytic cleavage takes place}$ $\text{Step 1:}$ $\text{The bond cleavage using curved arrows to show the electron flow of the given reaction can be represented as}$ $\text{CH}_3\text{O}-\text{OCH}_3 \rightarrow \text{CH}_3\text{O}^\bullet + \text{OCH}_3^\bullet$ $\text{Step 2:}$ $\text{It is an example of homolytic cleavage as one of the shared pairs in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical. The bond cleavage using curved arrows to show the electron flow of the given reaction can be represented as}$ $\text{H}_3\text{C}\text{COCH}_3 + \text{OH}^- \longrightarrow \text{H}_3\text{C}\text{COCH}_2^- + \text{H}_2\text{O}$ $\text{It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The reaction intermediate formed is carbanion.}$ $\text{(c) The bond cleavage using curved arrows to show the electron flow of the given reaction can be represented as}$ $\text{(CH}_3\text{)}_3\text{C}-\text{Br} \longrightarrow \text{(CH}_3\text{)}_3\text{C}^+ + \text{Br}^-$ $\text{It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromide ion. The reaction intermediate formed is a carbocation.}$ $\text{(d) The bond cleavage using curved arrows to show the electron flow of the given reaction can be represented as}$ $\text{C}_6\text{H}_6 + \text{E}^+ \longrightarrow \text{C}_6\text{H}_5\text{E} + \text{H}^+$ $\text{It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.}$

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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
}