Import Question JSON

Current Question (ID: 21246)

Question:

$\text{The order of stability of the following carbocations is:}$ $\begin{array}{ccc} \text{H}_2\text{C} = \text{C}^+ - \text{CH}_2 & \text{H}_3\text{C} - \overset{+}{\text{CH}} - \text{CH}_2 & \overset{+}{\text{CH}_2} \text{(benzene ring)} \\ \text{(I)} & \text{(II)} & \text{(III)} \end{array}$

Options:

1. $\text{I} > \text{II} > \text{III}$
2. $\text{III} > \text{I} > \text{II}$
3. $\text{III} > \text{II} > \text{I}$
4. $\text{II} > \text{III} > \text{I}$

Solution:

$\text{Hint: Resonance is more powerful than hyperconjugation and inductive effect.}$ $\text{Explanation:}$ $\text{The stability of carbocation is explained by inductive, hyperconjugation and resonance effect. The power of effect is:}$ $\text{resonance} > \text{hyperconjugation} > \text{inductive effect.}$ $\text{The first and third carbocation is resonance stabilized but II carbocation is stabilized by hyperconjugation.}$ $\text{The resonance effect is more powerful than hyperconjugation, hence, I and III is more stable than II.}$ $\text{The III structure has more resonating structures, hence it is more stable than I.}$ $\begin{array}{l} (1) \ \text{CH}_2 = \text{CH} - \overset{+}{\text{CH}}_2; \ \text{(I) Less conjugation, less stable.} \\ (2) \ \text{CH}_3 - \text{CH}_2 - \overset{+}{\text{CH}}_2; \ \text{(II) Non conjugated, so least stable.} \\ (3) \ \overset{+}{\text{CH}_2} \text{(benzene ring)}; \ \text{(III) More conjugation, thus more stable.} \end{array}$ $\text{Hence, the stability order is III > I > II. The correct answer is option 2.}$

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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
}