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Current Question (ID: 21217)

Question:

$\text{The order of stability of given carbocations is:}$

Options:

1. $\text{A > B > C}$
2. $\text{C > B > A}$
3. $\text{B > A > C}$
4. $\text{B > C > A}$

Solution:

$\text{Hint: Resonance is a more powerful effect than inductive effect.}$ $\text{Carbocations are stabilized if the positive charge can be delocalized through resonance with adjacent atoms or groups. It can also be stabilized by +M effect of the oxygen atom.}$ $\text{Electron-donating groups (such as alkyl groups, -OH, or -OR) stabilize the carbocation by donating electron density to the positively charged carbon.}$ $\text{Electron-withdrawing groups (such as -NO}_2 \text{ or -CF}_3\text{) destabilize the carbocation because they pull electron density away from the positively charged carbon, increasing the electron deficiency.}$ $\text{Among the given compounds, B is most stable due to the resonance effect caused by the lone pair of oxygen atom. A is more stable than C because in compound C, -I effect of oxygen destabilizes the carbocation.}$ $\text{So, the correct order of stability of the compounds is: B > A > C.}$

Import JSON File

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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
}