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Current Question (ID: 10904)

Question:

$\text{The most stable carbocation among the following is-}$

Options:

1. $(\text{CH}_3)_3\text{C}^+\text{CH}_2$
2. $(\text{CH}_3)_3\text{C}^+$ (Correct)
3. $\text{CH}_3\text{CH}_2\text{CH}_2^+$
4. $(\text{CH}_3)\text{CH}^+\text{CH}_2\text{CH}_3$

Solution:

$\text{Hint: Carbocation stability order : Tertiary > Secondary > Primary}$ $\text{Explanation:}$ $\text{The order of stability of carbocation is as follows:}$ $\text{Tertiary carbocation > Secondary carbocation > Primary carbocation > Methyl carbocation}$ $\text{A tertiary carbocation is the most stable carbocation due to the hyperconjugation effect of alpha hydrogen. The tertiary carbocation given in option 2 has 9 alpha hydrogen. Hence, it can form 9 hyperconjugation structures.}$ $\text{Hence, option second is the correct answer.}$

Import JSON File

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
}