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Current Question (ID: 10940)

Question:

$\text{The double bond in the above-mentioned compound that accepts proton (H}^+\text{) fastest is:}$ $\text{The compound shows a bicyclic structure with methyl groups at specific positions, with double bonds labeled A, B, C, and D.}$

Options:

1. $\text{A}$ (Correct)
2. $\text{B}$
3. $\text{C}$
4. $\text{D}$

Solution:

$\text{The structure of carbocations are as follows:}$ $\text{When the compound undergoes protonation at different double bonds, various carbocations are formed:}$ $\text{Protonation at position A: Forms a 3° allylic carbocation, stabilized by hyperconjugation effect}$ $\text{Protonation at position B: Forms a tertiary carbocation stabilized by hyperconjugation}$ $\text{Protonation at position C: Forms a tertiary carbocation stabilized by hyperconjugation}$ $\text{Protonation at position D: Forms a secondary carbocation}$ $\text{Analysis of carbocation stability:}$ $\text{The carbocation formed by protonation at position A is the most stable because it is stabilized by both resonance effect and hyperconjugation effect.}$ $\text{The 3° allylic carbocation (from position A) has maximum stability due to:}$ $\text{1. Tertiary nature of the carbocation}$ $\text{2. Allylic stabilization through resonance}$ $\text{3. Hyperconjugation with adjacent methyl groups}$ $\text{Therefore, position A accepts proton fastest as it forms the most stable carbocation intermediate.}$

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
}