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Current Question (ID: 10895)

Question:
$\text{Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. The major intermediate formed in the first step is -}$ $\text{H}_3\text{C} - \text{HC} = \text{CH}_2 + \text{H}^+ \rightarrow ?$
Options:
  • 1. $2^\circ \text{ carbanion}$
  • 2. $1^\circ \text{ carbocation}$
  • 3. $2^\circ \text{ carbocation}$ (Correct)
  • 4. $1^\circ \text{ carbanion}$
Solution:
$\text{Hint: Stability of carbocation}$ $\text{When electrophile attacks } \text{CH}_3 - \text{CH} = \text{CH}_2 \text{ delocalisation of electrons can take place, in two possible ways}$ $\text{Carbocation X is } 2^\circ \text{ and carbocation Y is } 1^\circ \text{. As } 2^\circ \text{ carbocation is more stable than } 1^\circ \text{ carbocation. Thus, option third correct.}$

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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
}