Import Question JSON

Current Question (ID: 10899)

Question:

$\text{The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H}^+ \text{ ion on the double bond portion. The same can be shown as-}$

Options:

1. $\text{H}^+ \text{ attacking one carbon of the double bond with arrow pointing from H}^+ \text{ to the carbon}$
2. $\text{H}^+ \text{ attacking one carbon of the double bond with arrow pointing from the double bond to H}^+$ (Correct)
3. $\text{H}^+ \text{ attacking the center of the double bond with arrow pointing from H}^+ \text{ to the bond}$
4. $\text{All of these are possible}$

Solution:

$\text{Hint: The addition of electrophile on the double bond of alkene occur in such a manner so that it can make more stable carbocation.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\pi\text{-bonds create an electron cloud. The }\pi\text{-bond acts as a nucleophile and attacks the electrophile (H}^+\text{) from H-Cl. The arrow always start from nucleophile and always end at electrophile. Here, carbocation is formed. The reaction is as follows:}$ $\text{[Chemical reaction showing }\pi\text{-bond electrons attacking H}^+\text{ to form carbocation]}$ $\text{Step 2:}$ $\text{The chloride anion attacks the carbocation. The reaction is as follows:}$ $\text{[Chemical reaction showing Cl}^-\text{ attacking the carbocation to form the final product]}$

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
}