Import Question JSON

Current Question (ID: 21294)

Question:
$\text{The major product of the following reaction is:}$ $\text{(image of reaction with } \text{NO}_2 \text{ and } \text{H}_2\text{SO}_4\text{)}$
Options:
  • 1. $\text{(image of option 1)}$
  • 2. $\text{(image of option 2)}$
  • 3. $\text{(image of option 3)}$
  • 4. $\text{(image of option 4)}$
Solution:
$\text{Hint: Electrophile reacts with a benzene ring and gives an electrophilic aromatic substitution reaction.}$ $\text{Step 1:}$ $\text{In the first step, nitroalkane reacts with an } \text{H}^+ \text{ ion and generate an electrophile. The reaction is as follows:}$ $\text{(image of reaction)}$ $\text{Step 2:}$ $\text{In the second step, electrophile reacts with a benzene ring and an electrophilic aromatic substitution reaction takes place. The reaction is known as the Friedel-Crafts alkylation reaction.}$ $\text{The reaction is as follows:}$ $\text{(image of reaction)}$ $\text{The reaction is depends on the stability of carbocation. In the reaction, two possible intermediate are formed. The structures are as follows:}$ $\text{(image of structures)}$ $\text{The primary carbocation is less stable than secondary carbocation because secondary carbocation has more hyperconjugation structure than primary carbocation. Here, stability of carbocation is explained by hyperconjugation effect not by inductive effect.}$

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
}