Import Question JSON

Current Question (ID: 17678)

Question:

$\text{Benzene} + \text{CH}_3\text{Cl} \overset{\text{AlCl}_3}{\longrightarrow} \text{C}_6\text{H}_5\text{CH}_3$ $\text{The mechanism \& intermediate involved in the above reaction are:}$

Options:

1. $\text{Aromatic electrophilic substitution \& carbocation}$
2. $\text{Aromatic Nucleophilic substitution \& carbanion}$
3. $\text{Aromatic free radical substitution \& Free radical}$
4. $\text{Carbene based substitution reaction \& Carbene}$

Solution:

$\text{Hint: The reaction is an example of Friedel craft alkylation reaction}$ $\text{Explanation:}$ $\text{Friedel craft alkylation is an example of an electrophilic aromatic}$ $\text{substitution reaction. In this reaction, benzene reacts with methyl chloride in}$ $\text{the presence of lewis acid and a toluene is obtained as a product. The}$ $\text{reaction is as follows:}$ $\text{Step 1: Removal of Cl from CH}_3\text{Cl to form the}$ $\text{electrophilic species CH}_3^+\text{ and AlCl}_4^-\text{.}$ $\text{Step 2: CH}_3^+\text{ attacks C=C bond in benzene which loses}$ $\text{its aromaticity to form a carbocation.}$ $\text{Step 3: Loss of proton (H}^+\text{) from the carbocation to}$ $\text{restore the aromaticity.}$ $\text{Step 4: H}^+\text{ reacts with AlCl}_4^-\text{ to give back AlCl}_3\text{.}$

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
}