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Current Question (ID: 21295)

Question:
$\text{The correct sequence of reagents used in the preparation of 4-Bromo-2-nitro-1-ethyl benzene from benzene is:}$
Options:
  • 1. $\text{HNO}_3/\text{H}_2\text{SO}_4, \text{Br}_2/\text{AlCl}_3, \text{CH}_3\text{COCl}/\text{AlCl}_3, \text{Zn} - \text{Hg}/\text{HCl}$
  • 2. $\text{Br}_2/\text{AlBr}_3, \text{CH}_3\text{COCl}/\text{AlCl}_3, \text{HNO}_3/\text{H}_2\text{SO}_4, \text{Zn}/\text{HCl}$
  • 3. $\text{CH}_3\text{COCl}/\text{AlCl}_3, \text{Br}_2/\text{AlBr}_3, \text{HNO}_3/\text{H}_2\text{SO}_4, \text{Zn}/\text{HCl}$
  • 4. $\text{CH}_3\text{COCl}/\text{AlCl}_3, \text{Zn} - \text{Hg}/\text{HCl}, \text{Br}_2/\text{AlBr}_3, \text{HNO}_3/\text{H}_2\text{SO}_4}$
Solution:
$\text{Hint: First acyl group is introduced and convert into ethyl group}$ $\text{Step 1:}$ $\text{The desired transformation is as follows:}$ $\text{benzene} \rightarrow \text{4-Bromo-2-nitro-1-ethyl benzene}$ $\text{Step 2:}$ $\text{The desired transformation occurs in four steps. In the first step, Friedel-Crafts acylation reaction occurs. In this reaction,}$ $\text{the acyl group is attached to the benzene ring. In the second step, Clemmensen reduction takes place and reduction of C=O}$ $\text{group takes place. In the third step, bromination occurs at the para position because the alkyl group is ortho/para directing.}$ $\text{In the fourth step, nitration occurs at the ortho position with respect to the alkyl group.}$ $\text{The overall reactions are as follows:}$ $\text{benzene} \xrightarrow{\text{CH}_3\text{COCl}/\text{AlCl}_3} \text{acylated benzene} \xrightarrow{\text{Zn}-\text{Hg}/\text{HCl}} \text{ethyl benzene} \xrightarrow{\text{Br}_2/\text{AlBr}_3} \text{bromoethyl benzene} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{4-Bromo-2-nitro-1-ethyl benzene}$

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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
}