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Current Question (ID: 21347)

Question:

$\text{In the following reaction sequence, identify the major product 'C':}$ $\text{C}_8\text{H}_{10} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{A} \xrightarrow{\text{Br}_2/\Delta} \text{B} \xrightarrow{\text{alcoholic KOH}} \text{C}$

Options:

1. $\begin{array}{c} \text{NO}_2 \\ \text{I} \\ \text{C} = \text{CH}_2 \end{array}$
2. $\begin{array}{c} \text{O}_2\text{N} \\ \text{CH} = \text{CH}_2 \end{array}$
3. $\begin{array}{c} \text{O}_2\text{N} \\ \text{C} = \text{CH}_2 \\ \text{NO}_2 \end{array}$
4. $\begin{array}{c} \text{NO}_2 \\ \text{CH} = \text{CH}_2 \end{array}$

Solution:

$\text{Hint: Bromination of para-nitroethylbenzene with bromine under heat (A) will occur at the benzylic position.}$ $\text{Step 1: The molecular formula of the given compound is C}_8\text{H}_{10}. \text{ So, the degree of unsaturation for the given compound will be:}$ $\frac{2 \times 8 - 10 + 2}{2} = 4$ $\text{Thus, the compound must have a benzene ring. The reactant is ethylbenzene.}$ $\text{Step 2: Nitration of ethyl benzene: Nitration of ethylbenzene (C}_8\text{H}_{10}) \text{ with HNO}_3 \text{ and H}_2\text{SO}_4 \text{ yields a mixture of ortho-nitroethylbenzene and para-nitroethylbenzene. The major product will be para-nitroethylbenzene due to less steric hindrance.}$ $\text{Step 3: Bromination of A: Bromination of para-nitroethylbenzene with bromine under heat (}\Delta\text{) will occur at the benzylic position. This yields 1-(bromoethyl)-4-nitrobenzene.}$ $\text{Step 4: Treatment of 1-(bromoethyl)-4-nitrobenzene with alcoholic KOH leads to elimination. The major product will be 4-nitrostyrene.}$ $\text{The complete reaction is given as follows:}$

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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
}