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Current Question (ID: 21296)

Question:

$\text{Given the following reaction:}$ $\text{HC} = \text{CHBr}$ $\begin{array}{l} \text{CH}_3 \\ \end{array}$ $\begin{array}{l} 1. \text{NaN}_2 \\ 2. \text{Red hot iron tube, 873 K} \end{array} \rightarrow (\text{A}) \text{(major product)}$ $\text{The major product A is:}$

Options:

1. $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$
2. $\begin{array}{c} \text{H} \\ \text{HC} = \text{C} - \text{NH}_2 \\ \text{CH}_3 \end{array}$
3. $\text{Benzene}$
4. $\begin{array}{c} \text{H}_3\text{C} \\ \text{CH}_3 \end{array} \text{C}_6\text{H}_3$

Solution:

$\text{Hint: In the first reaction, alkyne is obtained as product}$ $\text{The first reaction is an example of a dehydrohalogenation reaction.}$ $\text{This reaction is also known as an elimination reaction.}$ $\text{In this reaction, bromoalkene reacts with sodamide and propynes is obtained as a product.}$ $\text{In the second reaction, propyne derivative on passing through a red hot iron tube at 873K undergoes cyclic polymerization.}$ $\text{Three molecules polymerise to form an alkyl derivative of benzene.}$ $\text{The reactions are as follows:}$ $\text{CH} = \text{CHBr} \xrightarrow{\text{NaNH}_2} \text{C} \equiv \text{CH}$ $\begin{array}{l} \text{CH}_3 \\ \end{array}$ $\xrightarrow{\text{Red hot iron tube 873 K}} \text{H}_3\text{C} \begin{array}{c} \text{CH}_3 \\ \end{array} \text{C}_6\text{H}_3$ $\text{pseudocumene}$

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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
}