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Current Question (ID: 10867)

Question:

$\text{The pair that represents chain isomers is:-}$

Options:

1. $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{C}(=\text{O})-\text{OH} \text{ and } \text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{OCH}_3$
2. $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{CH}_3 \text{ and } \text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$
3. $\text{CH}_3-\text{CH}_2-\text{CH}(\text{CN})-\text{CH}_3 \text{ and } \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CN}$ (Correct)
4. $\text{CH}_3-\text{CH}_2-\text{CH}(\text{Cl})-\text{CH}_3 \text{ and } \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Cl}$

Solution:

$\text{Hint: Molecules containing the same parent chain and having the same molecular formula.}$ $\text{In the first option, both compounds contain the same parent chain that is pentanoic acid.}$ $\text{In the second option, both compounds contain the same parent chain that is pentane acid.}$ $\text{In the third option, both compounds contain different parent chains. The first structure parent chain name is butanenitrile and the second structure parent chain name is pantanenitrile.}$ $\text{In the fourth option, both compounds contain the same parent chain that is butane acid.}$ $\text{Chain isomers are compounds with the same molecular formula but different carbon chain arrangements. Option 3 shows butanenitrile and pentanenitrile, which have different parent chain lengths, making them chain isomers.}$

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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
}