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Current Question (ID: 10858)

Question:

$\text{Which types of isomerism can be exhibited by compounds with the molecular formula } \text{C}_4\text{H}_{11}\text{N}\text{?}$

Options:

1. $\text{Position isomerism}$
2. $\text{Metamerism}$
3. $\text{Functional isomerism}$
4. $\text{All of the above}$ (Correct)

Solution:

$\text{Hint: Amine general formula is } \text{RNH}_2$ $\text{Amine can show position isomerism, metamerism, and functional isomerism.}$ $\text{In position isomerism, the position of } \text{NH}_2 \text{ group is different but the parent chain is the same. For example, butan-1-amine and butan-2-amine.}$ $\text{In metamerism, different alkyl groups on either side of the functional group are present. For example, secondary amines with different alkyl groups attached to nitrogen.}$ $\text{It can also show functional group isomerism. } \text{C}_4\text{H}_{11}\text{N} \text{ can form primary, secondary and tertiary amines, and it is an example of functional group isomerism.}$

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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
}