Import Question JSON

Current Question (ID: 10873)

Question:

$\text{The number of chain isomers for } \text{C}_5\text{H}_{12} \text{ and } \text{C}_6\text{H}_{14} \text{ are, respectively:}$

Options:

1. $3, 3$ (Correct)
2. $3, 5$
3. $4, 4$
4. $3, 4$

Solution:

$\text{Hint: Chain isomers are molecules with the same molecular formula, but different arrangements of the carbon 'skeleton'}$ $\text{The chain isomers of } \text{C}_5\text{H}_{12} \text{ are as follows:}$ $\text{i. } \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \text{ (Pentane)}$ $\text{ii. } \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \text{ (2-Methylbutane)}$ $\text{iii. } \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}_3 \text{ (2,2-Dimethylpropane)}$ $\text{Total chain isomers for } \text{C}_5\text{H}_{12} = 3$ $\text{The chain isomers of } \text{C}_6\text{H}_{14} \text{ are as follows:}$ $\text{i. } \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \text{ (Hexane)}$ $\text{ii. } \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{CH}_3 \text{ (2-Methylpentane)}$ $\text{iii. } \text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \text{ (3-Methylpentane)}$ $\text{iv. } \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CH}_3 \text{ (2,3-Dimethylbutane)}$ $\text{v. } \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}_2-\text{CH}_3 \text{ (2,2-Dimethylbutane)}$ $\text{In the case of } \text{C}_6\text{H}_{14}\text{, only three chain isomers are possible. Consider only one chain isomer from ii and iii, and iv and v respectively, out of these pairs, another is the positional isomer, not the chain isomer.}$ $\text{Therefore, for } \text{C}_6\text{H}_{14}\text{, there are 3 distinct chain isomers based on different carbon skeleton arrangements.}$ $\text{The answer is 3 chain isomers for } \text{C}_5\text{H}_{12} \text{ and 3 chain isomers for } \text{C}_6\text{H}_{14}$

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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
}