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Current Question (ID: 17594)

Question:

$\text{An unsaturated hydrocarbon 'A' reacts with two molecules of H}_2 \text{ and upon}$ $\text{reductive ozonolysis A gives butane-1,4-dial, ethanal, and propanone.}$ $\text{The IUPAC name of A is:}$

Options:

1. $\text{2-Methylocta-2,6-diene}$
2. $\text{2-Methylocta-1,5-diene}$
3. $\text{3-Methylocta-2,6-diene}$
4. $\text{2-Methylocta-1,6-diene}$

Solution:

$\text{Hint: Reductive ozonolysis: Alkene reacts with ozone followed by Zn.}$ $\text{Explanation:}$ $\text{Step 1: The addition of H-atom on alkene is called hydrogenation. The}$ $\text{addition of ozone to alkene molecules to form}$ $\text{ozonide is called ozonolysis.}$ $\text{This reaction is useful in detecting the position of a double bond in alkenes.}$ $\text{Use the properties that 'A' should have to go}$ $\text{through hydrogenation and ozonolysis.}$ $\text{Step 2: See the reaction process}$ $\text{A + 2H}_2$ $\begin{array}{cccccccccccccccc} \text{H} & & \text{O} & & \text{O} & & \text{O} & & \text{O} & & \text{O} & & \text{O} & & \text{O} \\ \text{C} & = & \text{C} & + & \text{C} & = & \text{C} & - & \text{C} & - & \text{H} & + & \text{C} & - & \text{C} \\ & & & & & & & & & & & & & & & \\ \text{H} & & & & & & & & & & & & & & & \\ \text{H} & & & & & & & & & & & & & & & \\ \text{C} & = & \text{C} & & & & & & & & & & & & & \\ \text{8} & & \text{7} & & \text{6} & & \text{5} & & \text{4} & & \text{3} & & \text{2} & & \text{1} \\ \end{array}$ $\text{Hence, the IUPAC name of the compound is 2-Methylocta-2,6-diene}$

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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
}