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Current Question (ID: 10845)

Question:

$\text{The IUPAC name of } \text{CH}_3\text{CH}=\text{CHC}\equiv\text{CH} \text{ is:}$

Options:

1. $\text{Pent - 3 - en - 1 - yne}$ (Correct)
2. $\text{Pent - 2 - en - 4 - yne}$
3. $\text{Pent - 1 - yn - 3 - ene}$
4. $\text{Pent - 4 - yn - 2 - ene}$

Solution:

$\text{Hint: Triple bond is preferred over double bond.}$ $\text{Step 1:}$ $\text{If a molecule contains both carbon-carbon double or triple bonds, the numbering starts from the end which is nearer to the double bond followed by the triple bond, and according to the lowest sum of locant rule the numbering or sum rule will follow the alphabetical order of the substituent. However, if the sum of numbers turns out to be the same starting from either of the carbon chain, then the lowest number is given to the double bond. Here, the parent chain is having 5 carbon atoms and using the lowest sum of the locant rule the triple bond will get priority over the double bond.}$ $\text{Step 2:}$ $\text{The number of the parent carbon chain from right to left.}$ $\text{CH}_3\text{CH} = \text{CHC} \equiv \text{CH}$ $\text{with carbons numbered 5, 4, 3, 2, 1 respectively}$ $\text{Step 3:}$ $\text{The parent chain contains 5 carbon hence pent is used, for double bond "en" is used and for the triple bond "yne" is used}$ $\text{Thus, the correct name of this compound is Pent - 3 - en - 1 - yne.}$

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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
}