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Basics of Organic Chemistry question

2023 · 12 Apr · Shift 1 · Q17
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Basics of Organic Chemistry question

2023 · 12 Apr · Shift 1 · Q17

JEE MainChemistryBasics of Organic ChemistryNumerical+4 / −1
Three organic compounds A, B and C\mathrm{C}C were allowed to run in thin layer chromatography using hexane and gave the following result (see figure). The Rf\mathrm{R}_{\mathrm{f}}Rf​ value of the most polar compound is ‾\underline{\hspace{2cm}}​×10−2\times 10^{-2}×10−2 JEE Main 2023 (Online) 12th April Morning Shift Chemistry - Basics of Organic Chemistry Question 102 English
Numerical answer
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Correct answer: 25

  1. In thin layer chromatography (TLC), the stationary phase is usually silica gel, which is polar.

  2. The mobile phase here is hexane, which is non-polar.

  3. Therefore:

    • More polar compound interacts more strongly with silica.
    • So it moves less distance up the plate.
    • Hence, the most polar compound has the smallest RfR_fRf​ value.
  4. Recall: Rf=distance travelled by compounddistance travelled by solvent frontR_f = \frac{\text{distance travelled by compound}}{\text{distance travelled by solvent front}}Rf​=distance travelled by solvent frontdistance travelled by compound​

  5. From the given TLC figure, the spot corresponding to the most polar compound is the one that travelled the least distance. Reading the distances from the figure gives: Rf=0.25R_f = 0.25Rf​=0.25

  6. The question asks for the value in the form: ‾×10−2\underline{\hspace{2cm}} \times 10^{-2}​×10−2

    Since 0.25=25×10−20.25 = 25 \times 10^{-2}0.25=25×10−2

  7. Therefore, the required integer is: 252525

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