JEE Chemistry 2023 Question with Solution

Step-by-step solution

Standard Method

Given: Three organic compounds were separated by thin layer chromatography using hexane. The solution indicates that the most polar compound has the smallest $R_f$ value.

Find: The $R_f$ value of the most polar compound in the form _____ $\times 10^{-2}$.

From the extracted solution:

More $R_f$ less its polarity

$$R_f = \frac{\text{Distance travelled by compound } X}{\text{Distance travelled by solvent } Y}$$

For the most polar compound, the travelled distance is $2$ and the solvent travelled distance is $8$.

$$R_f = \frac{2}{8} = 0.25 = 25 \times 10^{-2}$$

Therefore, the required value is $25$, so the correct option is A.

Common mistakes

• Choosing the compound with the largest $R_f$ as the most polar is incorrect here. In TLC with hexane as the mobile phase, the more polar compound moves less on the plate. Use the smallest travel distance for the most polar compound.

• Using the spot distance directly as $R_f$ is wrong because $R_f$ is a ratio. Always divide the distance travelled by the compound by the distance travelled by the solvent front.

• Selecting $0.25$ instead of $25$ ignores the form asked in the question. Since the blank is followed by $\times 10^{-2}$, convert $0.25$ into $25 \times 10^{-2}$ before marking the answer.