Question

A mixture of 4-methylphenol and 4-methylcyclohexanol was run on a TLC plate using 3:1 hexane:ethyl acetate as the eluent. Only one spot was seen using a UV lamp. Does this mean that both spots have the same R_f or is there another explanation? How might you solve this problem?

Answer 1

4-methylphenol is more acid than 4-methylcyclohexanol which in turn should lead us to conclude that 4-methylphenol is more polar. This is actually expected, since the lone pair of O-atom can participate in resonance with the ring. No such resonance stabilization is offered in 4-methylcyclohexanol. However, on the TLC plate with 3:1 hexane:ethyl acetate, both the alcohols appear as a single spot. However, this may not necessarily mean that both the alcohols have the same R_f value (though it may be admitted that retention factor values are not universal constants, but constant for a particular plate and solvent composition usd).

The choice of solvent is crucial in TLC chromatography and in this case, it appears that the solvent choice is possibly not accurate. It appears that the solvent here is less polar to distinguish between the two solutes (the alcohols). That is why, 4-methylphenol traverses on the TLC plate more or less similar to 4-methylcyclohexanol, giving a single spot. An option is to increase the solvent polarity, possibly use 1:1 mixture of hexane:ethyl acetate. Such a solvent composition will allow to distinguish between the two alcohols of different polarity and we can expect to see distinct spots for both on the TLC plate.

Another option may be to repeat the experiment with a different TLC plate, made of a different material. Suppose, the first experiment was run on a silica plate (non-polar) with the given solvent compostion. Now, we may run the same experiment on an alumina plate (more polar) with the same solvent composition. It may be possible to obtain distinct spots since the retention of a solute depends on how the solute interacts with the stationary phase (TLC plate) and the solvent.